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A FIRST COURSE IN PROBABILITY

Eighth Edition

Sheldon Ross University of Southern California

Upper Saddle River, New Jersey 07458

Library of Congress Cataloging-in-Publication Data Ross, Sheldon M. A ﬁrst course in probability / Sheldon Ross. — 8th ed. p. cm. Includes bibliographical references and index. ISBN-13: 978-0-13-603313-4 ISBN-10: 0-13-603313-X 1. Probabilities—Textbooks. I. Title. QA273.R83 2010 519.2—dc22 2008033720 Editor in Chief, Mathematics and Statistics: Deirdre Lynch Senior Project Editor: Rachel S. Reeve Assistant Editor: Christina Lepre Editorial Assistant: Dana Jones Project Manager: Robert S. Merenoff Associate Managing Editor: Bayani Mendoza de Leon Senior Managing Editor: Linda Mihatov Behrens Senior Operations Supervisor: Diane Peirano Marketing Assistant: Kathleen DeChavez Creative Director: Jayne Conte Art Director/Designer: Bruce Kenselaar AV Project Manager: Thomas Benfatti Compositor: Integra Software Services Pvt. Ltd, Pondicherry, India Cover Image Credit: Getty Images, Inc.

© 2010, 2006, 2002, 1998, 1994, 1988, 1984, 1976 by Pearson Education, Inc., Pearson Prentice Hall Pearson Education, Inc. Upper Saddle River, NJ 07458

All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher. Pearson Prentice Hall™ is a trademark of Pearson Education, Inc. Printed in the United States of America 10 9 8 7 6 5 4 3 2 1

ISBN-13: 978-0-13-603313-4 ISBN-10: 0-13-603313-X Pearson Education, Ltd., London Pearson Education Australia PTY. Limited, Sydney Pearson Education Singapore, Pte. Ltd Pearson Education North Asia Ltd, Hong Kong Pearson Education Canada, Ltd., Toronto ´ de Mexico, S.A. de C.V. Pearson Educacion Pearson Education – Japan, Tokyo Pearson Education Malaysia, Pte. Ltd Pearson Education Upper Saddle River, New Jersey

For Rebecca

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Contents Preface 1

2

3

4

xi

Combinatorial Analysis 1.1 Introduction . . . . . . . . . . . . . . . . . . . . 1.2 The Basic Principle of Counting . . . . . . . . . 1.3 Permutations . . . . . . . . . . . . . . . . . . . . 1.4 Combinations . . . . . . . . . . . . . . . . . . . 1.5 Multinomial Coefﬁcients . . . . . . . . . . . . . 1.6 The Number of Integer Solutions of Equations Summary . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . Theoretical Exercises . . . . . . . . . . . . . . . Self-Test Problems and Exercises . . . . . . . .

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1 1 1 3 5 9 12 15 16 18 20

Axioms of Probability 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . 2.2 Sample Space and Events . . . . . . . . . . . . . . 2.3 Axioms of Probability . . . . . . . . . . . . . . . . 2.4 Some Simple Propositions . . . . . . . . . . . . . 2.5 Sample Spaces Having Equally Likely Outcomes 2.6 Probability as a Continuous Set Function . . . . . 2.7 Probability as a Measure of Belief . . . . . . . . . Summary . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . Theoretical Exercises . . . . . . . . . . . . . . . . Self-Test Problems and Exercises . . . . . . . . .

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22 22 22 26 29 33 44 48 49 50 54 56

Conditional Probability and Independence 3.1 Introduction . . . . . . . . . . . . . . 3.2 Conditional Probabilities . . . . . . . 3.3 Bayes’s Formula . . . . . . . . . . . . 3.4 Independent Events . . . . . . . . . . 3.5 P(·|F) Is a Probability . . . . . . . . . Summary . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . Theoretical Exercises . . . . . . . . . Self-Test Problems and Exercises . .

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58 58 58 65 79 93 101 102 110 114

Random Variables 4.1 Random Variables . . . . . . . . . . . . . . . . . . . . . 4.2 Discrete Random Variables . . . . . . . . . . . . . . . 4.3 Expected Value . . . . . . . . . . . . . . . . . . . . . . 4.4 Expectation of a Function of a Random Variable . . . 4.5 Variance . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 The Bernoulli and Binomial Random Variables . . . . 4.6.1 Properties of Binomial Random Variables . . . 4.6.2 Computing the Binomial Distribution Function

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Contents

4.7

The Poisson Random Variable . . . . . . . . . . . . . 4.7.1 Computing the Poisson Distribution Function 4.8 Other Discrete Probability Distributions . . . . . . . 4.8.1 The Geometric Random Variable . . . . . . . 4.8.2 The Negative Binomial Random Variable . . 4.8.3 The Hypergeometric Random Variable . . . 4.8.4 The Zeta (or Zipf) Distribution . . . . . . . . 4.9 Expected Value of Sums of Random Variables . . . 4.10 Properties of the Cumulative Distribution Function . Summary . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . Theoretical Exercises . . . . . . . . . . . . . . . . . . Self-Test Problems and Exercises . . . . . . . . . . . 5

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143 154 155 155 157 160 163 164 168 170 172 179 183

Continuous Random Variables 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Expectation and Variance of Continuous Random Variables . . 5.3 The Uniform Random Variable . . . . . . . . . . . . . . . . . . . 5.4 Normal Random Variables . . . . . . . . . . . . . . . . . . . . . . 5.4.1 The Normal Approximation to the Binomial Distribution 5.5 Exponential Random Variables . . . . . . . . . . . . . . . . . . . 5.5.1 Hazard Rate Functions . . . . . . . . . . . . . . . . . . . . 5.6 Other Continuous Distributions . . . . . . . . . . . . . . . . . . . 5.6.1 The Gamma Distribution . . . . . . . . . . . . . . . . . . 5.6.2 The Weibull Distribution . . . . . . . . . . . . . . . . . . 5.6.3 The Cauchy Distribution . . . . . . . . . . . . . . . . . . . 5.6.4 The Beta Distribution . . . . . . . . . . . . . . . . . . . . 5.7 The Distribution of a Function of a Random Variable . . . . . . Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Theoretical Exercises . . . . . . . . . . . . . . . . . . . . . . . . . Self-Test Problems and Exercises . . . . . . . . . . . . . . . . . .

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186 186 190 194 198 204 208 212 215 215 216 217 218 219 222 224 227 229

Jointly Distributed Random Variables 6.1 Joint Distribution Functions . . . . . . . . . . . . . . . . . . . . . 6.2 Independent Random Variables . . . . . . . . . . . . . . . . . . . 6.3 Sums of Independent Random Variables . . . . . . . . . . . . . . 6.3.1 Identically Distributed Uniform Random Variables . . . 6.3.2 Gamma Random Variables . . . . . . . . . . . . . . . . . 6.3.3 Normal Random Variables . . . . . . . . . . . . . . . . . 6.3.4 Poisson and Binomial Random Variables . . . . . . . . . 6.3.5 Geometric Random Variables . . . . . . . . . . . . . . . . 6.4 Conditional Distributions: Discrete Case . . . . . . . . . . . . . . 6.5 Conditional Distributions: Continuous Case . . . . . . . . . . . . 6.6 Order Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7 Joint Probability Distribution of Functions of Random Variables 6.8 Exchangeable Random Variables . . . . . . . . . . . . . . . . . . Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Theoretical Exercises . . . . . . . . . . . . . . . . . . . . . . . . . Self-Test Problems and Exercises . . . . . . . . . . . . . . . . . .

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232 232 240 252 252 254 256 259 260 263 266 270 274 282 285 287 291 293

Contents

7

8

9

Properties of Expectation 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 7.2 Expectation of Sums of Random Variables . . . . . . 7.2.1 Obtaining Bounds from Expectations via the Probabilistic Method . . . . . . . . . . 7.2.2 The Maximum–Minimums Identity . . . . . . 7.3 Moments of the Number of Events that Occur . . . . 7.4 Covariance, Variance of Sums, and Correlations . . . 7.5 Conditional Expectation . . . . . . . . . . . . . . . . 7.5.1 Deﬁnitions . . . . . . . . . . . . . . . . . . . . 7.5.2 Computing Expectations by Conditioning . . 7.5.3 Computing Probabilities by Conditioning . . 7.5.4 Conditional Variance . . . . . . . . . . . . . . 7.6 Conditional Expectation and Prediction . . . . . . . 7.7 Moment Generating Functions . . . . . . . . . . . . . 7.7.1 Joint Moment Generating Functions . . . . . 7.8 Additional Properties of Normal Random Variables 7.8.1 The Multivariate Normal Distribution . . . . 7.8.2 The Joint Distribution of the Sample Mean and Sample Variance . . . . . . . . . . . . . . 7.9 General Deﬁnition of Expectation . . . . . . . . . . . Summary . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . Theoretical Exercises . . . . . . . . . . . . . . . . . . Self-Test Problems and Exercises . . . . . . . . . . .

297 . . . . . . . . . . 297 . . . . . . . . . . 298 . . . . . . . . . . . . . .

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311 313 315 322 331 331 333 344 347 349 354 363 365 365

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367 369 370 373 380 384

Limit Theorems 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Chebyshev’s Inequality and the Weak Law of Large Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 The Central Limit Theorem . . . . . . . . . . . . . . . . . . . . 8.4 The Strong Law of Large Numbers . . . . . . . . . . . . . . . . 8.5 Other Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6 Bounding the Error Probability When Approximating a Sum of Independent Bernoulli Random Variables by a Poisson Random Variable . . . . . . . . . . . . . . . . . . . . . . . . . . Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Theoretical Exercises . . . . . . . . . . . . . . . . . . . . . . . . Self-Test Problems and Exercises . . . . . . . . . . . . . . . . . Additional Topics in Probability 9.1 The Poisson Process . . . . . . . . . 9.2 Markov Chains . . . . . . . . . . . . 9.3 Surprise, Uncertainty, and Entropy 9.4 Coding Theory and Entropy . . . . Summary . . . . . . . . . . . . . . . Problems and Theoretical Exercises Self-Test Problems and Exercises . References . . . . . . . . . . . . . .

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x

Contents

10 Simulation 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 General Techniques for Simulating Continuous Random Variables 10.2.1 The Inverse Transformation Method . . . . . . . . . . . . . 10.2.2 The Rejection Method . . . . . . . . . . . . . . . . . . . . . 10.3 Simulating from Discrete Distributions . . . . . . . . . . . . . . . . 10.4 Variance Reduction Techniques . . . . . . . . . . . . . . . . . . . . 10.4.1 Use of Antithetic Variables . . . . . . . . . . . . . . . . . . 10.4.2 Variance Reduction by Conditioning . . . . . . . . . . . . . 10.4.3 Control Variates . . . . . . . . . . . . . . . . . . . . . . . . Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Self-Test Problems and Exercises . . . . . . . . . . . . . . . . . . . Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Answers to Selected Problems

457

Solutions to Self-Test Problems and Exercises

461

Index

521

Preface

“We see that the theory of probability is at bottom only common sense reduced to calculation; it makes us appreciate with exactitude what reasonable minds feel by a sort of instinct, often without being able to account for it. . . . It is remarkable that this science, which originated in the consideration of games of chance, should have become the most important object of human knowledge. . . . The most important questions of life are, for the most part, really only problems of probability.” So said the famous French mathematician and astronomer (the “Newton of France”) PierreSimon, Marquis de Laplace. Although many people feel that the famous marquis, who was also one of the great contributors to the development of probability, might have exaggerated somewhat, it is nevertheless true that probability theory has become a tool of fundamental importance to nearly all scientists, engineers, medical practitioners, jurists, and industrialists. In fact, the enlightened individual had learned to ask not “Is it so?” but rather “What is the probability that it is so?” This book is intended as an elementary introduction to the theory of probability for students in mathematics, statistics, engineering, and the sciences (including computer science, biology, the social sciences, and management science) who possess the prerequisite knowledge of elementary calculus. It attempts to present not only the mathematics of probability theory, but also, through numerous examples, the many diverse possible applications of this subject. Chapter 1 presents the basic principles of combinatorial analysis, which are most useful in computing probabilities. Chapter 2 handles the axioms of probability theory and shows how they can be applied to compute various probabilities of interest. Chapter 3 deals with the extremely important subjects of conditional probability and independence of events. By a series of examples, we illustrate how conditional probabilities come into play not only when some partial information is available, but also as a tool to enable us to compute probabilities more easily, even when no partial information is present. This extremely important technique of obtaining probabilities by “conditioning” reappears in Chapter 7, where we use it to obtain expectations. The concept of random variables is introduced in Chapters 4, 5, and 6. Discrete random variables are dealt with in Chapter 4, continuous random variables in Chapter 5, and jointly distributed random variables in Chapter 6. The important concepts of the expected value and the variance of a random variable are introduced in Chapters 4 and 5, and these quantities are then determined for many of the common types of random variables. Additional properties of the expected value are considered in Chapter 7. Many examples illustrating the usefulness of the result that the expected value of a sum of random variables is equal to the sum of their expected values are presented. Sections on conditional expectation, including its use in prediction, and on moment-generating functions are contained in this chapter. In addition, the ﬁnal section introduces the multivariate normal distribution and presents a simple proof concerning the joint distribution of the sample mean and sample variance of a sample from a normal distribution. xi

xii

Preface

Chapter 8 presents the major theoretical results of probability theory. In particular, we prove the strong law of large numbers and the central limit theorem. Our proof of the strong law is a relatively simple one which assumes that the random variables have a ﬁnite fourth moment, and our proof of the central limit theorem assumes Levy’s continuity theorem. This chapter also presents such probability inequalities as Markov’s inequality, Chebyshev’s inequality, and Chernoff bounds. The ﬁnal section of Chapter 8 gives a bound on the error involved when a probability concerning a sum of independent Bernoulli random variables is approximated by the corresponding probability of a Poisson random variable having the same expected value. Chapter 9 presents some additional topics, such as Markov chains, the Poisson process, and an introduction to information and coding theory, and Chapter 10 considers simulation. As in the previous edition, three sets of exercises are given at the end of each chapter. They are designated as Problems, Theoretical Exercises, and Self-Test Problems and Exercises. This last set of exercises, for which complete solutions appear in Solutions to Self-Test Problems and Exercises, is designed to help students test their comprehension and study for exams. CHANGES IN THE NEW EDITION The eighth edition continues the evolution and ﬁne tuning of the text. It includes new problems, exercises, and text material chosen both for its inherent interest and for its use in building student intuition about probability. Illustrative of these goals are Example 5d of Chapter 1 on knockout tournaments, and Examples 4k and 5i of Chapter 7 on multiple player gambler’s ruin problems. A key change in the current edition is that the important result that the expectation of a sum of random variables is equal to the sum of the expectations is now ﬁrst presented in Chapter 4 (rather than Chapter 7 as in previous editions). A new and elementary proof of this result when the sample space of the probability experiment is ﬁnite is given in this chapter. Another change is the expansion of Section 6.3, which deals with the sum of independent random variables. Section 6.3.1 is a new section in which we derive the distribution of the sum of independent and identically distributed uniform random variables, and then use our results to show that the expected number of random numbers that needs to be added for their sum to exceed 1 is equal to e. Section 6.3.5 is a new section in which we derive the distribution of the sum of independent geometric random variables with different means. ACKNOWLEDGMENTS I am grateful for the careful work of Hossein Hamedani in checking the text for accuracy. I also appreciate the thoughtfulness of the following people that have taken the time to contact me with comments for improving the text: Amir Ardestani, Polytechnic University of Teheran; Joe Blitzstein, Harvard University; Peter Nuesch, University of Lausaunne; Joseph Mitchell, SUNY, Stony Brook; Alan Chambless, actuary; Robert Kriner; Israel David, Ben-Gurion University; T. Lim, George Mason University; Wei Chen, Rutgers; D. Monrad, University of Illinois; W. Rosenberger, George Mason University; E. Ionides, University of Michigan; J. Corvino, Lafayette College; T. Seppalainen, University of Wisconsin. Finally, I would like to thank the following reviewers for their many helpful comments. Reviewers of the eighth edition are marked with an asterisk.

Acknowledgments

K. B. Athreya, Iowa State University Richard Bass, University of Connecticut Robert Bauer, University of Illinois at Urbana-Champaign Phillip Beckwith, Michigan Tech Arthur Benjamin, Harvey Mudd College Geoffrey Berresford, Long Island University Baidurya Bhattacharya, University of Delaware Howard Bird, St. Cloud State University Shahar Boneh, Metropolitan State College of Denver Jean Cadet, State University of New York at Stony Brook Steven Chiappari, Santa Clara University Nicolas Christou, University of California, Los Angeles James Clay, University of Arizona at Tucson Francis Conlan, University of Santa Clara *Justin Corvino, Lafayette College Jay DeVore, California Polytechnic University, San Luis Obispo Scott Emerson, University of Washington Thomas R. Fischer, Texas A & M University Anant Godbole, Michigan Technical University Zakkula Govindarajulu, University of Kentucky Richard Groeneveld, Iowa State University Mike Hardy, Massachusetts Institute of Technology Bernard Harris, University of Wisconsin Larry Harris, University of Kentucky David Heath, Cornell University Stephen Herschkorn, Rutgers University Julia L. Higle, University of Arizona Mark Huber, Duke University

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*Edward Ionides, University of Michigan Anastasia Ivanova, University of North Carolina Hamid Jafarkhani, University of California, Irvine Chuanshu Ji, University of North Carolina, Chapel Hill Robert Keener, University of Michigan Fred Leysieffer, Florida State University Thomas Liggett, University of California, Los Angeles Helmut Mayer, University of Georgia Bill McCormick, University of Georgia Ian McKeague, Florida State University R. Miller, Stanford University *Ditlev Monrad, University of Illinois Robb J. Muirhead, University of Michigan Joe Naus, Rutgers University Nhu Nguyen, New Mexico State University Ellen O’Brien, George Mason University N. U. Prabhu, Cornell University Kathryn Prewitt, Arizona State University Jim Propp, University of Wisconsin *William F. Rosenberger, George Mason University Myra Samuels, Purdue University I. R. Savage, Yale University Art Schwartz, University of Michigan at Ann Arbor Therese Shelton, Southwestern University Malcolm Sherman, State University of New York at Albany Murad Taqqu, Boston University Eli Upfal, Brown University Ed Wheeler, University of Tennessee Allen Webster, Bradley University S. R. [email protected]

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C H A P T E R

1

Combinatorial Analysis 1.1 1.2 1.3 1.4 1.5 1.6

INTRODUCTION THE BASIC PRINCIPLE OF COUNTING PERMUTATIONS COMBINATIONS MULTINOMIAL COEFFICIENTS THE NUMBER OF INTEGER SOLUTIONS OF EQUATIONS

1.1

INTRODUCTION Here is a typical problem of interest involving probability: A communication system is to consist of n seemingly identical antennas that are to be lined up in a linear order. The resulting system will then be able to receive all incoming signals—and will be called functional—as long as no two consecutive antennas are defective. If it turns out that exactly m of the n antennas are defective, what is the probability that the resulting system will be functional? For instance, in the special case where n = 4 and m = 2, there are 6 possible system conﬁgurations, namely, 0 0 1 0 1 1

1 1 0 0 0 1

1 0 1 1 0 0

0 1 0 1 1 0

where 1 means that the antenna is working and 0 that it is defective. Because the resulting system will be functional in the ﬁrst 3 arrangements and not functional in the remaining 3, it seems reasonable to take 36 = 12 as the desired probability. In the case of general n and m, we could compute the probability that the system is functional in a similar fashion. That is, we could count the number of conﬁgurations that result in the system’s being functional and then divide by the total number of all possible conﬁgurations. From the preceding discussion, we see that it would be useful to have an effective method for counting the number of ways that things can occur. In fact, many problems in probability theory can be solved simply by counting the number of different ways that a certain event can occur. The mathematical theory of counting is formally known as combinatorial analysis. 1.2

THE BASIC PRINCIPLE OF COUNTING The basic principle of counting will be fundamental to all our work. Loosely put, it states that if one experiment can result in any of m possible outcomes and if another experiment can result in any of n possible outcomes, then there are mn possible outcomes of the two experiments. 1

2

Chapter 1

Combinatorial Analysis

The basic principle of counting Suppose that two experiments are to be performed. Then if experiment 1 can result in any one of m possible outcomes and if, for each outcome of experiment 1, there are n possible outcomes of experiment 2, then together there are mn possible outcomes of the two experiments. Proof of the Basic Principle: The basic principle may be proven by enumerating all the possible outcomes of the two experiments; that is, (1, 1), (1, 2), . . . , (1, n) (2, 1), (2, 2), . . . , (2, n) # # # (m, 1), (m, 2), . . . , (m, n) where we say that the outcome is (i, j) if experiment 1 results in its ith possible outcome and experiment 2 then results in its jth possible outcome. Hence, the set of possible outcomes consists of m rows, each containing n elements. This proves the result. EXAMPLE 2a A small community consists of 10 women, each of whom has 3 children. If one woman and one of her children are to be chosen as mother and child of the year, how many different choices are possible? Solution. By regarding the choice of the woman as the outcome of the ﬁrst experiment and the subsequent choice of one of her children as the outcome of the second experiment, we see from the basic principle that there are 10 * 3 = 30 possible choices. . When there are more than two experiments to be performed, the basic principle can be generalized.

The generalized basic principle of counting If r experiments that are to be performed are such that the ﬁrst one may result in any of n1 possible outcomes; and if, for each of these n1 possible outcomes, there are n2 possible outcomes of the second experiment; and if, for each of the possible outcomes of the ﬁrst two experiments, there are n3 possible outcomes of the third experiment; and if . . . , then there is a total of n1 · n2 · · · nr possible outcomes of the r experiments.

EXAMPLE 2b A college planning committee consists of 3 freshmen, 4 sophomores, 5 juniors, and 2 seniors. A subcommittee of 4, consisting of 1 person from each class, is to be chosen. How many different subcommittees are possible?

Section 1.3

Permutations

3

Solution. We may regard the choice of a subcommittee as the combined outcome of the four separate experiments of choosing a single representative from each of the classes. It then follows from the generalized version of the basic principle that there are 3 * 4 * 5 * 2 = 120 possible subcommittees. . EXAMPLE 2c How many different 7-place license plates are possible if the ﬁrst 3 places are to be occupied by letters and the ﬁnal 4 by numbers? Solution. By the generalized version of the basic principle, the answer is 26 · 26 · 26 · 10 · 10 · 10 · 10 = 175,760,000. . EXAMPLE 2d How many functions deﬁned on n points are possible if each functional value is either 0 or 1? Solution. Let the points be 1, 2, . . . , n. Since f (i) must be either 0 or 1 for each i = 1, 2, . . . , n, it follows that there are 2n possible functions. . EXAMPLE 2e In Example 2c, how many license plates would be possible if repetition among letters or numbers were prohibited? Solution. In this case, there would be 26 · 25 · 24 · 10 · 9 · 8 · 7 = 78,624,000 possible license plates. . 1.3

PERMUTATIONS How many different ordered arrangements of the letters a, b, and c are possible? By direct enumeration we see that there are 6, namely, abc, acb, bac, bca, cab, and cba. Each arrangement is known as a permutation. Thus, there are 6 possible permutations of a set of 3 objects. This result could also have been obtained from the basic principle, since the ﬁrst object in the permutation can be any of the 3, the second object in the permutation can then be chosen from any of the remaining 2, and the third object in the permutation is then the remaining 1. Thus, there are 3 · 2 · 1 = 6 possible permutations. Suppose now that we have n objects. Reasoning similar to that we have just used for the 3 letters then shows that there are n(n − 1)(n − 2) · · · 3 · 2 · 1 = n! different permutations of the n objects. EXAMPLE 3a How many different batting orders are possible for a baseball team consisting of 9 players? Solution. There are 9! = 362,880 possible batting orders.

.

4

Chapter 1

Combinatorial Analysis

EXAMPLE 3b A class in probability theory consists of 6 men and 4 women. An examination is given, and the students are ranked according to their performance. Assume that no two students obtain the same score. (a) How many different rankings are possible? (b) If the men are ranked just among themselves and the women just among themselves, how many different rankings are possible? Solution. (a) Because each ranking corresponds to a particular ordered arrangement of the 10 people, the answer to this part is 10! = 3,628,800. (b) Since there are 6! possible rankings of the men among themselves and 4! possible rankings of the women among themselves, it follows from the basic principle that there are (6!)(4!) = (720)(24) = 17,280 possible rankings in this case. . EXAMPLE 3c Ms. Jones has 10 books that she is going to put on her bookshelf. Of these, 4 are mathematics books, 3 are chemistry books, 2 are history books, and 1 is a language book. Ms. Jones wants to arrange her books so that all the books dealing with the same subject are together on the shelf. How many different arrangements are possible? Solution. There are 4! 3! 2! 1! arrangements such that the mathematics books are ﬁrst in line, then the chemistry books, then the history books, and then the language book. Similarly, for each possible ordering of the subjects, there are 4! 3! 2! 1! possible arrangements. Hence, as there are 4! possible orderings of the subjects, the desired answer is 4! 4! 3! 2! 1! = 6912. . We shall now determine the number of permutations of a set of n objects when certain of the objects are indistinguishable from each other. To set this situation straight in our minds, consider the following example. EXAMPLE 3d How many different letter arrangements can be formed from the letters PEPPER? Solution. We ﬁrst note that there are 6! permutations of the letters P1 E1 P2 P3 E2 R when the 3P’s and the 2E’s are distinguished from each other. However, consider any one of these permutations—for instance, P1 P2 E1 P3 E2 R. If we now permute the P’s among themselves and the E’s among themselves, then the resultant arrangement would still be of the form PPEPER. That is, all 3! 2! permutations P1 P2 E1 P3 E2 R P1 P3 E1 P2 E2 R P2 P1 E1 P3 E2 R P2 P3 E1 P1 E2 R P3 P1 E1 P2 E2 R P3 P2 E1 P1 E2 R

P1 P2 E2 P3 E1 R P1 P3 E2 P2 E1 R P2 P1 E2 P3 E1 R P2 P3 E2 P1 E1 R P3 P1 E2 P2 E1 R P3 P2 E2 P1 E1 R

are of the form PPEPER. Hence, there are 6!/(3! 2!) = 60 possible letter arrangements of the letters PEPPER. .

Section 1.4

Combinations

5

In general, the same reasoning as that used in Example 3d shows that there are n! n1 ! n2 ! · · · nr ! different permutations of n objects, of which n1 are alike, n2 are alike, . . . , nr are alike. EXAMPLE 3e A chess tournament has 10 competitors, of which 4 are Russian, 3 are from the United States, 2 are from Great Britain, and 1 is from Brazil. If the tournament result lists just the nationalities of the players in the order in which they placed, how many outcomes are possible? Solution. There are

10! = 12,600 4! 3! 2! 1! .

possible outcomes. EXAMPLE 3f

How many different signals, each consisting of 9 ﬂags hung in a line, can be made from a set of 4 white ﬂags, 3 red ﬂags, and 2 blue ﬂags if all ﬂags of the same color are identical? Solution. There are

9! = 1260 4! 3! 2! .

different signals. 1.4

COMBINATIONS We are often interested in determining the number of different groups of r objects that could be formed from a total of n objects. For instance, how many different groups of 3 could be selected from the 5 items A, B, C, D, and E? To answer this question, reason as follows: Since there are 5 ways to select the initial item, 4 ways to then select the next item, and 3 ways to select the ﬁnal item, there are thus 5 · 4 · 3 ways of selecting the group of 3 when the order in which the items are selected is relevant. However, since every group of 3—say, the group consisting of items A, B, and C—will be counted 6 times (that is, all of the permutations ABC, ACB, BAC, BCA, CAB, and CBA will be counted when the order of selection is relevant), it follows that the total number of groups that can be formed is 5 · 4 · 3 = 10 3 · 2 · 1 In general, as n(n − 1) · · · (n − r + 1) represents the number of different ways that a group of r items could be selected from n items when the order of selection is relevant, and as each group of r items will be counted r! times in this count, it follows that the number of different groups of r items that could be formed from a set of n items is n! n(n − 1) · · · (n − r + 1) = r! (n − r)! r!

6

Chapter 1

Combinatorial Analysis

Notation and terminology n , for r … n, by We deﬁne r

n r

=

n! (n − r)! r!

n represents the number of possible combinations of n objects r taken r at a time.†

and say that

n Thus, represents the number of different groups of size r that could be r selected from a set of n objects when the order of selection is not considered relevant. EXAMPLE 4a A committee of 3 is to be formed from a group of 20 people. How many different committees are possible? Solution. There are

20 3

=

20 · 19 · 18 = 1140 possible committees. 3 · 2 · 1

.

EXAMPLE 4b From a group of 5 women and 7 men, how many different committees consisting of 2 women and 3 men can be formed? What if 2 of the men are feuding and refuse to serve on the committee together? 7 possible 3 5 7 groups of 3 men, it follows from the basic principle that there are = 2 3 5 · 4 7 · 6 · 5 = 350 possible committees consisting of 2 women and 3 men. 2 · 1 3 · 2 · 1 that 2 of the refuse to serve together. Because a total of men suppose Now 7 5 2 = 35 possible groups of 3 men contain both of = 5 out of the 3 1 2 the feuding men, it follows that there are 35 − 5= 30 groups that do not contain 5 both of the feuding men. Because there are still = 10 ways to choose the 2 2 women, there are 30 · 10 = 300 possible committees in this case. .

Solution. As there are

5 2

possible groups of 2 women, and

† By convention, 0! is deﬁned to be 1. Thus,

to 0 when either i < 0 or i > n.

n 0

=

n n

= 1. We also take

n i

to be equal

Section 1.4

Combinations

7

EXAMPLE 4c Consider a set of n antennas of which m are defective and n − m are functional and assume that all of the defectives and all of the functionals are considered indistinguishable. How many linear orderings are there in which no two defectives are consecutive? Solution. Imagine that the n − m functional antennas are lined up among themselves. Now, if no two defectives are to be consecutive, then the spaces between the functional antennas must each contain at most one defective antenna. That is, in the n − m + 1 possible positions—represented in Figure 1.1 by carets—between the n − m functional antennas,we must selectm of these in which to put the defective n − m + 1 possible orderings in which there is at antennas. Hence, there are m least one functional antenna between any two defective ones. .

^1^1^1...^1^1^ 1 ⫽ functional ^ ⫽ place for at most one defective

FIGURE 1.1: No consecutive defectives

A useful combinatorial identity is n − 1 n − 1 n + = r r − 1 r

1 … r … n

(4.1)

Equation (4.1) may be proved analytically or by the following combinatorial argument: Consider a group of n objects, andﬁx attention on some particular one of these n − 1 groups of size r that contain object objects—call it object 1. Now, there are r − 1 1 (since each such groupis formed by selecting r − 1 from the remaining n − 1 n − 1 groups of size r that do not contain object 1. As objects). Also, there are r n there is a total of groups of size r, Equation (4.1) follows. r n The values are often referred to as binomial coefﬁcients because of their r prominence in the binomial theorem. The binomial theorem n

(x + y) =

n n k=0

k

xk yn−k

(4.2)

We shall present two proofs of the binomial theorem. The ﬁrst is a proof by mathematical induction, and the second is a proof based on combinatorial considerations.

8

Chapter 1

Combinatorial Analysis

Proof of the Binomial Theorem by Induction: When n = 1, Equation (4.2) reduces to 1 1 0 1 x y + x1 y0 = y + x x + y= 0 1 Assume Equation (4.2) for n − 1. Now, (x + y)n = (x + y)(x + y)n−1 n−1 n − 1 = (x + y) xk yn−1−k k =

n−1 k=0

k=0

n − 1 k

xk+1 yn−1−k +

n−1 n − 1 xk yn−k k k=0

Letting i = k + 1 in the ﬁrst sum and i = k in the second sum, we ﬁnd that n n−1 n − 1 n − 1 xi yn−i + xi yn−i i − 1 i i=0 i=1 n−1 n − 1 n − 1 xi yn−i + yn + = xn + i i − 1

(x + y)n =

= xn +

i=1 n−i i=1

n i

xi yn−i + yn

n n xi yn−i = i i=0

where the next-to-last equality follows by Equation (4.1). By induction, the theorem is now proved. Combinatorial Proof of the Binomial Theorem:

Consider the product

(x1 + y1 )(x2 + y2 ) · · · (xn + yn ) Its expansion consists of the sum of 2n terms, each term being the product of n factors. Furthermore, each of the 2n terms in the sum will contain as a factor either xi or yi for each i = 1, 2, . . . , n. For example, (x1 + y1 )(x2 + y2 ) = x1 x2 + x1 y2 + y1 x2 + y1 y2 Now, how many of the 2n terms in the sum will have k of the xi ’s and (n − k) of the yi ’s as factors? As each term consisting of k of the xi ’s and (n − k) of theyi ’scorresponds n such terms. to a choice of a group of k from the n values x1 , x2 , . . . , xn , there are k Thus, letting xi = x, yi = y, i = 1, . . . , n, we see that n n xk yn−k (x + y)n = k k=0

Section 1.5

Multinomial Coefﬁcients

9

EXAMPLE 4d Expand (x + y)3 . Solution. 3

(x + y) =

3 0

0 3

x y +

3 1

1 2

x y +

3 2

2

x y +

3 3

x3 y0

= y3 + 3xy2 + 3x2 y + x3

.

EXAMPLE 4e How many subsets are there of a set consisting of n elements? n subsets of size k, the desired answer is Solution. Since there are k n n = (1 + 1)n = 2n k k=0

This result could also have been obtained by assigning either the number 0 or the number 1 to each element in the set. To each assignment of numbers, there corresponds, in a one-to-one fashion, a subset, namely, that subset consisting of all elements that were assigned the value 1. As there are 2n possible assignments, the result follows. Note that we have included the set consisting of 0 elements (that is, the null set) as a subset of the original set. Hence, the number of subsets that contain at least one . element is 2n − 1. 1.5

MULTINOMIAL COEFFICIENTS In this section, we consider the following problem: A set of n distinct items is to be divided into r distinct groups of respective sizes n1 , n2 , . . . , nr , where ri=1 ni = n. How many divisions are possible? To answer this question, we note that different n there are possible choices for the ﬁrst group; for each choice of the ﬁrst group, n1 n − n1 there are possible choices for the second group; for each choice of the n2 n − n1 − n2 ﬁrst two groups, there are possible choices for the third group; and n3 so on. It then follows from the generalized version of the basic counting principle that there are n − n1 n − n1 − n2 − · · · − nr−1 n ··· n2 nr n1 n! (n − n1 )! (n − n1 − n2 − · · · − nr−1 )! = ··· (n − n1 )! n1 ! (n − n1 − n2 )! n2 ! 0! nr ! n! = n1 ! n2 ! · · · nr ! possible divisions.

10

Chapter 1

Combinatorial Analysis

Another way to see this result is to consider the n values 1, 1, . . . , 1, 2, . . . , 2, . . . , r, . . . , r, where i appears ni times, for i = 1, . . . , r. Every permutation of these values corresponds to a division of the n items into the r groups in the following manner: Let the permutation i1 , i2 , . . . , in correspond to assigning item 1 to group i1 , item 2 to group i2 , and so on. For instance, if n = 8 and if n1 = 4, n2 = 3, and n3 = 1, then the permutation 1, 1, 2, 3, 2, 1, 2, 1 corresponds to assigning items 1, 2, 6, 8 to the ﬁrst group, items 3, 5, 7 to the second group, and item 4 to the third group. Because every permutation yields a division of the items and every possible division results from some permutation, it follows that the number of divisions of n items into r distinct groups of sizes n1 , n2 , . . . , nr is the same as the number of permutations of n items of which n1 are alike, and n2 are alike, . . ., and nr are alike, which was shown in Section n! . 1.3 to equal n1 !n2 ! · · · nr !

Notation

If n1 + n2 + · · · + nr = n, we deﬁne

n n1 , n2 , . . . , nr

=

n n1 , n2 , . . . , nr

by

n! n1 ! n2 ! · · · nr !

n Thus, represents the number of possible divisions of n distinct n1 , n2 , . . . , nr objects into r distinct groups of respective sizes n1 , n2 , . . . , nr .

EXAMPLE 5a A police department in a small city consists of 10 ofﬁcers. If the department policy is to have 5 of the ofﬁcers patrolling the streets, 2 of the ofﬁcers working full time at the station, and 3 of the ofﬁcers on reserve at the station, how many different divisions of the 10 ofﬁcers into the 3 groups are possible? Solution. There are

10! = 2520 possible divisions. 5! 2! 3!

.

EXAMPLE 5b Ten children are to be divided into an A team and a B team of 5 each. The A team will play in one league and the B team in another. How many different divisions are possible? Solution. There are

10! = 252 possible divisions. 5! 5!

.

EXAMPLE 5c In order to play a game of basketball, 10 children at a playground divide themselves into two teams of 5 each. How many different divisions are possible?

Section 1.5

Multinomial Coefﬁcients

11

Solution. Note that this example is different from Example 5b because now the order of the two teams is irrelevant. That is, there is no A and B team, but just a division consisting of 2 groups of 5 each. Hence, the desired answer is 10!/(5! 5!) = 126 2!

.

The proof of the following theorem, which generalizes the binomial theorem, is left as an exercise. The multinomial theorem

n

(x1 + x2 + · · · + xr ) =

(n1 , . . . , nr ) : n1 + · · · + nr = n

n n1 , n2 , . . . , nr

n

n

x1 1 x2 2 · · · xnr r

That is, the sum is over all nonnegative integer-valued vectors (n1 , n2 , . . . , nr ) such that n1 + n2 + · · · + nr = n. The numbers

n n1 , n2 , . . . , nr

are known as multinomial coefﬁcients.

EXAMPLE 5d In the ﬁrst round of a knockout tournament involving n = 2m players, the n players are divided into n/2 pairs, with each of these pairs then playing a game. The losers of the games are eliminated while the winners go on to the next round, where the process is repeated until only a single player remains. Suppose we have a knockout tournament of 8 players. (a) How many possible outcomes are there for the initial round? (For instance, one outcome is that 1 beats 2, 3 beats 4, 5 beats 6, and 7 beats 8. ) (b) How many outcomes of the tournament are possible, where an outcome gives complete information for all rounds? Solution. One way to determine the number of possible outcomes for the initial round is to ﬁrst determine the number of possible pairings for that round. To do so, note that the number of waysto dividethe 8 players into a ﬁrst pair, a second pair, 8 8! a third pair, and a fourth pair is = 4 . Thus, the number of possible pair2, 2, 2, 2 2 8! ings when there is no ordering of the 4 pairs is 4 . For each such pairing, there are 2 4! 2 possible choices from each pair as to the winner of that game, showing that there 8! 8!24 possible results of round 1. (Another way to see this is to note that = are 4 4! 2 4! 8 there are possible choices of the 4 winners and, for each such choice, there are 4 8 8! 4! ways to pair the 4 winners with the 4 losers, showing that there are 4! = 4! 4 possible results for the ﬁrst round.)

12

Chapter 1

Combinatorial Analysis

4! possible outcomes of round 2, 2! 2! possible outcomes and for each of the outcomes of the ﬁrst two rounds, there are 1! of round 3. Consequently, by the generalized basic principle of counting, there are 8! 4! 2! = 8! possible outcomes of the tournament. Indeed, the same argument 4! 2! 1! can be used to show that a knockout tournament of n = 2m players has n! possible outcomes. Knowing the preceding result, it is not difﬁcult to come up with a more direct argument by showing that there is a one-to-one correspondence between the set of possible tournament results and the set of permutations of 1, . . . , n. To obtain such a correspondence, rank the players as follows for any tournament result: Give the tournament winner rank 1, and give the ﬁnal-round loser rank 2. For the two players who lost in the next-to-last round, give rank 3 to the one who lost to the player ranked 1 and give rank 4 to the one who lost to the player ranked 2. For the four players who lost in the second-to-last round, give rank 5 to the one who lost to player ranked 1, rank 6 to the one who lost to the player ranked 2, rank 7 to the one who lost to the player ranked 3, and rank 8 to the one who lost to the player ranked 4. Continuing on in this manner gives a rank to each player. (A more succinct description is to give the winner of the tournament rank 1 and let the rank of a player who lost in a round having 2k matches be 2k plus the rank of the player who beat him, for k = 0, . . . , m − 1.) In this manner, the result of the tournament can be represented by a permutation i1 , i2 , . . . , in , where ij is the player who was given rank j. Because different tournament results give rise to different permutations, and because there is a tournament result for each permutation, it follows that there are the same number of possible tournament results as there are permutations of 1, . . . , n. . Similarly, for each result of round 1, there are

EXAMPLE 5e

(x1 + x2 + x3 )2 =

2 2 x21 x02 x03 + x01 x22 x03 2, 0, 0 0, 2, 0 2 2 0 0 2 + x1 x2 x3 + x11 x12 x03 0, 0, 2 1, 1, 0 2 2 1 0 1 x1 x2 x3 + x01 x12 x13 + 1, 0, 1 0, 1, 1

= x21 + x22 + x23 + 2x1 x2 + 2x1 x3 + 2x2 x3 ∗ 1.6

.

THE NUMBER OF INTEGER SOLUTIONS OF EQUATIONS There are rn possible outcomes when n distinguishable balls are to be distributed into r distinguishable urns. This result follows because each ball may be distributed into any of r possible urns. Let us now, however, suppose that the n balls are indistinguishable from each other. In this case, how many different outcomes are possible? As the balls are indistinguishable, it follows that the outcome of the experiment of distributing the n balls into r urns can be described by a vector (x1 , x2 , . . . , xr ), where xi denotes the number of balls that are distributed into the ith urn. Hence, the problem reduces to ﬁnding the number of distinct nonnegative integer-valued vectors (x1 , x2 , . . . , xr ) such that x1 + x2 + · · · + xr = n ∗ Asterisks denote material that is optional.

Section 1.6

The Number of Integer Solutions of Equations

13

To compute this number, let us start by considering the number of positive integervalued solutions. Toward that end, imagine that we have n indistinguishable objects lined up and that we want to divide them into r nonempty groups. To do so, we can select r − 1 of the n − 1 spaces between adjacent objects as our dividing points. (See Figure 1.2.) For instance, if we have n = 8 and r = 3 and we choose the 2 divisors so as to obtain ooo|ooo|oo

0^0^0^...^0^0 n objects 0 Choose r ⫺ 1 of the spaces ^.

FIGURE 1.2: Number of positive solutions

then the resulting vector is x1 = 3, x2 = 3, x3 = 2. As there are

n − 1 r − 1

possible

selections, we have the following proposition. n − 1 distinct positive integer-valued vectors (x1 , Proposition 6.1. There are r − 1 x2 , . . . , xr ) satisfying the equation x1 + x2 + · · · + xr = n xi > 0, i = 1, . . . , r To obtain the number of nonnegative (as opposed to positive) solutions, note that the number of nonnegative solutions of x1 + x2 + · · · + xr = n is the same as the number of positive solutions of y1 + · · · + yr = n + r (seen by letting yi = xi + 1, i = 1, . . . , r). Hence, from Proposition 6.1, we obtain the following proposition. n + r − 1 Proposition 6.2. There are distinct nonnegative integer-valued vecr − 1 tors (x1 , x2 , . . . , xr ) satisfying the equation x1 + x2 + · · · + xr = n

(6.1)

EXAMPLE 6a How many distinct nonnegative integer-valued solutions of x1 + x2 = 3 are possible? 3 + 2 − 1 Solution. There are = 4 such solutions: (0, 3), (1, 2), (2, 1), (3, 0). . 2 − 1 EXAMPLE 6b An investor has 20 thousand dollars to invest among 4 possible investments. Each investment must be in units of a thousand dollars. If the total 20 thousand is to be

14

Chapter 1

Combinatorial Analysis

invested, how many different investment strategies are possible? What if not all the money need be invested? Solution. If we let xi , i = 1, 2, 3, 4, denote the number of thousands invested in investment i, then, when all is to be invested, x1 , x2 , x3 , x4 are integers satisfying the equation x1 + x2 + x3 + x4 = 20

xi Ú 0

23 = 1771 possible investment strategies. If Hence, by Proposition 6.2, there are 3 not all of the money need be invested, then if we let x5 denote the amount kept in reserve, a strategy is a nonnegative integer-valued vector (x1 , x2 , x3 , x4 , x5 ) satisfying the equation

x1 + x2 + x3 + x4 + x5 = 20 Hence, by Proposition 6.2, there are now

24 4

= 10,626 possible strategies.

.

EXAMPLE 6c How many terms are there in the multinomial expansion of (x1 + x2 + · · · + xr )n ? Solution. n

(x1 + x2 + · · · + xr ) =

n n1 , . . . , nr

n

x1 1 · · · xnr r

where the sum is over all nonnegative integer-valued (n1 , . .. , nr ) such that n1 + · · · + n + r − 1 such terms. . nr = n. Hence, by Proposition 6.2, there are r − 1 EXAMPLE 6d Let us consider again Example 4c, in which we have a set of n items, of which m are (indistinguishable and) defective and the remaining n − m are (also indistinguishable and) functional. Our objective is to determine the number of linear orderings in which no two defectives are next to each other. To determine this number, let us imagine that the defective items are lined up among themselves and the functional ones are now to be put in position. Let us denote x1 as the number of functional items to the left of the ﬁrst defective, x2 as the number of functional items between the ﬁrst two defectives, and so on. That is, schematically, we have x1 0 x2 0 · · · xm 0 xm+1 Now, there will be at least one functional item between any pair of defectives as long as xi > 0, i = 2, . . . , m. Hence, the number of outcomes satisfying the condition is the number of vectors x1 , . . . , xm+1 that satisfy the equation x1 + · · · + xm+1 = n − m x1 Ú 0, xm+1 Ú 0, xi > 0, i = 2, . . . , m

Summary

15

But, on letting y1 = x1 + 1, yi = xi , i = 2, . . . , m, ym+1 = xm+1 + 1, we see that this number is equal to the number of positive vectors (y1 , . . . , ym+1 ) that satisfy the equation y1 + y2 + · · · + ym+1 = n − m + 2 Hence, by Proposition 6.1, there are

n − m + 1 m

such outcomes, in agreement

with the results of Example 4c. Suppose now that we are interested in the number of outcomes in which each pair of defective items is separated by at least 2 functional items. By the same reasoning as that applied previously, this would equal the number of vectors satisfying the equation x1 + · · · + xm+1 = n − m

x1 Ú 0, xm+1 Ú 0, xi Ú 2, i = 2, . . . , m

Upon letting y1 = x1 + 1, yi = xi − 1, i = 2, . . . , m, ym+1 = xm+1 + 1, we see that this is the same as the number of positive solutions of the equation y1 + · · · + ym+1 = n − 2m + 3 Hence, from Proposition 6.1, there are

n − 2m + 2 m

such outcomes.

.

SUMMARY The basic principle of counting states that if an experiment consisting of two phases is such that there are n possible outcomes of phase 1 and, for each of these n outcomes, there are m possible outcomes of phase 2, then there are nm possible outcomes of the experiment. There are n! = n(n − 1) · · · 3 · 2 · 1 possible linear orderings of n items. The quantity 0! is deﬁned to equal 1. Let n! n = i (n − i)! i! when 0 … i … n, and let it equal 0 otherwise. This quantity represents the number of different subgroups of size i that can be chosen from a set of size n. It is often called a binomial coefﬁcient because of its prominence in the binomial theorem, which states that n n n (x + y) = xi yn−i i i=0

For nonnegative integers n1 , . . . , nr summing to n,

n n1 , n2 , . . . , nr

=

n! n1 !n2 ! · · · nr !

is the number of divisions of n items into r distinct nonoverlapping subgroups of sizes n1 , n2 , . . . , nr .

16

Chapter 1

Combinatorial Analysis

PROBLEMS 1. (a) How many different 7-place license plates are possible if the ﬁrst 2 places are for letters and the other 5 for numbers? (b) Repeat part (a) under the assumption that no letter or number can be repeated in a single license plate. 2. How many outcome sequences are possible when a die is rolled four times, where we say, for instance, that the outcome is 3, 4, 3, 1 if the ﬁrst roll landed on 3, the second on 4, the third on 3, and the fourth on 1? 3. Twenty workers are to be assigned to 20 different jobs, one to each job. How many different assignments are possible? 4. John, Jim, Jay, and Jack have formed a band consisting of 4 instruments. If each of the boys can play all 4 instruments, how many different arrangements are possible? What if John and Jim can play all 4 instruments, but Jay and Jack can each play only piano and drums? 5. For years, telephone area codes in the United States and Canada consisted of a sequence of three digits. The ﬁrst digit was an integer between 2 and 9, the second digit was either 0 or 1, and the third digit was any integer from 1 to 9. How many area codes were possible? How many area codes starting with a 4 were possible? 6. A well-known nursery rhyme starts as follows: “As I was going to St. Ives I met a man with 7 wives. Each wife had 7 sacks. Each sack had 7 cats. Each cat had 7 kittens. . .” How many kittens did the traveler meet? 7. (a) In how many ways can 3 boys and 3 girls sit in a row? (b) In how many ways can 3 boys and 3 girls sit in a row if the boys and the girls are each to sit together? (c) In how many ways if only the boys must sit together? (d) In how many ways if no two people of the same sex are allowed to sit together? 8. How many different letter arrangements can be made from the letters (a) Fluke? (b) Propose? (c) Mississippi? (d) Arrange? 9. A child has 12 blocks, of which 6 are black, 4 are red, 1 is white, and 1 is blue. If the child puts the blocks in a line, how many arrangements are possible?

10. In how many ways can 8 people be seated in a row if (a) there are no restrictions on the seating arrangement? (b) persons A and B must sit next to each other? (c) there are 4 men and 4 women and no 2 men or 2 women can sit next to each other? (d) there are 5 men and they must sit next to each other? (e) there are 4 married couples and each couple must sit together? 11. In how many ways can 3 novels, 2 mathematics books, and 1 chemistry book be arranged on a bookshelf if (a) the books can be arranged in any order? (b) the mathematics books must be together and the novels must be together? (c) the novels must be together, but the other books can be arranged in any order? 12. Five separate awards (best scholarship, best leadership qualities, and so on) are to be presented to selected students from a class of 30. How many different outcomes are possible if (a) a student can receive any number of awards? (b) each student can receive at most 1 award? 13. Consider a group of 20 people. If everyone shakes hands with everyone else, how many handshakes take place? 14. How many 5-card poker hands are there? 15. A dance class consists of 22 students, of which 10 are women and 12 are men. If 5 men and 5 women are to be chosen and then paired off, how many results are possible? 16. A student has to sell 2 books from a collection of 6 math, 7 science, and 4 economics books. How many choices are possible if (a) both books are to be on the same subject? (b) the books are to be on different subjects? 17. Seven different gifts are to be distributed among 10 children. How many distinct results are possible if no child is to receive more than one gift? 18. A committee of 7, consisting of 2 Republicans, 2 Democrats, and 3 Independents, is to be chosen from a group of 5 Republicans, 6 Democrats, and 4 Independents. How many committees are possible? 19. From a group of 8 women and 6 men, a committee consisting of 3 men and 3 women is to be formed. How many different committees are possible if (a) 2 of the men refuse to serve together? (b) 2 of the women refuse to serve together? (c) 1 man and 1 woman refuse to serve together?

Problems 20. A person has 8 friends, of whom 5 will be invited to a party. (a) How many choices are there if 2 of the friends are feuding and will not attend together? (b) How many choices if 2 of the friends will only attend together? 21. Consider the grid of points shown here. Suppose that, starting at the point labeled A, you can go one step up or one step to the right at each move. This procedure is continued until the point labeled B is reached. How many different paths from A to B are possible? Hint: Note that to reach B from A, you must take 4 steps to the right and 3 steps upward.

24. 25.

26. 27.

28.

29.

B

30.

A

22. In Problem 21, how many different paths are there from A to B that go through the point circled in the following lattice?

∗ 31.

∗ 32.

B

∗ 33.

A

23. A psychology laboratory conducting dream research contains 3 rooms, with 2 beds in each room. If 3 sets of identical twins are to be assigned to these 6 beds so that each set of twins sleeps

17

in different beds in the same room, how many assignments are possible? Expand (3x2 + y)5 . The game of bridge is played by 4 players, each of whom is dealt 13 cards. How many bridge deals are possible? Expand (x1 + 2x2 + 3x3 )4 . If 12 people are to be divided into 3 committees of respective sizes 3, 4, and 5, how many divisions are possible? If 8 new teachers are to be divided among 4 schools, how many divisions are possible? What if each school must receive 2 teachers? Ten weight lifters are competing in a team weightlifting contest. Of the lifters, 3 are from the United States, 4 are from Russia, 2 are from China, and 1 is from Canada. If the scoring takes account of the countries that the lifters represent, but not their individual identities, how many different outcomes are possible from the point of view of scores? How many different outcomes correspond to results in which the United States has 1 competitor in the top three and 2 in the bottom three? Delegates from 10 countries, including Russia, France, England, and the United States, are to be seated in a row. How many different seating arrangements are possible if the French and English delegates are to be seated next to each other and the Russian and U.S. delegates are not to be next to each other? If 8 identical blackboards are to be divided among 4 schools, how many divisions are possible? How many if each school must receive at least 1 blackboard? An elevator starts at the basement with 8 people (not including the elevator operator) and discharges them all by the time it reaches the top ﬂoor, number 6. In how many ways could the operator have perceived the people leaving the elevator if all people look alike to him? What if the 8 people consisted of 5 men and 3 women and the operator could tell a man from a woman? We have 20 thousand dollars that must be invested among 4 possible opportunities. Each investment must be integral in units of 1 thousand dollars, and there are minimal investments that need to be made if one is to invest in these opportunities. The minimal investments are 2, 2, 3, and 4 thousand dollars. How many different investment strategies are available if (a) an investment must be made in each opportunity? (b) investments must be made in at least 3 of the 4 opportunities?

18

Chapter 1

Combinatorial Analysis

THEORETICAL EXERCISES 1. Prove the generalized version of the basic counting principle. 2. Two experiments are to be performed. The ﬁrst can result in any one of m possible outcomes. If the ﬁrst experiment results in outcome i, then the second experiment can result in any of ni possible outcomes, i = 1, 2, . . . , m. What is the number of possible outcomes of the two experiments? 3. In how many ways can r objects be selected from a set of n objects if the order of selection is considered relevant? n 4. There are different linear arrangements of n r balls of which r are black and n − r are white. Give a combinatorial explanation of this fact. 5. Determine the number of vectors (x1 , . . . , xn ), such that each xi is either 0 or 1 and n

xi Ú k

6. How many vectors x1 , . . . , xk are there for which each xi is a positive integer such that 1 … xi … n and x1 < x2 < · · · < xk ? 7. Give an analytic proof of Equation (4.1). 8. Prove that n + m r

=

n 0

m r

+··· +

+

n r

n 1

m 0

m r − 1

Hint: Consider a group of n men and m women. How many groups of size r are possible? 9. Use Theoretical Exercise 8 to prove that

2n n

=

k

n k

2 n n k k=0

10. From a group of n people, suppose that we want to choose a committee of k, k … n, one of whom is to be designated as chairperson. (a) By focusing ﬁrst on the choice of the committee and then on the choice of the chair, argue n that there are k possible choices. k (b) By focusing ﬁrst on the choice of the nonchair committee members and then on

= (n − k + 1)

n k − 1

=n

(e) Use the factorial deﬁnition of

m r

n − 1 k − 1

to verify

the identity in part (d). 11. The following identity is known as Fermat’s combinatorial identity:

i=1

the choice of the chair, argue that there are n (n − k + 1) possible choices. k − 1 (c) By focusing ﬁrst on the choice of the chair and then on the choice of the other committee n − 1 members, argue that there are n k − 1 possible choices. (d) Conclude from parts (a), (b), and (c) that

n k

=

n i − 1 k − 1

n Ú k

i=k

Give a combinatorial argument (no computations are needed) to establish this identity. Hint: Consider the set of numbers 1 through n. How many subsets of size k have i as their highestnumbered member? 12. Consider the following combinatorial identity: n n k = n · 2n−1 k k=1

(a) Present a combinatorial argument for this identity by considering a set of n people and determining, in two ways, the number of possible selections of a committee of any size and a chairperson for the committee. Hint: (i) How many possible selections are there of a committee of size k and its chairperson? (ii) How many possible selections are there of a chairperson and the other committee members? (b) Verify the following identity for n = 1, 2, 3, 4, 5: n n k2 = 2n−2 n(n + 1) k k=1

Theoretical Exercises For a combinatorial proof of the preceding, consider a set of n people and argue that both sides of the identity represent the number of different selections of a committee, its chairperson, and its secretary (possibly the same as the chairperson). Hint: (i) How many different selections result in the committee containing exactly k people? (ii) How many different selections are there in which the chairperson and the secretary are the same? (ANSWER: n2n−1 .) (iii) How many different selections result in the chairperson and the secretary being different? (c) Now argue that n n k3 = 2n−3 n2 (n + 3) k k=1

13. Show that, for n > 0, n n (−1)i =0 i i=0

Hint: Use the binomial theorem. 14. From a set of n people, a committee of size j is to be chosen, and from this committee, a subcommittee of size i, i … j, is also to be chosen. (a) Derive a combinatorial identity by computing, in two ways, the number of possible choices of the committee and subcommittee— ﬁrst by supposing that the committee is chosen ﬁrst and then the subcommittee is chosen, and second by supposing that the subcommittee is chosen ﬁrst and then the remaining members of the committee are chosen. (b) Use part (a) to prove the following combinatorial identity: n n j n = 2n−i j i i

i … n

j=i

(c) Use part (a) and Theoretical Exercise 13 to show that n n j (−1)n−j = 0 i < n j i j=i

19

15. Let Hk (n) be the number of vectors x1 , . . . , xk for which each xi is a positive integer satisfying 1 … xi … n and x1 … x2 … · · · … xk . (a) Without any computations, argue that H1 (n) = n n Hk (n) = Hk−1 (j)

k > 1

j=1

Hint: How many vectors are there in which xk = j? (b) Use the preceding recursion to compute H3 (5). Hint: First compute H2 (n) for n = 1, 2, 3, 4, 5. 16. Consider a tournament of n contestants in which the outcome is an ordering of these contestants, with ties allowed. That is, the outcome partitions the players into groups, with the ﬁrst group consisting of the players that tied for ﬁrst place, the next group being those that tied for the next-best position, and so on. Let N(n) denote the number of different possible outcomes. For instance, N(2) = 3, since, in a tournament with 2 contestants, player 1 could be uniquely ﬁrst, player 2 could be uniquely ﬁrst, or they could tie for ﬁrst. (a) List all the possible outcomes when n = 3. (b) With N(0) deﬁned to equal 1, argue, without any computations, that

N(n) =

n n N(n − i) i i=1

Hint: How many outcomes are there in which i players tie for last place? (c) Show that the formula of part (b) is equivalent to the following:

N(n) =

n−1 n N(i) i i=0

(d) Use the recursion to ﬁnd N(3) and N(4). 17. Present explanation of why a combinatorial n n = . r r, n − r

20

Chapter 1

Combinatorial Analysis

18. Argue that

n n1 , n 2 , . . . , n r

=

n − 1 n1 − 1, n2 , . . . , nr n − 1 + ··· + n1 , n2 − 1, . . . , nr n − 1 + n1 , n 2 , . . . , n r − 1

Hint: Use an argument similar to the one used to establish Equation (4.1). 19. Prove the multinomial theorem. ∗ 20. In how many ways can n identical balls be distributed into r urns so that the ith urn contains at least mi balls, for each i = 1, . . . , r? Assume that n Ú ri=1 mi .

∗ 21.

Argue that there are exactly

r k

n − 1 n − r + k

solutions of x1 + x2 + · · · + x r = n for which exactly k of the xi are equal to 0. Consider a function f (x1 , . . . , xn ) of n variables. How many different partial derivatives of order r does f possess? ∗ 23. Determine the number of vectors (x , . . . , x ) such n 1 that each xi is a nonnegative integer and ∗ 22.

n

xi … k

i=1

SELF-TEST PROBLEMS AND EXERCISES 1. How many different linear arrangements are there of the letters A, B, C, D, E, F for which (a) A and B are next to each other? (b) A is before B? (c) A is before B and B is before C? (d) A is before B and C is before D? (e) A and B are next to each other and C and D are also next to each other? (f) E is not last in line? 2. If 4 Americans, 3 French people, and 3 British people are to be seated in a row, how many seating arrangements are possible when people of the same nationality must sit next to each other? 3. A president, treasurer, and secretary, all different, are to be chosen from a club consisting of 10 people. How many different choices of ofﬁcers are possible if (a) there are no restrictions? (b) A and B will not serve together? (c) C and D will serve together or not at all? (d) E must be an ofﬁcer? (e) F will serve only if he is president? 4. A student is to answer 7 out of 10 questions in an examination. How many choices has she? How many if she must answer at least 3 of the ﬁrst 5 questions? 5. In how many ways can a man divide 7 gifts among his 3 children if the eldest is to receive 3 gifts and the others 2 each? 6. How many different 7-place license plates are possible when 3 of the entries are letters and 4 are digits? Assume that repetition of letters and numbers is allowed and that there is no restriction on where the letters or numbers can be placed.

7. Give a combinatorial explanation of the identity

n r

=

n n − r

8. Consider n-digit numbers where each digit is one of the 10 integers 0, 1, . . . , 9. How many such numbers are there for which (a) no two consecutive digits are equal? (b) 0 appears as a digit a total of i times, i = 0, . . . , n? 9. Consider three classes, each consisting of n students. From this group of 3n students, a group of 3 students is to be chosen. (a) How many choices are possible? (b) How many choices are there in which all 3 students are in the same class? (c) How many choices are there in which 2 of the 3 students are in the same class and the other student is in a different class? (d) How many choices are there in which all 3 students are in different classes? (e) Using the results of parts (a) through (d), write a combinatorial identity. 10. How many 5-digit numbers can be formed from the integers 1, 2, . . . , 9 if no digit can appear more than twice? (For instance, 41434 is not allowed.) 11. From 10 married couples, we want to select a group of 6 people that is not allowed to contain a married couple. (a) How many choices are there? (b) How many choices are there if the group must also consist of 3 men and 3 women?

Self-Test Problems and Exercises 12. A committee of 6 people is to be chosen from a group consisting of 7 men and 8 women. If the committee must consist of at least 3 women and at least 2 men, how many different committees are possible? ∗ 13. An art collection on auction consisted of 4 Dalis, 5 van Goghs, and 6 Picassos. At the auction were 5 art collectors. If a reporter noted only the number of Dalis, van Goghs, and Picassos acquired by each collector, how many different results could have been recorded if all of the works were sold? ∗ 14. Determine the number of vectors (x , . . . , x ) such n 1 that each xi is a positive integer and n

xi … k

i=1

where k Ú n. 15. A total of n students are enrolled in a review course for the actuarial examination in probability. The posted results of the examination will list the names of those who passed, in decreasing order of their scores. For instance, the posted result will be “Brown, Cho” if Brown and Cho are the only ones to pass, with Brown receiving the higher score.

21

Assuming that all scores are distinct (no ties), how many posted results are possible? 16. How many subsets of size 4 of the set S = {1, 2, . . . , 20} contain at least one of the elements 1, 2, 3, 4, 5? 17. Give an analytic veriﬁcation of n k n − k = + k(n − k) + , 2 2 2

1…k…n

Now, give a combinatorial argument for this identity. 18. In a certain community, there are 3 families consisting of a single parent and 1 child, 3 families consisting of a single parent and 2 children, 5 families consisting of 2 parents and a single child, 7 families consisting of 2 parents and 2 children, and 6 families consisting of 2 parents and 3 children. If a parent and child from the same family are to be chosen, how many possible choices are there? 19. If there are no restrictions on where the digits and letters are placed, how many 8-place license plates consisting of 5 letters and 3 digits are possible if no repetitions of letters or digits are allowed. What if the 3 digits must be consecutive?

C H A P T E R

2

Axioms of Probability 2.1 2.2 2.3 2.4 2.5 2.6 2.7

INTRODUCTION SAMPLE SPACE AND EVENTS AXIOMS OF PROBABILITY SOME SIMPLE PROPOSITIONS SAMPLE SPACES HAVING EQUALLY LIKELY OUTCOMES PROBABILITY AS A CONTINUOUS SET FUNCTION PROBABILITY AS A MEASURE OF BELIEF

2.1

INTRODUCTION In this chapter, we introduce the concept of the probability of an event and then show how probabilities can be computed in certain situations. As a preliminary, however, we need the concept of the sample space and the events of an experiment.

2.2

SAMPLE SPACE AND EVENTS Consider an experiment whose outcome is not predictable with certainty. However, although the outcome of the experiment will not be known in advance, let us suppose that the set of all possible outcomes is known. This set of all possible outcomes of an experiment is known as the sample space of the experiment and is denoted by S. Following are some examples: 1. If the outcome of an experiment consists in the determination of the sex of a newborn child, then S = {g, b} where the outcome g means that the child is a girl and b that it is a boy. 2. If the outcome of an experiment is the order of ﬁnish in a race among the 7 horses having post positions 1, 2, 3, 4, 5, 6, and 7, then S = {all 7! permutations of (1, 2, 3, 4, 5, 6, 7)} The outcome (2, 3, 1, 6, 5, 4, 7) means, for instance, that the number 2 horse comes in ﬁrst, then the number 3 horse, then the number 1 horse, and so on. 3. If the experiment consists of ﬂipping two coins, then the sample space consists of the following four points: S = {(H, H), (H, T), (T, H), (T, T)} The outcome will be (H, H) if both coins are heads, (H, T) if the ﬁrst coin is heads and the second tails, (T, H) if the ﬁrst is tails and the second heads, and (T, T) if both coins are tails.

22

Section 2.2

Sample Space and Events

23

4. If the experiment consists of tossing two dice, then the sample space consists of the 36 points S = {(i, j): i, j = 1, 2, 3, 4, 5, 6} where the outcome (i, j) is said to occur if i appears on the leftmost die and j on the other die. 5. If the experiment consists of measuring (in hours) the lifetime of a transistor, then the sample space consists of all nonnegative real numbers; that is, S = {x: 0 … x < q} Any subset E of the sample space is known as an event. In other words, an event is a set consisting of possible outcomes of the experiment. If the outcome of the experiment is contained in E, then we say that E has occurred. Following are some examples of events. In the preceding Example 1, if E = {g}, then E is the event that the child is a girl. Similarly, if F = {b}, then F is the event that the child is a boy. In Example 2, if E = {all outcomes in S starting with a 3} then E is the event that horse 3 wins the race. In Example 3, if E = {(H, H), (H, T)}, then E is the event that a head appears on the ﬁrst coin. In Example 4, if E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}, then E is the event that the sum of the dice equals 7. In Example 5, if E = {x: 0 … x … 5}, then E is the event that the transistor does not last longer than 5 hours. For any two events E and F of a sample space S, we deﬁne the new event E ∪ F to consist of all outcomes that are either in E or in F or in both E and F. That is, the event E ∪ F will occur if either E or F occurs. For instance, in Example 1, if event E = {g} and F = {b}, then E ∪ F = {g, b} That is, E ∪ F is the whole sample space S. In Example 3, if E = {(H, H), (H, T)} and F = {(T, H)}, then E ∪ F = {(H, H), (H, T), (T, H)} Thus, E ∪ F would occur if a head appeared on either coin. The event E ∪ F is called the union of the event E and the event F. Similarly, for any two events E and F, we may also deﬁne the new event EF, called the intersection of E and F, to consist of all outcomes that are both in E and in F. That is, the event EF (sometimes written E ∩ F) will occur only if both E and F occur. For instance, in Example 3, if E = {(H, H), (H, T), (T, H)} is the event that at least 1 head occurs and F = {(H, T), (T, H), (T, T)} is the event that at least 1 tail occurs, then EF = {(H, T), (T, H)} is the event that exactly 1 head and 1 tail occur. In example 4, if E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} is the event that the sum of the dice is 7 and F = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)} is the event that the sum is 6, then the event EF does not contain

24

Chapter 2

Axioms of Probability

any outcomes and hence could not occur. To give such an event a name, we shall refer to it as the null event and denote it by Ø. (That is, Ø refers to the event consisting of no outcomes.) If EF = Ø, then E and F are said to be mutually exclusive. We deﬁne unions and intersections of more than two events in a similar manner. q

En , is deﬁned If E1 , E2 , . . . are events, then the union of these events, denoted by n=1

to be that event which consists of all outcomes that are in En for at least one value q En , is of n = 1, 2, . . .. Similarly, the intersection of the events En , denoted by n=1

deﬁned to be the event consisting of those outcomes which are in all of the events En , n = 1, 2, . . .. Finally, for any event E, we deﬁne the new event Ec , referred to as the complement of E, to consist of all outcomes in the sample space S that are not in E. That is, Ec will occur if and only if E does not occur. In Example 4, if event E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}, then Ec will occur when the sum of the dice does not equal 7. Note that because the experiment must result in some outcome, it follows that Sc = Ø. For any two events E and F, if all of the outcomes in E are also in F, then we say that E is contained in F, or E is a subset of F, and write E ( F (or equivalently, F ) E, which we sometimes say as F is a superset of E). Thus, if E ( F, then the occurrence of E implies the occurrence of F. If E ( F and F ( E, we say that E and F are equal and write E = F. A graphical representation that is useful for illustrating logical relations among events is the Venn diagram. The sample space S is represented as consisting of all the outcomes in a large rectangle, and the events E, F, G, . . . are represented as consisting of all the outcomes in given circles within the rectangle. Events of interest can then be indicated by shading appropriate regions of the diagram. For instance, in the three Venn diagrams shown in Figure 2.1, the shaded areas represent, respectively, the events E ∪ F, EF, and Ec . The Venn diagram in Figure 2.2 indicates that E ( F.

S E

F

S E

(a) Shaded region: E 傼 F.

F

(b) Shaded region: EF. S E

(c) Shaded region: E c. FIGURE 2.1: Venn Diagrams

Section 2.2

Sample Space and Events

25

S F E

FIGURE 2.2: E ( F

The operations of forming unions, intersections, and complements of events obey certain rules similar to the rules of algebra. We list a few of these rules: Commutative laws E∪F = F ∪E EF = FE Associative laws (E ∪ F) ∪ G = E ∪ (F ∪ G) (EF)G = E(FG) Distributive laws (E ∪ F)G = EG ∪ FG EF ∪ G = (E ∪ G)(F ∪ G) These relations are veriﬁed by showing that any outcome that is contained in the event on the left side of the equality sign is also contained in the event on the right side, and vice versa. One way of showing this is by means of Venn diagrams. For instance, the distributive law may be veriﬁed by the sequence of diagrams in Figure 2.3.

E

F

G (a) Shaded region: EG. E

E

F

G (b) Shaded region: FG. F

G (c) Shaded region: (E 傼 F )G. FIGURE 2.3: (E ∪ F)G = EG ∪ FG

26

Chapter 2

Axioms of Probability

The following useful relationships between the three basic operations of forming unions, intersections, and complements are known as DeMorgan’s laws: ⎛ ⎞c n n ⎝ Ei ⎠ = Eic ⎛ ⎝

i=1 n

i=1

⎞c Ei ⎠ =

i=1

n

Eic

i=1

To prove DeMorgan’s laws, suppose ﬁrst that x is an outcome of x is not contained in

n

n

c Ei

. Then

i=1

Ei , which means that x is not contained in any of the events

i=1

Ei , i = 1, 2, . . . , n, implying that x is contained in Eic for all i = 1, 2, . . . , n and thus is n n Eic . To go the other way, suppose that x is an outcome of Eic . Then contained in i=1

i=1

x is contained in Eic for all i = 1, 2, . . . , n, which means that x is not contained in Ei for n

any i = 1, 2, . . . , n, implying that x is not contained in Ei , in turn implying that x is i c n

Ei . This proves the ﬁrst of DeMorgan’s laws. contained in 1

To prove the second of DeMorgan’s laws, we use the ﬁrst law to obtain ⎛ ⎞c n n ⎝ Ec ⎠ = (Eic )c i i=1

which, since

(Ec )c

i=1

= E, is equivalent to ⎞c ⎛ n n ⎝ Ec ⎠ = Ei i 1

1

Taking complements of both sides of the preceding equation yields the result we seek, namely, ⎞c ⎛ n n Ec = ⎝ Ei ⎠ i

1

2.3

1

AXIOMS OF PROBABILITY One way of deﬁning the probability of an event is in terms of its relative frequency. Such a deﬁnition usually goes as follows: We suppose that an experiment, whose sample space is S, is repeatedly performed under exactly the same conditions. For each event E of the sample space S, we deﬁne n(E) to be the number of times in the ﬁrst n repetitions of the experiment that the event E occurs. Then P(E), the probability of the event E, is deﬁned as n(E) P(E) = lim n→q n

Section 2.3

Axioms of Probability

27

That is, P(E) is deﬁned as the (limiting) proportion of time that E occurs. It is thus the limiting frequency of E. Although the preceding deﬁnition is certainly intuitively pleasing and should always be kept in mind by the reader, it possesses a serious drawback: How do we know that n(E)/n will converge to some constant limiting value that will be the same for each possible sequence of repetitions of the experiment? For example, suppose that the experiment to be repeatedly performed consists of ﬂipping a coin. How do we know that the proportion of heads obtained in the ﬁrst n ﬂips will converge to some value as n gets large? Also, even if it does converge to some value, how do we know that, if the experiment is repeatedly performed a second time, we shall obtain the same limiting proportion of heads? Proponents of the relative frequency deﬁnition of probability usually answer this objection by stating that the convergence of n(E)/n to a constant limiting value is an assumption, or an axiom, of the system. However, to assume that n(E)/n will necessarily converge to some constant value seems to be an extraordinarily complicated assumption. For, although we might indeed hope that such a constant limiting frequency exists, it does not at all seem to be a priori evident that this need be the case. In fact, would it not be more reasonable to assume a set of simpler and more self-evident axioms about probability and then attempt to prove that such a constant limiting frequency does in some sense exist? The latter approach is the modern axiomatic approach to probability theory that we shall adopt in this text. In particular, we shall assume that, for each event E in the sample space S, there exists a value P(E), referred to as the probability of E. We shall then assume that all these probabilities satisfy a certain set of axioms, which, we hope the reader will agree, is in accordance with our intuitive notion of probability. Consider an experiment whose sample space is S. For each event E of the sample space S, we assume that a number P(E) is deﬁned and satisﬁes the following three axioms: Axiom 1 0 … P(E) … 1 Axiom 2 P(S) = 1 Axiom 3 For any sequence of mutually exclusive events E1 , E2 , . . . (that is, events for which Ei Ej = Ø when i Z j), ⎞ ⎛ q q ⎠ ⎝ Ei = P(Ei ) P i=1

i=1

We refer to P(E) as the probability of the event E. Thus, Axiom 1 states that the probability that the outcome of the experiment is an outcome in E is some number between 0 and 1. Axiom 2 states that, with probability 1, the outcome will be a point in the sample space S. Axiom 3 states that, for any sequence of mutually exclusive events, the probability of at least one of these events occurring is just the sum of their respective probabilities.

28

Chapter 2

Axioms of Probability

If we consider a sequence of events E1 , E2 , . . ., where E1 = S and Ei = Ø for i > 1, q

Ei , we have, from then, because the events are mutually exclusive and because S = i=1

Axiom 3, P(S) =

q

P(Ei ) = P(S) +

q

P(Ø)

i=2

i=1

implying that

P(Ø) = 0

That is, the null event has probability 0 of occurring. Note that it follows that, for any ﬁnite sequence of mutually exclusive events E1 , E2 , . . . , En , ⎞ ⎛ n n P(Ei ) (3.1) P ⎝ Ei ⎠ = 1

i=1

This equation follows from Axiom 3 by deﬁning Ei as the null event for all values of i greater than n. Axiom 3 is equivalent to Equation (3.1) when the sample space is ﬁnite. (Why?) However, the added generality of Axiom 3 is necessary when the sample space consists of an inﬁnite number of points. EXAMPLE 3a If our experiment consists of tossing a coin and if we assume that a head is as likely to appear as a tail, then we would have P({H}) = P({T}) =

1 2

On the other hand, if the coin were biased and we felt that a head were twice as likely to appear as a tail, then we would have P({H}) =

2 3

P({T}) =

1 3

.

EXAMPLE 3b If a die is rolled and we suppose that all six sides are equally likely to appear, then we would have P({1}) = P({2}) = P({3}) = P({4}) = P({5}) = P({6}) = 16 . From Axiom 3, it would thus follow that the probability of rolling an even number would equal 1 . P({2, 4, 6}) = P({2}) + P({4}) + P({6}) = 2 The assumption of the existence of a set function P, deﬁned on the events of a sample space S and satisfying Axioms 1, 2, and 3, constitutes the modern mathematical approach to probability theory. Hopefully, the reader will agree that the axioms are natural and in accordance with our intuitive concept of probability as related to chance and randomness. Furthermore, using these axioms we shall be able to prove that if an experiment is repeated over and over again, then, with probability 1, the proportion of time during which any speciﬁc event E occurs will equal P(E). This result, known as the strong law of large numbers, is presented in Chapter 8. In addition, we present another possible interpretation of probability—as being a measure of belief—in Section 2.7.

Section 2.4

Some Simple Propositions

29

Technical Remark. We have supposed that P(E) is deﬁned for all the events E of the sample space. Actually, when the sample space is an uncountably inﬁnite set, P(E) is deﬁned only for a class of events called measurable. However, this restriction need not concern us, as all events of any practical interest are measurable. 2.4

SOME SIMPLE PROPOSITIONS In this section, we prove some simple propositions regarding probabilities. We ﬁrst note that, since E and Ec are always mutually exclusive and since E ∪ Ec = S, we have, by Axioms 2 and 3, 1 = P(S) = P(E ∪ Ec ) = P(E) + P(Ec ) Or, equivalently, we have Proposition 4.1. Proposition 4.1.

P(Ec ) = 1 − P(E)

In words, Proposition 4.1 states that the probability that an event does not occur is 1 minus the probability that it does occur. For instance, if the probability of obtaining a head on the toss of a coin is 38 , then the probability of obtaining a tail must be 58 . Our second proposition states that if the event E is contained in the event F, then the probability of E is no greater than the probability of F. Proposition 4.2. If E ( F, then P(E) … P(F). Proof. Since E ( F, it follows that we can express F as F = E ∪ Ec F Hence, because E and Ec F are mutually exclusive, we obtain, from Axiom 3, P(F) = P(E) + P(Ec F) which proves the result, since P(Ec F) Ú 0. Proposition 4.2 tells us, for instance, that the probability of rolling a 1 with a die is less than or equal to the probability of rolling an odd value with the die. The next proposition gives the relationship between the probability of the union of two events, expressed in terms of the individual probabilities, and the probability of the intersection of the events. Proposition 4.3.

P(E ∪ F) = P(E) + P(F) − P(EF)

Proof. To derive a formula for P(E ∪ F), we ﬁrst note that E ∪ F can be written as the union of the two disjoint events E and Ec F. Thus, from Axiom 3, we obtain P(E ∪ F) = P(E ∪ Ec F) = P(E) + P(Ec F) Furthermore, since F = EF ∪ Ec F, we again obtain from Axiom 3 P(F) = P(EF) + P(Ec F)

30

Chapter 2

Axioms of Probability

E

F

FIGURE 2.4: Venn Diagram

E

F

I

II

III

FIGURE 2.5: Venn Diagram in Sections

or, equivalently,

P(Ec F) = P(F) − P(EF)

thereby completing the proof. Proposition 4.3 could also have been proved by making use of the Venn diagram in Figure 2.4. Let us divide E ∪ F into three mutually exclusive sections, as shown in Figure 2.5. In words, section I represents all the points in E that are not in F (that is, EF c ), section II represents all points both in E and in F (that is, EF), and section III represents all points in F that are not in E (that is, Ec F). From Figure 2.5, we see that E ∪ F = I ∪ II ∪ III E = I ∪ II F = II ∪ III As I, II, and III are mutually exclusive, it follows from Axiom 3 that P(E ∪ F) = P(I) + P(II) + P(III) P(E) = P(I) + P(II) P(F) = P(II) + P(III) which shows that

P(E ∪ F) = P(E) + P(F) − P(II)

and Proposition 4.3 is proved, since II = EF. EXAMPLE 4a J is taking two books along on her holiday vacation. With probability .5, she will like the ﬁrst book; with probability .4, she will like the second book; and with probability .3, she will like both books. What is the probability that she likes neither book?

Section 2.4

Some Simple Propositions

31

Solution. Let Bi denote the event that J likes book i, i = 1, 2. Then the probability that she likes at least one of the books is P(B1 ∪ B2 ) = P(B1 ) + P(B2 ) − P(B1 B2 ) = .5 + .4 − .3 = .6 Because the event that J likes neither book is the complement of the event that she likes at least one of them, we obtain the result . P(Bc1 Bc2 ) = P (B1 ∪ B2 )c = 1 − P(B1 ∪ B2 ) = .4 We may also calculate the probability that any one of the three events E, F, and G occurs, namely, P(E ∪ F ∪ G) = P[(E ∪ F) ∪ G] which, by Proposition 4.3, equals P(E ∪ F) + P(G) − P[(E ∪ F)G] Now, it follows from the distributive law that the events (E ∪ F)G and EG ∪ FG are equivalent; hence, from the preceding equations, we obtain P(E ∪ F ∪ G) = P(E) + P(F) − P(EF) + P(G) − P(EG ∪ FG) = P(E) + P(F) − P(EF) + P(G) − P(EG) − P(FG) + P(EGFG) = P(E) + P(F) + P(G) − P(EF) − P(EG) − P(FG) + P(EFG) In fact, the following proposition, known as the inclusion–exclusion identity, can be proved by mathematical induction: Proposition 4.4. P(E1 ∪ E2 ∪ · · · ∪ En ) =

n

P(Ei ) −

P(Ei1 Ei2 ) + · · ·

i1

i=1

+ (−1)

r+1

P(Ei1 Ei2 · · · Eir )

i1

P(E1 E2 · · · En ) n P(Ei1 Ei2 · · · Eir ) is taken over all of the possible subThe summation r i1

In words, Proposition 4.4 states that the probability of the union of n events equals the sum of the probabilities of these events taken one at a time, minus the sum of the probabilities of these events taken two at a time, plus the sum of the probabilities of these events taken three at a time, and so on. Remarks. 1. For a noninductive argument for Proposition 4.4, note ﬁrst that if an outcome of the sample space is not a member of any of the sets Ei , then its probability does not contribute anything to either side of the equality. Now, suppose that an outcome is in exactly m of the events Ei , where m > 0. Then, since it is in Ei , its i

32

Chapter 2

Axioms of Probability

probability is counted once in P

Ei ; also, as this outcome is contained in

i

m k

subsets of the type Ei1 Ei2 · · · Eik , its probability is counted

m 1

−

m 2

+

m 3

− ··· ;

m m

times on the right of the equality sign in Proposition 4.4. Thus, for m > 0, we must show that m m m m 1= − + − ··· ; 1 2 3 m However, since 1 =

m , the preceding equation is equivalent to 0 m m i

i=0

(−1)i = 0

and the latter equation follows from the binomial theorem, since m

0 = (−1 + 1) =

m m i

i=0

(−1)i (1)m−i

2. The following is a succinct way of writing the inclusion–exclusion identity: P(∪ni=1 Ei ) =

n

(−1)r+1

P(Ei1 · · · Eir )

i1 <···

r=1

3. In the inclusion–exclusion identity, going out one term results in an upper bound on the probability of the union, going out two terms results in a lower bound on the probability, going out three terms results in an upper bound on the probability, going out four terms results in a lower bound, and so on. That is, for events E1 , . . . , En , we have P(∪ni=1 Ei ) … P(∪ni=1 Ei ) Ú P(∪ni=1 Ei ) …

n i=1 n i=1 n i=1

P(Ei ) P(Ei ) −

(4.1)

P(Ei Ej )

(4.2)

j

P(Ei ) −

j

P(Ei Ej ) +

P(Ei Ej Ek )

k

and so on. To prove the validity of these bounds, note the identity c En ∪ni=1 Ei = E1 ∪ E1c E2 ∪ E1c E2c E3 ∪ · · · ∪ E1c · · · En−1

(4.3)

Section 2.5

Sample Spaces Having Equally Likely Outcomes

33

That is, at least one of the events Ei occurs if E1 occurs, or if E1 does not occur but E2 does, or if E1 and E2 do not occur but E3 does, and so on. Because the right-hand side is the union of disjoint events, we obtain c P(∪ni=1 Ei ) = P(E1 ) + P(E1c E2 ) + P(E1c E2c E3 ) + . . . + P(E1c · · · En−1 En ) n c = P(E1 ) + P(E1c · · · Ei−1 Ei ) (4.4) i=2 c Now, let Bi = E1c · · · Ei−1 = (∪j

P(Ei ) = P(Bi Ei ) + P(Bci Ei ) shows that or, equivalently,

c Ei ) + P(Ei ∪j

Substituting this equation into (4.4) yields P(∪ni=1 Ei ) = P(Ei ) − P(∪j

(4.5)

i

Because probabilities are always nonnegative, Inequality (4.1) follows directly from Equation (4.5). Now, ﬁxing i and applying Inequality (4.1) to P(∪j

which, by Equation (4.5), gives Inequality (4.2). Similarly, ﬁxing i and applying Inequality (4.2) to P(∪j

=

k

P(Ei Ej ) −

j

P(Ei Ej Ek )

k

which, by Equation (4.5), gives Inequality (4.3). The next inclusion–exclusion inequality is now obtained by ﬁxing i and applying Inequality (4.3) to P(∪j

SAMPLE SPACES HAVING EQUALLY LIKELY OUTCOMES In many experiments, it is natural to assume that all outcomes in the sample space are equally likely to occur. That is, consider an experiment whose sample space S is a ﬁnite set, say, S = {1, 2, . . . , N}. Then it is often natural to assume that P({1}) = P({2}) = · · · = P({N}) which implies, from Axioms 2 and 3 (why?), that P({i}) =

1 N

i = 1, 2, . . . , N

34

Chapter 2

Axioms of Probability

From this equation, it follows from Axiom 3 that, for any event E, P(E) =

number of outcomes in E number of outcomes in S

In words, if we assume that all outcomes of an experiment are equally likely to occur, then the probability of any event E equals the proportion of outcomes in the sample space that are contained in E. EXAMPLE 5a If two dice are rolled, what is the probability that the sum of the upturned faces will equal 7? Solution. We shall solve this problem under the assumption that all of the 36 possible outcomes are equally likely. Since there are 6 possible outcomes—namely, (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1)—that result in the sum of the dice being equal to 7, 6 = 16 . . the desired probability is 36 EXAMPLE 5b If 3 balls are “randomly drawn” from a bowl containing 6 white and 5 black balls, what is the probability that one of the balls is white and the other two black? Solution. If we regard the order in which the balls are selected as being relevant, then the sample space consists of 11 · 10 · 9 = 990 outcomes. Furthermore, there are 6 · 5 · 4 = 120 outcomes in which the ﬁrst ball selected is white and the other two are black; 5 · 6 · 4 = 120 outcomes in which the ﬁrst is black, the second is white, and the third is black; and 5 · 4 · 6 = 120 in which the ﬁrst two are black and the third is white. Hence, assuming that “randomly drawn” means that each outcome in the sample space is equally likely to occur, we see that the desired probability is 4 120 + 120 + 120 = 990 11 This problem could also have been solved by regarding the outcome of the experi11 = ment as the unordered set of drawn balls. From this point of view, there are 3 165 outcomes in the sample space. Now, each set of 3 balls corresponds to 3! outcomes when the order of selection is noted. As a result, if all outcomes are assumed equally likely when the order of selection is noted, then it follows that they remain equally likely when the outcome is taken to be the unordered set of selected balls. Hence, using the latter representation of the experiment, we see that the desired probability is 5 6 2 1 4 = 11 11 3 which, of course, agrees with the answer obtained previously. When the experiment consists of a random selection of k items from a set of n items, we have the ﬂexibility of either letting the outcome of the experiment be the ordered selection of the k items or letting it be the unordered set of items selected. In the former case we would assume that each new selection is equally likely to be

Section 2.5

Sample Spaces Having Equally Likely Outcomes

35

any of the so far unselected items of the set, and in the latter case we would assume that all nk possible subsets of k items are equally likely to be the set selected. For instance, suppose 5 people are to be randomly selected from a group of 20 individuals consisting of 10 married couples, and we want to determine P(N), the probability that the 5 chosen are all unrelated. (That is, no two are married to each If other.) equally we regard the sample space as the set of 5 people chosen, then there are 20 5 likely outcomes. An outcome that does not contain a married couple can be thought of as being the result of a six-stage experiment: In the ﬁrst stage, 5 of the 10 couples to have a member in the group are chosen; in the next 55 stages, 1 of the 2 members of each of these couples is selected. Thus, there are 10 5 2 possible outcomes in which the 5 members selected are unrelated, yielding the desired probability of 10 5 2 5 P(N) = 20 5 In contrast, we could let the outcome of the experiment be the ordered selection of the 5 individuals. In this setting, there are 20 · 19 · 18 · 17 · 16 equally likely outcomes, of which 20 · 18 · 16 · 14 · 12 outcomes result in a group of 5 unrelated individuals, yielding the result 20 · 18 · 16 · 14 · 12 20 · 19 · 18 · 17 · 16 We leave it for the reader to verify that the two answers are identical. P(N) =

.

EXAMPLE 5c A committee of 5 is to be selected from a group of 6 men and 9 women. If the selection is made randomly, what is the probability that the committee consists of 3 men and 2 women? Solution. Because each of the 15 5 possible committees is equally likely to be selected, the desired probability is 9 6 2 3 240 = . 1001 15 5 EXAMPLE 5d An urn contains n balls, one of which is special. If k of these balls are withdrawn one at a time, with each selection being equally likely to be any of the balls that remain at the time, what is the probability that the special ball is chosen? Solution. Since all of the balls are treated in an identical manner, it follows that the n sets of k balls. Therefore, set of k balls selected is equally likely to be any of the k 1 n − 1 1 k − 1 k = P{special ball is selected} = n n k

36

Chapter 2

Axioms of Probability

We could also have obtained this result by letting Ai denote the event that the special ball is the ith ball to be chosen, i = 1, . . . , k. Then, since each one of the n balls is equally likely to be the ith ball chosen, it follows that P(Ai ) = 1/n. Hence, because these events are clearly mutually exclusive, we have ⎛ ⎞ k k k P(Ai ) = P{special ball is selected} = P ⎝ Ai ⎠ = n i=1

i=1

We could also have argued that P(Ai ) = 1/n, by noting that there are n(n − 1) · · · (n − k + 1) = n!/(n − k)! equally likely outcomes of the experiment, of which (n − 1)(n − 2) · · · (n − i + 1)(1)(n − i) · · · (n − k + 1) = (n − 1)!/(n − k)! result in the special ball being the ith one chosen. From this reasoning, it follows that P(Ai ) =

1 (n − 1)! = n! n

.

EXAMPLE 5e Suppose that n + m balls, of which n are red and m are blue, are arranged in a linear order in such a way that all (n + m)! possible orderings are equally likely. If we record the result of this experiment by listing only the colors of the successive balls, show that all the possible results remain equally likely. Solution. Consider any one of the (n + m)! possible orderings, and note that any permutation of the red balls among themselves and of the blue balls among themselves does not change the sequence of colors. As a result, every ordering of colorings corresponds to n! m! different orderings of the n + m balls, so every ordering of the n!m! colors has probability (n+m)! of occurring. For example, suppose that there are 2 red balls, numbered r1 , r2 , and 2 blue balls, numbered b1 , b2 . Then, of the 4! possible orderings, there will be 2! 2! orderings that result in any speciﬁed color combination. For instance, the following orderings result in the successive balls alternating in color, with a red ball ﬁrst: r1 , b1 , r2 , b2

r1 , b2 , r2 , b1

r2 , b1 , r1 , b2

r2 , b2 , r1 , b1

Therefore, each of the possible orderings of the colors has probability occurring.

4 24

=

1 6

of .

EXAMPLE 5f A poker hand consists of 5 cards. If the cards have distinct consecutive values and are not all of the same suit, we say that the hand is a straight. For instance, a hand consisting of the ﬁve of spades, six of spades, seven of spades, eight of spades, and nine of hearts is a straight. What is the probability that one is dealt a straight? 52 possible poker hands are equally Solution. We start by assuming that all 5 likely. To determine the number of outcomes that are straights, let us ﬁrst determine the number of possible outcomes for which the poker hand consists of an ace, two, three, four, and ﬁve (the suits being irrelevant). Since the ace can be any 1 of the 4 possible aces, and similarly for the two, three, four, and ﬁve, it follows that there are 45 outcomes leading to exactly one ace, two, three, four, and ﬁve. Hence, since in 4 of these outcomes all the cards will be of the same suit (such a hand is called a straight

Section 2.5

Sample Spaces Having Equally Likely Outcomes

37

ﬂush), it follows that there are 45 − 4 hands that make up a straight of the form ace, two, three, four, and ﬁve. Similarly, there are 45 − 4 hands that make up a straight of the form ten, jack, queen, king, and ace. Thus, there are 10(45 − 4) hands that are straights, and it follows that the desired probability is 10(45 − 4) L .0039 52 5

.

EXAMPLE 5g A 5-card poker hand is said to be a full house if it consists of 3 cards of the same denomination and 2 other cards of the same denomination (of course, different from the ﬁrst denomination). Thus, one kind of full house is three of a kind plus a pair. What is the probability that one is dealt a full house? 52 Solution. Again, we assume that all possible hands are equally likely. To 5 4 4 determine the number of possible full houses, we ﬁrst note that there are 2 3 different combinations of, say, 2 tens and 3 jacks. Because there are 13 different choices for the kind of pair and, after a pair has been chosen, there are 12 other choices for the denomination of the remaining 3 cards, it follows that the probability of a full house is 4 4 13 · 12 · 2 3 L .0014 . 52 5 EXAMPLE 5h In the game of bridge, the entire deck of 52 cards is dealt out to 4 players. What is the probability that (a) one of the players receives all 13 spades; (b) each player receives 1 ace? Solution. (a) Letting Ei be the event that hand i has all 13 spades, then 1 P(Ei ) = 52 ,

i = 1, 2, 3, 4

13

Because the events Ei , i = 1, 2, 3, 4, are mutually exclusive, the probability that one of the hands is dealt all 13 spades is P(∪4i=1 Ei ) =

4 i=1

52 P(Ei ) = 4/ L 6.3 * 10−12 13

(b) To determine the number of outcomes in which each of the distinct players 48 posreceives exactly 1 ace, put aside the aces and note that there are 12, 12, 12, 12 sible divisions of the other 48 cards when each player is to receive 12. Because there

38

Chapter 2

Axioms of Probability

are 4! ways of dividing the 4 aces so that each player receives 1, we see that the num- 48 . ber of possible outcomes in which each player receives exactly 1 ace is 4! 12, 12, 12, 12 52 As there are 13,13,13,13 possible hands, the desired probability is thus 48 4! 12,12,12,12 52 L .1055

.

13,13,13,13

Some results in probability are quite surprising when initially encountered. Our next two examples illustrate this phenomenon. EXAMPLE 5i If n people are present in a room, what is the probability that no two of them celebrate their birthday on the same day of the year? How large need n be so that this probability is less than 12 ? Solution. As each person can celebrate his or her birthday on any one of 365 days, there is a total of (365)n possible outcomes. (We are ignoring the possibility of someone’s having been born on February 29.) Assuming that each outcome is equally likely, we see that the desired probability is (365)(364)(363) . . . (365 − n + 1)/(365)n . It is a rather surprising fact that when n Ú 23, this probability is less than 12 . That is, if there are 23 or more people in a room, then the probability that at least two of them have the same birthday exceeds 12 . Many people are initially surprised by this result, since 23 seems so small in relation to 365, the number of days of the year. However, 1 365 of having the same birthday, = every pair of individuals has probability 2 365 (365) 23 and in a group of 23 people there are = 253 different pairs of individuals. 2 Looked at this way, the result no longer seems so surprising. When there are 50 people in the room, the probability that at least two share the same birthday is approximately .970, and with 100 persons in the room, the odds are 3 * 106 better than 3,000,000:1. (That is, the probability is greater than that at 3 * 106 + 1 least two people have the same birthday.) . EXAMPLE 5j A deck of 52 playing cards is shufﬂed, and the cards are turned up one at a time until the ﬁrst ace appears. Is the next card—that is, the card following the ﬁrst ace—more likely to be the ace of spades or the two of clubs? Solution. To determine the probability that the card following the ﬁrst ace is the ace of spades, we need to calculate how many of the (52)! possible orderings of the cards have the ace of spades immediately following the ﬁrst ace. To begin, note that each ordering of the 52 cards can be obtained by ﬁrst ordering the 51 cards different from the ace of spades and then inserting the ace of spades into that ordering. Furthermore, for each of the (51)! orderings of the other cards, there is only one place where the ace of spades can be placed so that it follows the ﬁrst ace. For instance, if the ordering of the other 51 cards is 4c, 6h, Jd, 5s, Ac, 7d, . . . , Kh

Section 2.5

Sample Spaces Having Equally Likely Outcomes

39

then the only insertion of the ace of spades into this ordering that results in its following the ﬁrst ace is 4c, 6h, Jd, 5s, Ac, As, 7d, . . . , Kh Therefore, there are (51)! orderings that result in the ace of spades following the ﬁrst ace, so (51)! 1 = P{the ace of spades follows the ﬁrst ace} = (52)! 52 In fact, by exactly the same argument, it follows that the probability that the two of 1 . In other words, each clubs (or any other speciﬁed card) follows the ﬁrst ace is also 52 of the 52 cards of the deck is equally likely to be the one that follows the ﬁrst ace! Many people ﬁnd this result rather surprising. Indeed, a common reaction is to suppose initially that it is more likely that the two of clubs (rather than the ace of spades) follows the ﬁrst ace, since that ﬁrst ace might itself be the ace of spades. This reaction is often followed by the realization that the two of clubs might itself appear before the ﬁrst ace, thus negating its chance of immediately following the ﬁrst ace. However, as there is one chance in four that the ace of spades will be the ﬁrst ace (because all 4 aces are equally likely to be ﬁrst) and only one chance in ﬁve that the two of clubs will appear before the ﬁrst ace (because each of the set of 5 cards consisting of the two of clubs and the 4 aces is equally likely to be the ﬁrst of this set to appear), it again appears that the two of clubs is more likely. However, this is not the case, and a more complete analysis shows that they are equally likely. . EXAMPLE 5k A football team consists of 20 offensive and 20 defensive players. The players are to be paired in groups of 2 for the purpose of determining roommates. If the pairing is done at random, what is the probability that there are no offensive–defensive roommate pairs? What is the probability that there are 2i offensive–defensive roommate pairs, i = 1, 2, . . . , 10? Solution. There are

40 2, 2, . . . , 2

=

(40)! (2!)20

ways of dividing the 40 players into 20 ordered pairs of two each. [That is, there are (40)!/220 ways of dividing the players into a ﬁrst pair, a second pair, and so on.] Hence, there are (40)!/220 (20)! ways of dividing the players into (unordered) pairs of 2 each. Furthermore, since a division will result in no offensive–defensive pairs if the offensive (and defensive) players are paired among themselves, it follows that there are [(20)!/210 (10)!]2 such divisions. Hence, the probability of no offensive–defensive roommate pairs, call it P0 , is given by (20)! 2 [(20)!]3 210 (10)! P0 = = (40)! [(10)!]2 (40)! 20 2 (20)! To determine P2i , the probability that there are 2i offensive–defensive pairs, we ﬁrst 2 20 ways of selecting the 2i offensive players and the 2i defennote that there are 2i sive players who are to be in the offensive–defensive pairs. These 4i players can then

40

Chapter 2

Axioms of Probability

be paired up into (2i)! possible offensive–defensive pairs. (This is so because the ﬁrst offensive player can be paired with any of the 2i defensive players, the second offensive player with any of the remaining 2i − 1 defensive players, and so on.) As the remaining 20 − 2i offensive (and defensive) players must be paired among themselves, it follows that there are 2 (20 − 2i)! 2 20 (2i)! 10−i 2i 2 (10 − i)! divisions which lead to 2i offensive–defensive pairs. Hence, 2 (20 − 2i)! 2 20 (2i)! 10−i 2i 2 (10 − i)! P2i = i = 0, 1, . . . , 10 (40)! 220 (20)! The P2i , i = 0, 1, . . . , 10, can now be computed, or they can be approximated by mak√ ing use of a result of Stirling which shows that n! can be approximated by nn+1/2 e−n 2π. For instance, we obtain P0 L 1.3403 * 10−6 P10 L .345861 P20 L 7.6068 * 10−6

.

Our next three examples illustrate the usefulness of Proposition 4.4. In Example 5l, the introduction of probability enables us to obtain a quick solution to a counting problem. EXAMPLE 5l A total of 36 members of a club play tennis, 28 play squash, and 18 play badminton. Furthermore, 22 of the members play both tennis and squash, 12 play both tennis and badminton, 9 play both squash and badminton, and 4 play all three sports. How many members of this club play at least one of three sports? Solution. Let N denote the number of members of the club, and introduce probability by assuming that a member of the club is randomly selected. If, for any subset C of members of the club, we let P(C) denote the probability that the selected member is contained in C, then number of members inC P(C) = N Now, with T being the set of members that plays tennis, S being the set that plays squash, and B being the set that plays badminton, we have, from Proposition 4.4, P(T ∪ S ∪ B) = P(T) + P(S) + P(B) − P(TS) − P(TB) − P(SB) + P(TSB) 36 + 28 + 18 − 22 − 12 − 9 + 4 = N 43 = N Hence, we can conclude that 43 members play at least one of the sports.

.

Section 2.5

Sample Spaces Having Equally Likely Outcomes

41

The next example in this section not only possesses the virtue of giving rise to a somewhat surprising answer, but is also of theoretical interest. EXAMPLE 5m The matching problem Suppose that each of N men at a party throws his hat into the center of the room. The hats are ﬁrst mixed up, and then each man randomly selects a hat. What is the probability that none of the men selects his own hat? Solution. We ﬁrst calculate the complementary probability of at least one man’s selecting his own hat. Let us denote by Ei , i = 1, 2, . . ., N the event that the ith man N

Ei , the probability that at least selects his own hat. Now, by Proposition 4.4 P i=1

one of the men selects his own hat is given by ⎛ ⎞ N N P(Ei ) − P(Ei1 Ei2 ) + · · · P ⎝ Ei ⎠ = i=1

i1

i=1

+ (−1)n+1

P(Ei1 Ei2 · · · Ein )

i1

P(E1 E2 · · · EN )

+ · · · + (−1)

If we regard the outcome of this experiment as a vector of N numbers, where the ith element is the number of the hat drawn by the ith man, then there are N! possible outcomes. [The outcome (1, 2, 3, . . . , N) means, for example, that each man selects his own hat.] Furthermore, Ei1 Ei2 . . . Ein , the event that each of the n men i1 , i2 , . . . , in selects his own hat, can occur in any of (N − n)(N − n − 1) · · · 3 · 2 · 1 = (N − n)! possible ways; for, of the remaining N − n men, the ﬁrst can select any of N − n hats, the second can then select any of N − n − 1 hats, and so on. Hence, assuming that all N! possible outcomes are equally likely, we see that P(Ei1 Ei2 · · · Ein ) = Also, as there are

N n

terms in

i1

Thus,

⎛ P⎝

N

P(Ei1 Ei2 · · · Ein ), it follows that

P(Ei1 Ei2 · · · Ein ) =

i1

(N − n)! N!

1 N!(N − n)! = (N − n)!n!N! n!

⎞ Ei ⎠ = 1 −

i=1

1 1 1 + − · · · + (−1)N+1 2! 3! N!

Hence, the probability that none of the men selects his own hat is 1 − 1 +

1 1 (−1)N − + ··· + 2! 3! N!

42

Chapter 2

Axioms of Probability

which is approximately equal to e−1 L .36788 for N large. In other words, for N large, the probability that none of the men selects his own hat is approximately .37. (How many readers would have incorrectly thought that this probability would go to 1 as N →q?) . For another illustration of the usefulness of Proposition 4.4, consider the following example. EXAMPLE 5n Compute the probability that 10 married couples are seated at random at a round table, then no wife sits next to her husband. sit next Solution. If we let Ei , i = 1, 2, . . . , 10 denote the event that the ith couple 10

Ei . Now, from to each other, it follows that the desired probability is 1 − P i=1

Proposition 4.4, ⎛ P⎝

10

⎞ Ei ⎠ =

10

1

P(Ei ) − · · · + (−1)n+1

P(Ei1 Ei2 · · · Ein )

i1

1

+ · · · − P(E1 E2 · · · E10 ) To compute P(Ei1 Ei2 · · · Ein ), we ﬁrst note that there are 19! ways of arranging 20 people around a round table. (Why?) The number of arrangements that result in a speciﬁed set of n men sitting next to their wives can most easily be obtained by ﬁrst thinking of each of the n married couples as being single entities. If this were the case, then we would need to arrange 20 − n entities around a round table, and there are clearly (20 − n − 1)! such arrangements. Finally, since each of the n married couples can be arranged next to each other in one of two possible ways, it follows that there are 2n (20 − n − 1)! arrangements that result in a speciﬁed set of n men each sitting next to their wives. Therefore, P(Ei1 Ei2 · · · Ein ) =

2n (19 − n)! (19)!

Thus, from Proposition 4.4, we obtain that the probability that at least one married couple sits together, namely,

10 1

21

(18)! − (19)!

10 2

22

(17)! + (19)!

10 3

23

(16)! − ··· − (19)!

and the desired probability is approximately .3395. ∗ EXAMPLE

10 10

210

9! L .6605 (19)! .

5o Runs

Consider an athletic team that had just ﬁnished its season with a ﬁnal record of n wins and m losses. By examining the sequence of wins and losses, we are hoping to determine whether the team had stretches of games in which it was more likely to win than at other times. One way to gain some insight into this question is to count the number of runs of wins and then see how likely that result would be when all

Section 2.5

Sample Spaces Having Equally Likely Outcomes

43

(n + m)!/(n! m!) orderings of the n wins and m losses are assumed equally likely. By a run of wins, we mean a consecutive sequence of wins. For instance, if n = 10, m = 6, and the sequence of outcomes was WWLLWWWLWLLLWWWW, then there would be 4 runs of wins—the ﬁrst run being of size 2, the second of size 3, the third of size 1, and the fourth of size 4. Supposenow thata team has n wins and m losses. Assuming that all (n + m)!/ n + m (n! m!) = orderings are equally likely, let us determine the probability n that there will be exactly r runs of wins. To do so, consider ﬁrst any vector of positive integers x1 , x2 , . . . , xr with x1 + · · · + xr = n, and let us see how many outcomes result in r runs of wins in which the ith run is of size xi , i = 1, . . . , r. For any such outcome, if we let y1 denote the number of losses before the ﬁrst run of wins, y2 the number of losses between the ﬁrst 2 runs of wins, . . . , yr+1 the number of losses after the last run of wins, then the yi satisfy y1 + y2 + · · · + yr+1 = m

y1 Ú 0, yr+1 Ú 0, yi > 0, i = 2, . . . , r

and the outcome can be represented schematically as LL . . . L WW . . . W L . . L WW . . . W · · · WW . . L . . L y1

x1

y2

x2

xr

yr+1

Hence, the number of outcomes that result in r runs of wins—the ith of size xi , i = 1, . . . r—is equal to the number of integers y1 , . . . , yr+1 that satisfy the foregoing, or, equivalently, to the number of positive integers y1 = y1 + 1 yi = yi , i = 2, . . . , r, yr+1 = yr+1 + 1 that satisfy y1 + y2 + · · · + yr+1 = m + 2 m + 1 By Proposition 6.1 in Chapter 1, there are such outcomes. Hence, the r m + 1 , multiplied by total number of outcomes that result in r runs of wins is r the number of positive integral solutions of x1 + · · · + xr = n. Thus, again from m + 1 n − 1 Proposition 6.1, there are outcomes resulting in r runs of wins. r r − 1 n + m equally likely outcomes, it follows that As there are n n − 1 m + 1 r − 1 r P({r runs of wins}) = r Ú 1 m + n n 7 7 For example, if n = 8 and m = 6, then the probability of 7 runs is 7 6 14 14 = 1/429 if all outcomes are equally likely. Hence, if the outcome 8 8 was WLWLWLWLWWLWLW, then we might suspect that the team’s probability of winning was changing over time. (In particular, the probability that the team wins

44

Chapter 2

Axioms of Probability

seems to be quite high when it lost its last game and quite low when it won its last game.) On the other extreme, if the outcome were WWWWWWWWLLLLLL, then 7 7 14 there would have been only 1 run, and as P({1 run}) = = 1/429, 1 0 8 it would thus again seem unlikely that the team’s probability of winning remained unchanged over its 14 games. . ∗ 2.6

PROBABILITY AS A CONTINUOUS SET FUNCTION A sequence of events {En , n Ú 1} is said to be an increasing sequence if E1 ( E2 ( · · · ( En ( En+1 ( · · · whereas it is said to be a decreasing sequence if E1 ) E2 ) · · · ) En ) En+1 ) · · · If {En , n Ú 1} is an increasing sequence of events, then we deﬁne a new event, denoted by lim En , by n→q q Ei lim En = n→q i=1

Similarly, if {En , n Ú 1} is a decreasing sequence of events, we deﬁne lim En by n

lim En = n→q

q

Ei

i=1

We now prove the following Proposition 1: Proposition 6.1. If {En , n Ú 1} is either an increasing or a decreasing sequence of events, then lim P(En ) = P( lim En ) n→q

n→q

Proof. Suppose, ﬁrst, that {En , n Ú 1} is an increasing sequence, and deﬁne the events Fn , n Ú 1, by F1 = E1

⎛

Fn = En ⎝

n−1

⎞c c Ei ⎠ = En En−1

n > 1

1

where we have used the fact that

n−1

Ei = En−1 , since the events are increasing.

1

In words, Fn consists of those outcomes in En which are not in any of the earlier Ei , i < n. It is easy to verify that the Fn are mutually exclusive events such that q i=1

Fi =

q i=1

Ei

and

n i=1

Fi =

n i=1

Ei

for all n Ú 1

Section 2.6

Probability as a Continuous Set Function

45

Thus, ⎛ P⎝

q

⎞

⎛

Ei ⎠ = P ⎝

1

q

⎞ Fi ⎠

1

=

q

P(Fi )

(by Axiom 3)

1

= lim n→q

n

P(Fi )

1

⎛

= lim P ⎝ n→q ⎛ = lim P ⎝ n→q

n 1 n

⎞ Fi ⎠ ⎞ Ei ⎠

1

= lim P(En ) n→q which proves the result when {En , n Ú 1} is increasing. If {En , n Ú 1} is a decreasing sequence, then {Enc , n Ú 1} is an increasing sequence; hence, from the preceding equations, ⎛ P⎝

q 1

However, because

q

1

Eic =

q

⎞ Eic ⎠ = lim P(Enc ) n→q

c Ei

, it follows that

1

⎛⎛ ⎞c ⎞ q ⎟ ⎜ P ⎝⎝ Ei ⎠ ⎠ = lim P(Enc ) n→q 1

or, equivalently, ⎛ 1 − P⎝

q 1

or

⎞ Ei ⎠ = lim [1 − P(En )] = 1 − lim P(En ) n→q n→q

⎛ P⎝

q 1

which proves the result.

⎞ Ei ⎠ = lim P(En ) n→q

46

Chapter 2

Axioms of Probability

EXAMPLE 6a Probability and a paradox Suppose that we possess an inﬁnitely large urn and an inﬁnite collection of balls labeled ball number 1, number 2, number 3, and so on. Consider an experiment performed as follows: At 1 minute to 12 P.M., balls numbered 1 through 10 are placed in the urn and ball number 10 is withdrawn. (Assume that the withdrawal takes no time.) At 12 minute to 12 P.M., balls numbered 11 through 20 are placed in the urn and ball number 20 is withdrawn. At 14 minute to 12 P.M., balls numbered 21 through 30 are placed in the urn and ball number 30 is withdrawn. At 18 minute to 12 P.M., and so on. The question of interest is, How many balls are in the urn at 12 P.M.? The answer to this question is clearly that there is an inﬁnite number of balls in the urn at 12 P.M., since any ball whose number is not of the form 10n, n Ú 1, will have been placed in the urn and will not have been withdrawn before 12 P.M. Hence, the problem is solved when the experiment is performed as described. However, let us now change the experiment and suppose that at 1 minute to 12 P.M. balls numbered 1 through 10 are placed in the urn and ball number 1 is withdrawn; at 1 2 minute to 12 P.M., balls numbered 11 through 20 are placed in the urn and ball number 2 is withdrawn; at 14 minute to 12 P.M., balls numbered 21 through 30 are placed in the urn and ball number 3 is withdrawn; at 18 minute to 12 P.M., balls numbered 31 through 40 are placed in the urn and ball number 4 is withdrawn, and so on. For this new experiment, how many balls are in the urn at 12 P.M.? Surprisingly enough, the answer now is that the urn is empty at 12 P.M. For, consider n−1 any ball—say, ball number n. At some time prior to 12 P.M. [in particular, at 12 minutes to 12 P.M.], this ball would have been withdrawn from the urn. Hence, for each n, ball number n is not in the urn at 12 P.M.; therefore, the urn must be empty at that time. We see then, from the preceding discussion that the manner in which the balls are withdrawn makes a difference. For, in the ﬁrst case only balls numbered 10n, n Ú 1, are ever withdrawn, whereas in the second case all of the balls are eventually withdrawn. Let us now suppose that whenever a ball is to be withdrawn, that ball is randomly selected from among those present. That is, suppose that at 1 minute to 12 P.M. balls numbered 1 through 10 are placed in the urn and a ball is randomly selected and withdrawn, and so on. In this case, how many balls are in the urn at 12 P.M.? Solution. We shall show that, with probability 1, the urn is empty at 12 P.M. Let us ﬁrst consider ball number 1. Deﬁne En to be the event that ball number 1 is still in the urn after the ﬁrst n withdrawals have been made. Clearly, P(En ) =

9 · 18 · 27 · · · (9n) 10 · 19 · 28 · · · (9n + 1)

[To understand this equation, just note that if ball number 1 is still to be in the urn after the ﬁrst n withdrawals, the ﬁrst ball withdrawn can be any one of 9, the second any one of 18 (there are 19 balls in the urn at the time of the second withdrawal, one of which must be ball number 1), and so on. The denominator is similarly obtained.]

Section 2.6

Probability as a Continuous Set Function

Now, the event that ball number 1 is in the urn at 12 P.M. is just the event

q

47

En .

n=1

Because the events En , n Ú 1, are decreasing events, it follows from Proposition 6.1 that P{ball number 1 is in the urn at 12 P.M.} ⎞ ⎛ q En ⎠ = P⎝ n=1

= lim P(En ) n→q q 9n = 9n + 1 n=1

We now show that q n=1

Since q n=1

9n 9n + 1

9n =0 9n + 1 ⎡

⎤ −1 q 9n + 1 ⎦ =⎣ 9n n=1

this is equivalent to showing that q n=1

1 +

1 9n

=q

Now, for all m Ú 1, q n=1

1 +

1 9n

m 1 1 + 9n n=1 1 1 1 1 1 + 1 + ··· 1 + = 1 + 9 18 27 9m 1 1 1 1 + + + ··· + > 9 18 27 9m m 11 = 9 i Ú

i=1

Hence, letting m→q and using the fact that

q

1/i = q yields

i=1 q n=1

1 1 + 9n

=q

Thus, letting Fi denote the event that ball number i is in the urn at 12 P.M., we have shown that P(F1 ) = 0. Similarly, we can show that P(Fi ) = 0 for all i.

48

Chapter 2

Axioms of Probability

(For instance, the same reasoning shows that P(Fi ) =

q $

[9n/(9n + 1)] for i = q

11, 12, . . . , 20.) Therefore, the probability that the urn is not empty at 12 P.M., P Fi , n=2

1

satisﬁes

⎛ P⎝

q 1

⎞ Fi ⎠ …

q

P(Fi ) = 0

1

by Boole’s inequality. (See Self-Test Exercise 14.) Thus, with probability 1, the urn will be empty at 12 P.M.

2.7

.

PROBABILITY AS A MEASURE OF BELIEF Thus far we have interpreted the probability of an event of a given experiment as being a measure of how frequently the event will occur when the experiment is continually repeated. However, there are also other uses of the term probability. For instance, we have all heard such statements as “It is 90 percent probable that Shakespeare actually wrote Hamlet” or “The probability that Oswald acted alone in assassinating Kennedy is .8.” How are we to interpret these statements? The most simple and natural interpretation is that the probabilities referred to are measures of the individual’s degree of belief in the statements that he or she is making. In other words, the individual making the foregoing statements is quite certain that Oswald acted alone and is even more certain that Shakespeare wrote Hamlet. This interpretation of probability as being a measure of the degree of one’s belief is often referred to as the personal or subjective view of probability. It seems logical to suppose that a “measure of the degree of one’s belief” should satisfy all of the axioms of probability. For example, if we are 70 percent certain that Shakespeare wrote Julius Caesar and 10 percent certain that it was actually Marlowe, then it is logical to suppose that we are 80 percent certain that it was either Shakespeare or Marlowe. Hence, whether we interpret probability as a measure of belief or as a long-run frequency of occurrence, its mathematical properties remain unchanged. EXAMPLE 7a Suppose that, in a 7-horse race, you feel that each of the ﬁrst 2 horses has a 20 percent chance of winning, horses 3 and 4 each have a 15 percent chance, and the remaining 3 horses have a 10 percent chance each. Would it be better for you to wager at even money that the winner will be one of the ﬁrst three horses or to wager, again at even money, that the winner will be one of the horses 1, 5, 6, and 7? Solution. On the basis of your personal probabilities concerning the outcome of the race, your probability of winning the ﬁrst bet is .2 + .2 + .15 = .55, whereas it is .2 + .1 + .1 + .1 = .5 for the second bet. Hence, the ﬁrst wager is more attractive. . Note that, in supposing that a person’s subjective probabilities are always consistent with the axioms of probability, we are dealing with an idealized rather than an

Summary

49

actual person. For instance, if we were to ask someone what he thought the chances were of (a) (b) (c) (d)

rain today, rain tomorrow, rain both today and tomorrow, rain either today or tomorrow,

it is quite possible that, after some deliberation, he might give 30 percent, 40 percent, 20 percent, and 60 percent as answers. Unfortunately, such answers (or such subjective probabilities) are not consistent with the axioms of probability. (Why not?) We would of course hope that, after this was pointed out to the respondent, she would change his answers. (One possibility we could accept is 30 percent, 40 percent, 10 percent, and 60 percent.)

SUMMARY Let S denote the set of all possible outcomes of an experiment. S is called the sample space of the experiment. An event is a subset of S. If Ai , i = 1, . . . , n, are events, then n

Ai , called the union of these events, consists of all outcomes that are in at least i=1

n

one of the events Ai , i = 1, . . . , n. Similarly,

Ai , sometimes written as A1 · · · An , is

i=1

called the intersection of the events Ai and consists of all outcomes that are in all of the events Ai , i = 1, . . . , n. For any event A, we deﬁne Ac to consist of all outcomes in the sample space that are not in A. We call Ac the complement of the event A. The event Sc , which is empty of outcomes, is designated by Ø and is called the null set. If AB = Ø, then we say that A and B are mutually exclusive. For each event A of the sample space S, we suppose that a number P(A), called the probability of A, is deﬁned and is such that (i) 0 … P(A) … 1 (ii) P(S) = 1 (iii) For mutually exclusive events Ai , i Ú 1, ⎛ P⎝

q i=1

⎞ Ai ⎠ =

q

P(Ai )

i=1

P(A) represents the probability that the outcome of the experiment is in A. It can be shown that P(Ac ) = 1 − P(A) A useful result is that P(A ∪ B) = P(A) + P(B) − P(AB)

50

Chapter 2

Axioms of Probability

which can be generalized to give ⎞ ⎛ n n P(Ai ) − P(Ai Aj ) + P(Ai Aj Ak ) P ⎝ Ai ⎠ = i=1

i

i=1

+ · · · + (−1)

n+1

i

P(A1 · · · An )

If S is ﬁnite and each one point set is assumed to have equal probability, then P(A) =

|A| |S|

where |E| denotes the number of outcomes in the event E. P(A) can be interpreted either as a long-run relative frequency or as a measure of one’s degree of belief. PROBLEMS 1. A box contains 3 marbles: 1 red, 1 green, and 1 blue. Consider an experiment that consists of taking 1 marble from the box and then replacing it in the box and drawing a second marble from the box. Describe the sample space. Repeat when the second marble is drawn without replacing the ﬁrst marble. 2. In an experiment, die is rolled continually until a 6 appears, at which point the experiment stops. What is the sample space of this experiment? Let En denote the event that n rolls are necessary to complete the experiment. What points of the samc q

ple space are contained in En ? What is En ? 1

3. Two dice are thrown. Let E be the event that the sum of the dice is odd, let F be the event that at least one of the dice lands on 1, and let G be the event that the sum is 5. Describe the events EF, E ∪ F, FG, EF c , and EFG. 4. A, B, and C take turns ﬂipping a coin. The ﬁrst one to get a head wins. The sample space of this experiment can be deﬁned by % 1, 01, 001, 0001, . . . , S= 0000 · · · (a) Interpret the sample space. (b) Deﬁne the following events in terms of S: (i) A wins = A. (ii) B wins = B. (iii) (A ∪ B)c . Assume that A ﬂips ﬁrst, then B, then C, then A, and so on. 5. A system is comprised of 5 components, each of which is either working or failed. Consider an experiment that consists of observing the status of

each component, and let the outcome of the experiment be given by the vector (x1 , x2 , x3 , x4 , x5 ), where xi is equal to 1 if component i is working and is equal to 0 if component i is failed. (a) How many outcomes are in the sample space of this experiment? (b) Suppose that the system will work if components 1 and 2 are both working, or if components 3 and 4 are both working, or if components 1, 3, and 5 are all working. Let W be the event that the system will work. Specify all the outcomes in W. (c) Let A be the event that components 4 and 5 are both failed. How many outcomes are contained in the event A? (d) Write out all the outcomes in the event AW. 6. A hospital administrator codes incoming patients suffering gunshot wounds according to whether they have insurance (coding 1 if they do and 0 if they do not) and according to their condition, which is rated as good (g), fair (f), or serious (s). Consider an experiment that consists of the coding of such a patient. (a) Give the sample space of this experiment. (b) Let A be the event that the patient is in serious condition. Specify the outcomes in A. (c) Let B be the event that the patient is uninsured. Specify the outcomes in B. (d) Give all the outcomes in the event Bc ∪ A. 7. Consider an experiment that consists of determining the type of job—either blue-collar or whitecollar—and the political afﬁliation—Republican, Democratic, or Independent—of the 15 members of an adult soccer team. How many outcomes are (a) in the sample space? (b) in the event that at least one of the team members is a blue-collar worker?

Problems

8.

9.

10.

11.

12.

13.

(c) in the event that none of the team members considers himself or herself an Independent? Suppose that A and B are mutually exclusive events for which P(A) = .3 and P(B) = .5. What is the probability that (a) either A or B occurs? (b) A occurs but B does not? (c) both A and B occur? A retail establishment accepts either the American Express or the VISA credit card. A total of 24 percent of its customers carry an American Express card, 61 percent carry a VISA card, and 11 percent carry both cards. What percentage of its customers carry a credit card that the establishment will accept? Sixty percent of the students at a certain school wear neither a ring nor a necklace. Twenty percent wear a ring and 30 percent wear a necklace. If one of the students is chosen randomly, what is the probability that this student is wearing (a) a ring or a necklace? (b) a ring and a necklace? A total of 28 percent of American males smoke cigarettes, 7 percent smoke cigars, and 5 percent smoke both cigars and cigarettes. (a) What percentage of males smokes neither cigars nor cigarettes? (b) What percentage smokes cigars but not cigarettes? An elementary school is offering 3 language classes: one in Spanish, one in French, and one in German. The classes are open to any of the 100 students in the school. There are 28 students in the Spanish class, 26 in the French class, and 16 in the German class. There are 12 students that are in both Spanish and French, 4 that are in both Spanish and German, and 6 that are in both French and German. In addition, there are 2 students taking all 3 classes. (a) If a student is chosen randomly, what is the probability that he or she is not in any of the language classes? (b) If a student is chosen randomly, what is the probability that he or she is taking exactly one language class? (c) If 2 students are chosen randomly, what is the probability that at least 1 is taking a language class? A certain town with a population of 100,000 has 3 newspapers: I, II, and III. The proportions of townspeople who read these papers are as follows: I: 10 percent

I and II: 8 percent

II: 30 percent I and III: 2 percent III: 5 percent II and III: 4 percent

I and II and III: 1 percent

14.

15.

16.

17.

51

(The list tells us, for instance, that 8000 people read newspapers I and II.) (a) Find the number of people who read only one newspaper. (b) How many people read at least two newspapers? (c) If I and III are morning papers and II is an evening paper, how many people read at least one morning paper plus an evening paper? (d) How many people do not read any newspapers? (e) How many people read only one morning paper and one evening paper? The following data were given in a study of a group of 1000 subscribers to a certain magazine: In reference to job, marital status, and education, there were 312 professionals, 470 married persons, 525 college graduates, 42 professional college graduates, 147 married college graduates, 86 married professionals, and 25 married professional college graduates. Show that the numbers reported in the study must be incorrect. Hint: Let M, W, and G denote, respectively, the set of professionals, married persons, and college graduates. Assume that one of the 1000 persons is chosen at random, and use Proposition 4.4 to show that if the given numbers are correct, then P(M ∪ W ∪ G) > 1. 52 If it is assumed that all poker hands are 5 equally likely, what is the probability of being dealt (a) a ﬂush? (A hand is said to be a ﬂush if all 5 cards are of the same suit.) (b) one pair? (This occurs when the cards have denominations a, a, b, c, d, where a, b, c, and d are all distinct.) (c) two pairs? (This occurs when the cards have denominations a, a, b, b, c, where a, b, and c are all distinct.) (d) three of a kind? (This occurs when the cards have denominations a, a, a, b, c, where a, b, and c are all distinct.) (e) four of a kind? (This occurs when the cards have denominations a, a, a, a, b.) Poker dice is played by simultaneously rolling 5 dice. Show that (a) P{no two alike} = .0926; (b) P{one pair} = .4630; (c) P{two pair} = .2315; (d) P{three alike} = .1543; (e) P{full house} = .0386; (f) P{four alike} = .0193; (g) P{ﬁve alike} = .0008. If 8 rooks (castles) are randomly placed on a chessboard, compute the probability that none of the rooks can capture any of the others. That is,

52

Chapter 2

18.

19.

20.

21.

22.

23.

24.

25.

Axioms of Probability

compute the probability that no row or ﬁle contains more than one rook. Two cards are randomly selected from an ordinary playing deck. What is the probability that they form a blackjack? That is, what is the probability that one of the cards is an ace and the other one is either a ten, a jack, a queen, or a king? Two symmetric dice have both had two of their sides painted red, two painted black, one painted yellow, and the other painted white. When this pair of dice is rolled, what is the probability that both dice land with the same color face up? Suppose that you are playing blackjack against a dealer. In a freshly shufﬂed deck, what is the probability that neither you nor the dealer is dealt a blackjack? A small community organization consists of 20 families, of which 4 have one child, 8 have two children, 5 have three children, 2 have four children, and 1 has ﬁve children. (a) If one of these families is chosen at random, what is the probability it has i children, i = 1, 2, 3, 4, 5? (b) If one of the children is randomly chosen, what is the probability that child comes from a family having i children, i = 1, 2, 3, 4, 5? Consider the following technique for shufﬂing a deck of n cards: For any initial ordering of the cards, go through the deck one card at a time and at each card, ﬂip a fair coin. If the coin comes up heads, then leave the card where it is; if the coin comes up tails, then move that card to the end of the deck. After the coin has been ﬂipped n times, say that one round has been completed. For instance, if n = 4 and the initial ordering is 1, 2, 3, 4, then if the successive ﬂips result in the outcome h, t, t, h, then the ordering at the end of the round is 1, 4, 2, 3. Assuming that all possible outcomes of the sequence of n coin ﬂips are equally likely, what is the probability that the ordering after one round is the same as the initial ordering? A pair of fair dice is rolled. What is the probability that the second die lands on a higher value than does the ﬁrst? If two dice are rolled, what is the probability that the sum of the upturned faces equals i? Find it for i = 2, 3, . . . , 11, 12. A pair of dice is rolled until a sum of either 5 or 7 appears. Find the probability that a 5 occurs ﬁrst. Hint: Let En denote the event that a 5 occurs on the nth roll and no 5 or 7 occurs on the ﬁrst n − 1 q rolls. Compute P(En ) and argue that P(En ) is n=1

the desired probability. 26. The game of craps is played as follows: A player rolls two dice. If the sum of the dice is either a 2,

3, or 12, the player loses; if the sum is either a 7 or an 11, the player wins. If the outcome is anything else, the player continues to roll the dice until she rolls either the initial outcome or a 7. If the 7 comes ﬁrst, the player loses, whereas if the initial outcome reoccurs before the 7 appears, the player wins. Compute the probability of a player winning at craps. Hint: Let Ei denote the event that the initial outcome is i and the player wins. The desired prob12 ability is P(Ei ). To compute P(Ei ), deﬁne the i=2

events Ei,n to be the event that the initial sum is i and the player wins on the nth roll. Argue that q P(Ei,n ). P(Ei ) = n=1

27. An urn contains 3 red and 7 black balls. Players A and B withdraw balls from the urn consecutively until a red ball is selected. Find the probability that A selects the red ball. (A draws the ﬁrst ball, then B, and so on. There is no replacement of the balls drawn.) 28. An urn contains 5 red, 6 blue, and 8 green balls. If a set of 3 balls is randomly selected, what is the probability that each of the balls will be (a) of the same color? (b) of different colors? Repeat under the assumption that whenever a ball is selected, its color is noted and it is then replaced in the urn before the next selection. This is known as sampling with replacement. 29. An urn contains n white and m black balls, where n and m are positive numbers. (a) If two balls are randomly withdrawn, what is the probability that they are the same color? (b) If a ball is randomly withdrawn and then replaced before the second one is drawn, what is the probability that the withdrawn balls are the same color? (c) Show that the probability in part (b) is always larger than the one in part (a). 30. The chess clubs of two schools consist of, respectively, 8 and 9 players. Four members from each club are randomly chosen to participate in a contest between the two schools. The chosen players from one team are then randomly paired with those from the other team, and each pairing plays a game of chess. Suppose that Rebecca and her sister Elise are on the chess clubs at different schools. What is the probability that (a) Rebecca and Elise will be paired? (b) Rebecca and Elise will be chosen to represent their schools but will not play each other? (c) either Rebecca or Elise will be chosen to represent her school?

Problems 31. A 3-person basketball team consists of a guard, a forward, and a center. (a) If a person is chosen at random from each of three different such teams, what is the probability of selecting a complete team? (b) What is the probability that all 3 players selected play the same position? 32. A group of individuals containing b boys and g girls is lined up in random order; that is, each of the (b + g)! permutations is assumed to be equally likely. What is the probability that the person in the ith position, 1 … i … b + g, is a girl? 33. A forest contains 20 elk, of which 5 are captured, tagged, and then released. A certain time later, 4 of the 20 elk are captured. What is the probability that 2 of these 4 have been tagged? What assumptions are you making? 34. The second Earl of Yarborough is reported to have bet at odds of 1000 to 1 that a bridge hand of 13 cards would contain at least one card that is ten or higher. (By ten or higher we mean that a card is either a ten, a jack, a queen, a king, or an ace.) Nowadays, we call a hand that has no cards higher than 9 a Yarborough. What is the probability that a randomly selected bridge hand is a Yarborough? 35. Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that (a) 3 red, 2 blue, and 2 green balls are withdrawn; (b) at least 2 red balls are withdrawn; (c) all withdrawn balls are the same color; (d) either exactly 3 red balls or exactly 3 blue balls are withdrawn. 36. Two cards are chosen at random from a deck of 52 playing cards. What is the probability that they (a) are both aces? (b) have the same value? 37. An instructor gives her class a set of 10 problems with the information that the ﬁnal exam will consist of a random selection of 5 of them. If a student has ﬁgured out how to do 7 of the problems, what is the probability that he or she will answer correctly (a) all 5 problems? (b) at least 4 of the problems? 38. There are n socks, 3 of which are red, in a drawer. What is the value of n if, when 2 of the socks are chosen randomly, the probability that they are both red is 12 ? 39. There are 5 hotels in a certain town. If 3 people check into hotels in a day, what is the probability that they each check into a different hotel? What assumptions are you making? 40. A town contains 4 people who repair televisions. If 4 sets break down, what is the probability that

53

exactly i of the repairers are called? Solve the problem for i = 1, 2, 3, 4. What assumptions are you making? 41. If a die is rolled 4 times, what is the probability that 6 comes up at least once? 42. Two dice are thrown n times in succession. Compute the probability that double 6 appears at least once. How large need n be to make this probability at least 12 ? 43. (a) If N people, including A and B, are randomly arranged in a line, what is the probability that A and B are next to each other? (b) What would the probability be if the people were randomly arranged in a circle? 44. Five people, designated as A, B, C, D, E, are arranged in linear order. Assuming that each possible order is equally likely, what is the probability that (a) there is exactly one person between A and B? (b) there are exactly two people between A and B? (c) there are three people between A and B? 45. A woman has n keys, of which one will open her door. (a) If she tries the keys at random, discarding those that do not work, what is the probability that she will open the door on her kth try? (b) What if she does not discard previously tried keys? 46. How many people have to be in a room in order that the probability that at least two of them celebrate their birthday in the same month is at least 1 2 ? Assume that all possible monthly outcomes are equally likely. 47. If there are 12 strangers in a room, what is the probability that no two of them celebrate their birthday in the same month? 48. Given 20 people, what is the probability that, among the 12 months in the year, there are 4 months containing exactly 2 birthdays and 4 containing exactly 3 birthdays? 49. A group of 6 men and 6 women is randomly divided into 2 groups of size 6 each. What is the probability that both groups will have the same number of men? 50. In a hand of bridge, ﬁnd the probability that you have 5 spades and your partner has the remaining 8. 51. Suppose that n balls are randomly distributed into N compartments. Find the probability that m balls will fall into the ﬁrst compartment. Assume that all N n arrangements are equally likely.

54

Chapter 2

Axioms of Probability

52. A closet contains 10 pairs of shoes. If 8 shoes are randomly selected, what is the probability that there will be (a) no complete pair? (b) exactly 1 complete pair?

(Why not?) Hint: Use Proposition 4.4. 55. Compute the probability that a hand of 13 cards contains (a) the ace and king of at least one suit; (b) all 4 of at least 1 of the 13 denominations. 56. Two players play the following game: Player A chooses one of the three spinners pictured in Figure 2.6, and then player B chooses one of the remaining two spinners. Both players then spin their spinner, and the one that lands on the higher number is declared the winner. Assuming that each spinner is equally likely to land in any of its 3 regions, would you rather be player A or player B? Explain your answer!

53. If 4 married couples are arranged in a row, ﬁnd the probability that no husband sits next to his wife. 54. Compute the probability that a bridge hand is void in at least one suit. Note that the answer is not 4 39 1 13 52 13

9

3

5

8

a

b

1

4

7

6 c

2 FIGURE 2.6: Spinners

THEORETICAL EXERCISES

Prove the following relations: 1. EF ( E ( E ∪ F. 2. If E ( F, then F c ( Ec . 3. F = FE ∪ FEc and E ∪ F = E ∪ Ec F.

4.

q

Ei F =

1 q 1

Ei

q

Ei F and

1

∪ F=

q 1

(Ei ∪ F).

Theoretical Exercises 5. For any sequence of events E1 , E2 , . . ., deﬁne a new sequence F1 , F2 , . . . of disjoint events (that is, events such that Fi Fj = Ø whenever i Z j) such that for all n Ú 1, n

Fi =

n

1

Ei

1

6. Let E, F, and G be three events. Find expressions for the events so that, of E, F, and G, (a) only E occurs; (b) both E and G, but not F, occur; (c) at least one of the events occurs; (d) at least two of the events occur; (e) all three events occur; (f) none of the events occurs; (g) at most one of the events occurs; (h) at most two of the events occur; (i) exactly two of the events occur; (j) at most three of the events occur. 7. Find the simplest expression for the following events: (a) (E ∪ F)(E ∪ F c ); (b) (E ∪ F)(Ec ∪ F)(E ∪ F c ); (c) (E ∪ F)(F ∪ G). 8. Let S be a given set. If, for some k > 0, S1 , S2 , . . . , Sk are mutually exclusive nonempty k

subsets of S such that Si = S, then we i=1

call the set {S1 , S2 , . . . , Sk } a partition of S. Let Tn denote the number of different partitions of {1, 2, . . . , n}. Thus, T1 = 1 (the only partition being S1 = {1}) and T2 = 2 (the two partitions being {{1, 2, }}, {{1}, {2}}). (a) Show, by computing all partitions, that T3 = 5, T4 = 15. (b) Show that Tn+1 = 1 +

n n Tk k k=1

and use this equation to compute T10 . Hint: One way of choosing a partition of n + 1 items is to call one of the items special. Then we obtain different partitions by ﬁrst choosing k, k = 0, 1, . . . , n, then a subset of size n − k of the nonspecial items, and then any of the Tk partitions of the remaining k nonspecial items. By adding the special item to the subset of size n − k, we obtain a partition of all n + 1 items. 9. Suppose that an experiment is performed n times. For any event E of the sample space, let n(E) denote the number of times that event E occurs and deﬁne f (E) = n(E)/n. Show that f (·) satisﬁes Axioms 1, 2, and 3.

55

10. Prove that P(E ∪ F ∪ G) = P(E) + P(F) + P(G) − P(Ec FG) − P(EF c G) − P(EFGc ) − 2P(EFG). 11. If P(E) = .9 and P(F) = .8, show that P(EF) Ú .7. In general, prove Bonferroni’s inequality, namely, P(EF) Ú P(E) + P(F) − 1 12. Show that the probability that exactly one of the events E or F occurs equals P(E) + P(F) − 2P(EF). 13. Prove that P(EF c ) = P(E) − P(EF). 14. Prove Proposition 4.4 by mathematical induction. 15. An urn contains M white and N black balls. If a random sample of size r is chosen, what is the probability that it contains exactly k white balls? 16. Use induction to generalize Bonferroni’s inequality to n events. That is, show that P(E1 E2 · · · En ) Ú P(E1 ) + · · · + P(En ) − (n − 1) 17. Consider the matching problem, Example 5m, and deﬁne AN to be the number of ways in which the N men can select their hats so that no man selects his own. Argue that AN = (N − 1)(AN−1 + AN−2 ) This formula, along with the boundary conditions A1 = 0, A2 = 1, can then be solved for AN , and the desired probability of no matches would be AN /N!. Hint: After the ﬁrst man selects a hat that is not his own, there remain N − 1 men to select among a set of N − 1 hats that does not contain the hat of one of these men. Thus, there is one extra man and one extra hat. Argue that we can get no matches either with the extra man selecting the extra hat or with the extra man not selecting the extra hat. 18. Let fn denote the number of ways of tossing a coin n times such that successive heads never appear. Argue that fn = fn−1 + fn−2

n Ú 2, where f0 K 1, f1 K 2

Hint: How many outcomes are there that start with a head, and how many start with a tail? If Pn denotes the probability that successive heads never appear when a coin is tossed n times, ﬁnd Pn (in terms of fn ) when all possible outcomes of the n tosses are assumed equally likely. Compute P10 . 19. An urn contains n red and m blue balls. They are withdrawn one at a time until a total of r, r … n,

56

Chapter 2

Axioms of Probability

red balls have been withdrawn. Find the probability that a total of k balls are withdrawn. Hint: A total of k balls will be withdrawn if there are r − 1 red balls in the ﬁrst k − 1 withdrawals and the kth withdrawal is a red ball. 20. Consider an experiment whose sample space consists of a countably inﬁnite number of points. Show that not all points can be equally likely. Can all points have a positive probability of occurring? ∗ 21. Consider Example 5o, which is concerned with the number of runs of wins obtained when n wins and m losses are randomly permuted. Now

consider the total number of runs—that is, win runs plus loss runs—and show that m − 1 n − 1 k − 1 k − 1 P{2k runs} = 2 m + n n P{2k + 1 runs} m − 1 n − 1 m − 1 n − 1 + k − 1 k k k − 1 = m + n n

SELF-TEST PROBLEMS AND EXERCISES 1. A cafeteria offers a three-course meal consisting of an entree, a starch, and a dessert. The possible choices are given in the following table:

Course Entree Starch Dessert

Choices Chicken or roast beef Pasta or rice or potatoes Ice cream or Jello or apple pie or a peach

A person is to choose one course from each category. (a) How many outcomes are in the sample space? (b) Let A be the event that ice cream is chosen. How many outcomes are in A? (c) Let B be the event that chicken is chosen. How many outcomes are in B? (d) List all the outcomes in the event AB. (e) Let C be the event that rice is chosen. How many outcomes are in C? (f) List all the outcomes in the event ABC. 2. A customer visiting the suit department of a certain store will purchase a suit with probability .22, a shirt with probability .30, and a tie with probability .28. The customer will purchase both a suit and a shirt with probability .11, both a suit and a tie with probability .14, and both a shirt and a tie with probability .10. A customer will purchase all 3 items with probability .06. What is the probability that a customer purchases (a) none of these items? (b) exactly 1 of these items? 3. A deck of cards is dealt out. What is the probability that the 14th card dealt is an ace? What is the probability that the ﬁrst ace occurs on the 14th card?

4. Let A denote the event that the midtown temperature in Los Angeles is 70◦ F, and let B denote the event that the midtown temperature in New York is 70◦ F. Also, let C denote the event that the maximum of the midtown temperatures in New York and in Los Angeles is 70◦ F. If P(A) = .3, P(B) = .4, and P(C) = .2, ﬁnd the probability that the minimum of the two midtown temperatures is 70◦ F. 5. An ordinary deck of 52 cards is shufﬂed. What is the probability that the top four cards have (a) different denominations? (b) different suits? 6. Urn A contains 3 red and 3 black balls, whereas urn B contains 4 red and 6 black balls. If a ball is randomly selected from each urn, what is the probability that the balls will be the same color? 7. In a state lottery, a player must choose 8 of the numbers from 1 to 40. The lottery commission then performs an experiment that selects 8 of these 40 numbers. Assuming that the choice of the lottery is equally likely to be any of the commission 40 combinations, what is the probability that a 8 player has (a) all 8 of the numbers selected by the lottery commission? (b) 7 of the numbers selected by the lottery commission? (c) at least 6 of the numbers selected by the lottery commission? 8. From a group of 3 freshmen, 4 sophomores, 4 juniors, and 3 seniors a committee of size 4 is randomly selected. Find the probability that the committee will consist of (a) 1 from each class; (b) 2 sophomores and 2 juniors; (c) only sophomores or juniors.

Self -Test Problems and Exercises 9. For a ﬁnite set A, let N(A) denote the number of elements in A. (a) Show that N(A ∪ B) = N(A) + N(B) − N(AB) (b) More generally, show that ⎞ ⎛ n N(Ai ) − N(Ai Aj ) N ⎝ Ai ⎠ = i

i=1

15. Show that if P(Ai ) = 1 for all i Ú 1, then q Ai = 1. P i=1

16. Let Tk (n) denote the number of partitions of the set {1, . . . , n} into k nonempty subsets, where 1 … k … n. (See Theoretical Exercise 8 for the deﬁnition of a partition.) Argue that Tk (n) = kTk (n − 1) + Tk−1 (n − 1)

i

+ · · · + (−1)

N(A1 · · · An )

10. Consider an experiment that consists of six horses, numbered 1 through 6, running a race, and suppose that the sample space consists of the 6! possible orders in which the horses ﬁnish. Let A be the event that the number-1 horse is among the top three ﬁnishers, and let B be the event that the number-2 horse comes in second. How many outcomes are in the event A ∪ B?

17.

18.

11. A 5-card hand is dealt from a well-shufﬂed deck of 52 playing cards. What is the probability that the hand contains at least one card from each of the four suits? 12. A basketball team consists of 6 frontcourt and 4 backcourt players. If players are divided into roommates at random, what is the probability that there will be exactly two roommate pairs made up of a backcourt and a frontcourt player?

19.

13. Suppose that a person chooses a letter at random from R E S E R V E and then chooses one at random from V E R T I C A L. What is the probability that the same letter is chosen? 14. Prove Boole’s inequality: ⎞ ⎛ q q P ⎝ Ai ⎠ … P(Ai ) i=1

i=1

57

20.

Hint: In how many partitions is {1} a subset, and in how many is 1 an element of a subset that contains other elements? Five balls are randomly chosen, without replacement, from an urn that contains 5 red, 6 white, and 7 blue balls. Find the probability that at least one ball of each color is chosen. Four red, 8 blue, and 5 green balls are randomly arranged in a line. (a) What is the probability that the ﬁrst 5 balls are blue? (b) What is the probability that none of the ﬁrst 5 balls are blue? (c) What is the probability that the ﬁnal 3 balls are differently colored. (d) What is the probability that all the red balls are together? Ten cards are randomly chosen from a deck of 52 cards that consists of 13 cards of each of 4 different suits. Each of the selected cards is put in one of 4 piles, depending on the suit of the card. (a) What is the probability that the largest pile has 4 cards, the next largest has 3, the next largest has 2, and the smallest has 1 card? (b) What is the probability that two of the piles have 3 cards, one has 4 cards, and one has no cards? Balls are randomly removed from an urn initially containing 20 red and 10 blue balls. What is the probability that all of the red balls are removed before all of the blue ones have been removed?

C H A P T E R

3

Conditional Probability and Independence 3.1 3.2 3.3 3.4 3.5

INTRODUCTION CONDITIONAL PROBABILITIES BAYES’S FORMULA INDEPENDENT EVENTS P(·|F ) IS A PROBABILITY

3.1

INTRODUCTION In this chapter, we introduce one of the most important concepts in probability theory, that of conditional probability. The importance of this concept is twofold. In the ﬁrst place, we are often interested in calculating probabilities when some partial information concerning the result of an experiment is available; in such a situation, the desired probabilities are conditional. Second, even when no partial information is available, conditional probabilities can often be used to compute the desired probabilities more easily.

3.2

CONDITIONAL PROBABILITIES Suppose that we toss 2 dice, and suppose that each of the 36 possible outcomes is 1 . Suppose further that we observe equally likely to occur and hence has probability 36 that the ﬁrst die is a 3. Then, given this information, what is the probability that the sum of the 2 dice equals 8? To calculate this probability, we reason as follows: Given that the initial die is a 3, there can be at most 6 possible outcomes of our experiment, namely, (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), and (3, 6). Since each of these outcomes originally had the same probability of occurring, the outcomes should still have equal probabilities. That is, given that the ﬁrst die is a 3, the (conditional) probability of each of the outcomes (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), and (3, 6) is 16 , whereas the (conditional) probability of the other 30 points in the sample space is 0. Hence, the desired probability will be 16 . If we let E and F denote, respectively, the event that the sum of the dice is 8 and the event that the ﬁrst die is a 3, then the probability just obtained is called the conditional probability that E occurs given that F has occurred and is denoted by P(E|F) A general formula for P(E|F) that is valid for all events E and F is derived in the same manner: If the event F occurs, then, in order for E to occur, it is necessary that the actual occurrence be a point both in E and in F; that is, it must be in EF. Now, since we know that F has occurred, it follows that F becomes our new, or reduced, sample space; hence, the probability that the event EF occurs will equal the probability of EF relative to the probability of F. That is, we have the following deﬁnition.

58

Section 3.2

Conditional Probabilities

59

Deﬁnition If P(F) > 0, then P(E|F) =

P(EF) P(F)

(2.1)

EXAMPLE 2a A student is taking a one-hour-time-limit makeup examination. Suppose the probability that the student will ﬁnish the exam in less than x hours is x/2, for all 0 … x … 1. Then, given that the student is still working after .75 hour, what is the conditional probability that the full hour is used? Solution. Let Lx denote the event that the student ﬁnishes the exam in less than x hours, 0 … x … 1, and let F be the event that the student uses the full hour. Because F is the event that the student is not ﬁnished in less than 1 hour, P(F) = P(Lc1 ) = 1 − P(L1 ) = .5 Now, the event that the student is still working at time .75 is the complement of the event L.75 , so the desired probability is obtained from P(FLc.75 ) P(Lc.75 ) P(F) = 1 − P(L.75 ) .5 = = .8 .625

P(F|Lc.75 ) =

.

If each outcome of a ﬁnite sample space S is equally likely, then, conditional on the event that the outcome lies in a subset F ( S, all outcomes in F become equally likely. In such cases, it is often convenient to compute conditional probabilities of the form P(E|F) by using F as the sample space. Indeed, working with this reduced sample space often results in an easier and better understood solution. Our next few examples illustrate this point. EXAMPLE 2b A coin is ﬂipped twice. Assuming that all four points in the sample space S = {(h, h), (h, t), (t, h), (t, t)} are equally likely, what is the conditional probability that both ﬂips land on heads, given that (a) the ﬁrst ﬂip lands on heads? (b) at least one ﬂip lands on heads? Solution. Let B = {(h, h)} be the event that both ﬂips land on heads; let F = {(h, h), (h, t)} be the event that the ﬁrst ﬂip lands on heads; and let A = {(h, h), (h, t), (t, h)} be the event that at least one ﬂip lands on heads. The probability for (a) can be obtained from

60

Chapter 3

Conditional Probability and Independence

P(BF) P(F) P({(h, h)}) = P({(h, h), (h, t)}) 1/4 = 1/2 = 2/4

P(B|F) =

For (b), we have P(BA) P(A) P({(h, h)}) = P({(h, h), (h, t), (t, h)}) 1/4 = 1/3 = 3/4

P(B|A) =

Thus, the conditional probability that both ﬂips land on heads given that the ﬁrst one does is 1/2, whereas the conditional probability that both ﬂips land on heads given that at least one does is only 1/3. Many students initially ﬁnd this latter result surprising. They reason that, given that at least one ﬂip lands on heads, there are two possible results: Either they both land on heads or only one does. Their mistake, however, is in assuming that these two possibilities are equally likely. For, initially, there are 4 equally likely outcomes. Because the information that at least one ﬂip lands on heads is equivalent to the information that the outcome is not (t, t), we are left with the 3 equally likely outcomes (h, h), (h, t), (t, h), only one of which results in both ﬂips landing on heads. . EXAMPLE 2c In the card game bridge, the 52 cards are dealt out equally to 4 players—called East, West, North, and South. If North and South have a total of 8 spades among them, what is the probability that East has 3 of the remaining 5 spades? Solution. Probably the easiest way to compute the desired probability is to work with the reduced sample space. That is, given that North–South have a total of 8 spades among their 26 cards, there remains a total of 26 cards, exactly 5 of them being spades, to be distributed among the East–West hands. Since each distribution is equally likely, it follows that the conditional probability that East will have exactly 3 spades among his or her 13 cards is 21 5 10 3 L .339 . 26 13 EXAMPLE 2d A total of n balls are sequentially and randomly chosen, without replacement, from an urn containing r red and b blue balls (n … r + b). Given that k of the n balls are blue, what is the conditional probability that the ﬁrst ball chosen is blue?

Section 3.2

Conditional Probabilities

61

Solution. If we imagine that the balls are numbered, with the blue balls having numbers 1 through b and the red balls b + 1 through b + r, then the outcome of the experiment of selecting n balls without replacement is a vector of distinct integers x1 , . . . , xn , where each xi is between 1 and r + b. Moreover, each such vector is equally likely to be the outcome. So, given that the vector contains k blue balls (that is, it contains k values between 1 and b), it follows that each of these outcomes is equally likely. But because the ﬁrst ball chosen is, therefore, equally likely to be any of the n chosen balls, of which k are blue, it follows that the desired probability is k/n. If we did not choose to work with the reduced sample space, we could have solved the problem by letting B be the event that the ﬁrst ball chosen is blue and Bk be the event that a total of k blue balls are chosen. Then P(BBk ) P(Bk ) P(Bk |B)P(B) = P(Bk )

P(B|Bk ) =

Now, P(Bk |B) is the probability that a random choice of n − 1 balls from an urn containing r red and b − 1 blue balls results in a total of k − 1 blue balls being chosen; consequently, b−1 r P(Bk |B) =

k−1 n−k r+b−1 n−1

Using the preceding formula along with P(B) =

b r + b

and the hypergeometric probability b P(Bk ) =

r k n−k r+b n

again yields the result that P(B|Bk ) =

k n

.

Multiplying both sides of Equation (2.1) by P(F ), we obtain P(EF ) = P(F)P(E|F )

(2.2)

In words, Equation (2.2) states that the probability that both E and F occur is equal to the probability that F occurs multiplied by the conditional probability of E given that F occurred. Equation (2.2) is often quite useful in computing the probability of the intersection of events. EXAMPLE 2e Celine is undecided as to whether to take a French course or a chemistry course. She estimates that her probability of receiving an A grade would be 12 in a French course and 23 in a chemistry course. If Celine decides to base her decision on the ﬂip of a fair coin, what is the probability that she gets an A in chemistry?

62

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Solution. (a) Let the event that Celine takes chemistry and A denote the event that she receives an A in whatever course she takes, then the desired probability is P(CA), which is calculated by using Equation (2.2) as follows: P(CA) = P(C)P(A|C) 2 1 1 = = 2 3 3

.

EXAMPLE 2f Suppose that an urn contains 8 red balls and 4 white balls. We draw 2 balls from the urn without replacement. (a) If we assume that at each draw each ball in the urn is equally likely to be chosen, what is the probability that both balls drawn are red? (b) Now suppose that the balls have different weights, with each red ball having weight r and each white ball having weight w. Suppose that the probability that a given ball in the urn is the next one selected is its weight divided by the sum of the weights of all balls currently in the urn. Now what is the probability that both balls are red? Solution. Let R1 and R2 denote, respectively, the events that the ﬁrst and second balls drawn are red. Now, given that the ﬁrst ball selected is red, there are 7 remaining 7 8 red balls and 4 white balls, so P(R2 |R1 ) = 11 . As P(R1 ) is clearly 12 , the desired probability is P(R1 R2 ) = P(R1 )P(R2 |R1 ) 2 7 14 = = 3 11 33 Of course, this probability could have been computed by P(R1 R2 ) = 82 / 12 2 . For part (b), we again let Ri be the event that the ith ball chosen is red and use P(R1 R2 ) = P(R1 )P(R2 |R1 ) Now, number the red balls, and let Bi , i = 1, . . . , 8 be the event that the ﬁrst ball drawn is red ball number i. Then P(R1 ) = P(∪8i=1 Bi ) =

8

P(Bi ) = 8

i=1

r 8r + 4w

Moreover, given that the ﬁrst ball is red, the urn then contains 7 red and 4 white balls. Thus, by an argument similar to the preceding one, P(R2 |R1 ) =

7r 7r + 4w

Hence, the probability that both balls are red is P(R1 R2 ) =

7r 8r 8r + 4w 7r + 4w

.

A generalization of Equation (2.2), which provides an expression for the probability of the intersection of an arbitrary number of events, is sometimes referred to as the multiplication rule.

Section 3.2

Conditional Probabilities

63

The multiplication rule P(E1 E2 E3 · · · En ) = P(E1 )P(E2 |E1 )P(E3 |E1 E2 ) · · · P(En |E1 · · · En−1 )

To prove the multiplication rule, just apply the deﬁnition of conditional probability to its right-hand side, giving P(E1 )

P(E1 E2 ) P(E1 E2 E3 ) P(E1 E2 · · · En ) ··· = P(E1 E2 · · · En ) P(E1 ) P(E1 E2 ) P(E1 E2 · · · En−1 )

EXAMPLE 2g In the match problem stated in Example 5m of Chapter 2, it was shown that PN , the probability that there are no matches when N people randomly select from among their own N hats, is given by N (−1)i /i! PN = i=0

What is the probability that exactly k of the N people have matches? Solution. Let us ﬁx our attention on a particular set of k people and determine the probability that these k individuals have matches and no one else does. Letting E denote the event that everyone in this set has a match, and letting G be the event that none of the other N − k people have a match, we have P(EG) = P(E )P(G|E ) Now, let Fi , i = 1, . . . , k, be the event that the ith member of the set has a match. Then P(E ) = P(F1 F2 · · · Fk ) = P(F1 )P(F2 |F1 )P(F3 |F1 F2 ) · · · P(Fk |F1 · · · Fk−1 ) 1 1 1 1 = ··· N N − 1N − 2 N − k + 1 (N − k)! = N! Given that everyone in the set of k has a match, the other N − k people will be randomly choosing among their own N − k hats, so the probability that none of them has a match is equal to the probability of no matches in a problem having N − k people choosing among their own N − k hats. Therefore, P(G|E ) = PN−k =

N−k

(−1)i /i!

i=0

showing that the probability that a speciﬁed set of k people have matches and no one else does is (N − k)! PN−k P(EG) = N!

64

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Because there will be exactly k matches if the preceding is true for any of the of k individuals, the desired probability is

N k

sets

P(exactly k matches) = PN−k /k! L e−1 /k!

when N is large

.

We will now employ the multiplication rule to obtain a second approach to solving Example 5h(b) of Chapter 2. EXAMPLE 2h An ordinary deck of 52 playing cards is randomly divided into 4 piles of 13 cards each. Compute the probability that each pile has exactly 1 ace. Solution. Deﬁne events Ei , i = 1, 2, 3, 4, as follows: E1 E2 E3 E4

= {the ace of spades is in any one of the piles} = {the ace of spades and the ace of hearts are in different piles} = {the aces of spades, hearts, and diamonds are all in different piles} = {all 4 aces are in different piles}

The desired probability is P(E1 E2 E3 E4 ), and by the multiplication rule, P(E1 E2 E3 E4 ) = P(E1 )P(E2 |E1 )P(E3 |E1 E2 )P(E4 |E1 E2 E3 ) Now, P(E1 ) = 1 since E1 is the sample space S. Also, P(E2 |E1 ) =

39 51

since the pile containing the ace of spades will receive 12 of the remaining 51 cards, and 26 P(E3 |E1 E2 ) = 50 since the piles containing the aces of spades and hearts will receive 24 of the remaining 50 cards. Finally, 13 P(E4 |E1 E2 E3 ) = 49 Therefore, the probability that each pile has exactly 1 ace is P(E1 E2 E3 E4 ) =

39 · 26 · 13 L .105 51 · 50 · 49

That is, there is approximately a 10.5 percent chance that each pile will contain an ace. (Problem 13 gives another way of using the multiplication rule to solve this problem.) . Remarks. Our deﬁnition of P(E|F) is consistent with the interpretation of probability as being a long-run relative frequency. To see this, suppose that n repetitions of the experiment are to be performed, where n is large. We claim that if

Section 3.3

Bayes’s Formula

65

we consider only those experiments in which F occurs, then P(E|F) will equal the long-run proportion of them in which E also occurs. To verify this statement, note that, since P(F) is the long-run proportion of experiments in which F occurs, it follows that in the n repetitions of the experiment F will occur approximately nP(F) times. Similarly, in approximately nP(EF) of these experiments both E and F will occur. Hence, out of the approximately nP(F) experiments in which F occurs, the proportion of them in which E also occurs is approximately equal to P(EF) nP(EF) = nP(F) P(F) Because this approximation becomes exact as n becomes larger and larger, we have the appropriate deﬁnition of P(E|F). 3.3

BAYES’S FORMULA Let E and F be events. We may express E as E = EF ∪ EF c for, in order for an outcome to be in E, it must either be in both E and F or be in E but not in F. (See Figure 3.1.) As EF and EF c are clearly mutually exclusive, we have, by Axiom 3, P(E ) = P(EF ) + P(EF c ) = P(E|F )P(F ) + P(E|F c )P(F c ) = P(E|F )P(F ) + P(E|F c )[1 − P(F)]

(3.1)

Equation (3.1) states that the probability of the event E is a weighted average of the conditional probability of E given that F has occurred and the conditional probability of E given that F has not occurred—each conditional probability being given as much weight as the event on which it is conditioned has of occurring. This is an extremely useful formula, because its use often enables us to determine the probability of an event by ﬁrst “conditioning” upon whether or not some second event has occurred. That is, there are many instances in which it is difﬁcult to compute the probability of an event directly, but it is straightforward to compute it once we know whether or not some second event has occurred. We illustrate this idea with some examples.

E

F EF c

EF

FIGURE 3.1: E = EF ∪ EF c . EF = Shaded Area; EF c = Striped Area

66

Chapter 3

Conditional Probability and Independence

EXAMPLE 3a (Part 1) An insurance company believes that people can be divided into two classes: those who are accident prone and those who are not. The company’s statistics show that an accident-prone person will have an accident at some time within a ﬁxed 1-year period with probability .4, whereas this probability decreases to .2 for a person who is not accident prone. If we assume that 30 percent of the population is accident prone, what is the probability that a new policyholder will have an accident within a year of purchasing a policy? Solution. We shall obtain the desired probability by ﬁrst conditioning upon whether or not the policyholder is accident prone. Let A1 denote the event that the policyholder will have an accident within a year of purchasing the policy, and let A denote the event that the policyholder is accident prone. Hence, the desired probability is given by P(A1 ) = P(A1 |A)P(A) + P(A1 |Ac )P(Ac ) = (.4)(.3) + (.2)(.7) = .26

.

EXAMPLE 3a (Part 2) Suppose that a new policyholder has an accident within a year of purchasing a policy. What is the probability that he or she is accident prone? Solution. The desired probability is P(AA1 ) P(A1 ) P(A)P(A1 |A) = P(A1 ) 6 (.3)(.4) = = .26 13

P(A|A1 ) =

.

EXAMPLE 3b Consider the following game played with an ordinary deck of 52 playing cards: The cards are shufﬂed and then turned over one at a time. At any time, the player can guess that the next card to be turned over will be the ace of spades; if it is, then the player wins. In addition, the player is said to win if the ace of spades has not yet appeared when only one card remains and no guess has yet been made. What is a good strategy? What is a bad strategy? Solution. Every strategy has probability 1/52 of winning! To show this, we will use induction to prove the stronger result that, for an n card deck, one of whose cards is the ace of spades, the probability of winning is 1/n, no matter what strategy is employed. Since this is clearly true for n = 1, assume it to be true for an n − 1 card deck, and now consider an n card deck. Fix any strategy, and let p denote the probability that the strategy guesses that the ﬁrst card is the ace of spades. Given that it does, the player’s probability of winning is 1/n. If, however, the strategy does not guess that the ﬁrst card is the ace of spades, then the probability that the player wins is the probability that the ﬁrst card is not the ace of spades, namely, (n − 1)/n, multiplied by the conditional probability of winning given that the ﬁrst card is not the ace of spades. But this latter conditional probability is equal to the probability of winning

Section 3.3

Bayes’s Formula

67

when using an n − 1 card deck containing a single ace of spades; it is thus, by the induction hypothesis, 1/(n − 1). Hence, given that the strategy does not guess the ﬁrst card, the probability of winning is n − 1 1 1 = n n − 1 n Thus, letting G be the event that the ﬁrst card is guessed, we obtain P{win} = P{win|G}P(G) + P{win|Gc }(1 − P(G)) = =

1 1 p + (1 − p) n n

1 n

.

EXAMPLE 3c In answering a question on a multiple-choice test, a student either knows the answer or guesses. Let p be the probability that the student knows the answer and 1 − p be the probability that the student guesses. Assume that a student who guesses at the answer will be correct with probability 1/m, where m is the number of multiple-choice alternatives. What is the conditional probability that a student knew the answer to a question given that he or she answered it correctly? Solution. Let C and K denote, respectively, the events that the student answers the question correctly and the event that he or she actually knows the answer. Now, P(K|C ) =

P(KC ) P(C )

P(C|K )P(K ) P(C|K )P(K ) + P(C|Kc )P(Kc ) p = p + (1/m)(1 − p) mp = 1 + (m − 1)p

=

For example, if m = 5, p = 12 , then the probability that the student knew the answer to a question he or she answered correctly is 56 . .

EXAMPLE 3d A laboratory blood test is 95 percent effective in detecting a certain disease when it is, in fact, present. However, the test also yields a “false positive” result for 1 percent of the healthy persons tested. (That is, if a healthy person is tested, then, with probability .01, the test result will imply that he or she has the disease.) If .5 percent of the population actually has the disease, what is the probability that a person has the disease given that the test result is positive?

68

Chapter 3

Conditional Probability and Independence

Solution. Let D be the event that the person tested has the disease and E the event that the test result is positive. Then the desired probability is P(D|E) =

P(DE) P(E)

P(E|D)P(D) P(E|D)P(D) + P(E|Dc )P(Dc ) (.95)(.005) = (.95)(.005) + (.01)(.995) 95 = L .323 294 =

Thus, only 32 percent of those persons whose test results are positive actually have the disease. Many students are often surprised at this result (they expect the percentage to be much higher, since the blood test seems to be a good one), so it is probably worthwhile to present a second argument that, although less rigorous than the preceding one, is probably more revealing. We now do so. Since .5 percent of the population actually has the disease, it follows that, on the average, 1 person out of every 200 tested will have it. The test will correctly conﬁrm that this person has the disease with probability .95. Thus, on the average, out of every 200 persons tested, the test will correctly conﬁrm that .95 person has the disease. On the other hand, however, out of the (on the average) 199 healthy people, the test will incorrectly state that (199)(.01) of these people have the disease. Hence, for every .95 diseased person that the test correctly states is ill, there are (on the average) (199)(.01) healthy persons that the test incorrectly states are ill. Thus, the proportion of time that the test result is correct when it states that a person is ill is 95 .95 = L .323 .95 + (199)(.01) 294

.

Equation (3.1) is also useful when one has to reassess one’s personal probabilities in the light of additional information. For instance, consider the examples that follow. EXAMPLE 3e Consider a medical practitioner pondering the following dilemma: “If I’m at least 80 percent certain that my patient has this disease, then I always recommend surgery, whereas if I’m not quite as certain, then I recommend additional tests that are expensive and sometimes painful. Now, initially I was only 60 percent certain that Jones had the disease, so I ordered the series A test, which always gives a positive result when the patient has the disease and almost never does when he is healthy. The test result was positive, and I was all set to recommend surgery when Jones informed me, for the ﬁrst time, that he was diabetic. This information complicates matters because, although it doesn’t change my original 60 percent estimate of his chances of having the disease in question, it does affect the interpretation of the results of the A test. This is so because the A test, while never yielding a positive result when the patient is healthy, does unfortunately yield a positive result 30 percent of the time in the case of diabetic patients who are not suffering from the disease. Now what do I do? More tests or immediate surgery?” Solution. In order to decide whether or not to recommend surgery, the doctor should ﬁrst compute her updated probability that Jones has the disease given that the A test

Section 3.3

Bayes’s Formula

69

result was positive. Let D denote the event that Jones has the disease and E the event that the A test result is positive. The desired conditional probability is then P(D|E) =

P(DE) P(E)

P(D)P(E|D) P(E|D)P(D) + P(E|Dc )P(Dc ) (.6)1 = 1(.6) + (.3)(.4) = .833

=

Note that we have computed the probability of a positive test result by conditioning on whether or not Jones has the disease and then using the fact that, because Jones is a diabetic, his conditional probability of a positive result given that he does not have the disease, P(E|Dc ), equals .3. Hence, as the doctor should now be over 80 percent certain that Jones has the disease, she should recommend surgery. . EXAMPLE 3f At a certain stage of a criminal investigation, the inspector in charge is 60 percent convinced of the guilt of a certain suspect. Suppose, however, that a new piece of evidence which shows that the criminal has a certain characteristic (such as left-handedness, baldness, or brown hair) is uncovered. If 20 percent of the population possesses this characteristic, how certain of the guilt of the suspect should the inspector now be if it turns out that the suspect has the characteristic? Solution. Letting G denote the event that the suspect is guilty and C the event that he possesses the characteristic of the criminal, we have P(G|C) =

P(GC) P(C)

P(C|G)P(G) P(C|G)P(G) + P(C|Gc )P(Gc ) 1(.6) = 1(.6) + (.2)(.4) L .882 =

where we have supposed that the probability of the suspect having the characteristic if he is, in fact, innocent is equal to .2, the proportion of the population possessing the characteristic. . EXAMPLE 3g In the world bridge championships held in Buenos Aires in May 1965, the famous British bridge partnership of Terrence Reese and Boris Schapiro was accused of cheating by using a system of ﬁnger signals that could indicate the number of hearts held by the players. Reese and Schapiro denied the accusation, and eventually a hearing was held by the British bridge league. The hearing was in the form of a legal proceeding with prosecution and defense teams, both having the power to call and cross-examine witnesses. During the course of the proceeding, the prosecutor examined speciﬁc hands played by Reese and Schapiro and claimed that their playing these

70

Chapter 3

Conditional Probability and Independence

hands was consistent with the hypothesis that they were guilty of having illicit knowledge of the heart suit. At this point, the defense attorney pointed out that their play of these hands was also perfectly consistent with their standard line of play. However, the prosecution then argued that, as long as their play was consistent with the hypothesis of guilt, it must be counted as evidence toward that hypothesis. What do you think of the reasoning of the prosecution? Solution. The problem is basically one of determining how the introduction of new evidence (in the preceding example, the playing of the hands) affects the probability of a particular hypothesis. If we let H denote a particular hypothesis (such as the hypothesis that Reese and Schapiro are guilty) and E the new evidence, then P(H|E) = =

P(HE) P(E) P(E|H)P(H) P(E|H)P(H) + P(E|H c )[1 − P(H)]

(3.2)

where P(H) is our evaluation of the likelihood of the hypothesis before the introduction of the new evidence. The new evidence will be in support of the hypothesis whenever it makes the hypothesis more likely—that is, whenever P(H|E) Ú P(H). From Equation (3.2), this will be the case whenever P(E|H) Ú P(E|H)P(H) + P(E|H c )[1 − P(H)] or, equivalently, whenever

P(E|H) Ú P(E|H c )

In other words, any new evidence can be considered to be in support of a particular hypothesis only if its occurrence is more likely when the hypothesis is true than when it is false. In fact, the new probability of the hypothesis depends on its initial probability and the ratio of these conditional probabilities, since, from Equation (3.2), P(H|E) =

P(H) P(H) + [1 − P(H)]

P(E|H c ) P(E|H)

Hence, in the problem under consideration, the play of the cards can be considered to support the hypothesis of guilt only if such play would have been more likely if the partnership were cheating than if they were not. As the prosecutor never made this claim, his assertion that the evidence is in support of the guilt hypothesis is invalid. . When the author of this text drinks iced tea at a coffee shop, he asks for a glass of water along with the (same-sized) glass of tea. As he drinks the tea, he continuously reﬁlls the tea glass with water. Assuming a perfect mixing of water and tea, he wondered about the probability that his ﬁnal gulp would be tea. This wonderment led to part (a) of the following problem and to a very interesting answer. EXAMPLE 3h Urn 1 initially has n red molecules and urn 2 has n blue molecules. Molecules are randomly removed from urn 1 in the following manner: After each removal from urn 1, a molecule is taken from urn 2 (if urn 2 has any molecules) and placed in urn 1. The process continues until all the molecules have been removed. (Thus, there are 2n removals in all.)

Section 3.3

Bayes’s Formula

71

(a) Find P(R), where R is the event that the ﬁnal molecule removed from urn 1 is red. (b) Repeat the problem when urn 1 initially has r1 red molecules and b1 blue molecules and urn 2 initially has r2 red molecules and b2 blue molecules. Solution. (a) Focus attention on any particular red molecule, and let F be the event that this molecule is the ﬁnal one selected. Now, in order for F to occur, the molecule in question must still be in the urn after the ﬁrst n molecules have been removed (at which time urn 2 is empty). So, letting Ni be the event that this molecule is not the ith molecule to be removed, we have P(F) = P(N1 · · · Nn F) = P(N1 )P(N2 |N1 ) · · · P(Nn |N1 · · · Nn−1 )P(F|N1 · · · Nn ) 1 1 1 = 1 − ··· 1 − n n n where the preceding formula uses the fact that the conditional probability that the molecule under consideration is the ﬁnal molecule to be removed, given that it is still in urn 1 when only n molecules remain, is, by symmetry, 1/n. Therefore, if we number the n red molecules and let Rj be the event that red molecule number j is the ﬁnal molecule removed, then it follows from the preceding formula that 1 n 1 P(Rj ) = 1 − n n Because the events Rj are mutually exclusive, we obtain ⎞ ⎛ n n 1 ⎜ ⎟ P(Rj ) = (1 − )n L e−1 P(R) = P ⎝ Rj ⎠ = n j=1

j=1

(b) Suppose now that urn i initially has ri red and bi blue molecules, for i = 1, 2. To ﬁnd P(R), the probability that the ﬁnal molecule removed is red, focus attention on any molecule that is initially in urn 1. As in part (a), it follows that the probability that this molecule is the ﬁnal one removed is r2 +b2 1 1 p= 1 − r1 + b1 r1 + b1 r2 +b2 1 That is, 1 − r +b is the probability that the molecule under consideration 1

1

1 is the conditional probability, is still in urn 1 when urn 2 becomes empty, and r +b 1 1 given the preceding event, that the molecule under consideration is the ﬁnal molecule removed. Hence, if we let O be the event that the last molecule removed is one of the molecules originally in urn 1, then r2 +b2 1 P(O) = (r1 + b1 )p = 1 − r1 + b1

To determine P(R), we condition on whether O occurs, to obtain P(R) = P(R|O)P(O) + P(R|Oc )P(Oc ) r2 +b2 r2 +b2 r1 r2 1 1 = + 1 − 1 − 1 − r1 + b1 r1 + b1 r2 + b2 r1 + b1

72

Chapter 3

Conditional Probability and Independence

If r1 + b1 = r2 + b2 = n, so that both urns initially have n molecules, then, when n is large, r1 r2 . P(L) L e−1 + (1 − e−1 ) r1 + b1 r2 + b2 The change in the probability of a hypothesis when new evidence is introduced can be expressed compactly in terms of the change in the odds of that hypothesis, where the concept of odds is deﬁned as follows. Deﬁnition The odds of an event A are deﬁned by P(A) P(A) = P(Ac ) 1 − P(A) That is, the odds of an event A tell how much more likely it is that the event A occurs than it is that it does not occur. For instance, if P(A) = 23 , then P(A) = 2P(Ac ), so the odds are 2. If the odds are equal to α, then it is common to say that the odds are “α to 1” in favor of the hypothesis. Consider now a hypothesis H that is true with probability P(H), and suppose that new evidence E is introduced. Then the conditional probabilities, given the evidence E, that H is true and that H is not true are respectively given by P(H|E) =

P(E|H)P(H) P(E)

P(H c |E) =

P(E|H c )P(H c ) P(E)

Therefore, the new odds after the evidence E has been introduced are P(H) P(E|H) P(H|E) = P(H c |E) P(H c ) P(E|H c )

(3.3)

That is, the new value of the odds of H is the old value, multiplied by the ratio of the conditional probability of the new evidence given that H is true to the conditional probability given that H is not true. Thus, Equation (3.3) veriﬁes the result of Example 3f, since the odds, and thus the probability of H, increase whenever the new evidence is more likely when H is true than when it is false. Similarly, the odds decrease whenever the new evidence is more likely when H is false than when it is true. EXAMPLE 3i An urn contains two type A coins and one type B coin. When a type A coin is ﬂipped, it comes up heads with probability 1/4, whereas when a type B coin is ﬂipped, it comes up heads with probability 3/4. A coin is randomly chosen from the urn and ﬂipped. Given that the ﬂip landed on heads, what is the probability that it was a type A coin? Solution. Let A be the event that a type A coin was ﬂipped, and let B = Ac be the event that a type B coin was ﬂipped. We want P(A|heads), where heads is the event that the ﬂip landed on heads. From Equation (3.3), we see that P(A) P(heads|A) P(A|heads) = P(Ac |heads) P(B) P(heads|B) 2/3 1/4 = 1/3 3/4 = 2/3

Section 3.3

Bayes’s Formula

73

Hence, the odds are 2/3 : 1, or, equivalently, the probability is 2/5 that a type A coin was ﬂipped. . Equation (3.1) may be generalized as follows: Suppose that F1 , F2 , . . . , Fn are mutually exclusive events such that n

Fi = S

i=1

In other words, exactly one of the events F1 , F2 , . . . , Fn must occur. By writing E=

n

EFi

i=1

and using the fact that the events EFi , i = 1, . . . , n are mutually exclusive, we obtain P(E) =

n

P(EFi )

i=1

=

n

P(E|Fi )P(Fi )

(3.4)

i=1

Thus, Equation (3.4) shows how, for given events F1 , F2 , . . . , Fn , of which one and only one must occur, we can compute P(E) by ﬁrst conditioning on which one of the Fi occurs. That is, Equation (3.4) states that P(E) is equal to a weighted average of P(E|Fi ), each term being weighted by the probability of the event on which it is conditioned. EXAMPLE 3j In Example 5j of Chapter 2, we considered the probability that, for a randomly shufﬂed deck, the card following the ﬁrst ace is some speciﬁed card, and we gave a combi1 natorial argument to show that this probability is 52 . Here is a probabilistic argument based on conditioning: Let E be the event that the card following the ﬁrst ace is some speciﬁed card, say, card x. To compute P(E), we ignore card x and condition on the relative ordering of the other 51 cards in the deck. Letting O be the ordering gives P(E|O)P(O) P(E) = O

Now, given O, there are 52 possible orderings of the cards, corresponding to having card x being the ith card in the deck, i = 1,..., 52. But because all 52! possible orderings were initially equally likely, it follows that, conditional on O, each of the 52 remaining possible orderings is equally likely. Because card x will follow the ﬁrst ace for only one of these orderings, we have P(E|O) = 1/52, implying that P(E) = 1/52. . Again, let F1 ,..., Fn be a set of mutually exclusive and exhaustive events (meaning that exactly one of these events must occur). Suppose now that E has occurred and we are interested in determining which one of the Fj also occurred. Then, by Equation (3.4), we have the following proposition.

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Proposition 3.1. P(EFj ) P(E) P(E|Fj )P(Fj ) = n P(E|Fi )P(Fi )

P(Fj |E) =

(3.5)

i=1

Equation (3.5) is known as Bayes’s formula, after the English philosopher Thomas Bayes. If we think of the events Fj as being possible “hypotheses” about some subject matter, then Bayes’s formula may be interpreted as showing us how opinions about these hypotheses held before the experiment was carried out [that is, the P(Fj )] should be modiﬁed by the evidence produced by the experiment. EXAMPLE 3k A plane is missing, and it is presumed that it was equally likely to have gone down in any of 3 possible regions. Let 1 − βi , i = 1, 2, 3, denote the probability that the plane will be found upon a search of the ith region when the plane is, in fact, in that region. (The constants βi are called overlook probabilities, because they represent the probability of overlooking the plane; they are generally attributable to the geographical and environmental conditions of the regions.) What is the conditional probability that the plane is in the ith region given that a search of region 1 is unsuccessful? Solution. Let Ri , i = 1, 2, 3, be the event that the plane is in region i, and let E be the event that a search of region 1 is unsuccessful. From Bayes’s formula, we obtain P(R1 |E) = =

P(ER1 ) P(E) P(E|R1 )P(R1 ) 3

P(E|Ri )P(Ri )

i=1

=

(β1 ) 13

(β1 ) 13 + (1) 13 + (1) 13 β1 = β1 + 2 For j = 2, 3, P(Rj |E) = =

P(E|Rj )P(Rj ) P(E) (1) 13

(β1 ) 13 + 1 = β1 + 2

1 3

+

1 3

j = 2, 3

Note that the updated (that is, the conditional) probability that the plane is in region j, given the information that a search of region 1 did not ﬁnd it, is greater than

Section 3.3

Bayes’s Formula

75

the initial probability that it was in region j when j Z 1 and is less than the initial probability when j = 1. This statement is certainly intuitive, since not ﬁnding the plane in region 1 would seem to decrease its chance of being in that region and increase its chance of being elsewhere. Further, the conditional probability that the plane is in region 1 given an unsuccessful search of that region is an increasing function of the overlook probability β1 . This statement is also intuitive, since the larger β1 is, the more it is reasonable to attribute the unsuccessful search to “bad luck” as opposed to the plane’s not being there. Similarly, P(Rj |E), j Z 1, is a decreasing function . of β1 . The next example has often been used by unscrupulous probability students to win money from their less enlightened friends. EXAMPLE 3l Suppose that we have 3 cards that are identical in form, except that both sides of the ﬁrst card are colored red, both sides of the second card are colored black, and one side of the third card is colored red and the other side black. The 3 cards are mixed up in a hat, and 1 card is randomly selected and put down on the ground. If the upper side of the chosen card is colored red, what is the probability that the other side is colored black? Solution. Let RR, BB, and RB denote, respectively, the events that the chosen card is all red, all black, or the red–black card. Also, let R be the event that the upturned side of the chosen card is red. Then the desired probability is obtained by P(RB|R) =

P(RB ∩ R) P(R)

P(R|RB)P(RB) P(R|RR)P(RR) + P(R|RB)P(RB) + P(R|BB)P(BB) 1 1 2 3 1 = = 1 1 1 1 3 + + 0 (1)

=

3

2

3

3

Hence, the answer is 13 . Some students guess 12 as the answer by incorrectly reasoning that, given that a red side appears, there are two equally likely possibilities: that the card is the all-red card or the red–black card. Their mistake, however, is in assuming that these two possibilities are equally likely. For, if we think of each card as consisting of two distinct sides, then we see that there are 6 equally likely outcomes of the experiment—namely, R1 , R2 , B1 , B2 , R3 , B3 —where the outcome is R1 if the ﬁrst side of the all-red card is turned face up, R2 if the second side of the all-red card is turned face up, R3 if the red side of the red–black card is turned face up, and so on. Since the other side of the upturned red side will be black only if the outcome is R3 , we see that the desired probability is the conditional probability of R3 given that either R1 or R2 or R3 occurred, which obviously equals 13 . . EXAMPLE 3m A new couple, known to have two children, has just moved into town. Suppose that the mother is encountered walking with one of her children. If this child is a girl, what is the probability that both children are girls?

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Solution. Let us start by deﬁning the following events: G1 : the ﬁrst (that is, the oldest) child is a girl. G2 : the second child is a girl. G: the child seen with the mother is a girl. Also, let B1 , B2 , and B denote similar events, except that “girl” is replaced by “boy.” Now, the desired probability is P(G1 G2 |G), which can be expressed as follows: P(G1 G2 G) P(G) P(G1 G2 ) = P(G)

P(G1 G2 |G) =

Also, P(G) = P(G|G1 G2 )P(G1 G2 ) + P(G|G1 B2 )P(G1 B2 ) + P(G|B1 G2 )P(B1 G2 ) + P(G|B1 B2 )P(B1 B2 ) = P(G1 G2 ) + P(G|G1 B2 )P(G1 B2 ) + P(G|B1 G2 )P(B1 G2 ) where the ﬁnal equation used the results P(G|G1 G2 ) = 1 and P(G|B1 B2 ) = 0. If we now make the usual assumption that all 4 gender possibilities are equally likely, then we see that P(G1 G2 |G) =

1 4

+ P(G|G1 B2 )/4 + P(G|B1 G2 )/4 1 = 1 + P(G|G1 B2 ) + P(G|B1 G2 ) 1 4

Thus, the answer depends on whatever assumptions we want to make about the conditional probabilities that the child seen with the mother is a girl given the event G1 B2 and that the child seen with the mother is a girl given the event G2 B1 . For instance, if we want to assume, on the one hand, that, independently of the genders of the children, the child walking with the mother is the elder child with some probability p, then it would follow that P(G|G1 B2 ) = p = 1 − P(G|B1 G2 ) implying under this scenario that P(G1 G2 |G) =

1 2

If, on the other hand, we were to assume that if the children are of different genders, then the mother would choose to walk with the girl with probability q, independently of the birth order of the children, then we would have P(G|G1 B2 ) = P(G|B1 G2 ) = q implying that P(G1 G2 |G) =

1 1 + 2q

For instance, if we took q = 1, meaning that the mother would always choose to walk with a daughter, then the conditional probability the she has two daughters would be

Section 3.3

Bayes’s Formula

77

1 3,

which is in accord with Example 2b because seeing the mother with a daughter is now equivalent to the event that she has at least one daughter. Hence, as stated, the problem is incapable of solution. Indeed, even when the usual assumption about equally likely gender probabilities is made, we still need to make additional assumptions before a solution can be given. This is because the sample space of the experiment consists of vectors of the form s1 , s2 , i, where s1 is the gender of the older child, s2 is the gender of the younger child, and i identiﬁes the birth order of the child seen with the mother. As a result, to specify the probabilities of the events of the sample space, it is not enough to make assumptions only about the genders of the children; it is also necessary to assume something about the conditional probabilities as to which child is with the mother given the genders of the children. . EXAMPLE 3n A bin contains 3 different types of disposable ﬂashlights. The probability that a type 1 ﬂashlight will give over 100 hours of use is .7, with the corresponding probabilities for type 2 and type 3 ﬂashlights being .4 and .3, respectively. Suppose that 20 percent of the ﬂashlights in the bin are type 1, 30 percent are type 2, and 50 percent are type 3. (a) What is the probability that a randomly chosen ﬂashlight will give more than 100 hours of use? (b) Given that a ﬂashlight lasted over 100 hours, what is the conditional probability that it was a type j ﬂashlight, j = 1, 2, 3? Solution. (a) Let A denote the event that the ﬂashlight chosen will give over 100 hours of use, and let Fj be the event that a type j ﬂashlight is chosen, j = 1, 2, 3. To compute P(A), we condition on the type of the ﬂashlight, to obtain P(A) = P(A|F1 )P(F1 ) + P(A|F2 )P(F2 ) + P(A|F3 )P(F3 ) = (.7)(.2) + (.4)(.3) + (.3)(.5) = .41 There is a 41 percent chance that the ﬂashlight will last for over 100 hours. (b) The probability is obtained by using Bayes’s formula: P(AFj ) P(A) P(A|Fj )P(Fj ) = .41

P(Fj |A) =

Thus, P(F1 |A) = (.7)(.2)/.41 = 14/41 P(F2 |A) = (.4)(.3)/.41 = 12/41 P(F3 |A) = (.3)(.5)/.41 = 15/41 For instance, whereas the initial probability that a type 1 ﬂashlight is chosen is only .2, the information that the ﬂashlight has lasted over 100 hours raises the probability of this event to 14/41 L .341. . EXAMPLE 3o A crime has been committed by a solitary individual, who left some DNA at the scene of the crime. Forensic scientists who studied the recovered DNA noted that

78

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only ﬁve strands could be identiﬁed and that each innocent person, independently, would have a probability of 10−5 of having his or her DNA match on all ﬁve strands. The district attorney supposes that the perpetrator of the crime could be any of the one million residents of the town. Ten thousand of these residents have been released from prison within the past 10 years; consequently, a sample of their DNA is on ﬁle. Before any checking of the DNA ﬁle, the district attorney feels that each of the ten thousand ex-criminals has probability α of being guilty of the new crime, while each of the remaining 990,000 residents has probability β, where α = cβ. (That is, the district attorney supposes that each recently released convict is c times as likely to be the crime’s perpetrator as is each town member who is not a recently released convict.) When the DNA that is analyzed is compared against the database of the ten thousand ex-convicts, it turns out that A. J. Jones is the only one whose DNA matches the proﬁle. Assuming that the district attorney’s estimate of the relationship between α and β is accurate, what is the probability that A. J. is guilty? Solution. To begin, note that, because probabilities must sum to 1, we have 1 = 10,000α + 990,000β = (10,000c + 990,000)β Thus, β=

1 , 10,000c + 990,000

α=

c 10,000c + 990,000

Now, let G be the event that A. J. is guilty, and let M denote the event that A. J. is the only one of the ten thousand on ﬁle to have a match. Then P(G|M) = =

P(GM) P(M) P(G)P(M|G) P(M|G)P(G) + P(M|Gc )P(Gc )

On the one hand, if A. J. is guilty, then he will be the only one to have a DNA match if none of the others on ﬁle have a match. Therefore, P(M|G) = (1 − 10−5 )9999 On the other hand, if A. J. is innocent, then in order for him to be the only match, his DNA must match (which will occur with probability 10−5 ), all others in the database must be innocent, and none of these others can have a match. Now, given that A. J. is innocent, the conditional probability that all the others in the database are also innocent is P(all in database innocent) P(AJ innocent) 1 − 10,000α = 1 − α

P(all others innocent|AJ innocent) =

Also, the conditional probability, given their innocence, that none of the others in the database will have a match is (1 − 10−5 )9999 . Therefore, 1 − 10,000α (1 − 10−5 )9999 P(M|Gc ) = 10−5 1 − α

Section 3.4

Independent Events

79

Because P(G) = α, the preceding formula gives P(G|M) =

α α +

10−5 (1

− 10,000α)

=

1 .9 +

10−5 α

Thus, if the district attorney’s initial feelings were that an arbitrary ex-convict was 100 times more likely to have committed the crime than was a nonconvict (that is, 1 and c = 100), then α = 19,900 P(G|M) =

1 L 0.9099 1.099

If the district attorney initially felt that the appropriate ratio was c = 10, then α = 1 and 109, 000 1 P(G|M) = L 0.5025 1.99 If the district attorney initially felt that the criminal was equally likely to be any of the members of the town (c = 1), then α = 10−6 and P(G|M) =

1 L 0.0917 10.9

Thus, the probability ranges from approximately 9 percent when the district attorney’s initial assumption is that all the members of the population have the same chance of being the perpetrator to approximately 91 percent when she assumes that each ex-convict is 100 times more likely to be the criminal than is a speciﬁed townsperson who is not an ex-convict. . 3.4

INDEPENDENT EVENTS The previous examples of this chapter show that P(E|F), the conditional probability of E given F, is not generally equal to P(E), the unconditional probability of E. In other words, knowing that F has occurred generally changes the chances of E’s occurrence. In the special cases where P(E|F) does in fact equal P(E), we say that E is independent of F. That is, E is independent of F if knowledge that F has occurred does not change the probability that E occurs. Since P(E|F) = P(EF)/P(F), it follows that E is independent of F if P(EF) = P(E)P(F)

(4.1)

The fact that Equation (4.1) is symmetric in E and F shows that whenever E is independent of F, F is also independent of E. We thus have the following deﬁnition. Deﬁnition Two events E and F are said to be independent if Equation (4.1) holds. Two events E and F that are not independent are said to be dependent. EXAMPLE 4a A card is selected at random from an ordinary deck of 52 playing cards. If E is the event that the selected card is an ace and F is the event that it is a spade, then E

80

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and F are independent. This follows because P(EF) = P(F) = 13 52 .

1 52 ,

whereas P(E) =

4 52

and .

EXAMPLE 4b Two coins are ﬂipped, and all 4 outcomes are assumed to be equally likely. If E is the event that the ﬁrst coin lands on heads and F the event that the second lands on tails, then E and F are independent, since P(EF) = P({(H, T)}) = 14 , whereas . P(E) = P({(H, H), (H, T)}) = 12 and P(F) = P({(H, T), (T, T)}) = 12 . EXAMPLE 4c Suppose that we toss 2 fair dice. Let E1 denote the event that the sum of the dice is 6 and F denote the event that the ﬁrst die equals 4. Then P(E1 F) = P({(4, 2)}) = whereas

P(E1 )P(F) =

5 36

1 36

1 5 = 6 216

Hence, E1 and F are not independent. Intuitively, the reason for this is clear because if we are interested in the possibility of throwing a 6 (with 2 dice), we shall be quite happy if the ﬁrst die lands on 4 (or, indeed, on any of the numbers 1, 2, 3, 4, and 5), for then we shall still have a possibility of getting a total of 6. If, however, the ﬁrst die landed on 6, we would be unhappy because we would no longer have a chance of getting a total of 6. In other words, our chance of getting a total of 6 depends on the outcome of the ﬁrst die; thus, E1 and F cannot be independent. Now, suppose that we let E2 be the event that the sum of the dice equals 7. Is E2 independent of F? The answer is yes, since P(E2 F) = P({(4, 3)}) = whereas P(E2 )P(F) =

1 36

1 1 1 = 6 6 36

We leave it for the reader to present the intuitive argument why the event that the sum of the dice equals 7 is independent of the outcome on the ﬁrst die. . EXAMPLE 4d If we let E denote the event that the next president is a Republican and F the event that there will be a major earthquake within the next year, then most people would probably be willing to assume that E and F are independent. However, there would probably be some controversy over whether it is reasonable to assume that E is independent of G, where G is the event that there will be a recession within two years after the election. . We now show that if E is independent of F, then E is also independent of F c .

Section 3.4

Independent Events

81

Proposition 4.1. If E and F are independent, then so are E and F c . Proof. Assume that E and F are independent. Since E = EF ∪ EF c and EF and EF c are obviously mutually exclusive, we have P(E) = P(EF) + P(EF c ) = P(E)P(F) + P(EF c ) or, equivalently, P(EF c ) = P(E)[1 − P(F)] = P(E)P(F c ) and the result is proved. Thus, if E is independent of F, then the probability of E’s occurrence is unchanged by information as to whether or not F has occurred. Suppose now that E is independent of F and is also independent of G. Is E then necessarily independent of FG? The answer, somewhat surprisingly, is no, as the following example demonstrates. EXAMPLE 4e Two fair dice are thrown. Let E denote the event that the sum of the dice is 7. Let F denote the event that the ﬁrst die equals 4 and G denote the event that the second die equals 3. From Example 4c, we know that E is independent of F, and the same reasoning as applied there shows that E is also independent of G; but clearly, E is not independent of FG [since P(E|FG) = 1]. . It would appear to follow from Example 4e that an appropriate deﬁnition of the independence of three events E, F, and G would have to go further than merely 3 assuming that all of the pairs of events are independent. We are thus led to the 2 following deﬁnition. Deﬁnition Three events E, F, and G are said to be independent if P(EFG) = P(E)P(F)P(G) P(EF) = P(E)P(F) P(EG) = P(E)P(G) P(FG) = P(F)P(G) Note that if E, F, and G are independent, then E will be independent of any event formed from F and G. For instance, E is independent of F ∪ G, since P[E(F ∪ G)] = P(EF ∪ EG) = P(EF) + P(EG) − P(EFG) = P(E)P(F) + P(E)P(G) − P(E)P(FG) = P(E)[P(F) + P(G) − P(FG)] = P(E)P(F ∪ G)

82

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Of course, we may also extend the deﬁnition of independence to more than three events. The events E1 , E2 , . . . , En are said to be independent if, for every subset E1 , E2 , . . . , Er , r … n, of these events, P(E1 E2 · · · Er ) = P(E1 )P(E2 ) · · · P(Er ) Finally, we deﬁne an inﬁnite set of events to be independent if every ﬁnite subset of those events is independent. Sometimes, a probability experiment under consideration consists of performing a sequence of subexperiments. For instance, if the experiment consists of continually tossing a coin, we may think of each toss as being a subexperiment. In many cases, it is reasonable to assume that the outcomes of any group of the subexperiments have no effect on the probabilities of the outcomes of the other subexperiments. If such is the case, we say that the subexperiments are independent. More formally, we say that the subexperiments are independent if E1 , E2 , . . . , En , . . . is necessarily an independent sequence of events whenever Ei is an event whose occurrence is completely determined by the outcome of the ith subexperiment. If each subexperiment has the same set of possible outcomes, then the subexperiments are often called trials. EXAMPLE 4f An inﬁnite sequence of independent trials is to be performed. Each trial results in a success with probability p and a failure with probability 1 − p. What is the probability that (a) at least 1 success occurs in the ﬁrst n trials; (b) exactly k successes occur in the ﬁrst n trials; (c) all trials result in successes? Solution. In order to determine the probability of at least 1 success in the ﬁrst n trials, it is easiest to compute ﬁrst the probability of the complementary event: that of no successes in the ﬁrst n trials. If we let Ei denote the event of a failure on the ith trial, then the probability of no successes is, by independence, P(E1 E2 · · · En ) = P(E1 )P(E2 ) · · · P(En ) = (1 − p)n Hence, the answer to part (a) is 1 − (1 − p)n . To compute the answer to part (b), consider any particular sequence of the ﬁrst n outcomes containing k successes and n − k failures. Each one of these sequences k n−k will, by the assumed independence of trials, occur with probability p (1 − p) . n such sequences (there are n!/k!(n − k)! permutations of k Since there are k successes and n − k failures), the desired probability in part (b) is n pk (1 − p)n−k P{exactly k successes} = k To answer part (c), we note that, by part (a), the probability of the ﬁrst n trials all resulting in success is given by P(E1c E2c · · · Enc ) = pn

Section 3.4

Independent Events

83

Thus, using the continuity property of probabilities (Section 2.6), we see that the desired probability is given by ⎛ ⎞ ⎛ ⎞ q n Eic ⎠ P ⎝ Eic ⎠ = P ⎝ lim n→q i=1 i=1 ⎞ ⎛ n = lim P ⎝ Eic ⎠ n→q i=1 % 0 if p < 1 . = lim pn = 1 if p = 1 n EXAMPLE 4g A system composed of n separate components is said to be a parallel system if it functions when at least one of the components functions. (See Figure 3.2.) For such a system, if component i, which is independent of the other components, functions with probability pi , i = 1, . . . , n, what is the probability that the system functions? Solution. Let Ai denote the event that component i functions. Then P{system functions} = 1 − P{system does not function} = 1 − P{all components do not function} ⎞ ⎛ = 1 − P ⎝ Aci ⎠ i

=1 −

n

(1 − pi ) by independence

.

i=1

1 2 A

B 3 n

FIGURE 3.2: Parallel System: Functions if Current Flows from A to B

EXAMPLE 4h Independent trials consisting of rolling a pair of fair dice are performed. What is the probability that an outcome of 5 appears before an outcome of 7 when the outcome of a roll is the sum of the dice? Solution. If we let En denote the event that no 5 or 7 appears on the ﬁrst n − 1 trials and a 5 appears on the nth trial, then the desired probability is ⎛ ⎞ q q ⎝ ⎠ P En = P(En ) n=1

n=1

84

Chapter 3

Conditional Probability and Independence

Now, since P{5 on any trial} = pendence of trials,

4 36

and P{7 on any trial} =

10 P(En ) = 1 − 36 Thus,

⎛ P⎝

q

n=1

n−1

6 36 ,

we obtain, by the inde-

4 36

⎞

q 1 13 n−1 ⎠ En = 9 18 n=1

1 1 91 − 2 = 5 =

13 18

This result could also have been obtained by the use of conditional probabilities. If we let E be the event that a 5 occurs before a 7, then we can obtain the desired probability, P(E), by conditioning on the outcome of the ﬁrst trial, as follows: Let F be the event that the ﬁrst trial results in a 5, let G be the event that it results in a 7, and let H be the event that the ﬁrst trial results in neither a 5 nor a 7. Then, conditioning on which one of these events occurs gives P(E) = P(E|F)P(F) + P(E|G)P(G) + P(E|H)P(H) However, P(E|F) = 1 P(E|G) = 0 P(E|H) = P(E) The ﬁrst two equalities are obvious. The third follows because if the ﬁrst outcome results in neither a 5 nor a 7, then at that point the situation is exactly as it was when the problem ﬁrst started—namely, the experimenter will continually roll a pair of fair dice until either a 5 or 7 appears. Furthermore, the trials are independent; therefore, the outcome of the ﬁrst trial will have no effect on subsequent rolls of the dice. Since 4 6 , P(G) = 36 , and P(H) = 26 P(F) = 36 36 , it follows that P(E) =

13 1 + P(E) 9 18

or P(E) =

2 5

The reader should note that the answer is quite intuitive. That is, because a 5 occurs 4 6 on any roll with probability 36 and a 7 with probability 36 , it seems intuitive that the odds that a 5 appears before a 7 should be 6 to 4 against. The probability should then 4 , as indeed it is. be 10 The same argument shows that if E and F are mutually exclusive events of an experiment, then, when independent trials of the experiment are performed, the event E will occur before the event F with probability P(E) P(E) + P(F)

.

Section 3.4

Independent Events

85

EXAMPLE 4i There are n types of coupons, and each new one collected is independently of type i with probability pi , ni=1 pi = 1. Suppose k coupons are to be collected. If Ai is the event that there is at least one type i coupon among those collected, then, for i Z j, ﬁnd (a) P(Ai ) (b) P(Ai ∪ Aj ) (c) P(Ai |Aj ) Solution. P(Ai ) = 1 − P(Aci ) = 1 − P{no coupon is type i} = 1 − (1 − pi )k where the preceding used that each coupon is, independently, not of type i with probability 1 − pi . Similarly, P(Ai ∪ Aj ) = 1 − P( Ai ∪ Aj )c = 1 − P{no coupon is either type i or type j} = 1 − (1 − pi − pj )k where the preceding used that each coupon is, independently, neither of type i nor type j with probability 1 − pi − pj . To determine P(Ai |Aj ), we will use the identity P(Ai ∪ Aj ) = P(Ai ) + P(Aj ) − P(Ai Aj ) which, in conjunction with parts (a) and (b), yields P(Ai Aj ) = 1 − (1 − pi )k + 1 − (1 − pj )k − [1 − (1 − pi − pj )k ] = 1 − (1 − pi )k − (1 − pj )k + (1 − pi − pj )k Consequently, P(Ai |Aj ) =

1 − (1 − pi )k − (1 − pj )k + (1 − pi − pj )k P(Ai Aj ) = P(Aj ) 1 − (1 − pj )k

.

The next example presents a problem that occupies an honored place in the history of probability theory. This is the famous problem of the points. In general terms, the problem is this: Two players put up stakes and play some game, with the stakes to go to the winner of the game. An interruption requires them to stop before either has won and when each has some sort of a “partial score.” How should the stakes be divided? This problem was posed to the French mathematician Blaise Pascal in 1654 by ´ e, ´ who was a professional gambler at that time. In attacking the Chevalier de Mer the problem, Pascal introduced the important idea that the proportion of the prize deserved by the competitors should depend on their respective probabilities of winning if the game were to be continued at that point. Pascal worked out some special cases and, more importantly, initiated a correspondence with the famous Frenchman Pierre de Fermat, who had a great reputation as a mathematician. The resulting exchange of letters not only led to a complete solution to the problem of the points,

86

Chapter 3

Conditional Probability and Independence

but also laid the framework for the solution to many other problems connected with games of chance. This celebrated correspondence, dated by some as the birth date of probability theory, was also important in stimulating interest in probability among the mathematicians in Europe, for Pascal and Fermat were both recognized as being among the foremost mathematicians of the time. For instance, within a short time of their correspondence, the young Dutch mathematician Christiaan Huygens came to Paris to discuss these problems and solutions, and interest and activity in this new ﬁeld grew rapidly. EXAMPLE 4j The problem of the points Independent trials resulting in a success with probability p and a failure with probability 1 − p are performed. What is the probability that n successes occur before m failures? If we think of A and B as playing a game such that A gains 1 point when a success occurs and B gains 1 point when a failure occurs, then the desired probability is the probability that A would win if the game were to be continued in a position where A needed n and B needed m more points to win. Solution. We shall present two solutions. The ﬁrst is due to Pascal and the second to Fermat. Let us denote by Pn,m the probability that n successes occur before m failures. By conditioning on the outcome of the ﬁrst trial, we obtain Pn,m = pPn−1,m + (1 − p)Pn,m−1

n Ú 1, m Ú 1

(Why? Reason it out.) Using the obvious boundary conditions Pn, 0 = 0, P0, m = 1, we can solve these equations for Pn,m . Rather than go through the tedious details, let us instead consider Fermat’s solution. Fermat argued that, in order for n successes to occur before m failures, it is necessary and sufﬁcient that there be at least n successes in the ﬁrst m + n − 1 trials. (Even if the game were to end before a total of m + n − 1 trials were completed, we could still imagine that the necessary additional trials were performed.) This is true, for if there are at least n successes in the ﬁrst m + n − 1 trials, there could be at most m − 1 failures in those m + n − 1 trials; thus, n successes would occur before m failures. If, however, there were fewer than n successes in the ﬁrst m + n − 1 trials, there would have to be at least m failures in that same number of trials; thus, n successes would not occur before m failures. Hence, since, as shown in Example 4f, the probability of exactly k successes in m + n − 1 pk (1 − p)m+n−1−k , it follows that the desired m + n − 1 trials is k probability of n successes before m failures is m+n−1 m + n − 1 pk (1 − p)m+n−1−k Pn,m = k . k=n Our next two examples deal with gambling problems, with the ﬁrst having a surprisingly elegant analysis.∗ EXAMPLE 4k Suppose that initially there are r players, with player i having ni units, ni > 0, i = 1, . . . , r. At each stage, two of the players are chosen to play a game, with the winner ∗ The remainder of this section should be considered optional.

Section 3.4

Independent Events

87

of the game receiving 1 unit from the loser. Any player whose fortune drops to 0 is r eliminated, and this continues until a single player has all n K n i=1 i units, with that player designated as the victor. Assuming that the results of successive games are independent and that each game is equally likely to be won by either of its two players, ﬁnd Pi , the probability that player i is the victor? Solution. To begin, suppose that there are n players, with each player initially having 1 unit. Consider player i. Each stage she plays will be equally likely to result in her either winning or losing 1 unit, with the results from each stage being independent. In addition, she will continue to play stages until her fortune becomes either 0 or n. Because this is the same for all n players, it follows that each player has the same chance of being the victor, implying that each player has probability 1/n of being the victor. Now, suppose these n players are divided into r teams, with team i containing ni players, i = 1, . . . , r. Then the probability that the victor is a member of team i is ni /n. But because (a) team i initially has a total fortune of ni units, i = 1, . . . , r, and (b) each game played by members of different teams is equally likely to be won by either player and results in the fortune of members of the winning team increasing by 1 and the fortune of the members of the losing team decreasing by 1, it is easy to see that the probability that the victor is from team i is exactly the probability we desire. Thus, Pi = ni /n. Interestingly, our argument shows that this result does not depend on how the players in each stage are chosen. . In the gambler’s ruin problem, there are only 2 gamblers, but they are not assumed to be of equal skill. EXAMPLE 4l The gambler’s ruin problem Two gamblers, A and B, bet on the outcomes of successive ﬂips of a coin. On each ﬂip, if the coin comes up heads, A collects 1 unit from B, whereas if it comes up tails, A pays 1 unit to B. They continue to do this until one of them runs out of money. If it is assumed that the successive ﬂips of the coin are independent and each ﬂip results in a head with probability p, what is the probability that A ends up with all the money if he starts with i units and B starts with N − i units? Solution. Let E denote the event that A ends up with all the money when he starts with i and B starts with N − i, and to make clear the dependence on the initial fortune of A, let Pi = P(E). We shall obtain an expression for P(E) by conditioning on the outcome of the ﬁrst ﬂip as follows: Let H denote the event that the ﬁrst ﬂip lands on heads; then Pi = P(E) = P(E|H)P(H) + P(E|H c )P(H c ) = pP(E|H) + (1 − p)P(E|H c ) Now, given that the ﬁrst ﬂip lands on heads, the situation after the ﬁrst bet is that A has i + 1 units and B has N − (i + 1). Since the successive ﬂips are assumed to be independent with a common probability p of heads, it follows that, from that point on, A’s probability of winning all the money is exactly the same as if the game were just starting with A having an initial fortune of i + 1 and B having an initial fortune of N − (i + 1). Therefore, P(E|H) = Pi+1

88

Chapter 3

Conditional Probability and Independence

and similarly, P(E|H c ) = Pi−1 Hence, letting q = 1 − p, we obtain Pi = pPi+1 + qPi−1

i = 1, 2, . . . , N − 1

(4.2)

By making use of the obvious boundary conditions P0 = 0 and PN = 1, we shall now solve Equation (4.2). Since p + q = 1, these equations are equivalent to pPi + qPi = pPi+1 + qPi−1 or Pi+1 − Pi =

q (Pi − Pi−1 ) p

i = 1, 2, . . . , N − 1

(4.3)

Hence, since P0 = 0, we obtain, from Equation (4.3), q q (P1 − P0 ) = P1 p p 2 q q P3 − P2 = (P2 − P1 ) = P1 p p . . . i−1 q q Pi − Pi−1 = (Pi−1 − Pi−2 ) = P1 p p . . . N−1 q q PN − PN−1 = (PN−1 − PN−2 ) = P1 p p P2 − P1 =

Adding the ﬁrst i − 1 equations of (4.4) yields 2 i−1 q q q + Pi − P1 = P1 + ··· + p p p or

⎧ ⎪ 1 − (q/p)i ⎪ ⎪ P1 ⎨ 1 − (q/p) Pi = ⎪ ⎪ ⎪ ⎩ iP1 Using the fact that PN = 1, we obtain ⎧ ⎪ 1 − (q/p) ⎪ ⎨ N P1 = 1 − (q/p) ⎪ ⎪ ⎩ 1 N

q Z 1 p q if = 1 p if

if p Z

1 2

if p =

1 2

(4.4)

Section 3.4

Hence,

⎧ i ⎪ ⎪ ⎨ 1 − (q/p) N Pi = 1 − (q/p) ⎪ ⎪ ⎩ i N

Independent Events

if p Z

1 2

if p =

1 2

89

(4.5)

Let Qi denote the probability that B winds up with all the money when A starts with i and B starts with N − i. Then, by symmetry to the situation described, and on replacing p by q and i by N − i, it follows that ⎧ ⎪ 1 − (p/q)N−i ⎪ ⎪ if q Z 12 ⎨ N 1 − (p/q) Qi = ⎪ ⎪N − i ⎪ ⎩ if q = 12 N Moreover, since q =

1 2

is equivalent to p = 12 , we have, when q Z

Pi + Qi =

1 2,

1 − (q/p)i 1 − (p/q)N−i + 1 − (q/p)N 1 − (p/q)N

=

pN − pN (q/p)i qN − qN (p/q)N−i + pN − qN qN − pN

=

pN − pN−i qi − qN + qi pN−i pN − qN

=1 This result also holds when p = q = 12 , so Pi + Qi = 1 In words, this equation states that, with probability 1, either A or B will wind up with all of the money; in other words, the probability that the game continues indeﬁnitely with A’s fortune always being between 1 and N − 1 is zero. (The reader must be careful because, a priori, there are three possible outcomes of this gambling game, not two: Either A wins, or B wins, or the game goes on forever with nobody winning. We have just shown that this last event has probability 0.) As a numerical illustration of the preceding result, if A were to start with 5 units and B with 10, then the probability of A’s winning would be 13 if p were 12 , whereas it would jump to 5 1 − 23 15 L .87 1 − 23 if p were .6. A special case of the gambler’s ruin problem, which is also known as the problem of duration of play, was proposed to Huygens by Fermat in 1657. The version Huygens proposed, which he himself solved, was that A and B have 12 coins each. They play for these coins in a game with 3 dice as follows: Whenever 11 is thrown (by either—it makes no difference who rolls the dice), A gives a coin to B. Whenever 14 is thrown, B gives a coin to A. The person who ﬁrst wins all the coins wins the game. Since

90

Chapter 3

Conditional Probability and Independence 27 15 P{roll 11} = 216 and P{roll 14} = 216 , we see from Example 4h that, for A, this is just the gambler’s ruin problem with p = 15 42 , i = 12, and N = 24. The general form of the gambler’s ruin problem was solved by the mathematician James Bernoulli and published 8 years after his death in 1713. For an application of the gambler’s ruin problem to drug testing, suppose that two new drugs have been developed for treating a certain disease. Drug i has a cure rate Pi , i = 1, 2, in the sense that each patient treated with drug i will be cured with probability Pi . These cure rates are, however, not known, and we are interested in ﬁnding a method for deciding whether P1 > P2 or P2 > P1 . To decide on one of these alternatives, consider the following test: Pairs of patients are to be treated sequentially, with one member of the pair receiving drug 1 and the other drug 2. The results for each pair are determined, and the testing stops when the cumulative number of cures from one of the drugs exceeds the cumulative number of cures from the other by some ﬁxed, predetermined number. More formally, let % 1 if the patient in the jth pair that receives drug 1 is cured Xj = 0 otherwise % 1 if the patient in the jth pair that receives drug 2 is cured Yj = 0 otherwise

For a predetermined positive integer M, the test stops after pair N, where N is the ﬁrst value of n such that either X1 + · · · + Xn − (Y1 + · · · + Yn ) = M or

X1 + · · · + Xn − (Y1 + · · · + Yn ) = −M

In the former case, we assert that P1 > P2 and in the latter that P2 > P1 . In order to help ascertain whether the foregoing is a good test, one thing we would like to know is the probability that it leads to an incorrect decision. That is, for given P1 and P2 , where P1 > P2 , what is the probability that the test will incorrectly assert that P2 > P1 ? To determine this probability, note that after each pair is checked, the cumulative difference of cures using drug 1 versus drug 2 will go up by 1 with probability P1 (1 − P2 )—since this is the probability that drug 1 leads to a cure and drug 2 does not—or go down by 1 with probability (1 − P1 )P2 , or remain the same with probability P1 P2 + (1 − P1 )(1 − P2 ). Hence, if we consider only those pairs in which the cumulative difference changes, then the difference will go up by 1 with probability P = P{up 1|up 1 or down 1} P1 (1 − P2 ) = P1 (1 − P2 ) + (1 − P1 )P2 and down by 1 with probability 1 − P=

P2 (1 − P1 ) P1 (1 − P2 ) + (1 − P1 )P2

Thus, the probability that the test will assert that P2 > P1 is equal to the probability that a gambler who wins each (one-unit) bet with probability P will go down M before going up M. But Equation (4.5), with i = M, N = 2M, shows that this probability is given by

Section 3.4

Independent Events

91

P{test asserts that P2 > P1 } 1 − P M 1 − P =1 − 1 − P 2M 1 − P 1 =1 − 1 − P M 1 + P 1 = 1 + γM where γ =

P1 (1 − P2 ) P = 1 − P P2 (1 − P1 )

For instance, if P1 = .6 and P2 = .4, then the probability of an incorrect decision is .017 when M = 5 and reduces to .0003 when M = 10. . Suppose that we are presented with a set of elements and we want to determine whether at least one member of the set has a certain property. We can attack this question probabilistically by randomly choosing an element of the set in such a way that each element has a positive probability of being selected. Then the original question can be answered by a consideration of the probability that the randomly selected element does not have the property of interest. If this probability is equal to 1, then none of the elements of the set have the property; if it is less than 1, then at least one element of the set has the property. The ﬁnal example of this section illustrates this technique. EXAMPLE 4m The complete graph having n vertices is deﬁned to be a set of n points (called vertices) n lines (called edges) connecting each pair of vertices. The in the plane and the 2 complete graph having 3 vertices is shown in Figure 3.3. Suppose now that each edge in a complete graph having n vertices is to be colored either red or blue. For a ﬁxed integer k, a question of interest is, Is there a way of coloring the edges so that no set k connecting edges the same color? It can be shown by of k vertices has all of its 2 a probabilistic argument that if n is not too large, then the answer is yes.

FIGURE 3.3

92

Chapter 3

Conditional Probability and Independence

The argument runs as follows: Suppose that each edge is, independently, equally 1 likely to be colored 2. either red or blue. That is, each edge is red with probability n n as sets of k vertices and deﬁne the events Ei , i = 1, . . . , Number the k k follows: Ei = {all of the connecting edges of the ith set of k vertices are the same color} k connecting edges of a set of k vertices is equally likely to Now, since each of the 2 be either red or blue, it follows that the probability that they are all the same color is P(Ei ) = 2

k(k−1)/2 1 2

Therefore, because ⎛ P⎝

⎞

Ei ⎠ …

i

we ﬁnd that P

P(Ei )

(Boole’s inequality)

i

Ei , the probability that there is a set of k vertices all of whose

i

connecting edges are similarly colored, satisﬁes ⎛ ⎞ k(k−1)/2−1 1 n ⎝ ⎠ Ei … P k 2 i

Hence, if

or, equivalently, if

n k

k(k−1)/2−1 1 < 1 2

n k

< 2k(k−1)/2−1

n then the probability that at least one of the sets of k vertices has all of its k connecting edges the same color is less than 1. Consequently, under the preceding condition on n and k, it follows that there is a positive probability that no set of k vertices has all of its connecting edges the same color. But this conclusion implies that there is at least one way of coloring the edges for which no set of k vertices has all of its connecting edges the same color. . Remarks. (a) Whereas the preceding argument established a condition on n and k that guarantees the existence of a coloring scheme satisfying the desired property, it gives no information about how to obtain such a scheme (although one possibility would be simply to choose the colors at random, check to see if the resulting coloring satisﬁes the property, and repeat the procedure until it does).

Section 3.5

P(·|F ) Is a Probability

93

(b) The method of introducing probability into a problem whose statement is purely deterministic has been called the probabilistic method.† Other examples of this method are given in Theoretical Exercise 24 and Examples 2t and 2u of Chapter 7. 3.5

P(·|F) IS A PROBABILITY Conditional probabilities satisfy all of the properties of ordinary probabilities, as is proved by Proposition 5.1, which shows that P(E|F) satisﬁes the three axioms of a probability. Proposition 5.1. (a) 0 … P(E|F) … 1. (b) P(S|F) = 1. (c) If Ei , i = 1, 2, . . ., are mutually exclusive events, then ⎛ ⎞ q q P ⎝ Ei |F ⎠ = P(Ei |F) 1

1

Proof. To prove part (a), we must show that 0 … P(EF)/P(F) … 1. The left-side inequality is obvious, whereas the right side follows because EF ( F, which implies that P(EF) … P(F). Part (b) follows because P(S|F) = Part (c) follows from ⎞ ⎛ q P ⎝ Ei |F ⎠ = i=1

⎛ ⎞ q

P⎝ Ei F ⎠ i=1

P(F) q

Ei F P 1

=

P(F) q

= =

P(F) P(SF) = =1 P(F) P(F)

⎛ since

⎝

q

⎞ Ei ⎠ F =

1

q

Ei F

1

P(Ei F)

1

P(F) q

P(Ei |F)

1

where the next-to-last equality follows because Ei Ej Ei FEj F = Ø.

=

Ø implies that

† See N. Alon, J. Spencer, and P. Erdos, The Probabilistic Method (New York: John Wiley & Sons, Inc., 1992).

94

Chapter 3

Conditional Probability and Independence

If we deﬁne Q(E) = P(E|F), then, from Proposition 5.1, Q(E) may be regarded as a probability function on the events of S. Hence, all of the propositions previously proved for probabilities apply to Q(E). For instance, we have Q(E1 ∪ E2 ) = Q(E1 ) + Q(E2 ) − Q(E1 E2 ) or, equivalently, P(E1 ∪ E2 |F) = P(E1 |F) + P(E2 |F) − P(E1 E2 |F) Also, if we deﬁne the conditional probability Q(E1 |E2 ) by Q(E1 |E2 ) = Q(E1 E2 )/ Q(E2 ), then, from Equation (3.1), we have Q(E1 ) = Q(E1 |E2 )Q(E2 ) + Q(E1 |E2c )Q(E2c ) Since

(5.1)

Q(E1 E2 ) Q(E2 ) P(E1 E2 |F) = P(E2 |F) P(E1 E2 F) P(F) = P(E2 F) P(F) = P(E1 |E2 F)

Q(E1 |E2 ) =

Equation (5.1) is equivalent to P(E1 |F) = P(E1 |E2 F)P(E2 |F) + P(E1 |E2c F)P(E2c |F) EXAMPLE 5a Consider Example 3a, which is concerned with an insurance company which believes that people can be divided into two distinct classes: those who are accident prone and those who are not. During any given year, an accident-prone person will have an accident with probability .4, whereas the corresponding ﬁgure for a person who is not prone to accidents is .2. What is the conditional probability that a new policyholder will have an accident in his or her second year of policy ownership, given that the policyholder has had an accident in the ﬁrst year? Solution. If we let A be the event that the policyholder is accident prone and we let Ai , i = 1, 2, be the event that he or she has had an accident in the ith year, then the desired probability P(A2 |A1 ) may be obtained by conditioning on whether or not the policyholder is accident prone, as follows: P(A2 |A1 ) = P(A2 |AA1 )P(A|A1 ) + P(A2 |Ac A1 )P(Ac |A1 ) Now, P(A|A1 ) = However, P(A) is assumed to equal .26. Hence,

P(A1 |A)P(A) P(A1 A) = P(A1 ) P(A1 ) 3 10 ,

and it was shown in Example 3a that P(A1 ) =

P(A|A1 ) =

6 (.4)(.3) = .26 13

Section 3.5

P(·|F ) Is a Probability

95

Thus, P(Ac |A1 ) = 1 − P(A|A1 ) =

7 13

Since P(A2 |AA1 ) = .4 and P(A2 |Ac A1 ) = .2, it follows that P(A2 |A1 ) = (.4)

7 6 + (.2) L .29 13 13

.

EXAMPLE 5b A female chimp has given birth. It is not certain, however, which of two male chimps is the father. Before any genetic analysis has been performed, it is felt that the probability that male number 1 is the father is p and the probability that male number 2 is the father is 1 − p. DNA obtained from the mother, male number 1, and male number 2 indicate that, on one speciﬁc location of the genome, the mother has the gene pair (A, A), male number 1 has the gene pair (a, a), and male number 2 has the gene pair (A, a). If a DNA test shows that the baby chimp has the gene pair (A, a), what is the probability that male number 1 is the father? Solution. Let all probabilities be conditional on the event that the mother has the gene pair (A, A), male number 1 has the gene pair (a, a), and male number 2 has the gene pair (A, a). Now, let Mi be the event that male number i, i = 1, 2, is the father, and let BA,a be the event that the baby chimp has the gene pair (A, a). Then P(M1 |BA,a ) is obtained as follows: P(M1 |BA,a ) =

P(M1 BA,a ) P(BA,a )

P(BA,a |M1 )P(M1 ) P(BA,a |M1 )P(M1 ) + P(BA,a |M2 )P(M2 ) 1 · p = 1 · p + (1/2)(1 − p) 2p = 1 + p

=

2p > p when p < 1, the information that the baby’s gene pair is (A, a) Because 1+p increases the probability that male number 1 is the father. This result is intuitive because it is more likely that the baby would have gene pair (A, a) if M1 is true than . if M2 is true (the respective conditional probabilities being 1 and 1/2).

The next example deals with a problem in the theory of runs. EXAMPLE 5c Independent trials, each resulting in a success with probability p or a failure with probability q = 1 − p, are performed. We are interested in computing the probability that a run of n consecutive successes occurs before a run of m consecutive failures. Solution. Let E be the event that a run of n consecutive successes occurs before a run of m consecutive failures. To obtain P(E), we start by conditioning on the outcome of the ﬁrst trial. That is, letting H denote the event that the ﬁrst trial results in a success, we obtain (5.2) P(E) = pP(E|H) + qP(E|H c )

96

Chapter 3

Conditional Probability and Independence

Now, given that the ﬁrst trial was successful, one way we can get a run of n successes before a run of m failures would be to have the next n − 1 trials all result in successes. So, let us condition on whether or not that occurs. That is, letting F be the event that trials 2 through n all are successes, we obtain P(E|H) = P(E|FH)P(F|H) + P(E|F c H)P(F c |H)

(5.3)

On the one hand, clearly, P(E|FH) = 1; on the other hand, if the event F c H occurs, then the ﬁrst trial would result in a success, but there would be a failure some time during the next n − 1 trials. However, when this failure occurs, it would wipe out all of the previous successes, and the situation would be exactly as if we started out with a failure. Hence, P(E|F c H) = P(E|H c ) Because the independence of trials implies that F and H are independent, and because P(F) = pn−1 , it follows from Equation (5.3) that P(E|H) = pn−1 + (1 − pn−1 )P(E|H c )

(5.4)

We now obtain an expression for P(E|H c ) in a similar manner. That is, we let G denote the event that trials 2 through m are all failures. Then P(E|H c ) = P(E|GH c )P(G|H c ) + P(E|Gc H c )P(Gc |H c )

(5.5)

Now, GH c is the event that the ﬁrst m trials all result in failures, so P(E|GH c ) = 0. Also, if Gc H c occurs, then the ﬁrst trial is a failure, but there is at least one success in the next m − 1 trials. Hence, since this success wipes out all previous failures, we see that P(E|Gc H c ) = P(E|H) Thus, because P(Gc |H c ) = P(Gc ) = 1 − qm−1 , we obtain, from (5.5), P(E|H c ) = (1 − qm−1 )P(E|H)

(5.6)

Solving Equations (5.4) and (5.6) yields P(E|H) =

pn−1

pn−1 + qm−1 − pn−1 qm−1

pn−1

(1 − qm−1 )pn−1 + qm−1 − pn−1 qm−1

and P(E|H c ) = Thus, P(E) = pP(E|H) + qP(E|H c ) pn + qpn−1 (1 − qm−1 ) pn−1 + qm−1 − pn−1 qm−1 pn−1 (1 − qm ) = n−1 p + qm−1 − pn−1 qm−1

=

(5.7)

Section 3.5

P(·|F ) Is a Probability

97

It is interesting to note that, by the symmetry of the problem, the probability of obtaining a run of m failures before a run of n successes would be given by Equation (5.7) with p and q interchanged and n and m interchanged. Hence, this probability would equal P{run of m failures before a run of n successes} =

qm−1

qm−1 (1 − pn ) + pn−1 − qm−1 pn−1

(5.8)

Since Equations (5.7) and (5.8) sum to 1, it follows that, with probability 1, either a run of n successes or a run of m failures will eventually occur. As an example of Equation (5.7), we note that, in tossing a fair coin, the probability 7 . For 2 consecutive heads before that a run of 2 heads will precede a run of 3 tails is 10 5 4 consecutive tails, the probability rises to 6 . . In our next example, we return to the matching problem (Example 5m, Chapter 2) and this time obtain a solution by using conditional probabilities. EXAMPLE 5d At a party, n men take off their hats. The hats are then mixed up, and each man randomly selects one. We say that a match occurs if a man selects his own hat. What is the probability of (a) no matches? (b) exactly k matches? Solution. (a) Let E denote the event that no matches occur, and to make explicit the dependence on n, write Pn = P(E). We start by conditioning on whether or not the ﬁrst man selects his own hat—call these events M and Mc , respectively. Then Pn = P(E) = P(E|M)P(M) + P(E|Mc )P(Mc ) Clearly, P(E|M) = 0, so Pn = P(E|Mc )

n − 1 n

(5.9)

Now, P(E|Mc ) is the probability of no matches when n − 1 men select from a set of n − 1 hats that does not contain the hat of one of these men. This can happen in either of two mutually exclusive ways: Either there are no matches and the extra man does not select the extra hat (this being the hat of the man who chose ﬁrst), or there are no matches and the extra man does select the extra hat. The probability of the ﬁrst of these events is just Pn−1 , which is seen by regarding the extra hat as “belonging” to the extra man. Because the second event has probability [1/(n − 1)]Pn−2 , we have P(E|Mc ) = Pn−1 +

1 Pn−2 n − 1

Thus, from Equation (5.9), Pn =

n − 1 1 Pn−1 + Pn−2 n n

or, equivalently, 1 Pn − Pn−1 = − (Pn−1 − Pn−2 ) n

(5.10)

98

Chapter 3

Conditional Probability and Independence

However, since Pn is the probability of no matches when n men select among their own hats, we have 1 P1 = 0 P2 = 2 So, from Equation (5.10), (P2 − P1 ) 1 1 1 =− or P3 = − 3 3! 2! 3! 1 1 1 (P3 − P2 ) 1 P4 − P3 = − = or P4 = − + 4 4! 2! 3! 4! P3 − P2 = −

and, in general, Pn =

1 1 1 (−1)n − + − ··· + 2! 3! 4! n!

(b) To obtain the probability of exactly k matches, we consider any ﬁxed group of k men. The probability that they, and only they, select their own hats is 1 (n − k)! 1 1 ··· Pn−k = Pn−k nn − 1 n − (k − 1) n! where Pn−k is the conditional probability that theother n − k men, selecting among n their own hats, have no matches. Since there are choices of a set of k men, the k desired probability of exactly k matches is Pn−k = k!

1 2!

−

1 3!

+ ··· +

(−1)n−k (n − k)!

k!

.

An important concept in probability theory is that of the conditional independence of events. We say that the events E1 and E2 are conditionally independent given F if, given that F occurs, the conditional probability that E1 occurs is unchanged by information as to whether or not E2 occurs. More formally, E1 and E2 are said to be conditionally independent given F if

or, equivalently,

P(E1 |E2 F) = P(E1 |F)

(5.11)

P(E1 E2 |F) = P(E1 |F)P(E2 |F)

(5.12)

The notion of conditional independence can easily be extended to more than two events, and this extension is left as an exercise. The reader should note that the concept of conditional independence was implicitly employed in Example 5a, where it was assumed that the events that a policyholder had an accident in his or her ith year, i = 1, 2, . . ., were conditionally independent given whether or not the person was accident prone. The following example, sometimes referred to as Laplace’s rule of succession, further illustrates the concept of conditional independence. EXAMPLE 5e Laplace’s rule of succession There are k + 1 coins in a box. When ﬂipped, the ith coin will turn up heads with probability i/k, i = 0, 1, . . . , k. A coin is randomly selected from the box and is then

Section 3.5

P(·|F ) Is a Probability

99

repeatedly ﬂipped. If the ﬁrst n ﬂips all result in heads, what is the conditional probability that the (n + 1)st ﬂip will do likewise? Solution. Let Ci denote the event that the ith coin, i = 0, 1, . . . , k, is initially selected; let Fn denote the event that the ﬁrst n ﬂips all result in heads; and let H be the event that the (n + 1)st ﬂip is a head. The desired probability, P(H|Fn ), is now obtained as follows: k P(H|Fn ) = P(H|Fn Ci )P(Ci |Fn ) i=0

Now, given that the ith coin is selected, it is reasonable to assume that the outcomes will be conditionally independent, with each one resulting in a head with probability i/k. Hence, i P(H|Fn Ci ) = P(H|Ci ) = k Also, P(Ci |Fn ) =

P(Fn |Ci )P(Ci ) (i/k)n [1/(k + 1)] P(Ci Fn ) = = k k P(Fn ) P(Fn |Cj )P(Cj ) (j/k)n [1/(k + 1)] j=0

j=0

Thus, k

P(H|Fn ) =

(i/k)n+1

i=0 k

(j/k)n

j=0

But if k is large, we can use the integral approximations * 1 k 1 i n+1 1 L xn+1 dx = k k n + 2 0 i=0

* 1 k 1 j n 1 L xn dx = k k n + 1 0 j=0

So, for k large, P(H|Fn ) L

n + 1 n + 2

.

EXAMPLE 5f Updating information sequentially Suppose there are n mutually exclusive and exhaustive possible hypotheses, with initial (sometimes referred to as prior) probabilities P(Hi ), ni=1 P(Hi ) = 1. Now, if information that the event E has occurred is received, then the conditional probability that Hi is the true hypothesis (sometimes referred to as the updated or posterior probability of Hi ) is P(E|Hi )P(Hi ) (5.13) P(Hi |E) = j P(E|Hj )P(Hj )

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Suppose now that we learn ﬁrst that E1 has occurred and then that E2 has occurred. Then, given only the ﬁrst piece of information, the conditional probability that Hi is the true hypothesis is P(Hi |E1 ) =

P(E1 |Hi )P(Hi ) P(E1 |Hi )P(Hi ) = P(E1 ) j P(E1 |Hj )P(Hj )

whereas given both pieces of information, the conditional probability that Hi is the true hypothesis is P(Hi |E1 E2 ), which can be computed by P(E1 E2 |Hi )P(Hi ) P(Hi |E1 E2 ) = j P(E1 E2 |Hj )P(Hj ) One might wonder, however, when one can compute P(Hi |E1 E2 ) by using the right side of Equation (5.13) with E = E2 and with P(Hj ) replaced by P(Hj |E1 ), j = 1, . . . , n. That is, when is it legitimate to regard P(Hj |E1 ), j Ú 1, as the prior probabilities and then use (5.13) to compute the posterior probabilities? Solution. The answer is that the preceding is legitimate, provided that, for each j = 1, . . . , n, the events E1 and E2 are conditionally independent, given Hj . For if this is the case, then P(E1 E2 |Hj ) = P(E2 |Hj )P(E1 |Hj ), j = 1, . . . , n Therefore, P(E2 |Hi )P(E1 |Hi )P(Hi ) P(E1 E2 ) P(E2 |Hi )P(E1 Hi ) = P(E1 E2 ) P(E2 |Hi )P(Hi |E1 )P(E1 ) = P(E1 E2 ) P(E2 |Hi )P(Hi |E1 ) = Q(1, 2)

P(Hi |E1 E2 ) =

where Q(1, 2) = upon summing,

P(E1 E2 ) P(E1 ) .

1=

Since the preceding equation is valid for all i, we obtain,

n

P(Hi |E1 E2 ) =

i=1

showing that Q(1, 2) =

n P(E2 |Hi )P(Hi |E1 ) i=1

n

Q(1, 2)

P(E2 |Hi )P(Hi |E1 )

i=1

and yielding the result P(E2 |Hi )P(Hi |E1 ) P(Hi |E1 E2 ) = n i=1 P(E2 |Hi )P(Hi |E1 ) For instance, suppose that one of two coins is chosen to be ﬂipped. Let Hi be the event that coin i, i = 1, 2, is chosen, and suppose that when coin i is ﬂipped, it lands on heads with probability pi , i = 1, 2. Then the preceding equations show that, to sequentially

Summary

101

update the probability that coin 1 is the one being ﬂipped, given the results of the previous ﬂips, all that must be saved after each new ﬂip is the conditional probability that coin 1 is the coin being used. That is, it is not necessary to keep track of all earlier results. . SUMMARY For events E and F, the conditional probability of E given that F has occurred is denoted by P(E|F) and is deﬁned by P(E|F) =

P(EF) P(F)

The identity P(E1 E2 · · · En ) = P(E1 )P(E2 |E1 ) · · · P(En |E1 · · · En−1 ) is known as the multiplication rule of probability. A valuable identity is P(E) = P(E|F)P(F) + P(E|F c )P(F c ) which can be used to compute P(E) by “conditioning” on whether F occurs. P(H)/P(H c ) is called the odds of the event H. The identity P(H) P(E|H) P(H|E) = P(H c |E) P(H c )P(E|H c ) shows that when new evidence E is obtained, the value of the odds of H becomes its old value multiplied by the ratio of the conditional probability of the new evidence when H is true to the conditional probability when H is not true. Let Fi , i = 1, . . . , n, be mutually exclusive events whose union is the entire sample space. The identity P(Fj |E) =

P(E|Fj )P(Fj ) n

P(E|Fi )P(Fi )

i=1

is known as Bayes’s formula. If the events Fi , i = 1, . . . , n, are competing hypotheses, then Bayes’s formula shows how to compute the conditional probabilities of these hypotheses when additional evidence E becomes available. If P(EF) = P(E)P(F), then we say that the events E and F are independent. This condition is equivalent to P(E|F) = P(E) and to P(F|E) = P(F). Thus, the events E and F are independent if knowledge of the occurrence of one of them does not affect the probability of the other. The events E1 , . . . , En are said to be independent if, for any subset Ei1 , . . . , Eir of them, P(Ei1 · · · Eir ) = P(Ei1 ) · · · P(Eir ) For a ﬁxed event F, P(E|F) can be considered to be a probability function on the events E of the sample space.

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PROBLEMS 3.1. Two fair dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on different numbers? 3.2. If two fair dice are rolled, what is the conditional probability that the ﬁrst one lands on 6 given that the sum of the dice is i? Compute for all values of i between 2 and 12. 3.3. Use Equation (2.1) to compute, in a hand of bridge, the conditional probability that East has 3 spades given that North and South have a combined total of 8 spades. 3.4. What is the probability that at least one of a pair of fair dice lands on 6, given that the sum of the dice is i, i = 2, 3, . . . , 12? 3.5. An urn contains 6 white and 9 black balls. If 4 balls are to be randomly selected without replacement, what is the probability that the ﬁrst 2 selected are white and the last 2 black? 3.6. Consider an urn containing 12 balls, of which 8 are white. A sample of size 4 is to be drawn with replacement (without replacement). What is the conditional probability (in each case) that the ﬁrst and third balls drawn will be white given that the sample drawn contains exactly 3 white balls? 3.7. The king comes from a family of 2 children. What is the probability that the other child is his sister? 3.8. A couple has 2 children. What is the probability that both are girls if the older of the two is a girl? 3.9. Consider 3 urns. Urn A contains 2 white and 4 red balls, urn B contains 8 white and 4 red balls, and urn C contains 1 white and 3 red balls. If 1 ball is selected from each urn, what is the probability that the ball chosen from urn A was white given that exactly 2 white balls were selected? 3.10. Three cards are randomly selected, without replacement, from an ordinary deck of 52 playing cards. Compute the conditional probability that the ﬁrst card selected is a spade given that the second and third cards are spades. 3.11. Two cards are randomly chosen without replacement from an ordinary deck of 52 cards. Let B be the event that both cards are aces, let As be the event that the ace of spades is chosen, and let A be the event that at least one ace is chosen. Find (a) P(B|As ) (b) P(B|A) 3.12. A recent college graduate is planning to take the ﬁrst three actuarial examinations in the coming summer. She will take the ﬁrst actuarial exam in June. If she passes that exam, then she will take the second exam in July, and if she also passes that one, then she will take the third exam in September. If she fails an exam, then she is not allowed

3.13.

3.14.

3.15.

3.16.

3.17.

3.18.

to take any others. The probability that she passes the ﬁrst exam is .9. If she passes the ﬁrst exam, then the conditional probability that she passes the second one is .8, and if she passes both the ﬁrst and the second exams, then the conditional probability that she passes the third exam is .7. (a) What is the probability that she passes all three exams? (b) Given that she did not pass all three exams, what is the conditional probability that she failed the second exam? Suppose that an ordinary deck of 52 cards (which contains 4 aces) is randomly divided into 4 hands of 13 cards each. We are interested in determining p, the probability that each hand has an ace. Let Ei be the event that the ith hand has exactly one ace. Determine p = P(E1 E2 E3 E4 ) by using the multiplication rule. An urn initially contains 5 white and 7 black balls. Each time a ball is selected, its color is noted and it is replaced in the urn along with 2 other balls of the same color. Compute the probability that (a) the ﬁrst 2 balls selected are black and the next 2 are white; (b) of the ﬁrst 4 balls selected, exactly 2 are black. An ectopic pregnancy is twice as likely to develop when the pregnant woman is a smoker as it is when she is a nonsmoker. If 32 percent of women of childbearing age are smokers, what percentage of women having ectopic pregnancies are smokers? Ninety-eight percent of all babies survive delivery. However, 15 percent of all births involve Cesarean (C) sections, and when a C section is performed, the baby survives 96 percent of the time. If a randomly chosen pregnant woman does not have a C section, what is the probability that her baby survives? In a certain community, 36 percent of the families own a dog and 22 percent of the families that own a dog also own a cat. In addition, 30 percent of the families own a cat. What is (a) the probability that a randomly selected family owns both a dog and a cat? (b) the conditional probability that a randomly selected family owns a dog given that it owns a cat? A total of 46 percent of the voters in a certain city classify themselves as Independents, whereas 30 percent classify themselves as Liberals and 24 percent say that they are Conservatives. In a recent local election, 35 percent of the Independents, 62 percent of the Liberals, and 58 percent of the Conservatives voted. A voter is chosen at random.

Problems Given that this person voted in the local election, what is the probability that he or she is (a) an Independent? (b) a Liberal? (c) a Conservative? (d) What fraction of voters participated in the local election? 3.19. A total of 48 percent of the women and 37 percent of the men that took a certain “quit smoking” class remained nonsmokers for at least one year after completing the class. These people then attended a success party at the end of a year. If 62 percent of the original class was male, (a) what percentage of those attending the party were women? (b) what percentage of the original class attended the party? 3.20. Fifty-two percent of the students at a certain college are females. Five percent of the students in this college are majoring in computer science. Two percent of the students are women majoring in computer science. If a student is selected at random, ﬁnd the conditional probability that (a) the student is female given that the student is majoring in computer science; (b) this student is majoring in computer science given that the student is female. 3.21. A total of 500 married working couples were polled about their annual salaries, with the following information resulting:

Less than $25,000 More than $25,000

3.24.

3.25.

Husband Wife

3.23.

Less than $25,000

More than $25,000

212 36

198 54

For instance, in 36 of the couples, the wife earned more and the husband earned less than $25,000. If one of the couples is randomly chosen, what is (a) the probability that the husband earns less than $25,000? (b) the conditional probability that the wife earns more than $25,000 given that the husband earns more than this amount? (c) the conditional probability that the wife earns more than $25,000 given that the husband earns less than this amount? 3.22. A red die, a blue die, and a yellow die (all six sided) are rolled. We are interested in the probability that the number appearing on the blue die is less than that appearing on the yellow die, which is less than that appearing on the red die. That is,

3.26.

3.27.

103

with B, Y, and R denoting, respectively, the number appearing on the blue, yellow, and red die, we are interested in P(B < Y < R). (a) What is the probability that no two of the dice land on the same number? (b) Given that no two of the dice land on the same number, what is the conditional probability that B < Y < R? (c) What is P(B < Y < R)? Urn I contains 2 white and 4 red balls, whereas urn II contains 1 white and 1 red ball. A ball is randomly chosen from urn I and put into urn II, and a ball is then randomly selected from urn II. What is (a) the probability that the ball selected from urn II is white? (b) the conditional probability that the transferred ball was white given that a white ball is selected from urn II? Each of 2 balls is painted either black or gold and then placed in an urn. Suppose that each ball is colored black with probability 12 and that these events are independent. (a) Suppose that you obtain information that the gold paint has been used (and thus at least one of the balls is painted gold). Compute the conditional probability that both balls are painted gold. (b) Suppose now that the urn tips over and 1 ball falls out. It is painted gold. What is the probability that both balls are gold in this case? Explain. The following method was proposed to estimate the number of people over the age of 50 who reside in a town of known population 100,000: “As you walk along the streets, keep a running count of the percentage of people you encounter who are over 50. Do this for a few days; then multiply the percentage you obtain by 100,000 to obtain the estimate.” Comment on this method. Hint: Let p denote the proportion of people in the town who are over 50. Furthermore, let α1 denote the proportion of time that a person under the age of 50 spends in the streets, and let α2 be the corresponding value for those over 50. What quantity does the method suggested estimate? When is the estimate approximately equal to p? Suppose that 5 percent of men and .25 percent of women are color blind. A color-blind person is chosen at random. What is the probability of this person being male? Assume that there are an equal number of males and females. What if the population consisted of twice as many males as females? All the workers at a certain company drive to work and park in the company’s lot. The company

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is interested in estimating the average number of workers in a car. Which of the following methods will enable the company to estimate this quantity? Explain your answer. 1. Randomly choose n workers, ﬁnd out how many were in the cars in which they were driven, and take the average of the n values. 2. Randomly choose n cars in the lot, ﬁnd out how many were driven in those cars, and take the average of the n values. 3.28. Suppose that an ordinary deck of 52 cards is shufﬂed and the cards are then turned over one at a time until the ﬁrst ace appears. Given that the ﬁrst ace is the 20th card to appear, what is the conditional probability that the card following it is the (a) ace of spades? (b) two of clubs? 3.29. There are 15 tennis balls in a box, of which 9 have not previously been used. Three of the balls are randomly chosen, played with, and then returned to the box. Later, another 3 balls are randomly chosen from the box. Find the probability that none of these balls has ever been used. 3.30. Consider two boxes, one containing 1 black and 1 white marble, the other 2 black and 1 white marble. A box is selected at random, and a marble is drawn from it at random. What is the probability that the marble is black? What is the probability that the ﬁrst box was the one selected given that the marble is white? 3.31. Ms. Aquina has just had a biopsy on a possibly cancerous tumor. Not wanting to spoil a weekend family event, she does not want to hear any bad news in the next few days. But if she tells the doctor to call only if the news is good, then if the doctor does not call, Ms. Aquina can conclude that the news is bad. So, being a student of probability, Ms. Aquina instructs the doctor to ﬂip a coin. If it comes up heads, the doctor is to call if the news is good and not call if the news is bad. If the coin comes up tails, the doctor is not to call. In this way, even if the doctor doesn’t call, the news is not necessarily bad. Let α be the probability that the tumor is cancerous; let β be the conditional probability that the tumor is cancerous given that the doctor does not call. (a) Which should be larger, α or β? (b) Find β in terms of α, and prove your answer in part (a). 3.32. A family has j children with probability pj , where p1 = .1, p2 = .25, p3 = .35, p4 = .3. A child from this family is randomly chosen. Given that this child is the eldest child in the family, ﬁnd the conditional probability that the family has

3.33.

3.34.

3.35.

3.36.

3.37.

3.38.

3.39.

3.40.

(a) only 1 child; (b) 4 children. Redo (a) and (b) when the randomly selected child is the youngest child of the family. On rainy days, Joe is late to work with probability .3; on nonrainy days, he is late with probability .1. With probability .7, it will rain tomorrow. (a) Find the probability that Joe is early tomorrow. (b) Given that Joe was early, what is the conditional probability that it rained? In Example 3f, suppose that the new evidence is subject to different possible interpretations and in fact shows only that it is 90 percent likely that the criminal possesses the characteristic in question. In this case, how likely would it be that the suspect is guilty (assuming, as before, that he has the characteristic)? With probability .6, the present was hidden by mom; with probability .4, it was hidden by dad. When mom hides the present, she hides it upstairs 70 percent of the time and downstairs 30 percent of the time. Dad is equally likely to hide it upstairs or downstairs. (a) What is the probability that the present is upstairs? (b) Given that it is downstairs, what is the probability it was hidden by dad? Stores A, B, and C have 50, 75, and 100 employees, respectively, and 50, 60, and 70 percent of them respectively are women. Resignations are equally likely among all employees, regardless of sex. One woman employee resigns. What is the probability that she works in store C? (a) A gambler has a fair coin and a two-headed coin in his pocket. He selects one of the coins at random; when he ﬂips it, it shows heads. What is the probability that it is the fair coin? (b) Suppose that he ﬂips the same coin a second time and, again, it shows heads. Now what is the probability that it is the fair coin? (c) Suppose that he ﬂips the same coin a third time and it shows tails. Now what is the probability that it is the fair coin? Urn A has 5 white and 7 black balls. Urn B has 3 white and 12 black balls. We ﬂip a fair coin. If the outcome is heads, then a ball from urn A is selected, whereas if the outcome is tails, then a ball from urn B is selected. Suppose that a white ball is selected. What is the probability that the coin landed tails? In Example 3a, what is the probability that someone has an accident in the second year given that he or she had no accidents in the ﬁrst year? Consider a sample of size 3 drawn in the following manner: We start with an urn containing 5 white

Problems

3.41.

3.42.

3.43.

3.44.

3.45.

3.46.

and 7 red balls. At each stage, a ball is drawn and its color is noted. The ball is then returned to the urn, along with an additional ball of the same color. Find the probability that the sample will contain exactly (a) 0 white balls; (b) 1 white ball; (c) 3 white balls; (d) 2 white balls. A deck of cards is shufﬂed and then divided into two halves of 26 cards each. A card is drawn from one of the halves; it turns out to be an ace. The ace is then placed in the second half-deck. The half is then shufﬂed, and a card is drawn from it. Compute the probability that this drawn card is an ace. Hint: Condition on whether or not the interchanged card is selected. Three cooks, A, B, and C, bake a special kind of cake, and with respective probabilities .02, .03, and .05, it fails to rise. In the restaurant where they work, A bakes 50 percent of these cakes, B 30 percent, and C 20 percent. What proportion of “failures” is caused by A? There are 3 coins in a box. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of the time. When one of the 3 coins is selected at random and ﬂipped, it shows heads. What is the probability that it was the two-headed coin? Three prisoners are informed by their jailer that one of them has been chosen at random to be executed and the other two are to be freed. Prisoner A asks the jailer to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information because he already knows that at least one of the two will go free. The jailer refuses to answer the question, pointing out that if A knew which of his fellow prisoners were to be set free, then his own probability of being executed would rise from 13 to 12 because he would then be one of two prisoners. What do you think of the jailer’s reasoning? Suppose we have 10 coins such that if the ith coin is ﬂipped, heads will appear with probability i/10, i = 1, 2, . . . , 10. When one of the coins is randomly selected and ﬂipped, it shows heads. What is the conditional probability that it was the ﬁfth coin? In any given year, a male automobile policyholder will make a claim with probability pm and a female policyholder will make a claim with probability pf , where pf Z pm . The fraction of the policyholders that are male is α, 0 < α < 1. A policyholder is randomly chosen. If Ai denotes the event that this

105

policyholder will make a claim in year i, show that P(A2 |A1 ) > P(A1 )

3.47.

3.48.

3.49.

3.50.

3.51.

Give an intuitive explanation of why the preceding inequality is true. An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What is the probability that all of the balls selected are white? What is the conditional probability that the die landed on 3 if all the balls selected are white? Each of 2 cabinets identical in appearance has 2 drawers. Cabinet A contains a silver coin in each drawer, and cabinet B contains a silver coin in one of its drawers and a gold coin in the other. A cabinet is randomly selected, one of its drawers is opened, and a silver coin is found. What is the probability that there is a silver coin in the other drawer? Prostate cancer is the most common type of cancer found in males. As an indicator of whether a male has prostate cancer, doctors often perform a test that measures the level of the prostatespeciﬁc antigen (PSA) that is produced only by the prostate gland. Although PSA levels are indicative of cancer, the test is notoriously unreliable. Indeed, the probability that a noncancerous man will have an elevated PSA level is approximately .135, increasing to approximately .268 if the man does have cancer. If, on the basis of other factors, a physician is 70 percent certain that a male has prostate cancer, what is the conditional probability that he has the cancer given that (a) the test indicated an elevated PSA level? (b) the test did not indicate an elevated PSA level? Repeat the preceding calculation, this time assuming that the physician initially believes that there is a 30 percent chance that the man has prostate cancer. Suppose that an insurance company classiﬁes people into one of three classes: good risks, average risks, and bad risks. The company’s records indicate that the probabilities that good-, average-, and bad-risk persons will be involved in an accident over a 1-year span are, respectively, .05, .15, and .30. If 20 percent of the population is a good risk, 50 percent an average risk, and 30 percent a bad risk, what proportion of people have accidents in a ﬁxed year? If policyholder A had no accidents in 1997, what is the probability that he or she is a good or average risk? A worker has asked her supervisor for a letter of recommendation for a new job. She estimates that there is an 80 percent chance that she will get the

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job if she receives a strong recommendation, a 40 percent chance if she receives a moderately good recommendation, and a 10 percent chance if she receives a weak recommendation. She further estimates that the probabilities that the recommendation will be strong, moderate, and weak are .7, .2, and .1, respectively. (a) How certain is she that she will receive the new job offer? (b) Given that she does receive the offer, how likely should she feel that she received a strong recommendation? a moderate recommendation? a weak recommendation? (c) Given that she does not receive the job offer, how likely should she feel that she received a strong recommendation? a moderate recommendation? a weak recommendation? 3.52. A high school student is anxiously waiting to receive mail telling her whether she has been accepted to a certain college. She estimates that the conditional probabilities of receiving notiﬁcation on each day of next week, given that she is accepted and that she is rejected, are as follows: Day Monday Tuesday Wednesday Thursday Friday

P(mail|accepted)

P(mail|rejected)

.15 .20 .25 .15 .10

.05 .10 .10 .15 .20

She estimates that her probability of being accepted is .6. (a) What is the probability that she receives mail on Monday? (b) What is the conditional probability that she received mail on Tuesday given that she does not receive mail on Monday? (c) If there is no mail through Wednesday, what is the conditional probability that she will be accepted? (d) What is the conditional probability that she will be accepted if mail comes on Thursday? (e) What is the conditional probability that she will be accepted if no mail arrives that week? 3.53. A parallel system functions whenever at least one of its components works. Consider a parallel system of n components, and suppose that each component works independently with probability 12 . Find the conditional probability that component 1 works given that the system is functioning. 3.54. If you had to construct a mathematical model for events E and F, as described in parts (a) through

3.55.

3.56.

3.57.

3.58.

(e), would you assume that they were independent events? Explain your reasoning. (a) E is the event that a businesswoman has blue eyes, and F is the event that her secretary has blue eyes. (b) E is the event that a professor owns a car, and F is the event that he is listed in the telephone book. (c) E is the event that a man is under 6 feet tall, and F is the event that he weighs over 200 pounds. (d) E is the event that a woman lives in the United States, and F is the event that she lives in the Western Hemisphere. (e) E is the event that it will rain tomorrow, and F is the event that it will rain the day after tomorrow. In a class, there are 4 freshman boys, 6 freshman girls, and 6 sophomore boys. How many sophomore girls must be present if sex and class are to be independent when a student is selected at random? Suppose that you continually collect coupons and that there are m different types. Suppose also that each time a new coupon is obtained, it is a type i coupon with probability pi , i = 1, . . . , m. Suppose that you have just collected your nth coupon. What is the probability that it is a new type? Hint: Condition on the type of this coupon. A simpliﬁed model for the movement of the price of a stock supposes that on each day the stock’s price either moves up 1 unit with probability p or moves down 1 unit with probability 1 − p. The changes on different days are assumed to be independent. (a) What is the probability that after 2 days the stock will be at its original price? (b) What is the probability that after 3 days the stock’s price will have increased by 1 unit? (c) Given that after 3 days the stock’s price has increased by 1 unit, what is the probability that it went up on the ﬁrst day? Suppose that we want to generate the outcome of the ﬂip of a fair coin, but that all we have at our disposal is a biased coin which lands on heads with some unknown probability p that need not be equal to 12 . Consider the following procedure for accomplishing our task: 1. Flip the coin. 2. Flip the coin again. 3. If both ﬂips land on heads or both land on tails, return to step 1. 4. Let the result of the last ﬂip be the result of the experiment.

Problems

3.59.

3.60.

3.61.

3.62.

(a) Show that the result is equally likely to be either heads or tails. (b) Could we use a simpler procedure that continues to ﬂip the coin until the last two ﬂips are different and then lets the result be the outcome of the ﬁnal ﬂip? Independent ﬂips of a coin that lands on heads with probability p are made. What is the probability that the ﬁrst four outcomes are (a) H, H, H, H? (b) T, H, H, H? (c) What is the probability that the pattern T, H, H, H occurs before the pattern H, H, H, H? Hint for part (c): How can the pattern H, H, H, H occur ﬁrst? The color of a person’s eyes is determined by a single pair of genes. If they are both blue-eyed genes, then the person will have blue eyes; if they are both brown-eyed genes, then the person will have brown eyes; and if one of them is a blue-eyed gene and the other a brown-eyed gene, then the person will have brown eyes. (Because of the latter fact, we say that the brown-eyed gene is dominant over the blue-eyed one.) A newborn child independently receives one eye gene from each of its parents, and the gene it receives from a parent is equally likely to be either of the two eye genes of that parent. Suppose that Smith and both of his parents have brown eyes, but Smith’s sister has blue eyes. (a) What is the probability that Smith possesses a blue-eyed gene? (b) Suppose that Smith’s wife has blue eyes. What is the probability that their ﬁrst child will have blue eyes? (c) If their ﬁrst child has brown eyes, what is the probability that their next child will also have brown eyes? Genes relating to albinism are denoted by A and a. Only those people who receive the a gene from both parents will be albino. Persons having the gene pair A, a are normal in appearance and, because they can pass on the trait to their offspring, are called carriers. Suppose that a normal couple has two children, exactly one of whom is an albino. Suppose that the nonalbino child mates with a person who is known to be a carrier for albinism. (a) What is the probability that their ﬁrst offspring is an albino? (b) What is the conditional probability that their second offspring is an albino given that their ﬁrstborn is not? Barbara and Dianne go target shooting. Suppose that each of Barbara’s shots hits a wooden duck target with probability p1 , while each shot of

3.63.

3.64.

3.65.

3.66.

3.67.

107

Dianne’s hits it with probability p2 . Suppose that they shoot simultaneously at the same target. If the wooden duck is knocked over (indicating that it was hit), what is the probability that (a) both shots hit the duck? (b) Barbara’s shot hit the duck? What independence assumptions have you made? A and B are involved in a duel. The rules of the duel are that they are to pick up their guns and shoot at each other simultaneously. If one or both are hit, then the duel is over. If both shots miss, then they repeat the process. Suppose that the results of the shots are independent and that each shot of A will hit B with probability pA , and each shot of B will hit A with probability pB . What is (a) the probability that A is not hit? (b) the probability that both duelists are hit? (c) the probability that the duel ends after the nth round of shots? (d) the conditional probability that the duel ends after the nth round of shots given that A is not hit? (e) the conditional probability that the duel ends after the nth round of shots given that both duelists are hit? A true–false question is to be posed to a husbandand-wife team on a quiz show. Both the husband and the wife will independently give the correct answer with probability p. Which of the following is a better strategy for the couple? (a) Choose one of them and let that person answer the question. (b) Have them both consider the question, and then either give the common answer if they agree or, if they disagree, ﬂip a coin to determine which answer to give. In Problem 3.5, if p = .6 and the couple uses the strategy in part (b), what is the conditional probability that the couple gives the correct answer given that the husband and wife (a) agree? (b) disagree? The probability of the closing of the ith relay in the circuits shown in Figure 3.4 is given by pi , i = 1, 2, 3, 4, 5. If all relays function independently, what is the probability that a current ﬂows between A and B for the respective circuits? Hint for (b): Condition on whether relay 3 closes. An engineering system consisting of n components is said to be a k-out-of-n system (k … n) if the system functions if and only if at least k of the n components function. Suppose that all components function independently of each other. (a) If the ith component functions with probability Pi , i = 1, 2, 3, 4, compute the probability that a 2-out-of-4 system functions.

108

Chapter 3

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(a) 1

2

A

5 3

B

4

(b) 1

4 3

A 2

B 5

FIGURE 3.4: Circuits for Problem 3.66

(b) Repeat part (a) for a 3-out-of-5 system. (c) Repeat for a k-out-of-n system when all the Pi equal p (that is, Pi = p, i = 1, 2, . . . , n). 3.68. In Problem 3.65a, ﬁnd the conditional probability that relays 1 and 2 are both closed given that a current ﬂows from A to B. 3.69. A certain organism possesses a pair of each of 5 different genes (which we will designate by the ﬁrst 5 letters of the English alphabet). Each gene appears in 2 forms (which we designate by lowercase and capital letters). The capital letter will be assumed to be the dominant gene, in the sense that if an organism possesses the gene pair xX, then it will outwardly have the appearance of the X gene. For instance, if X stands for brown eyes and x for blue eyes, then an individual having either gene pair XX or xX will have brown eyes, whereas one having gene pair xx will have blue eyes. The characteristic appearance of an organism is called its phenotype, whereas its genetic constitution is called its genotype. (Thus, 2 organisms with respective genotypes aA, bB, cc, dD, ee and AA, BB, cc, DD, ee would have different genotypes but the same phenotype.) In a mating between 2 organisms, each one contributes, at random, one of its gene pairs of each type. The 5 contributions of an organism (one of each of the 5 types) are assumed to be independent and are also independent of the contributions of the organism’s mate. In a mating between organisms having genotypes aA, bB, cC, dD, eE and aa, bB, cc, Dd, ee what is the probability that the progeny will (i) phenotypically and (ii) genotypically resemble (a) the ﬁrst parent? (b) the second parent?

(c) either parent? (d) neither parent? 3.70. There is a 50–50 chance that the queen carries the gene for hemophilia. If she is a carrier, then each prince has a 50–50 chance of having hemophilia. If the queen has had three princes without the disease, what is the probability that the queen is a carrier? If there is a fourth prince, what is the probability that he will have hemophilia? 3.71. On the morning of September 30, 1982, the won– lost records of the three leading baseball teams in the Western Division of the National League were as follows: Team Atlanta Braves San Francisco Giants Los Angeles Dodgers

Won

Lost

87 86 86

72 73 73

Each team had 3 games remaining. All 3 of the Giants’ games were with the Dodgers, and the 3 remaining games of the Braves were against the San Diego Padres. Suppose that the outcomes of all remaining games are independent and each game is equally likely to be won by either participant. For each team, what is the probability that it will win the division title? If two teams tie for ﬁrst place, they have a playoff game, which each team has an equal chance of winning. 3.72. A town council of 7 members contains a steering committee of size 3. New ideas for legislation go ﬁrst to the steering committee and then on to the council as a whole if at least 2 of the 3 committee members approve the legislation. Once at the full council, the legislation requires a majority vote

Problems

3.73.

3.74.

3.75.

3.76.

3.77.

3.78.

(of at least 4) to pass. Consider a new piece of legislation, and suppose that each town council member will approve it, independently, with probability p. What is the probability that a given steering committee member’s vote is decisive in the sense that if that person’s vote were reversed, then the ﬁnal fate of the legislation would be reversed? What is the corresponding probability for a given council member not on the steering committee? Suppose that each child born to a couple is equally likely to be a boy or a girl, independently of the sex distribution of the other children in the family. For a couple having 5 children, compute the probabilities of the following events: (a) All children are of the same sex. (b) The 3 eldest are boys and the others girls. (c) Exactly 3 are boys. (d) The 2 oldest are girls. (e) There is at least 1 girl. A and B alternate rolling a pair of dice, stopping either when A rolls the sum 9 or when B rolls the sum 6. Assuming that A rolls ﬁrst, ﬁnd the probability that the ﬁnal roll is made by A. In a certain village, it is traditional for the eldest son (or the older son in a two-son family) and his wife to be responsible for taking care of his parents as they age. In recent years, however, the women of this village, not wanting that responsibility, have not looked favorably upon marrying an eldest son. (a) If every family in the village has two children, what proportion of all sons are older sons? (b) If every family in the village has three children, what proportion of all sons are eldest sons? Assume that each child is, independently, equally likely to be either a boy or a girl. Suppose that E and F are mutually exclusive events of an experiment. Show that if independent trials of this experiment are performed, then E will occur before F with probability P(E)/[P(E) + P(F)]. Consider an unending sequence of independent trials, where each trial is equally likely to result in any of the outcomes 1, 2, or 3. Given that outcome 3 is the last of the three outcomes to occur, ﬁnd the conditional probability that (a) the ﬁrst trial results in outcome 1; (b) the ﬁrst two trials both result in outcome 1. A and B play a series of games. Each game is independently won by A with probability p and by B with probability 1 − p. They stop when the total number of wins of one of the players is two greater than that of the other player. The player with the

109

greater number of total wins is declared the winner of the series. (a) Find the probability that a total of 4 games are played. (b) Find the probability that A is the winner of the series. 3.79. In successive rolls of a pair of fair dice, what is the probability of getting 2 sevens before 6 even numbers? 3.80. In a certain contest, the players are of equal skill and the probability is 12 that a speciﬁed one of the two contestants will be the victor. In a group of 2n players, the players are paired off against each other at random. The 2n−1 winners are again paired off randomly, and so on, until a single winner remains. Consider two speciﬁed contestants, A and B, and deﬁne the events Ai , i … n, E by Ai : E:

A plays in exactly i contests: A and B never play each other.

(a) Find P(Ai ), i = 1, . . . , n. (b) Find P(E). (c) Let Pn = P(E). Show that Pn =

1 2n − 2 + n n 2 − 1 2 − 1

2 1 Pn−1 2

and use this formula to check the answer you obtained in part (b). Hint: Find P(E) by conditioning on which of the events Ai , i = 1, . . . , n occur. In simplifying your answer, use the algebraic identity n−1 i=1

ixi−1 =

1 − nxn−1 + (n − 1)xn (1 − x)2

For another approach to solving this problem, note that there are a total of 2n − 1 games played. (d) Explain why 2n − 1 games are played. Number these games, and let Bi denote the event that A and B play each other in game i, i = 1, . . . , 2n − 1. (e) What is P(Bi )? (f) Use part (e) to ﬁnd P(E). 3.81. An investor owns shares in a stock whose present value is 25. She has decided that she must sell her stock if it goes either down to 10 or up to 40. If each change of price is either up 1 point with probability .55 or down 1 point with probability .45, and the successive changes are independent, what is the probability that the investor retires a winner? 3.82. A and B ﬂip coins. A starts and continues ﬂipping until a tail occurs, at which point B starts ﬂipping and continues until there is a tail. Then A takes

110

3.83.

3.84.

3.85.

3.86.

Chapter 3

Conditional Probability and Independence

over, and so on. Let P1 be the probability of the coin’s landing on heads when A ﬂips and P2 when B ﬂips. The winner of the game is the ﬁrst one to get (a) 2 heads in a row; (b) a total of 2 heads; (c) 3 heads in a row; (d) a total of 3 heads. In each case, ﬁnd the probability that A wins. Die A has 4 red and 2 white faces, whereas die B has 2 red and 4 white faces. A fair coin is ﬂipped once. If it lands on heads, the game continues with die A; if it lands on tails, then die B is to be used. (a) Show that the probability of red at any throw is 12 . (b) If the ﬁrst two throws result in red, what is the probability of red at the third throw? (c) If red turns up at the ﬁrst two throws, what is the probability that it is die A that is being used? An urn contains 12 balls, of which 4 are white. Three players—A, B, and C—successively draw from the urn, A ﬁrst, then B, then C, then A, and so on. The winner is the ﬁrst one to draw a white ball. Find the probability of winning for each player if (a) each ball is replaced after it is drawn; (b) the balls that are withdrawn are not replaced. Repeat Problem 3.84 when each of the 3 players selects from his own urn. That is, suppose that there are 3 different urns of 12 balls with 4 white balls in each urn. Let S = {1, 2, . . . , n} and suppose that A and B are, independently, equally likely to be any of the 2n subsets (including the null set and S itself) of S. (a) Show that n 3 P{A ( B} = 4

Hint: Let N(B) denote the number of elements in B. Use P{A ( B} =

n

P{A ( B|N(B) = i}P{N(B) = i}

i=0

Show that P{AB = Ø} =

n 3 4

.

3.87. In Example 5e, what is the conditional probability that the ith coin was selected given that the ﬁrst n trials all result in heads? 3.88. In Laplace’s rule of succession (Example 5e), are the outcomes of the successive ﬂips independent? Explain. 3.89. A person tried by a 3-judge panel is declared guilty if at least 2 judges cast votes of guilty. Suppose that when the defendant is in fact guilty, each judge will independently vote guilty with probability .7, whereas when the defendant is in fact innocent, this probability drops to .2. If 70 percent of defendants are guilty, compute the conditional probability that judge number 3 votes guilty given that (a) judges 1 and 2 vote guilty; (b) judges 1 and 2 cast 1 guilty and 1 not guilty vote; (c) judges 1 and 2 both cast not guilty votes. Let Ei , i = 1, 2, 3 denote the event that judge i casts a guilty vote. Are these events independent. Are they conditionally independent? Explain. 3.90. Suppose that n independent trials, each of which results in any of the outcomes 0, 1, or 2, with respective probabilities p0 , p1 , and p2 , 2i=0 pi = 1, are performed. Find the probability that outcomes 1 and 2 both occur at least once.

THEORETICAL EXERCISES 3.1. Show that if P(A) > 0, then P(AB|A) Ú P(AB|A ∪ B) 3.2. Let A ( B. Express the following probabilities as simply as possible: P(A|B),

P(A|Bc ),

P(B|A),

P(B|Ac )

3.3. Consider a school community of m families, with ni k of them having i children, i = 1, . . . , k, ni = m. i=1

Consider the following two methods for choosing a child:

1. Choose one of the m families at random and then randomly choose a child from that family. k ini children at random. 2. Choose one of the i=1

Show that method 1 is more likely than method 2 to result in the choice of a ﬁrstborn child. Hint: In solving this problem, you will need to show that k i=1

ini

k nj j=1

j

Ú

k i=1

ni

k j=1

nj

Theoretical Exercises To do so, multiply the sums and show that, for all pairs i, j, the coefﬁcient of the term ni nj is greater in the expression on the left than in the one on the right. 3.4. A ball is in any one of n boxes and is in the ith box with probability Pi . If the ball is in box i, a search of that box will uncover it with probability αi . Show that the conditional probability that the ball is in box j, given that a search of box i did not uncover it, is Pj if j Z i 1 − αi Pi

3.8. Let A, B, and C be events relating to the experiment of rolling a pair of dice. (a) If P(A|C) > P(B|C)

P(A|C) > P(A|Cc )

3.5. An event F is said to carry negative information about an event E, and we write F R E, if P(E|F) … P(E)

P(E1 ∪ E2 ∪ · · · ∪ En ) = 1 −

n

3.9.

3.10.

[1 − P(Ei )]

i=1

3.7. (a) An urn contains n white and m black balls. The balls are withdrawn one at a time until only those of the same color are left. Show that, with probability n/(n + m), they are all white. Hint: Imagine that the experiment continues until all the balls are removed, and consider the last ball withdrawn. (b) A pond contains 3 distinct species of ﬁsh, which we will call the Red, Blue, and Green ﬁsh. There are r Red, b Blue, and g Green ﬁsh. Suppose that the ﬁsh are removed from the pond in a random order. (That is, each selection is equally likely to be any of the remaining ﬁsh.) What is the probability that the Red ﬁsh are the ﬁrst species to become extinct in the pond? Hint: Write P{R} = P{RBG} + P{RGB}, and compute the probabilities on the right by ﬁrst conditioning on the last species to be removed.

and P(A|Cc ) > P(B|Cc )

either prove that P(A) > P(B) or give a counterexample by deﬁning events A, B, and C for which that relationship is not true. (b) If

(1 − αi )Pi if j = i 1 − αi Pi

Prove or give counterexamples to the following assertions: (a) If F R E, then E R F. (b) If F R E and E R G, then F R G. (c) If F R E and G R E, then FG R E. Repeat parts (a), (b), and (c) when R is replaced by Q, where we say that F carries positive information about E, written F Q E, when P(E|F) Ú P(E). 3.6. Prove that if E1 , E2 , . . . , En are independent events, then

111

3.11.

3.12.

and P(B|C) > P(B|Cc )

either prove that P(AB|C) > P(AB|Cc ) or give a counterexample by deﬁning events A, B, and C for which that relationship is not true. Hint: Let C be the event that the sum of a pair of dice is 10; let A be the event that the ﬁrst die lands on 6; let B be the event that the second die lands on 6. Consider two independent tosses of a fair coin. Let A be the event that the ﬁrst toss results in heads, let B be the event that the second toss results in heads, and let C be the event that in both tosses the coin lands on the same side. Show that the events A, B, and C are pairwise independent—that is, A and B are independent, A and C are independent, and B and C are independent—but not independent. Consider a collection of n individuals. Assume that each person’s birthday is equally likely to be any of the 365 days of the year and also that the birthdays are independent. Let Ai,j , i Z j, denote the event that persons i and j have the same birthday. Show that these events are pairwise independent, but not independent. That is,show that Ai,j and Ar,s are n independent, but the events Ai,j , i Z j are not 2 independent. In each of n independent tosses of a coin, the coin lands on heads with probability p. How large need n be so that the probability of obtaining at least one head is at least 12 ? Show that 0 … ai … 1, i = 1, 2, . . ., then ⎤ ⎡ i−1 q q ⎥ ⎢ (1 − a ) + (1 − ai ) = 1 a ⎣ i j ⎦ i=1

j=1

i=1

Hint: Suppose that an inﬁnite number of coins are to be ﬂipped. Let ai be the probability that the ith coin lands on heads, and consider when the ﬁrst head occurs. 3.13. The probability of getting a head on a single toss of a coin is p. Suppose that A starts and continues to ﬂip the coin until a tail shows up, at which point

112

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B starts ﬂipping. Then B continues to ﬂip until a tail comes up, at which point A takes over, and so on. Let Pn,m denote the probability that A accumulates a total of n heads before B accumulates m. Show that Pn,m = pPn−1,m + (1 − p)(1 − Pm,n ) ∗ 3.14.

Suppose that you are gambling against an inﬁnitely rich adversary and at each stage you either win or lose 1 unit with respective probabilities p and 1 − p. Show that the probability that you eventually go broke is 1

if p …

(q/p)i if p >

1 2 1 2

where q = 1 − p and where i is your initial fortune. 3.15. Independent trials that result in a success with probability p are successively performed until a total of r successes is obtained. Show that the probability that exactly n trials are required is n − 1 pr (1 − p)n−r r − 1 Use this result to solve the problem of the points (Example 4j). Hint: In order for it to take n trials to obtain r successes, how many successes must occur in the ﬁrst n − 1 trials? 3.16. Independent trials that result in a success with probability p and a failure with probability 1 − p are called Bernoulli trials. Let Pn denote the probability that n Bernoulli trials result in an even number of successes (0 being considered an even number). Show that Pn = p(1 − Pn−1 ) + (1 − p)Pn−1

n Ú 1

and use this formula to prove (by induction) that Pn =

1 + (1 − 2p)n 2

3.17. Suppose that n independent trials are performed, with trial i being a success with probability 1/(2i + 1). Let Pn denote the probability that the total number of successes that result is an odd number. (a) Find Pn for n = 1, 2, 3, 4, 5. (b) Conjecture a general formula for Pn . (c) Derive a formula for Pn in terms of Pn−1 . (d) Verify that your conjecture in part (b) satisﬁes the recursive formula in part (d). Because the recursive formula has a unique solution, this then proves that your conjecture is correct.

3.18. Let Qn denote the probability that no run of 3 consecutive heads appears in n tosses of a fair coin. Show that 1 1 1 Qn−1 + Qn−2 + Qn−3 2 4 8 Q0 = Q 1 = Q 2 = 1

Qn =

Find Q8 . Hint: Condition on the ﬁrst tail. 3.19. Consider the gambler’s ruin problem, with the exception that A and B agree to play no more than n games. Let Pn,i denote the probability that A winds up with all the money when A starts with i and B starts with N − i. Derive an equation for Pn,i in terms of Pn−1, i+1 and Pn−1, i−1 , and compute P7, 3 , N = 5. 3.20. Consider two urns, each containing both white and black balls. The probabilities of drawing white balls from the ﬁrst and second urns are, respectively, p and p . Balls are sequentially selected with replacement as follows: With probability α, a ball is initially chosen from the ﬁrst urn, and with probability 1 − α, it is chosen from the second urn. The subsequent selections are then made according to the rule that whenever a white ball is drawn (and replaced), the next ball is drawn from the same urn, but when a black ball is drawn, the next ball is taken from the other urn. Let αn denote the probability that the nth ball is chosen from the ﬁrst urn. Show that αn+1 = αn (p + p − 1) + 1 − p

n Ú 1

and use this formula to prove that 1 − p 1 − p + α − αn = 2 − p − p 2 − p − p * (p + p − 1)n−1 Let Pn denote the probability that the nth ball selected is white. Find Pn . Also, compute limn→q αn and limn→q Pn . 3.21. The Ballot Problem. In an election, candidate A receives n votes and candidate B receives m votes, where n > m. Assuming that all of the (n + m)!/n! m! orderings of the votes are equally likely, let Pn,m denote the probability that A is always ahead in the counting of the votes. (a) Compute P2,1 , P3,1 , P3,2 , P4,1 , P4,2 , P4,3 . (b) Find Pn,1 , Pn,2 . (c) On the basis of your results in parts (a) and (b), conjecture the value of Pn,m . (d) Derive a recursion for Pn,m in terms of Pn−1,m and Pn,m−1 by conditioning on who receives the last vote.

Theoretical Exercises (e) Use part (d) to verify your conjecture in part (c) by an induction proof on n + m. 3.22. As a simpliﬁed model for weather forecasting, suppose that the weather (either wet or dry) tomorrow will be the same as the weather today with probability p. Show that the weather is dry on January 1, then Pn , the probability that it will be dry n days later, satisﬁes Pn = (2p − 1)Pn−1 + (1 − p)

n Ú 1

P0 = 1 Prove that Pn =

1 1 + (2p − 1)n 2 2

n Ú 0

3.23. A bag contains a white and b black balls. Balls are chosen from the bag according to the following method: 1. A ball is chosen at random and is discarded. 2. A second ball is then chosen. If its color is different from that of the preceding ball, it is replaced in the bag and the process is repeated from the beginning. If its color is the same, it is discarded and we start from step 2. In other words, balls are sampled and discarded until a change in color occurs, at which point the last ball is returned to the urn and the process starts anew. Let Pa,b denote the probability that the last ball in the bag is white. Prove that Pa,b = ∗ 3.24.

1 2

Hint: Use induction on k K a + b. A round-robin tournament of n contestants is a n tournament in which each of the pairs of 2 contestants play each other exactly once, with the outcome of any play being that one of the contestants wins and the other loses. For a ﬁxed integer k, k < n, a question of interest is whether it is possible that the tournament outcome is such that, for every set of k players, there is a player who beat each member of that set. Show that if k n−k 1 n 1 − < 1 k 2 then such an outcome is possible. Hint: Suppose that the results of the games are independent and that each game is equally likely

113

n k sets of k contestants, and let Bi denote the event that no contestant beat all of the k players in the Then use Boole’s inequality to bound ith set.

Bi . P to be won by either contestant. Number the

i

3.25. Prove directly that P(E|F) = P(E|FG)P(G|F) + P(E|FGc )P(Gc |F) 3.26. Prove the equivalence of Equations (5.11) and (5.12). 3.27. Extend the deﬁnition of conditional independence to more than 2 events. 3.28. Prove or give a counterexample. If E1 and E2 are independent, then they are conditionally independent given F. 3.29. In Laplace’s rule of succession (Example 5e), show that if the ﬁrst n ﬂips all result in heads, then the conditional probability that the next m ﬂips also result in all heads is (n + 1)/(n + m + 1). 3.30. In Laplace’s rule of succession (Example 5e), suppose that the ﬁrst n ﬂips resulted in r heads and n − r tails. Show that the probability that the (n + 1)st ﬂip turns up heads is (r + 1)/(n + 2). To do so, you will have to prove and use the identity * 1 n!m! yn (1 − y)m dy = (n + m + 1)! 0 Hint: To prove the identity, let C(n, m) -1 n m 0 y (1 − y) dy. Integrating by parts yields C(n, m) =

=

m C(n + 1, m − 1) n + 1

Starting with C(n, 0) = 1/(n + 1), prove the identity by induction on m. 3.31. Suppose that a nonmathematical, but philosophically minded, friend of yours claims that Laplace’s rule of succession must be incorrect because it can lead to ridiculous conclusions. “For instance,” says he, “the rule states that if a boy is 10 years old, having lived 10 years, the boy has probability 11 12 of living another year. On the other hand, if the boy has an 80-year-old grandfather, then, by Laplace’s rule, the grandfather has probability 81 82 of surviving another year. However, this is ridiculous. Clearly, the boy is more likely to survive an additional year than the grandfather is.” How would you answer your friend?

114

Chapter 3

Conditional Probability and Independence

SELF-TEST PROBLEMS AND EXERCISES 3.1. In a game of bridge, West has no aces. What is the probability of his partner’s having (a) no aces? (b) 2 or more aces? (c) What would the probabilities be if West had exactly 1 ace? 3.2. The probability that a new car battery functions for over 10,000 miles is .8, the probability that it functions for over 20,000 miles is .4, and the probability that it functions for over 30,000 miles is .1. If a new car battery is still working after 10,000 miles, what is the probability that (a) its total life will exceed 20,000 miles? (b) its additional life will exceed 20,000 miles? 3.3. How can 20 balls, 10 white and 10 black, be put into two urns so as to maximize the probability of drawing a white ball if an urn is selected at random and a ball is drawn at random from it? 3.4. Urn A contains 2 white balls and 1 black ball, whereas urn B contains 1 white ball and 5 black balls. A ball is drawn at random from urn A and placed in urn B. A ball is then drawn from urn B. It happens to be white. What is the probability that the ball transferred was white? 3.5. An urn has r red and w white balls that are randomly removed one at a time. Let Ri be the event that the ith ball removed is red. Find (a) P(Ri ) (b) P(R5 |R3 ) (c) P(R3 |R5 ) 3.6. An urn contains b black balls and r red balls. One of the balls is drawn at random, but when it is put back in the urn, c additional balls of the same color are put in with it. Now, suppose that we draw another ball. Show that the probability that the ﬁrst ball was black, given that the second ball drawn was red, is b/(b + r + c). 3.7. A friend randomly chooses two cards, without replacement, from an ordinary deck of 52 playing cards. In each of the following situations, determine the conditional probability that both cards are aces. (a) You ask your friend if one of the cards is the ace of spades, and your friend answers in the afﬁrmative. (b) You ask your friend if the ﬁrst card selected is an ace, and your friend answers in the afﬁrmative. (c) You ask your friend if the second card selected is an ace, and your friend answers in the afﬁrmative. (d) You ask your friend if either of the cards selected is an ace, and your friend answers in the afﬁrmative.

3.8. Show that P(H|E) P(H) P(E|H) = P(G|E) P(G) P(E|G)

3.9.

3.10.

3.11.

3.12.

3.13.

Suppose that, before new evidence is observed, the hypothesis H is three times as likely to be true as is the hypothesis G. If the new evidence is twice as likely when G is true than it is when H is true, which hypothesis is more likely after the evidence has been observed? You ask your neighbor to water a sickly plant while you are on vacation. Without water, it will die with probability .8; with water, it will die with probability .15. You are 90 percent certain that your neighbor will remember to water the plant. (a) What is the probability that the plant will be alive when you return? (b) If the plant is dead upon your return, what is the probability that your neighbor forgot to water it? Six balls are to be randomly chosen from an urn containing 8 red, 10 green, and 12 blue balls. (a) What is the probability at least one red ball is chosen? (b) Given that no red balls are chosen, what is the conditional probability that there are exactly 2 green balls among the 6 chosen? A type C battery is in working condition with probability .7, whereas a type D battery is in working condition with probability .4. A battery is randomly chosen from a bin consisting of 8 type C and 6 type D batteries. (a) What is the probability that the battery works? (b) Given that the battery does not work, what is the conditional probability that it was a type C battery? Maria will take two books with her on a trip. Suppose that the probability that she will like book 1 is .6, the probability that she will like book 2 is .5, and the probability that she will like both books is .4. Find the conditional probability that she will like book 2 given that she did not like book 1. Balls are randomly removed from an urn that initially contains 20 red and 10 blue balls. (a) What is the probability that all of the red balls are removed before all of the blue ones have been removed? Now suppose that the urn initially contains 20 red, 10 blue, and 8 green balls. (b) Now what is the probability that all of the red balls are removed before all of the blue ones have been removed?

Self-Test Problems and Exercises

3.14.

3.15.

3.16.

3.17.

3.18.

3.19.

(c) What is the probability that the colors are depleted in the order blue, red, green? (d) What is the probability that the group of blue balls is the ﬁrst of the three groups to be removed? A coin having probability .8 of landing on heads is ﬂipped. A observes the result—either heads or tails—and rushes off to tell B. However, with probability .4, A will have forgotten the result by the time he reaches B. If A has forgotten, then, rather than admitting this to B, he is equally likely to tell B that the coin landed on heads or that it landed tails. (If he does remember, then he tells B the correct result.) (a) What is the probability that B is told that the coin landed on heads? (b) What is the probability that B is told the correct result? (c) Given that B is told that the coin landed on heads, what is the probability that it did in fact land on heads? In a certain species of rats, black dominates over brown. Suppose that a black rat with two black parents has a brown sibling. (a) What is the probability that this rat is a pure black rat (as opposed to being a hybrid with one black and one brown gene)? (b) Suppose that when the black rat is mated with a brown rat, all 5 of their offspring are black. Now what is the probability that the rat is a pure black rat? (a) In Problem 3.65b, ﬁnd the probability that a current ﬂows from A to B, by conditioning on whether relay 1 closes. (b) Find the conditional probability that relay 3 is closed given that a current ﬂows from A to B. For the k-out-of-n system described in Problem 3.67, assume that each component independently works with probability 12 . Find the conditional probability that component 1 is working, given that the system works, when (a) k = 1, n = 2; (b) k = 2, n = 3. Mr. Jones has devised a gambling system for winning at roulette. When he bets, he bets on red and places a bet only when the 10 previous spins of the roulette have landed on a black number. He reasons that his chance of winning is quite large because the probability of 11 consecutive spins resulting in black is quite small. What do you think of this system? Three players simultaneously toss coins. The coin tossed by A(B)[C] turns up heads with probability P1 (P2 )[P3 ]. If one person gets an outcome different from those of the other two, then he is the odd

115

man out. If there is no odd man out, the players ﬂip again and continue to do so until they get an odd man out. What is the probability that A will be the odd man? 3.20. Suppose that there are n possible outcomes of a trial, with outcome i resulting with probability n pi , i = 1, . . . , n, pi = 1. If two independent trii=1

3.21.

3.22.

3.23.

3.24.

als are observed, what is the probability that the result of the second trial is larger than that of the ﬁrst? If A ﬂips n + 1 and B ﬂips n fair coins, show that the probability that A gets more heads than B is 12 . Hint: Condition on which player has more heads after each has ﬂipped n coins. (There are three possibilities.) Prove or give counterexamples to the following statements: (a) If E is independent of F and E is independent of G, then E is independent of F ∪ G. (b) If E is independent of F, and E is independent of G, and FG = Ø, then E is independent of F ∪ G. (c) If E is independent of F, and F is independent of G, and E is independent of FG, then G is independent of EF. Let A and B be events having positive probability. State whether each of the following statements is (i) necessarily true, (ii) necessarily false, or (iii) possibly true. (a) If A and B are mutually exclusive, then they are independent. (b) If A and B are independent, then they are mutually exclusive. (c) P(A) = P(B) = .6, and A and B are mutually exclusive. (d) P(A) = P(B) = .6, and A and B are independent. Rank the following from most likely to least likely to occur: 1. A fair coin lands on heads. 2. Three independent trials, each of which is a success with probability .8, all result in successes. 3. Seven independent trials, each of which is a success with probability .9, all result in successes.

3.25. Two local factories, A and B, produce radios. Each radio produced at factory A is defective with probability .05, whereas each one produced at factory B is defective with probability .01. Suppose you purchase two radios that were produced at the same factory, which is equally likely to have been either factory A or factory B. If the ﬁrst radio that you check is defective, what is the conditional probability that the other one is also defective?

116

Chapter 3

Conditional Probability and Independence

3.26. Show that if P(A|B) = 1, then P(Bc |Ac ) = 1. 3.27. An urn initially contains 1 red and 1 blue ball. At each stage, a ball is randomly withdrawn and replaced by two other balls of the same color. (For instance, if the red ball is initially chosen, then there would be 2 red and 1 blue ball in the urn when the next selection occurs.) Show by mathematical induction that the probability that there are exactly i red balls in the urn 1 after n stages have been completed is n+1 ,1 … i … n + 1. 3.28. A total of 2n cards, of which 2 are aces, are to be randomly divided among two players, with each player receiving n cards. Each player is then to declare, in sequence, whether he or she has received any aces. What is the conditional probability that the second player has no aces, given that the ﬁrst player declares in the afﬁrmative, when (a) n = 2? (b) n = 10? (c) n = 100? To what does the probability converge as n goes to inﬁnity? Why?

3.29. There are n distinct types of coupons, and each coupon obtained is, independently of prior types n collected, of type i with probability pi , i=1 pi = 1. (a) If n coupons are collected, what is the probability that one of each type is obtained? (b) Now suppose that p1 = p2 = · · · = pn = 1/n. Let Ei be the event that there are no type i coupons among the n collected. Apply the inclusion–exclusion identity for the probability of the union of events to P(∪i Ei ) to prove the identity n! =

n n (−1)k (n − k)n k k=0

3.30. Show that, for any events E and F, P(E|E ∪ F) Ú P(E|F) Hint: Compute P(E|E ∪ F) by conditioning on whether F occurs.

C H A P T E R

4

Random Variables 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10

RANDOM VARIABLES DISCRETE RANDOM VARIABLES EXPECTED VALUE EXPECTATION OF A FUNCTION OF A RANDOM VARIABLE VARIANCE THE BERNOULLI AND BINOMIAL RANDOM VARIABLES THE POISSON RANDOM VARIABLE OTHER DISCRETE PROBABILITY DISTRIBUTIONS EXPECTED VALUE OF SUMS OF RANDOM VARIABLES PROPERTIES OF THE CUMULATIVE DISTRIBUTION FUNCTION

4.1

RANDOM VARIABLES Frequently, when an experiment is performed, we are interested mainly in some function of the outcome as opposed to the actual outcome itself. For instance, in tossing dice, we are often interested in the sum of the two dice and are not really concerned about the separate values of each die. That is, we may be interested in knowing that the sum is 7 and may not be concerned over whether the actual outcome was (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), or (6, 1). Also, in ﬂipping a coin, we may be interested in the total number of heads that occur and not care at all about the actual head–tail sequence that results. These quantities of interest, or, more formally, these real-valued functions deﬁned on the sample space, are known as random variables. Because the value of a random variable is determined by the outcome of the experiment, we may assign probabilities to the possible values of the random variable. EXAMPLE 1a Suppose that our experiment consists of tossing 3 fair coins. If we let Y denote the number of heads that appear, then Y is a random variable taking on one of the values 0, 1, 2, and 3 with respective probabilities P{Y = 0} = P{(T, T, T)} =

1 8

3 8 3 P{Y = 2} = P{(T, H, H), (H, T, H), (H, H, T)} = 8 1 P{Y = 3} = P{(H, H, H)} = 8 P{Y = 1} = P{(T, T, H), (T, H, T), (H, T, T)} =

117

118

Chapter 4

Random Variables

Since Y must take on one of the values 0 through 3, we must have ⎛ ⎞ 3 3 1 = P ⎝ {Y = i}⎠ = P{Y = i} i=0

i=0

which, of course, is in accord with the preceding probabilities.

.

EXAMPLE 1b Three balls are to be randomly selected without replacement from an urn containing 20 balls numbered 1 through 20. If we bet that at least one of the balls that are drawn has a number as large as or larger than 17, what is the probability that we win the bet? Solution. Let X denote the largest number selected. Then X is a random variable taking on one of the values 3, 4, . . . , 20. Furthermore, if we suppose that each of the 20 possible selections are equally likely to occur, then 3 i − 1 2 i = 3, . . . , 20 (1.1) P{X = i} = 20 3 Equation (1.1) follows because the number of selections that result in the event {X = i} is just the number of selections that result in the ball numbered i and two 1 of the balls numbered 1 through i − 1 being chosen. Because there are clearly 1 i − 1 such selections, we obtain the probabilities expressed in Equation (1.1), 2 from which we see that 19 2 3 = P{X = 20} = = .150 20 20 3 18 2 51 = P{X = 19} = L .134 380 20 3 17 2 34 = P{X = 18} = L .119 285 20 3 16 2 2 = P{X = 17} = L .105 19 20 3

Section 4.1

Random Variables

119

Hence, since the event {X Ú 17} is the union of the disjoint events {X = i}, i = 17, 18, 19, 20, it follows that the probability of our winning the bet is given by P{X Ú 17} L .105 + .119 + .134 + .150 = .508

.

EXAMPLE 1c Independent trials consisting of the ﬂipping of a coin having probability p of coming up heads are continually performed until either a head occurs or a total of n ﬂips is made. If we let X denote the number of times the coin is ﬂipped, then X is a random variable taking on one of the values 1, 2, 3, . . . , n with respective probabilities P{X = 1} = P{H} = p P{X = 2} = P{(T, H)} = (1 − p)p P{X = 3} = P{(T, T, H)} = (1 − p)2 p # # # P{X = n − 1} = P{(T, T, . . . , T, H)} = (1 − p)n−2 p n−2

P{X = n} = P{(T, T, . . . , T, T), (T, T, . . . , T, H)} = (1 − p)n−1 n−1

n−1

As a check, note that ⎛ ⎞ n n P ⎝ {X = i}⎠ = P{X = i} i=1

i=1

=

n−1

p(1 − p)i−1 + (1 − p)n−1

i=1

1 − (1 − p)n−1 =p 1 − (1 − p)

+ (1 − p)n−1

= 1 − (1 − p)n−1 + (1 − p)n−1 =1 . EXAMPLE 1d Three balls are randomly chosen from an urn containing 3 white, 3 red, and 5 black balls. Suppose that we win $1 for each white ball selected and lose $1 for each red ball

120

Chapter 4

Random Variables

selected. If we let X denote our total winnings from the experiment, then X is a random variable taking on the possible values 0, ;1, ;2, ;3 with respective probabilities P{X = 0} =

5 3

5 3 3 1 1 1 55 = 165 11 3

+

P{X = 1} = P{X = −1} = P{X = 2} = P{X = −2} = P{X = 3} = P{X = −3} =

3 1

3 2

3 3 11 3

11 3

5 2

5 1

+

11 3

3 2

3 1

=

39 165

=

15 165

=

1 165

These probabilities are obtained, for instance, by noting that in order for X to equal 0, either all 3 balls selected must be black or 1 ball of each color must be selected. Similarly, the event {X = 1} occurs either if 1 white and 2 black balls are selected or if 2 white and 1 red is selected. As a check, we note that 3

P{X = i} +

i=0

3

P{X = −i} =

i=1

55 + 39 + 15 + 1 + 39 + 15 + 1 =1 165

The probability that we win money is given by 3 i=1

P{X = i} =

1 55 = 165 3

.

EXAMPLE 1e Suppose that there are N distinct types of coupons and that each time one obtains a coupon, it is, independently of previous selections, equally likely to be any one of the N types. One random variable of interest is T, the number of coupons that needs to be collected until one obtains a complete set of at least one of each type. Rather than derive P{T = n} directly, let us start by considering the probability that T is greater than n. To do so, ﬁx n and deﬁne the events A1 , A2 , . . . , AN as follows: Aj is the event

Section 4.1

Random Variables

121

that no type j coupon is contained among the ﬁrst n coupons collected, j = 1, . . . , N. Hence, ⎛ ⎞ N ⎜ ⎟ P{T > n} = P ⎝ Aj ⎠ =

j=1

P(Aj ) −

P(Aj1 Aj2 ) + · · ·

j1

j

+ (−1)k+1

P(Aj1 Aj2 · · · Ajk ) · · ·

j1

+ (−1)N+1 P(A1 A2 · · · AN ) Now, Aj will occur if each of the n coupons collected is not of type j. Since each of the coupons will not be of type j with probability (N − 1)/N, we have, by the assumed independence of the types of successive coupons, N − 1 n P(Aj ) = N Also, the event Aj1 Aj2 will occur if none of the ﬁrst n coupons collected is of either type j1 or type j2 . Thus, again using independence, we see that N − 2 n P(Aj1 Aj2 ) = N The same reasoning gives P(Aj1 Aj2 · · · Ajk ) =

N − k N

n

and we see that, for n > 0, N − 2 n N − 3 n N − 1 n N N − + − ··· P{T > n} = N 2 3 N N N n 1 N + (−1)N N − 1 N N−1 n N N − i = (−1)i+1 (1.2) i N i=1

The probability that T equals n can now be obtained from the preceding formula by the use of P{T > n − 1} = P{T = n} + P{T > n} or, equivalently, P{T = n} = P{T > n − 1} − P{T > n} Another random variable of interest is the number of distinct types of coupons that are contained in the ﬁrst n selections—call this random variable Dn . To compute

122

Chapter 4

Random Variables

P{Dn = k}, let us start by ﬁxing attention on a particular set of k distinct types, and let us then determine the probability that this set constitutes the set of distinct types obtained in the ﬁrst n selections. Now, in order for this to be the situation, it is necessary and sufﬁcient that, of the ﬁrst n coupons obtained, A: B:

each is one of these k types. each of these k types is represented.

Now, each coupon selected will be one of the k types with probability k/N, so the probability that A will be valid is (k/N)n . Also, given that a coupon is of one of the k types under consideration, it is easy to see that it is equally likely to be of any one of these k types. Hence, the conditional probability of B given that A occurs is the same as the probability that a set of n coupons, each equally likely to be any of k possible types, contains a complete set of all k types. But this is just the probability that the number needed to amass a complete set, when choosing among k types, is less than or equal to n and is thus obtainable from Equation (1.2) with k replacing N. Thus, we have n k P(A) = N k−1 k − i n k P(B|A) = 1 − (−1)i+1 i k i=1

N possible choices for the set of k types, we arrive at k N P(AB) P{Dn = k} = k ⎡ ⎤ n n k−1 k k − i N k ⎣1 − = (−1)i+1 ⎦ k i N k

Finally, as there are

i=1

Remark. Since one must collect at least N coupons to obtain a compete set, it follows that P{T > n} = 1 if n < N. Therefore, from Equation (1.2), we obtain the interesting combinatorial identity that, for integers 1 … n < N, N−1 i=1

N i

N − i N

n (−1)i+1 = 1

which can be written as N−1 i=0

N i

N − i N

n (−1)i+1 = 0

or, upon multiplying by (−1)N N n and letting j = N − i, N N j=1

j

jn (−1)j−1 = 0

1 … n < N .

Section 4.2

Discrete Random Variables

123

For a random variable X, the function F deﬁned by F(x) = P{X … x}

− q < x < q

is called the cumulative distribution function, or, more simply, the distribution function, of X. Thus, the distribution function speciﬁes, for all real values x, the probability that the random variable is less than or equal to x. Now, suppose that a … b. Then, because the event {X … a} is contained in the event {X … b}, it follows that F(a), the probability of the former, is less than or equal to F(b), the probability of the latter. In other words, F(x) is a nondecreasing function of x. Other general properties of the distribution function are given in Section 4.10. 4.2

DISCRETE RANDOM VARIABLES A random variable that can take on at most a countable number of possible values is said to be discrete. For a discrete random variable X, we deﬁne the probability mass function p(a) of X by p(a) = P{X = a} The probability mass function p(a) is positive for at most a countable number of values of a. That is, if X must assume one of the values x1 , x2 , . . . , then p(xi ) Ú 0 p(x) = 0

for i = 1, 2, . . . for all other values of x

Since X must take on one of the values xi , we have q

p(xi ) = 1

i=1

It is often instructive to present the probability mass function in a graphical format by plotting p(xi ) on the y-axis against xi on the x-axis. For instance, if the probability mass function of X is p(0) =

1 4

p(1) =

1 2

p(2) =

p(x)

1 1– 2 1– 4

0

1 FIGURE 4.1

2

x

1 4

124

Chapter 4

Random Variables p(x)

6 — 36 5 — 36 4 — 36 3 — 36 2 — 36 1 — 36

0

1

2

3

4

5

6

7

8

9

10

11

12

x

FIGURE 4.2

we can represent this function graphically as shown in Figure 4.1. Similarly, a graph of the probability mass function of the random variable representing the sum when two dice are rolled looks like Figure 4.2. EXAMPLE 2a The probability mass function of a random variable X is given by p(i) = cλi /i!, i = 0, 1, 2, . . . , where λ is some positive value. Find (a) P{X = 0} and (b) P{X > 2}. Solution. Since

q

p(i) = 1, we have

i=0

c

q λi

i!

i=0

which, because e x =

q

=1

xi /i!, implies that

i=0

ceλ = 1

or c = e−λ

Hence, (a) P{X = 0} = e−λ λ0 /0! = e−λ (b) P{X > 2} = 1 − P{X … 2} = 1 − P{X = 0} − P{X = 1} − P{X = 2} λ2 e−λ = 1 − e−λ − λe−λ − 2 The cumulative distribution function F can be expressed in terms of p(a) by p(x) F(a) =

.

all x … a

If X is a discrete random variable whose possible values are x1 , x2 , x3 , . . . , where x1 < x2 < x3 < · · · , then the distribution function F of X is a step function. That is,

Section 4.3

Expected Value

125

the value of F is constant in the intervals [xi−1 , xi ) and then takes a step (or jump) of size p(xi ) at xi . For instance, if X has a probability mass function given by p(1) =

1 4

p(2) =

1 2

p(3) =

then its cumulative distribution function is ⎧ ⎪ 0 a ⎪ ⎪ ⎪ ⎪ 1 ⎪ 1 ⎪ ⎪ ⎨4 F(a) = 34 2 ⎪ ⎪ ⎪ 7 ⎪ ⎪ ⎪ 8 3 ⎪ ⎪ ⎩1 4

1 8

p(4) =

1 8

< 1 … a < 2 … a < 3 … a < 4 … a

This function is depicted graphically in Figure 4.3. F(a) 1 7– 8 3– 4

1– 4

1

2

3

4

a

FIGURE 4.3

Note that the size of the step at any of the values 1, 2, 3, and 4 is equal to the probability that X assumes that particular value. 4.3

EXPECTED VALUE One of the most important concepts in probability theory is that of the expectation of a random variable. If X is a discrete random variable having a probability mass function p(x), then the expectation, or the expected value, of X, denoted by E[X], is deﬁned by E[X] = xp(x) x:p(x)>0

In words, the expected value of X is a weighted average of the possible values that X can take on, each value being weighted by the probability that X assumes it. For instance, on the one hand, if the probability mass function of X is given by p(0) = then E[X] = 0

1 = p(1) 2

1 1 1 + 1 = 2 2 2

126

Chapter 4

Random Variables

is just the ordinary average of the two possible values, 0 and 1, that X can assume. On the other hand, if p(0) = then

1 3

p(1) =

2 3

2 2 1 + 1 = E[X] = 0 3 3 3

is a weighted average of the two possible values 0 and 1, where the value 1 is given twice as much weight as the value 0, since p(1) = 2p(0). Another motivation of the deﬁnition of expectation is provided by the frequency interpretation of probabilities. This interpretation (partially justiﬁed by the strong law of large numbers, to be presented in Chapter 8) assumes that if an inﬁnite sequence of independent replications of an experiment is performed, then, for any event E, the proportion of time that E occurs will be P(E). Now, consider a random variable X that must take on one of the values x1 , x2 , . . . xn with respective probabilities p(x1 ), p(x2 ), . . . , p(xn ), and think of X as representing our winnings in a single game of chance. That is, with probability p(xi ) we shall win xi units i = 1, 2, . . . , n. By the frequency interpretation, if we play this game continually, then the proportion of time that we win xi will be p(xi ). Since this is true for all i, i = 1, 2, . . . , n, it follows that our average winnings per game will be n

xi p(xi ) = E[X]

i=1

EXAMPLE 3a Find E[X], where X is the outcome when we roll a fair die. Solution. Since p(1) = p(2) = p(3) = p(4) = p(5) = p(6) = 16 , we obtain 1 1 1 1 1 1 7 + 2 + 3 + 4 + 5 + 6 = E[X] = 1 6 6 6 6 6 6 2

.

EXAMPLE 3b We say that I is an indicator variable for the event A if % 1 if A occurs I= 0 if Ac occurs Find E[I]. Solution. Since p(1) = P(A), p(0) = 1 − P(A), we have E[I] = P(A) That is, the expected value of the indicator variable for the event A is equal to the probability that A occurs. .

Section 4.3

Expected Value

127

EXAMPLE 3c A contestant on a quiz show is presented with two questions, questions 1 and 2, which he is to attempt to answer in some order he chooses. If he decides to try question i ﬁrst, then he will be allowed to go on to question j, j Z i, only if his answer to question i is correct. If his initial answer is incorrect, he is not allowed to answer the other question. The contestant is to receive Vi dollars if he answers question i correctly, i = 1, 2. For instance, he will receive V1 + V2 dollars if he answers both questions correctly. If the probability that he knows the answer to question i is Pi , i = 1, 2, which question should he attempt to answer ﬁrst so as to maximize his expected winnings? Assume that the events Ei , i = 1, 2, that he knows the answer to question i are independent events. Solution. On the one hand, if he attempts to answer question 1 ﬁrst, then he will win 0 V1 V1 + V2

with probability 1 − P1 with probability P1 (1 − P2 ) with probability P1 P2

Hence, his expected winnings in this case will be V1 P1 (1 − P2 ) + (V1 + V2 )P1 P2 On the other hand, if he attempts to answer question 2 ﬁrst, his expected winnings will be V2 P2 (1 − P1 ) + (V1 + V2 )P1 P2 Therefore, it is better to try question 1 ﬁrst if V1 P1 (1 − P2 ) Ú V2 P2 (1 − P1 ) or, equivalently, if V2 P2 V1 P1 Ú 1 − P1 1 − P2 For example, if he is 60 percent certain of answering question 1, worth $200, correctly and he is 80 percent certain of answering question 2, worth $100, correctly, then he should attempt to answer question 2 ﬁrst because 400 =

(200)(.6) (100)(.8) > = 300 .2 .4

.

EXAMPLE 3d A school class of 120 students is driven in 3 buses to a symphonic performance. There are 36 students in one of the buses, 40 in another, and 44 in the third bus. When the buses arrive, one of the 120 students is randomly chosen. Let X denote the number of students on the bus of that randomly chosen student, and ﬁnd E[X]. Solution. Since the randomly chosen student is equally likely to be any of the 120 students, it follows that P{X = 36} = Hence,

E[X] = 36

36 120

3 10

P{X = 40} =

+ 40

40 120

P{X = 44} =

44 120

1 11 1208 + 44 = = 40.2667 3 30 30

128

Chapter 4

Random Variables

However, the average number of students on a bus is 120/3 = 40, showing that the expected number of students on the bus of a randomly chosen student is larger than the average number of students on a bus. This is a general phenomenon, and it occurs because the more students there are on a bus, the more likely it is that a randomly chosen student would have been on that bus. As a result, buses with many students are given more weight than those with fewer students. (See Self-Test Problem 4.) . Remark. The probability concept of expectation is analogous to the physical concept of the center of gravity of a distribution of mass. Consider a discrete random variable X having probability mass function p(xi ), i Ú 1. If we now imagine a weightless rod in which weights with mass p(xi ), i Ú 1, are located at the points xi , i Ú 1 (see Figure 4.4), then the point at which the rod would be in balance is known as the center of gravity. For those readers acquainted with elementary statics, it is now a . simple matter to show that this point is at E[X].†

–1

0

p(–1) = .10, p(0) = .25,

^1

2

p(1) = .30,

p(2) = .35

^ = center of gravity = .9 FIGURE 4.4

4.4

EXPECTATION OF A FUNCTION OF A RANDOM VARIABLE Suppose that we are given a discrete random variable along with its probability mass function and that we want to compute the expected value of some function of X, say, g(X). How can we accomplish this? One way is as follows: Since g(X) is itself a discrete random variable, it has a probability mass function, which can be determined from the probability mass function of X. Once we have determined the probability mass function of g(X), we can compute E[g(X)] by using the deﬁnition of expected value. EXAMPLE 4a Let X denote a random variable that takes on any of the values −1, 0, and 1 with respective probabilities P{X = −1} = .2

P{X = 0} = .5 P{X = 1} = .3

Compute E[X 2 ]. Solution. Let Y = X 2 . Then the probability mass function of Y is given by P{Y = 1} = P{X = −1} + P{X = 1} = .5 P{Y = 0} = P{X = 0} = .5 Hence,

E[X 2 ] = E[Y] = 1(.5) + 0(.5) = .5

† To prove this, we must show that the sum of the torques tending to turn the point around E[X] is equal to 0. That is, we must show that 0 = (xi − E[X])p(xi ), which is immediate. i

Section 4.4

Expectation of a Function of a Random Variable

129

Note that .5 = E[X 2 ] Z (E[X])2 = .01

.

Although the preceding procedure will always enable us to compute the expected value of any function of X from a knowledge of the probability mass function of X, there is another way of thinking about E[g(X)]: Since g(X) will equal g(x) whenever X is equal to x, it seems reasonable that E[g(X)] should just be a weighted average of the values g(x), with g(x) being weighted by the probability that X is equal to x. That is, the following result is quite intuitive: Proposition 4.1. If X is a discrete random variable that takes on one of the values xi , i Ú 1, with respective probabilities p(xi ), then, for any real-valued function g, g(xi )p(xi ) E[g(X)] = i

Before proving this proposition, let us check that it is in accord with the results of Example 4a. Applying it to that example yields E{X 2 } = (−1)2 (.2) + 02 (.5) + 12 (.3) = 1(.2 + .3) + 0(.5) = .5 which is in agreement with the result given in Example 4a. Proof of Proposition 4.1: The proof of Proposition 4.1 proceeds, as in the preceding veriﬁcation, by grouping together all the terms in g(xi )p(xi ) having the same value i

of g(xi ). Speciﬁcally, suppose that yj , j Ú 1, represent the different values of g(xi ), i Ú 1. Then, grouping all the g(xi ) having the same value gives g(xi )p(xi ) = g(xi )p(xi ) i

= =

j

i:g(xi )=yj

j

i:g(xi )=yj

j

=

yj p(xi )

yj

p(xi )

i:g(xi )=yj

yj P{g(X) = yj }

j

= E[g(X)]

EXAMPLE 4b A product that is sold seasonally yields a net proﬁt of b dollars for each unit sold and a net loss of dollars for each unit left unsold when the season ends. The number of units of the product that are ordered at a speciﬁc department store during any season is a random variable having probability mass function p(i), i Ú 0. If the store must stock this product in advance, determine the number of units the store should stock so as to maximize its expected proﬁt.

130

Chapter 4

Random Variables

Solution. Let X denote the number of units ordered. If s units are stocked, then the proﬁt—call it P(s)—can be expressed as P(s) = bX − (s − X) = sb

if X … s if X > s

Hence, the expected proﬁt equals E[P(s)] =

s

[bi − (s − i)]p(i) +

i=0

q

sbp(i)

i=s+1

= (b + ) = (b + )

s i=0 s

ip(i) − s

s

⎡

p(i) + sb ⎣1 −

i=0

ip(i) − (b + )s

s

⎤ p(i)⎦

i=0 s

i=0

p(i) + sb

i=0

= sb + (b + )

s

(i − s)p(i)

i=0

To determine the optimum value of s, let us investigate what happens to the proﬁt when we increase s by 1 unit. By substitution, we see that the expected proﬁt in this case is given by E[P(s + 1)] = b(s + 1) + (b + ) = b(s + 1) + (b + )

s+1 (i − s − 1)p(i) i=0 s

(i − s − 1)p(i)

i=0

Therefore, E[P(s + 1)] − E[P(s)] = b − (b + )

s

p(i)

i=0

Thus, stocking s + 1 units will be better than stocking s units whenever s i=0

p(i) <

b b +

(4.1)

Because the left-hand side of Equation (4.1) is increasing in s while the right-hand side is constant, the inequality will be satisﬁed for all values of s … s∗ , where s∗ is the largest value of s satisfying Equation (4.1). Since E[P(0)] < · · · < E[P(s∗ )] < E[P(s∗ + 1)] > E[P(s∗ + 2)] > · · · it follows that stocking s∗ + 1 items will lead to a maximum expected proﬁt.

.

EXAMPLE 4c Utility Suppose that you must choose one of two possible actions, each of which can result in any of n consequences, denoted as C1 , . . . , Cn . Suppose that if the ﬁrst action is

Section 4.4

Expectation of a Function of a Random Variable

131

chosen, then consequence Ci will result with probability pi , i = 1, . . . , n, whereas if the second action is chosen, then consequence Ci will result with probability qi , n n i = 1, . . . , n, where pi = qi = 1. The following approach can be used to deteri=1

i=1

mine which action to choose: Start by assigning numerical values to the different consequences in the following manner: First, identify the least and the most desirable consequences—call them c and C, respectively; give consequence c the value 0 and give C the value 1. Now consider any of the other n − 2 consequences, say, Ci . To value this consequence, imagine that you are given the choice between either receiving Ci or taking part in a random experiment that either earns you consequence C with probability u or consequence c with probability 1 − u. Clearly, your choice will depend on the value of u. On the one hand, if u = 1, then the experiment is certain to result in consequence C, and since C is the most desirable consequence, you will prefer participating in the experiment to receiving Ci . On the other hand, if u = 0, then the experiment will result in the least desirable consequence—namely, c—so in this case you will prefer the consequence Ci to participating in the experiment. Now, as u decreases from 1 to 0, it seems reasonable that your choice will at some point switch from participating in the experiment to the certain return of Ci , and at that critical switch point you will be indifferent between the two alternatives. Take that indifference probability u as the value of the consequence Ci . In other words, the value of Ci is that probability u such that you are indifferent between either receiving the consequence Ci or taking part in an experiment that returns consequence C with probability u or consequence c with probability 1 − u. We call this indifference probability the utility of the consequence Ci , and we designate it as u(Ci ). To determine which action is superior, we need to evaluate each one. Consider the ﬁrst action, which results in consequence Ci with probability pi , i = 1, . . . , n. We can think of the result of this action as being determined by a two-stage experiment. In the ﬁrst stage, one of the values 1, . . . , n is chosen according to the probabilities p1 , . . . , pn ; if value i is chosen, you receive consequence Ci . However, since Ci is equivalent to obtaining consequence C with probability u(Ci ) or consequence c with probability 1 − u(Ci ), it follows that the result of the two-stage experiment is equivalent to an experiment in which either consequence C or consequence c is obtained, with C being obtained with probability n pi u(Ci ) i=1

Similarly, the result of choosing the second action is equivalent to taking part in an experiment in which either consequence C or consequence c is obtained, with C being obtained with probability n qi u(Ci ) i=1

Since C is preferable to c, it follows that the ﬁrst action is preferable to the second action if n n pi u(Ci ) > qi u(Ci ) i=1

i=1

In other words, the worth of an action can be measured by the expected value of the utility of its consequence, and the action with the largest expected utility is the most preferable. .

132

Chapter 4

Random Variables

A simple logical consequence of Proposition 4.1 is Corollary 4.1. Corollary 4.1. If a and b are constants, then E[aX + b] = aE[X] + b Proof.

E[aX + b] =

(ax + b)p(x)

x:p(x)>0

=a

xp(x) + b

x:p(x)>0

p(x)

x:p(x)>0

= aE[X] + b

The expected value of a random variable X, E[X], is also referred to as the mean or the ﬁrst moment of X. The quantity E[X n ], n Ú 1, is called the nth moment of X. By Proposition 4.1, we note that xn p(x) E[X n ] = x:p(x)>0

4.5

VARIANCE Given a random variable X along with its distribution function F, it would be extremely useful if we were able to summarize the essential properties of F by certain suitably deﬁned measures. One such measure would be E[X], the expected value of X. However, although E[X] yields the weighted average of the possible values of X, it does not tell us anything about the variation, or spread, of these values. For instance, although random variables W, Y, and Z having probability mass functions determined by W = 0 with probability 1 ⎧ ⎨−1 with probability 1 2 Y= ⎩+1 with probability 1 2 ⎧ ⎨−100 with probability Z= ⎩+100 with probability

1 2 1 2

all have the same expectation—namely, 0—there is a much greater spread in the possible values of Y than in those of W (which is a constant) and in the possible values of Z than in those of Y. Because we expect X to take on values around its mean E[X], it would appear that a reasonable way of measuring the possible variation of X would be to look at how far apart X would be from its mean, on the average. One possible way to measure this variation would be to consider the quantity E[|X − μ|], where μ = E[X]. However, it turns out to be mathematically inconvenient to deal with this quantity, so a more tractable quantity is usually considered—namely, the expectation of the square of the difference between X and its mean. We thus have the following deﬁnition.

Section 4.5

Variance

133

Deﬁnition If X is a random variable with mean μ, then the variance of X, denoted by Var(X), is deﬁned by Var(X) = E[(X − μ)2 ]

An alternative formula for Var(X) is derived as follows: Var(X) = E[(X − μ)2 ] (x − μ)2 p(x) = x

(x2 − 2μx + μ2 )p(x) = x

=

x2 p(x) − 2μ

x

xp(x) + μ2

x

p(x)

x

= E[X 2 ] − 2μ2 + μ2 = E[X 2 ] − μ2 That is, Var(X) = E[X 2 ] − (E[X])2 In words, the variance of X is equal to the expected value of X 2 minus the square of its expected value. In practice, this formula frequently offers the easiest way to compute Var(X). EXAMPLE 5a Calculate Var(X) if X represents the outcome when a fair die is rolled. Solution. It was shown in Example 3a that E[X] = 72 . Also, 1 1 1 1 1 1 + 22 + 32 + 42 + 52 + 62 E[X 2 ] = 12 6 6 6 6 6 6 1 = (91) 6 Hence, 91 Var(X) = − 6

2 35 7 = 2 12

A useful identity is that, for any constants a and b, Var(aX + b) = a2 Var(X)

.

134

Chapter 4

Random Variables

To prove this equality, let μ = E[X] and note from Corollary 4.1 that E[aX + b] = aμ + b. Therefore, Var(aX + b) = E[(aX + b − aμ − b)2 ] = E[a2 (X − μ)2 ] = a2 E[(X − μ)2 ] = a2 Var(X) Remarks. (a) Analogous to the means being the center of gravity of a distribution of mass, the variance represents, in the terminology of mechanics, the moment of inertia. (b) The square root of the Var(X) is called the standard deviation of X, and we denote it by SD(X). That is, . SD(X) = Var(X) Discrete random variables are often classiﬁed according to their probability mass functions. In the next few sections, we consider some of the more common types. 4.6

THE BERNOULLI AND BINOMIAL RANDOM VARIABLES Suppose that a trial, or an experiment, whose outcome can be classiﬁed as either a success or a failure is performed. If we let X = 1 when the outcome is a success and X = 0 when it is a failure, then the probability mass function of X is given by p(0) = P{X = 0} = 1 − p p(1) = P{X = 1} = p

(6.1)

where p, 0 … p … 1, is the probability that the trial is a success. A random variable X is said to be a Bernoulli random variable (after the Swiss mathematician James Bernoulli) if its probability mass function is given by Equations (6.1) for some p ∈ (0, 1). Suppose now that n independent trials, each of which results in a success with probability p and in a failure with probability 1 − p, are to be performed. If X represents the number of successes that occur in the n trials, then X is said to be a binomial random variable with parameters (n, p). Thus, a Bernoulli random variable is just a binomial random variable with parameters (1, p). The probability mass function of a binomial random variable having parameters (n, p) is given by n pi (1 − p)n−i i = 0, 1, . . . , n (6.2) p(i) = i The validity of Equation (6.2) may be veriﬁed by ﬁrst noting that the probability of any particular sequence of n outcomes containing i successes and n − i failures is, by the assumed independence of trials, pi (1 − p)n−i . Equation (6.2) then follows, n since there are different sequences of the n outcomes leading to i successes and i n n − i failures. This perhaps can most easily be seen by noting that there are i different choices of the i trials that result in successes. For instance, if n = 4, i = 2, 4 then there are = 6 ways in which the four trials can result in two successes, 2

Section 4.6

The Bernoulli and Binomial Random Variables

135

namely, any of the outcomes (s, s, f , f ), (s, f , s, f ), (s, f , f , s), (f , s, s, f ), (f , s, f , s), and (f , f , s, s), where the outcome (s, s, f , f ) means, for instance, that the ﬁrst two trials are successes and the last two failures. Since each of these outcomes has probability 2 (1 − p2 p) of occurring, the desired probability of two successes in the four trials 4 p2 (1 − p)2 . is 2 Note that, by the binomial theorem, the probabilities sum to 1; that is, q i=0

p(i) =

n

ni pi (1 − p)n−i = [p + (1 − p)]n = 1

i=0

EXAMPLE 6a Five fair coins are ﬂipped. If the outcomes are assumed independent, ﬁnd the probability mass function of the number of heads obtained. Solution. If we let X equal the number of heads (successes) that appear, then X 1 is a binomial random variable with parameters n = 5, p = 2 . Hence, by Equation (6.2), 0 5 1 1 1 5 P{X = 0} = = 0 2 2 32 1 4 1 5 1 5 P{X = 1} = = 1 2 2 32 2 3 10 1 1 5 = P{X = 2} = 2 2 2 32 3 2 1 10 1 5 P{X = 3} = = 3 2 2 32 4 1 1 5 1 5 P{X = 4} = = 4 2 2 32 5 0 1 1 1 5 P{X = 5} = = 5 . 2 2 32 EXAMPLE 6b It is known that screws produced by a certain company will be defective with probability .01, independently of each other. The company sells the screws in packages of 10 and offers a money-back guarantee that at most 1 of the 10 screws is defective. What proportion of packages sold must the company replace? Solution. If X is the number of defective screws in a package, then X is a binomial random variable with parameters (10, .01). Hence, the probability that a package will have to be replaced is 10 10 (.01)1 (.99)9 (.01)0 (.99)10 − 1 − P{X = 0} − P{X = 1} = 1 − 1 0 L .004 Thus, only .4 percent of the packages will have to be replaced.

.

136

Chapter 4

Random Variables

EXAMPLE 6c The following gambling game, known as the wheel of fortune (or chuck-a-luck), is quite popular at many carnivals and gambling casinos: A player bets on one of the numbers 1 through 6. Three dice are then rolled, and if the number bet by the player appears i times, i = 1, 2, 3, then the player wins i units; if the number bet by the player does not appear on any of the dice, then the player loses 1 unit. Is this game fair to the player? (Actually, the game is played by spinning a wheel that comes to rest on a slot labeled by three of the numbers 1 through 6, but this variant is mathematically equivalent to the dice version.) Solution. If we assume that the dice are fair and act independently of each other, then the number of times that the number bet appears is a binomial random variable 1 with parameters 3, 6 . Hence, letting X denote the player’s winnings in the game, we have 0 3 125 5 1 3 = P{X = −1} = 0 6 6 216 1 2 1 75 5 3 P{X = 1} = = 1 6 6 216 2 1 1 15 5 3 P{X = 2} = = 2 6 6 216 3 0 1 5 1 3 = P{X = 3} = 3 6 6 216 In order to determine whether or not this is a fair game for the player, let us calculate E[X]. From the preceding probabilities, we obtain −125 + 75 + 30 + 3 216 −17 = 216

E[X] =

Hence, in the long run, the player will lose 17 units per every 216 games he plays. . In the next example, we consider the simplest form of the theory of inheritance as developed by Gregor Mendel (1822–1884). EXAMPLE 6d Suppose that a particular trait (such as eye color or left-handedness) of a person is classiﬁed on the basis of one pair of genes, and suppose also that d represents a dominant gene and r a recessive gene. Thus, a person with dd genes is purely dominant, one with rr is purely recessive, and one with rd is hybrid. The purely dominant and the hybrid individuals are alike in appearance. Children receive 1 gene from each parent. If, with respect to a particular trait, 2 hybrid parents have a total of 4 children, what is the probability that 3 of the 4 children have the outward appearance of the dominant gene? Solution. If we assume that each child is equally likely to inherit either of 2 genes from each parent, the probabilities that the child of 2 hybrid parents will have dd, rr, and rd pairs of genes are, respectively, 14 , 14 , and 12 . Hence, since an offspring will

Section 4.6

The Bernoulli and Binomial Random Variables

137

have the outward appearance of the dominant gene if its gene pair is either dd or rd, it that the number of such children is binomially distributed with parameters follows 3 4, 4 . Thus, the desired probability is

4 3

3 1 3 27 1 = 4 4 64

.

EXAMPLE 6e Consider a jury trial in which it takes 8 of the 12 jurors to convict the defendant; that is, in order for the defendant to be convicted, at least 8 of the jurors must vote him guilty. If we assume that jurors act independently and that, whether or not the defendant is guilty, each makes the right decision with probability θ, what is the probability that the jury renders a correct decision? Solution. The problem, as stated, is incapable of solution, for there is not yet enough information. For instance, if the defendant is innocent, the probability of the jury’s rendering a correct decision is 12 12 θ i (1 − θ)12−i i i=5

whereas, if he is guilty, the probability of a correct decision is 12 12 i=8

i

θ i (1 − θ)12−i

Therefore, if α represents the probability that the defendant is guilty, then, by conditioning on whether or not he is guilty, we obtain the probability that the jury renders a correct decision: 12 12 12 12 i 12−i θ (1 − θ) θ i (1 − θ)12−i α + (1 − α) i i i=8

i=5

.

EXAMPLE 6f A communication system consists of n components, each of which will, independently, function with probability p. The total system will be able to operate effectively if at least one-half of its components function. (a) For what values of p is a 5-component system more likely to operate effectively than a 3-component system? (b) In general, when is a (2k + 1)-component system better than a (2k − 1)component system? Solution. (a) Because the number of functioning components is a binomial random variable with parameters (n, p), it follows that the probability that a 5-component system will be effective is 5 5 p3 (1 − p)2 + p4 (1 − p) + p5 3 4

138

Chapter 4

Random Variables

whereas the corresponding probability for a 3-component system is

3 2

p2 (1 − p) + p3

Hence, the 5-component system is better if 10p3 (1 − p)2 + 5p4 (1 − p) + p5 > 3p2 (1 − p) + p3 which reduces to 3(p − 1)2 (2p − 1) > 0 or p >

1 2

(b) In general, a system with 2k + 1 components will be better than one with 2k − 1 components if (and only if) p > 12 . To prove this, consider a system of 2k + 1 components and let X denote the number of the ﬁrst 2k − 1 that function. Then P2k+1 (effective) = P{X Ú k + 1} + P{X = k}(1 − (1 − p)2 ) + P{X = k − 1}p2 which follows because the (2k + 1)-component system will be effective if either (i) X Ú k + 1; (ii) X = k and at least one of the remaining 2 components function; or (iii) X = k − 1 and both of the next 2 components function. Since P2k−1 (effective) = P{X Ú k} = P{X = k} + P{X Ú k + 1} we obtain P2k+1 (effective) − P2k−1 (effective) = P{X = k − 1}p2 − (1 − p)2 P{X = k} 2k − 1 2k − 1 pk−1 (1 − p)k p2 − (1 − p)2 pk (1 − p)k−1 = k − 1 k 2k − 1 2k − 1 2k − 1 k k = p (1 − p) [ p − (1 − p)] since = k k − 1 k 1 > 0 3 p > . 2

Section 4.6

4.6.1

The Bernoulli and Binomial Random Variables

139

Properties of Binomial Random Variables

We will now examine the properties of a binomial random variable with parameters n and p. To begin, let us compute its expected value and variance. Now, E[X k ] =

n

ik

i=0

=

n

ik

i=1

Using the identity

i

n i

n i n i

=n

pi (1 − p)n−i pi (1 − p)n−i

n − 1 i − 1

gives E[X k ] = np

n

ik−1

i=1

= np

n−1

n − 1 i − 1

(j + 1)

k−1

j=0

pi−1 (1 − p)n−i n − 1 j

pj (1 − p)n−1−j

by letting j=i − 1

= npE[(Y + 1)k−1 ] where Y is a binomial random variable with parameters n − 1, p. Setting k = 1 in the preceding equation yields E[X] = np That is, the expected number of successes that occur in n independent trials when each is a success with probability p is equal to np. Setting k = 2 in the preceding equation, and using the preceding formula for the expected value of a binomial random variable yields E[X 2 ] = npE[Y + 1] = np[(n − 1)p + 1] Since E[X] = np, we obtain Var(X) = E[X 2 ] − (E[X])2 = np[(n − 1)p + 1] − (np)2 = np(1 − p) Summing up, we have shown the following: If X is a binomial random variable with parameters n and p, then E[X] = np Var(X) = np(1 − p)

140

Chapter 4

Random Variables

The following proposition details how the binomial probability mass function ﬁrst increases and then decreases. Proposition 6.1. If X is a binomial random variable with parameters (n, p), where 0 < p < 1, then as k goes from 0 to n, P{X = k} ﬁrst increases monotonically and then decreases monotonically, reaching its largest value when k is the largest integer less than or equal to (n + 1)p. Proof. We prove the proposition by considering P{X = k}/P{X = k − 1} and determining for what values of k it is greater or less than 1. Now, n! pk (1 − p)n−k (n − k)!k! n! pk−1 (1 − p)n−k+1 (n − k + 1)!(k − 1)! (n − k + 1)p = k(1 − p)

P{X = k} = P{X = k − 1}

Hence, P{X = k} Ú P{X = k − 1} if and only if (n − k + 1)p Ú k(1 − p) or, equivalently, if and only if k … (n + 1)p .

and the proposition is proved.

As an illustration of Proposition 6.1 consider Figure 4.5, the graph of the probability mass function of a binomial random variable with parameters (10, 12 ). EXAMPLE 6g In a U.S. presidential election, the candidate who gains the maximum number of votes in a state is awarded the total number of electoral college votes allocated to

1024 ⫻ p(k)

252 210 120 45 10 1 0

1

2

3

4

5

6

7

8

1 10 FIGURE 4.5 Graph of p(k) = 10 k 2

9

10

k

Section 4.6

The Bernoulli and Binomial Random Variables

141

that state. The number of electoral college votes of a given state is roughly proportional to the population of that state—that is, a state with population n has roughly nc electoral votes. (Actually, it is closer to nc + 2, as a state is given an electoral vote for each member it has in the House of Representatives, with the number of such representatives being roughly proportional to the population of the state, and one electoral college vote for each of its two senators.) Let us determine the average power of a citizen in a state of size n in a close presidential election, where, by average power in a close election, we mean that a voter in a state of size n = 2k + 1 will be decisive if the other n − 1 voters split their votes evenly between the two candidates. (We are assuming here that n is odd, but the case where n is even is quite similar.) Because the election is close, we shall suppose that each of the other n − 1 = 2k voters acts independently and is equally likely to vote for either candidate. Hence, the probability that a voter in a state of size n = 2k + 1 will make a difference to the outcome is the same as the probability that 2k tosses of a fair coin land heads and tails an equal number of times. That is, P{voter in state of size 2k + 1 makes a difference} k k 1 1 2k = k 2 2 (2k)! = k!k!22k To approximate the preceding equality, we make use of Stirling’s approximation, which says that, for k large, √ k! ∼ kk+1/2 e−k 2π where we say that ak ∼ bk when the ratio ak /bk approaches 1 as k approaches q. Hence, it follows that P{voter in state of size 2k + 1 makes a difference} √ (2k)2k+1/2 e−2k 2π 1 ∼ 2k+1 −2k =√ 2k k e (2π)2 kπ Because such a voter (if he or she makes a difference) will affect nc electoral votes, the expected number of electoral votes a voter in a state of size n will affect—or the voter’s average power—is given by average power = ncP{makes a difference} nc ∼. nπ/2 . = c 2n/π Thus, the average power of a voter in a state of size n is proportional to the square root of n, showing that, in presidential elections, voters in large states have more power than do those in smaller states. .

142

Chapter 4

Random Variables

4.6.2

Computing the Binomial Distribution Function

Suppose that X is binomial with parameters (n, p). The key to computing its distribution function i n pk (1 − p)n−k P{X … i} = i = 0, 1, . . . , n k k=0

is to utilize the following relationship between P{X = k + 1} and P{X = k}, which was established in the proof of Proposition 6.1: P{X = k + 1} =

p n − k P{X = k} 1 − pk + 1

(6.3)

EXAMPLE 6h Let X be a binomial random variable with parameters n = 6, p = .4. Then, starting with P{X = 0} = (.6)6 and recursively employing Equation (6.3), we obtain P{X = 0} = (.6)6 L .0467 46 P{X = 0} L .1866 61 45 P{X = 2} = P{X = 1} L .3110 62 44 P{X = 3} = P{X = 2} L .2765 63 43 P{X = 3} L .1382 P{X = 4} = 64 42 P{X = 4} L .0369 P{X = 5} = 65 41 . P{X = 6} = P{X = 5} L .0041 66 A computer program that utilizes the recursion (6.3) to compute the binomial distribution function is easily written. To compute P{X … i}, the program should ﬁrst compute P{X = i} and then use the recursion to successively compute P{X = i − 1}, P{X = i − 2}, and so on. P{X = 1} =

Historical Note Independent trials having a common probability of success p were ﬁrst studied by the Swiss mathematician Jacques Bernoulli (1654–1705). In his book Ars Conjectandi (The Art of Conjecturing), published by his nephew Nicholas eight years after his death in 1713, Bernoulli showed that if the number of such trials were large, then the proportion of them that were successes would be close to p with a probability near 1. Jacques Bernoulli was from the ﬁrst generation of the most famous mathematical family of all time. Altogether, there were between 8 and 12 Bernoullis, spread over three generations, who made fundamental contributions to probability, statistics, and mathematics. One difﬁculty in knowing their exact number is the fact that several had the same name. (For example, two of the sons of Jacques’s brother Jean

Section 4.7

The Poisson Random Variable

143

were named Jacques and Jean.) Another difﬁculty is that several of the Bernoullis were known by different names in different places. Our Jacques (sometimes written Jaques) was, for instance, also known as Jakob (sometimes written Jacob) and as James Bernoulli. But whatever their number, their inﬂuence and output were prodigious. Like the Bachs of music, the Bernoullis of mathematics were a family for the ages! EXAMPLE 6i If X is a binomial random variable with parameters n = 100 and p = .75, ﬁnd P{X = 70} and P{X … 70}. Solution. The answer is shown here in Figure 4.6.

Binomial Distribution

Start

Enter Value For p:.75 Enter Value For n:100

Quit

Enter Value For i:70

Probability (Number of Successes = i) = .04575381 Probability (Number of Successes < = i) = .14954105

FIGURE 4.6

4.7

THE POISSON RANDOM VARIABLE A random variable X that takes on one of the values 0, 1, 2, . . . is said to be a Poisson random variable with parameter λ if, for some λ > 0, λi i = 0, 1, 2, . . . i! Equation (7.1) deﬁnes a probability mass function, since p(i) = P{X = i} = e−λ q i=0

p(i) = e−λ

q λi i=0

i!

(7.1)

= e−λ eλ = 1

´ The Poisson probability distribution was introduced by Simeon Denis Poisson in a book he wrote regarding the application of probability theory to lawsuits, criminal trials, and the like. This book, published in 1837, was entitled Recherches sur la probabilit´e des jugements en mati`ere criminelle et en mati`ere civile (Investigations into the Probability of Verdicts in Criminal and Civil Matters). The Poisson random variable has a tremendous range of applications in diverse areas because it may be used as an approximation for a binomial random variable with parameters (n, p) when n is large and p is small enough so that np is of moderate

144

Chapter 4

Random Variables

size. To see this, suppose that X is a binomial random variable with parameters (n, p), and let λ = np. Then n! pi (1 − p)n−i (n − i)!i! i λ n! λ n−i = 1 − (n − i)!i! n n n(n − 1) · · · (n − i + 1) λi (1 − λ/n)n = i! (1 − λ/n)i ni

P{X = i} =

Now, for n large and λ moderate,

λ 1 − n

n

−λ

L e

n(n − 1) · · · (n − i + 1) L 1 ni

λ i 1 − L 1 n

Hence, for n large and λ moderate, P{X = i} L e−λ

λi i!

In other words, if n independent trials, each of which results in a success with probability p, are performed, then, when n is large and p is small enough to make np moderate, the number of successes occurring is approximately a Poisson random variable with parameter λ = np. This value λ (which will later be shown to equal the expected number of successes) will usually be determined empirically. Some examples of random variables that generally obey the Poisson probability law [that is, they obey Equation (7.1)] are as follows: 1. 2. 3. 4. 5. 6. 7.

The number of misprints on a page (or a group of pages) of a book The number of people in a community who survive to age 100 The number of wrong telephone numbers that are dialed in a day The number of packages of dog biscuits sold in a particular store each day The number of customers entering a post ofﬁce on a given day The number of vacancies occurring during a year in the federal judicial system The number of α-particles discharged in a ﬁxed period of time from some radioactive material

Each of the preceding, and numerous other random variables, are approximately Poisson for the same reason—namely, because of the Poisson approximation to the binomial. For instance, we can suppose that there is a small probability p that each letter typed on a page will be misprinted. Hence, the number of misprints on a page will be approximately Poisson with λ = np, where n is the number of letters on a page. Similarly, we can suppose that each person in a community has some small probability of reaching age 100. Also, each person entering a store may be thought of as having some small probability of buying a package of dog biscuits, and so on. EXAMPLE 7a Suppose that the number of typographical errors on a single page of this book has a Poisson distribution with parameter λ = 12 . Calculate the probability that there is at least one error on this page.

Section 4.7

The Poisson Random Variable

145

Solution. Letting X denote the number of errors on this page, we have P{X Ú 1} = 1 − P{X = 0} = 1 − e−1/2 L .393

.

EXAMPLE 7b Suppose that the probability that an item produced by a certain machine will be defective is .1. Find the probability that a sample of 10 items will contain at most 1 defective item. 10 10 0 10 (.1)1 (.9)9 = .7361, (.1) (.9) + Solution. The desired probability is 1 0 whereas the Poisson approximation yields the value e−1 + e−1 L .7358. . EXAMPLE 7c Consider an experiment that consists of counting the number of α particles given off in a 1-second interval by 1 gram of radioactive material. If we know from past experience that, on the average, 3.2 such α particles are given off, what is a good approximation to the probability that no more than 2 α particles will appear? Solution. If we think of the gram of radioactive material as consisting of a large number n of atoms, each of which has probability of 3.2/n of disintegrating and sending off an α particle during the second considered, then we see that, to a very close approximation, the number of α particles given off will be a Poisson random variable with parameter λ = 3.2. Hence, the desired probability is P{X … 2} = e−3.2 + 3.2e−3.2 + L .3799

(3.2)2 −3.2 e 2 .

Before computing the expected value and variance of the Poisson random variable with parameter λ, recall that this random variable approximates a binomial random variable with parameters n and p when n is large, p is small, and λ = np. Since such a binomial random variable has expected value np = λ and variance np(1 − p) = λ(1 − p) L λ (since p is small), it would seem that both the expected value and the variance of a Poisson random variable would equal its parameter λ. We now verify this result: E[X] =

q ie−λ λi i=0 q

=λ

i=1

= λe−λ

i! e−λ λi−1 (i − 1)! q λj j=0

=λ

j!

since

by letting j=i − 1 q λj j=0

j!

= eλ

146

Chapter 4

Random Variables

Thus, the expected value of a Poisson random variable X is indeed equal to its parameter λ. To determine its variance, we ﬁrst compute E[X 2 ]: E[X 2 ] =

q 2 −λ i i e λ i=0 q

=λ

i=1

i! ie−λ λi−1 (i − 1)!

q (j + 1)e−λ λj by letting =λ j=i − 1 j! j=0 ⎤ ⎡ q q −λ j e−λ λj ⎥ ⎢ je λ = λ⎣ + ⎦ j! j! j=0

j=0

= λ(λ + 1) where the ﬁnal equality follows because the ﬁrst sum is the expected value of a Poisson random variable with parameter λ and the second is the sum of the probabilities of this random variable. Therefore, since we have shown that E[X] = λ, we obtain Var(X) = E[X 2 ] − (E[X])2 =λ Hence, the expected value and variance of a Poisson random variable are both equal to its parameter λ. We have shown that the Poisson distribution with parameter np is a very good approximation to the distribution of the number of successes in n independent trials when each trial has probability p of being a success, provided that n is large and p small. In fact, it remains a good approximation even when the trials are not independent, provided that their dependence is weak. For instance, recall the matching problem (Example 5m of Chapter 2) in which n men randomly select hats from a set consisting of one hat from each person. From the point of view of the number of men who select their own hat, we may regard the random selection as the result of n trials where we say that trial i is a success if person i selects his own hat, i = 1, . . . , n. Deﬁning the events Ei , i = 1, . . . , n, by Ei = {trial i is a success} it is easy to see that P{Ei } =

1 n

and

P{Ei |Ej } =

1 , n − 1

j Z i

Thus, we see that, although the events Ei , i = 1, . . . , n are not independent, their dependence, for large n, appears to be weak. Because of this it seems reasonable to expect that the number of successes will approximately have a Poisson distribution with parameter n * 1/n = 1, and indeed this is veriﬁed in Example 5m of Chapter 2. For a second illustration of the strength of the Poisson approximation when the trials are weakly dependent, let us consider again the birthday problem presented in Example 5i of Chapter 2. In this example, we suppose that each of n people is equally likely to have any of the 365 days of the year as his or her birthday, and the problem

Section 4.7

The Poisson Random Variable

147

is to determine the probability that a set of n independent people all have different birthdays. A combinatorial argument was used to determine this probability, which was shown to be less than 12 when n = 23. We can approximate the preceding probability by using Poisson approximation the n pairs of individuals i and as follows: Imagine that we have a trial for each of the 2 j, i Z j, and say that trial i, j is a success if persons i and j have the samebirthday. If n we let Eij denote the event that trial i, j is a success, then, whereas the events 2 Eij , 1 … i < j … n, are not independent (see Theoretical Exercise 21), their dependence appears to be rather weak. (Indeed, these events are even pairwise independent, in that any 2 of the events Eij and Ekl are independent—again, see Theoretical Exercise 21). Since P(Eij ) = 1/365, it is reasonable to suppose that the of number n 365 = successes should approximately have a Poisson distribution with mean 2 n(n − 1)/730. Therefore, P{no 2 people have the same birthday} = P{0 successes} % / −n(n − 1) L exp 730 To determine the smallest integer n for which this probability is less than 12 , note that / % 1 −n(n − 1) … exp 730 2 is equivalent to

% exp

n(n − 1) 730

/ Ú 2

Taking logarithms of both sides, we obtain n(n − 1) Ú 730 log 2 L 505.997 which yields the solution n = 23, in agreement with the result of Example 5i of Chapter 2. Suppose now that we wanted the probability that, among the n people, no 3 of them have their birthday on the same day. Whereas this now becomes a difﬁcult combinatorial problem, it is a simple matter to a good approximation. To begin, obtain n triplets i, j, k, where 1 … i < j < k … n, imagine that we have a trial for each of the 3 and call the i, j, k trial a success if persons i, j, and k all have their birthday on the same day. As before, we can then conclude that the number of successes is approximately a Poisson random variable with parameter

n 3

1 2 n P{i, j, k have the same birthday} = 3 365 n(n − 1)(n − 2) = 6 * (365)2

148

Chapter 4

Random Variables

Hence, %

−n(n − 1)(n − 2) P{no 3 have the same birthday} L exp 799350 This probability will be less than

1 2

/

when n is such that

n(n − 1)(n − 2) Ú 799350 log 2 L 554067.1 which is equivalent to n Ú 84. Thus, the approximate probability that at least 3 people in a group of size 84 or larger will have the same birthday exceeds 12 . For the number of events to occur to approximately have a Poisson distribution, it is not essential that all the events have the same probability of occurrence, but only that all of these probabilities be small. The following is referred to as the Poisson paradigm. Poisson Paradigm. Consider n events, with pi equal to the probability that event i occurs, i = 1, . . . , n. If all the pi are “small” and the trials are either independent or at most “weakly dependent,” then the number of these events that occur approximately has a Poisson distribution with mean ni=1 pi . Our next example not only makes use of the Poisson paradigm, but also illustrates a variety of the techniques we have studied so far. EXAMPLE 7d Length of the longest run A coin is ﬂipped n times. Assuming that the ﬂips are independent, with each one coming up heads with probability p, what is the probability that there is a string of k consecutive heads? Solution. We will ﬁrst use the Poisson paradigm to approximate this probability. Now, if, for i = 1, . . . , n − k + 1, we let Hi denote the event that ﬂips i, i + 1, . . . , i + k − 1 all land on heads, then the desired probability is that at least one of the events Hi occur. Because Hi is the event that, starting with ﬂip i, the next k ﬂips all land on heads, it follows that P(Hi ) = pk . Thus, when pk is small, we might think that the number of the Hi that occur should have an approximate Poisson distribution. However, such is not the case, because, although the events all have small probabilities, some of their dependencies are too great for the Poisson distribution to be a good approximation. For instance, because the conditional probability that ﬂips 2, . . . , k + 1 are all heads given that ﬂips 1, . . . , k are all heads is equal to the probability that ﬂip k + 1 is a head, it follows that P(H2 |H1 ) = p which is far greater than the unconditional probability of H2 . The trick that enables us to use a Poisson approximation is to note that there will be a string of k consecutive heads either if there is such a string that is immediately followed by a tail or if the ﬁnal k ﬂips all land on heads. Consequently, for i = 1, . . . , n − k, let Ei be the event that ﬂips i, . . . , i + k − 1 are all heads and ﬂip i + k is a tail; also, let En−k+1 be the event that ﬂips n − k + 1, . . . , n are all heads. Note that P(Ei ) = pk (1 − p), k

P(En−k+1 ) = p

i … n − k

Section 4.7

The Poisson Random Variable

149

Thus, when pk is small, each of the events Ei has a small probability of occurring. Moreover, for i Z j, if the events Ei and Ej refer to nonoverlapping sequences of ﬂips, then P(Ei |Ej ) = P(Ei ); if they refer to overlapping sequences, then P(Ei |Ej ) = 0. Hence, in both cases, the conditional probabilities are close to the unconditional ones, indicating that N, the number of the events Ei that occur, should have an approximate Poisson distribution with mean E[N] =

n−k+1

P(Ei ) = (n − k)pk (1 − p) + pk

i=1

Because there will not be a run of k heads if (and only if) N = 0, thus the preceding gives P(no head strings of length k) = P(N = 0) L exp{−(n − k)pk (1 − p) − pk } If we let Ln denote the largest number of consecutive heads in the n ﬂips, then, because Ln will be less than k if (and only if) there are no head strings of length k, the preceding equation can be written as P{Ln < k} L exp{− (n − k)pk (1 − p) − pk } Now, let us suppose that the coin being ﬂipped is fair; that is, suppose that p = 1/2. Then the preceding gives % / % / n − k + 2 n P{Ln < k} L exp − L exp − 2k+1 2k+1 k−2

where the ﬁnal approximation supposes that e 2k+1 L 1 (that is, that j = log2 n, and assume that j is an integer. For k = j + i, n 2k+1

=

n 2j 2i+1

=

k−2 2k+1

1 2i+1

Consequently, P{Ln < j + i} L exp{−(1/2)i+1 } which implies that P{Ln = j + i} = P{Ln < j + i + 1} − P{Ln < j + i} L exp{−(1/2)i+2 } − exp{−(1/2)i+1 } For instance, P{Ln < j − 3} L e−4 L .0183 P{Ln = j − 3} L e−2 − e−4 L .1170 P{Ln = j − 2} L e−1 − e−2 L .2325 P{Ln = j − 1} L e−1/2 − e−1 L .2387 P{Ln = j} L e−1/4 − e−1/2 L .1723 P{Ln = j + 1} L e−1/8 − e−1/4 L .1037 P{Ln = j + 2} L e−1/16 − e−1/8 L .0569 P{Ln = j + 3} L e−1/32 − e−1/16 L .0298 P{Ln Ú j + 4} L 1 − e−1/32 L .0308

L 0). Let

150

Chapter 4

Random Variables

Thus, we observe the rather interesting fact that no matter how large n is, the length of the longest run of heads in a sequence of n ﬂips of a fair coin will be within 2 of log2 (n) − 1 with a probability approximately equal to .86. We now derive an exact expression for the probability that there is a string of k consecutive heads when a coin that lands on heads with probability p is ﬂipped n times. With the events Ei , i = 1, . . . , n − k + 1, as deﬁned earlier, and with Ln denoting, as before, the length of the longest run of heads, Ei ) P(Ln Ú k) = P(there is a string of k consecutive heads) = P(∪n−k+1 i=1 The inclusion–exclusion identity for the probability of a union can be written as n−k+1 P(∪i=1 Ei ) =

n−k+1 r=1

(−1)r+1

P(Ei1 · · · Eir )

i1 <···

Let Si denote the set of ﬂip numbers to which the event Ei refers. (So, for instance, S1 = {1, . . . , k + 1}.) Now, consider one of the r-way intersection probabilities that does not include the event En−k+1 . That is, consider P(Ei1 · · · Eir ) where i1 < · · · < ir < n − k + 1. On the one hand, if there is any overlap in the sets Si1 , . . . , Sir then this probability is 0. On the other hand, if there is no overlap, then the events Ei1 , . . . , Eir are independent. Therefore, 0 0, if there is any overlap in Si1 , . . . , Sir P(Ei1 · · · Eir ) = rk r p (1 − p) , if there is no overlap We must now determine the number of different choices of i1 < · · · < ir < n − k + 1 for which there is no overlap in the sets Si1 , . . . , Sir . To do so, note ﬁrst that each of the Sij , j = 1, . . . , r, refer to k + 1 ﬂips, so, without any overlap, they together refer to r(k + 1) ﬂips. Now consider any permutation of r identical letters a (one for each of the sets Si1 , . . . , Sir−1 ) and of n − r(k + 1) identical letters b (one for each of the trials that are not part of any of Si1 , . . . , Sir−1 , Sn−k+1 ). Interpret the number of b’s before the ﬁrst a as the number of ﬂips before Si1 , the number of b’s between the ﬁrst and second a as the number of ﬂips between Si1 and Si2 , and so on, with the number of b’s after the ﬁnal a representing the number of ﬂips after Sir . Because there are n−rk permutations of r letters a and of n − r(k + 1) letters b, with every such r permutation corresponding (in a one-to-one fashion) to a different nonoverlapping choice, it follows that n − rk rk P(Ei1 · · · Eir ) = p (1 − p)r r i1 <···

We must now consider r-way intersection probabilities of the form P(Ei1 · · · Eir−1 En−k+1 ), where i1 < · · · < ir−1 < n − k + 1. Now, this probability will equal 0 if there is any overlap in Si1 , . . . , Sir−1 , Sn−k ; if there is no overlap, then the events of the intersection will be independent, so P(Ei1 · · · Eir−1 En−k+1 ) = [pk (1 − p)]r−1 pk = pkr (1 − p)r−1 By a similar argument as before, the number of nonoverlapping sets Si1 , . . . , Sir−1 , Sn−k will equal the number of permutations of r − 1 letters a (one for each of the sets

Section 4.7

The Poisson Random Variable

151

Si1 , . . . , Sir−1 ) and of n − (r − 1)(k + 1) − k = n − rk − (r − 1) letters b (one for each of the trials that are not part of any of Si1 , . . . , Sir−1 , Sn−k+1 ). Since there are n−rk r−1 permutations of r − 1 letters a and of n − rk − (r − 1) letters b, we have n − rk kr P(Ei1 · · · Eir−1 En−k+1 ) = p (1 − p)r−1 r − 1 i1 <...

Putting it all together yields the exact expression, namely, n−k+1 n − rk n − rk 1 pkr (1 − p)r (−1)r+1 P(Ln Ú k) = + p r − 1 r r=1

where we utilize the convention that mj = 0 if m < j. From a computational point of view, a more efﬁcient method for computing the desired probability than the use of the preceding identity is to derive a set of recursive equations. To do so, let An be the event that there is a string of k consecutive heads in a sequence of n ﬂips of a fair coin, and let Pn = P(An ). We will derive a set of recursive equations for Pn by conditioning on when the ﬁrst tail appears. For j = 1, . . . , k, let Fj be the event that the ﬁrst tail appears on ﬂip j, and let H be the event that the ﬁrst k ﬂips are all heads. Because the events F1 , . . . , Fk , H are mutually exclusive and exhaustive (that is, exactly one of these events must occur), we have P(An ) =

k

P(An |Fj )P(Fj ) + P(An |H)P(H)

j=1

Now, given that the ﬁrst tail appears on ﬂip j, where j < k, it follows that those j ﬂips are wasted as far as obtaining a string of k heads in a row; thus, the conditional probability of this event is the probability that such a string will occur among the remaining n − j ﬂips. Therefore, P(An |Fj ) = Pn−j Because P(An |H) = 1, the preceding equation gives Pn = P(An ) =

k

Pn−j P(Fj ) + P(H)

j=1

=

k

Pn−j p j−1 (1 − p) + pk

j=1

Starting with Pj = 0, j < k, and Pk = pk , we can use the latter formula to recursively compute Pk+1 , Pk+2 , and so on, up to Pn . For instance, suppose we want the probability that there is a run of 2 consecutive heads when a fair coin is ﬂipped 4 times. Then, with k = 2, we have P1 = 0, P2 = (1/2)2 . Because, when p = 1/2, the recursion becomes k Pn−j (1/2)j + (1/2)k Pn = j=1

152

Chapter 4

Random Variables

we obtain P3 = P2 (1/2) + P1 (1/2)2 + (1/2)2 = 3/8 and P4 = P3 (1/2) + P2 (1/2)2 + (1/2)2 = 1/2 which is clearly true because there are 8 outcomes that result in a string of 2 consecutive heads: hhhh, hhht, hhth, hthh, thhh, hhtt, thht, and tthh. Each of these outcomes occurs with probability 1/16. . Another use of the Poisson probability distribution arises in situations where “events” occur at certain points in time. One example is to designate the occurrence of an earthquake as an event; another possibility would be for events to correspond to people entering a particular establishment (bank, post ofﬁce, gas station, and so on); and a third possibility is for an event to occur whenever a war starts. Let us suppose that events are indeed occurring at certain (random) points of time, and let us assume that, for some positive constant λ, the following assumptions hold true: 1. The probability that exactly 1 event occurs in a given interval of length h is equal to λh + o(h), where o(h) stands for any function f (h) for which lim f (h)/h = 0. h→0 [For instance, f (h) = h2 is o(h), whereas f (h) = h is not.] 2. The probability that 2 or more events occur in an interval of length h is equal to o(h). 3. For any integers n, j1 , j2 , . . . , jn and any set of n nonoverlapping intervals, if we deﬁne Ei to be the event that exactly ji of the events under consideration occur in the ith of these intervals, then events E1 , E2 , . . . , En are independent. Loosely put, assumptions 1 and 2 state that, for small values of h, the probability that exactly 1 event occurs in an interval of size h equals λh plus something that is small compared with h, whereas the probability that 2 or more events occur is small compared with h. Assumption 3 states that whatever occurs in one interval has no (probability) effect on what will occur in other, nonoverlapping intervals. We now show that, under assumptions 1, 2, and 3, the number of events occurring in any interval of length t is a Poisson random variable with parameter λt. To be precise, let us call the interval [0, t] and denote the number of events occurring in that interval by N(t). To obtain an expression for P{N(t) = k}, we start by breaking the interval [0, t] into n nonoverlapping subintervals, each of length t/n (Figure 4.7). Now, P{N(t) = k} = P{k of the n subintervals contain exactly 1 event and the other n − k contain 0 events} + P{N(t) = k and at least 1 subinterval contains 2 or more events}

(7.2)

The proceding equation holds because the event on the left side of Equation (7.2), that is, {N(t) = k}, is clearly equal to the union of the two mutually exclusive events 0 –t n

2t — n

3t — n

(n – 1) n–t FIGURE 4.7

nt t=— n

Section 4.7

The Poisson Random Variable

153

on the right side of the equation. Letting A and B denote the two mutually exclusive events on the right side of Equation (7.2), we have P(B) … P{at least one subinterval contains 2 or more events} ⎞ ⎛ n = P ⎝ {ith subinterval contains 2 or more events}⎠ i=1

…

n

P{ith subinterval contains 2 or more events}

i=1

n t o = n i=1 t = no n o(t/n) =t t/n

by Boole’s inequality by assumption 2

Now, in addition for any t, t/n→0 as n→q, so o(t/n)/(t/n)→0 as n→q, by the deﬁnition of o(h). Hence, P(B)→0 as n→q (7.3) Moreover, since assumptions 1 and 2 imply that† P{0 events occur in an interval of length h} = 1 − [λh + o(h) + o(h)] = 1 − λh − o(h) we see from the independence assumption (number 3) that P(A) = P{k of the subintervals contain exactly 1 event and the other n − k contain 0 events} k n−k λt t t λt n + o − o 1 − = k n n n n However, since

λt t o(t/n) n + o = λt + t →λt n n t/n

as

n→q

it follows, by the same argument that veriﬁed the Poisson approximation to the binomial, that (λt)k P(A)→e−λt as n→q (7.4) k! Thus, from Equations (7.2), (7.3), and (7.4), by letting n→q, we obtain P{N(t) = k} = e−λt

(λt)k k!

k = 0, 1, . . .

(7.5)

† The sum of two functions, both of which are o(h), is also o(h). This is so because if limh→0 f (h)/h = limh→0 g(h)/h = 0, then limh→0 [f (h) + g(h)]/h = 0.

154

Chapter 4

Random Variables

Hence, if assumptions 1, 2, and 3 are satisﬁed, then the number of events occurring in any ﬁxed interval of length t is a Poisson random variable with mean λt, and we say that the events occur in accordance with a Poisson process having rate λ. The value λ, which can be shown to equal the rate per unit time at which events occur, is a constant that must be empirically determined. The preceding discussion explains why a Poisson random variable is usually a good approximation for such diverse phenomena as the following: 1. The number of earthquakes occurring during some ﬁxed time span 2. The number of wars per year 3. The number of electrons emitted from a heated cathode during a ﬁxed time period 4. The number of deaths, in a given period of time, of the policyholders of a life insurance company EXAMPLE 7e Suppose that earthquakes occur in the western portion of the United States in accordance with assumptions 1, 2, and 3, with λ = 2 and with 1 week as the unit of time. (That is, earthquakes occur in accordance with the three assumptions at a rate of 2 per week.) (a) Find the probability that at least 3 earthquakes occur during the next 2 weeks. (b) Find the probability distribution of the time, starting from now, until the next earthquake. Solution. (a) From Equation (7.5), we have P{N(2) Ú 3} = 1 − P{N(2) = 0} − P{N(2) = 1} − P{N(2) = 2} = 1 − e−4 − 4e−4 − = 1 − 13e−4

42 −4 e 2

(b) Let X denote the amount of time (in weeks) until the next earthquake. Because X will be greater than t if and only if no events occur within the next t units of time, we have, from Equation (7.5), P{X > t} = P{N(t) = 0} = e−λt so the probability distribution function F of the random variable X is given by F(t) = P{X … t} = 1 − P{X > t} = 1 − e−λt = 1 − e−2t

4.7.1

.

Computing the Poisson Distribution Function

If X is Poisson with parameter λ, then P{X = i + 1} e−λ λi+1 /(i + 1)! λ = = P{X = i} i + 1 e−λ λi /i!

(7.6)

Section 4.8

Other Discrete Probability Distributions

155

Starting with P{X = 0} = e−λ , we can use (7.6) to compute successively P{X = 1} = λP{X = 0} λ P{X = 2} = P{X = 1} 2 . . . λ P{X = i} P{X = i + 1} = i + 1 The website includes a program that uses Equation (7.6) to compute Poisson probabilities. EXAMPLE 7f (a) Determine P{X … 90} when X is Poisson with mean 100. (b) Determine P{Y … 1075} when Y is Poisson with mean 1000. Solution. From the website, we obtain the solutions: (a) P{X … 90} L .1714; (b) P{Y … 1075} L .9894. 4.8

.

OTHER DISCRETE PROBABILITY DISTRIBUTIONS 4.8.1

The Geometric Random Variable

Suppose that independent trials, each having a probability p, 0 < p < 1, of being a success, are performed until a success occurs. If we let X equal the number of trials required, then n = 1, 2, . . . (8.1) P{X = n} = (1 − p)n−1 p Equation (8.1) follows because, in order for X to equal n, it is necessary and sufﬁcient that the ﬁrst n − 1 trials are failures and the nth trial is a success. Equation (8.1) then follows, since the outcomes of the successive trials are assumed to be independent. Since q q p =1 P{X = n} = p (1 − p)n−1 = 1 − (1 − p) n=1

n=1

it follows that, with probability 1, a success will eventually occur. Any random variable X whose probability mass function is given by Equation (8.1) is said to be a geometric random variable with parameter p. EXAMPLE 8a An urn contains N white and M black balls. Balls are randomly selected, one at a time, until a black one is obtained. If we assume that each ball selected is replaced before the next one is drawn, what is the probability that (a) exactly n draws are needed? (b) at least k draws are needed? Solution. If we let X denote the number of draws needed to select a black ball, then X satisﬁes Equation (8.1) with p = M/(M + N). Hence,

156

Chapter 4

Random Variables

(a) P{X = n} =

N M + N

n−1

MN n−1 M = M + N (M + N)n

(b) n−1 q N M M + N M + N n=k k−1 1 N M N = 1 − M + N M + N M + N k−1 N = M + N

P{X Ú k} =

Of course, part (b) could have been obtained directly, since the probability that at least k trials are necessary to obtain a success is equal to the probability that the ﬁrst k − 1 trials are all failures. That is, for a geometric random variable, P{X Ú k} = (1 − p)k−1 EXAMPLE 8b Find the expected value of a geometric random variable. Solution. With q = 1 − p, we have E[X] =

q

iqi−1 p

i=1 q (i − 1 + 1)qi−1 p = i=1 q q (i − 1)qi−1 p + qi−1 p = i=1

=

q

i=1

jqj p + 1

j=0 q

=q

jqj−1 p + 1

j=1

= qE[X] + 1 Hence, pE[X] = 1 yielding the result E[X] =

1 p

.

Section 4.8

Other Discrete Probability Distributions

157

In other words, if independent trials having a common probability p of being successful are performed until the ﬁrst success occurs, then the expected number of required trials equals 1/p. For instance, the expected number of rolls of a fair die that it takes to obtain the value 1 is 6. . EXAMPLE 8c Find the variance of a geometric random variable. Solution. To determine Var(X), let us ﬁrst compute E[X 2 ]. With q = 1 − p, we have E[X 2 ] =

q

i2 qi−1 p

i=1

=

q

(i − 1 + 1)2 qi−1 p

i=1

=

q

(i − 1)2 qi−1 p +

i=1

=

q

q

2(i − 1)qi−1 p +

i=1

j2 qj p + 2

j=0

q

q

qi−1 p

i=1

jqj p + 1

j=1 2

= qE[X ] + 2qE[X] + 1 Using E[X] = 1/p, the equation for E[X 2 ] yields pE[X 2 ] =

2q + 1 p

Hence, E[X 2 ] =

2q + p q + 1 = p2 p2

giving the result Var(X) = 4.8.2

1 q 1 − p q + 1 − 2 = 2 = 2 p p p p2

.

The Negative Binomial Random Variable

Suppose that independent trials, each having probability p, 0 < p < 1, of being a success are performed until a total of r successes is accumulated. If we let X equal the number of trials required, then n − 1 (8.2) P{X = n} = pr (1 − p)n−r n = r, r + 1, . . . r − 1 Equation (8.2) follows because, in order for the rth success to occur at the nth trial, there must be r − 1 successes in the ﬁrst n − 1 trials and the nth trial must be a success. The probability of the ﬁrst event is n − 1 pr−1 (1 − p)n−r r − 1

158

Chapter 4

Random Variables

and the probability of the second is p; thus, by independence, Equation (8.2) is established. To verify that a total of r successes must eventually be accumulated, either we can prove analytically that q q n − 1 pr (1 − p)n−r = 1 P{X = n} = (8.3) r − 1 n=r

n=r

or we can give a probabilistic argument as follows: The number of trials required to obtain r successes can be expressed as Y1 + Y2 + · · · + Yr , where Y1 equals the number of trials required for the ﬁrst success, Y2 the number of additional trials after the ﬁrst success until the second success occurs, Y3 the number of additional trials until the third success, and so on. Because the trials are independent and all have the same probability of success, it follows that Y1 , Y2 , . . . , Yr are all geometric r random variables. Hence, each is ﬁnite with probability 1, so Yi must also be ﬁnite, i=1

establishing Equation (8.3). Any random variable X whose probability mass function is given by Equation (8.2) is said to be a negative binomial random variable with parameters (r, p). Note that a geometric random variable is just a negative binomial with parameter (1, p). In the next example, we use the negative binomial to obtain another solution of the problem of the points. EXAMPLE 8d If independent trials, each resulting in a success with probability p, are performed, what is the probability of r successes occurring before m failures? Solution. The solution will be arrived at by noting that r successes will occur before m failures if and only if the rth success occurs no later than the (r + m − 1)th trial. This follows because if the rth success occurs before or at the (r + m − 1)th trial, then it must have occurred before the mth failure, and conversely. Hence, from Equation (8.2), the desired probability is r+m−1 n=r

n − 1 r − 1

pr (1 − p)n−r

.

EXAMPLE 8e The Banach match problem At all times, a pipe-smoking mathematician carries 2 matchboxes—1 in his left-hand pocket and 1 in his right-hand pocket. Each time he needs a match, he is equally likely to take it from either pocket. Consider the moment when the mathematician ﬁrst discovers that one of his matchboxes is empty. If it is assumed that both matchboxes initially contained N matches, what is the probability that there are exactly k matches, k = 0, 1, . . . , N, in the other box? Solution. Let E denote the event that the mathematician ﬁrst discovers that the righthand matchbox is empty and that there are k matches in the left-hand box at the time. Now, this event will occur if and only if the (N + 1)th choice of the right-hand matchbox is made at the (N + 1 + N − k)th trial. Hence, from Equation (8.2) (with p = 12 , r = N + 1, and n = 2N − k + 1), we see that P(E) =

2N − k N

2N−k+1 1 2

Section 4.8

Other Discrete Probability Distributions

159

Since there is an equal probability that it is the left-hand box that is ﬁrst discovered to be empty and there are k matches in the right-hand box at that time, the desired result is 2N−k 1 2N − k 2P(E) = N . 2 EXAMPLE 8f Compute the expected value and the variance of a negative binomial random variable with parameters r and p. Solution. We have n − 1 pr (1 − p)n−r r − 1 n=r q r k−1 n = n pr+1 (1 − p)n−r r p

E[X k ] =

q

r = p

nk

n=r q

(m − 1)

k−1

m=r+1

m − 1 r

since

n

n − 1 r − 1

=r

n r

by setting pr+1 (1 − p)m−(r+1) m = n + 1

r = E[(Y − 1)k−1 ] p where Y is a negative binomial random variable with parameters r + 1, p. Setting k = 1 in the preceding equation yields E[X] =

r p

Setting k = 2 in the equation for E[X k ] and using the formula for the expected value of a negative binomial random variable gives r E[Y − 1] p r r + 1 − 1 = p p

E[X 2 ] =

Therefore, 2 r r + 1 r − 1 − Var(X) = p p p r(1 − p) = p2

.

Thus, from Example 8f, if independent trials, each of which is a success with probability p, are performed, then the expected value and variance of the number of trials that it takes to amass r successes is r/p and r(1 − p)/p2 , respectively. Since a geometric random variable is just a negative binomial with parameter r = 1, it follows from the preceding example that the variance of a geometric random variable with parameter p is equal to (1 − p)/p2 , which checks with the result of Example 8c.

160

Chapter 4

Random Variables

EXAMPLE 8g Find the expected value and the variance of the number of times one must throw a die until the outcome 1 has occurred 4 times. Solution. Since the random variable of interest is a negative binomial with parameters r = 4 and p = 16 , it follows that E[X] = 24 4 56 Var(X) = 2 = 120 1 6

4.8.3

.

The Hypergeometric Random Variable

Suppose that a sample of size n is to be chosen randomly (without replacement) from an urn containing N balls, of which m are white and N − m are black. If we let X denote the number of white balls selected, then N − m m n − i i i = 0, 1, . . . , n (8.4) P{X = i} = N n A random variable X whose probability mass function is given by Equation (8.4) for some values of n, N, m is said to be a hypergeometric random variable. Remark. Although we have written the hypergeometric probability mass function with i going from 0 to n, P{X = i} will actually be 0, unless i satisﬁes the inequalities n − (N − m) … i … m). However, Equation (8.4) is always valid because min(n, r of our convention that is equal to 0 when either k < 0 or r < k. . k EXAMPLE 8h An unknown number, say, N, of animals inhabit a certain region. To obtain some information about the size of the population, ecologists often perform the following experiment: They ﬁrst catch a number, say, m, of these animals, mark them in some manner, and release them. After allowing the marked animals time to disperse throughout the region, a new catch of size, say, n, is made. Let X denote the number of marked animals in this second capture. If we assume that the population of animals in the region remained ﬁxed between the time of the two catches and that each time an animal was caught it was equally likely to be any of the remaining uncaught animals, it follows that X is a hypergeometric random variable such that m N − m n − i i P{X = i} = K Pi (N) N n Suppose now that X is observed to equal i. Then, since Pi (N) represents the probability of the observed event when there are actually N animals present in the region, it would appear that a reasonable estimate of N would be the value of N that maximizes Pi (N). Such an estimate is called a maximum likelihood estimate. (See Theoretical Exercises 13 and 18 for other examples of this type of estimation procedure.)

Section 4.8

Other Discrete Probability Distributions

161

The maximization of Pi (N) can be done most simply by ﬁrst noting that Pi (N) (N − m)(N − n) = Pi (N − 1) N(N − m − n + i) Now, the preceding ratio is greater than 1 if and only if (N − m)(N − n) Ú N(N − m − n + i) or, equivalently, if and only if N …

mn i

Thus, Pi (N) is ﬁrst increasing and then decreasing, and reaches its maximum value at the largest integral value not exceeding mn/i. This value is the maximum likelihood estimate of N. For example, suppose that the initial catch consisted of m = 50 animals, which are marked and then released. If a subsequent catch consists of n = 40 animals of which i = 4 are marked, then we would estimate that there are some 500 animals in the region. (Note that the preceding estimate could also have been obtained by assuming that the proportion of marked animals in the region, m/N, is approximately equal to the proportion of marked animals in our second catch, i/n.) . EXAMPLE 8i A purchaser of electrical components buys them in lots of size 10. It is his policy to inspect 3 components randomly from a lot and to accept the lot only if all 3 are nondefective. If 30 percent of the lots have 4 defective components and 70 percent have only 1, what proportion of lots does the purchaser reject? Solution. Let A denote the event that the purchaser accepts a lot. Now, 3 7 P(A) = P(A|lot has 4 defectives) + P(A|lot has 1 defective) 10 10 6 9 4 1 3 3 0 0 3 7 = + 10 10 10 10 3 3 54 = 100 Hence, 46 percent of the lots are rejected.

.

If n balls are randomly chosen without replacement from a set of N balls of which the fraction p = m/N is white, then the number of white balls selected is hypergeometric. Now, it would seem that when m and N are large in relation to n, it shouldn’t make much difference whether the selection is being done with or without replacement, because, no matter which balls have previously been selected, when m and N are large, each additional selection will be white with a probability approximately equal to p. In other words, it seems intuitive that when m and N are large in relation to n, the probability mass function of X should approximately be that of a binomial random variable with parameters n and p. To verify this intuition, note that if X is hypergeometric, then, for i … n,

162

Chapter 4

Random Variables

N − m n − i P{X = i} = N n m! (N − m)! (N − n)! n! = (m − i)! i! (N − m − n + i)! (n − i)! N! m − i + 1N − mN − m − 1 n mm − 1 ··· = i NN − 1 N − i + 1 N − i N − i − 1 N − m − (n − i − 1) ··· N − i − (n − i − 1) when p = m/N and m and N are n pi (1 − p)n−i L large in relation to n and i i m i

EXAMPLE 8j Determine the expected value and the variance of X, a hypergeometric random variable with parameters n, N, and m. Solution. E[X k ] = =

n i=0 n

ik P{X = i} k

i

i=1

m i

Using the identities m − 1 m =m i i − 1 i

N − m n − i

and

n

N n

1

N n

=N

N − 1 n − 1

we obtain n

m − 1 i − 1

1

N − 1 n − 1 i=1 1 n−1 nm m − 1 N − m N − 1 k−1 (j + 1) = j n − 1 − j n − 1 N

nm k−1 E[X ] = i N k

N − m n − i

j=0

nm E[(Y + 1)k−1 ] = N where Y is a hypergeometric random variable with parameters n − 1, N − 1, and m − 1. Hence, upon setting k = 1, we have E[X] =

nm N

In words, if n balls are randomly selected from a set of N balls, of which m are white, then the expected number of white balls selected is nm/N.

Section 4.8

Other Discrete Probability Distributions

163

Upon setting k = 2 in the equation for E[X k ], we obtain nm E[Y + 1] N nm (n − 1)(m − 1) = + 1 N N − 1

E[X 2 ] =

where the ﬁnal equality uses our preceding result to compute the expected value of the hypergeometric random variable Y. Because E[X] = nm/N, we can conclude that nm nm (n − 1)(m − 1) + 1 − Var(X) = N N − 1 N Letting p = m/N and using the identity m − 1 Np − 1 1 − p = =p − N − 1 N − 1 N − 1 shows that Var(X) = np[(n − 1)p − (n − 1) = np(1 − p)(1 −

n − 1 ) N − 1

1 − p + 1 − np] N − 1 .

Remark. We have shown in Example 8j that if n balls are randomly selected without replacement from a set of N balls, of which the fraction p are white, then the expected number of white balls chosen is np. In addition, if N is large in relation to n [so that (N − n)/(N − 1) is approximately equal to 1], then Var(X) L np(1 − p) In other words, E[X] is the same as when the selection of the balls is done with replacement (so that the number of white balls is binomial with parameters n and p), and if the total collection of balls is large, then Var(X) is approximately equal to what it would be if the selection were done with replacement. This is, of course, exactly what we would have guessed, given our earlier result that when the number of balls in the urn is large, the number of white balls chosen approximately has the mass function of a binomial random variable. . 4.8.4

The Zeta (or Zipf) Distribution

A random variable is said to have a zeta (sometimes called the Zipf) distribution if its probability mass function is given by P{X = k} =

C kα+1

k = 1, 2, . . .

for some value of α > 0. Since the sum of the foregoing probabilities must equal 1, it follows that ⎡ ⎤−1 q α+1 1 ⎦ C=⎣ k k=1

164

Chapter 4

Random Variables

The zeta distribution owes its name to the fact that the function s s s 1 1 1 ζ (s) = 1 + + + ··· + + ··· 2 3 k is known in mathematical disciplines as the Riemann zeta function (after the German mathematician G. F. B. Riemann). The zeta distribution was used by the Italian economist V. Pareto to describe the distribution of family incomes in a given country. However, it was G. K. Zipf who applied zeta distribution to a wide variety of problems in different areas and, in doing so, popularized its use. 4.9

EXPECTED VALUE OF SUMS OF RANDOM VARIABLES A very important property of expectations is that the expected value of a sum of random variables is equal to the sum of their expectations. In this section, we will prove this result under the assumption that the set of possible values of the probability experiment—that is, the sample space S—is either ﬁnite or countably inﬁnite. Although the result is true without this assumption (and a proof is outlined in the theoretical exercises), not only will the assumption simplify the argument, but it will also result in an enlightening proof that will add to our intuition about expectations. So, for the remainder of this section, suppose that the sample space S is either a ﬁnite or a countably inﬁnite set. For a random variable X, let X(s) denote the value of X when s ∈ S is the outcome of the experiment. Now, if X and Y are both random variables, then so is their sum. That is, Z = X + Y is also a random variable. Moreover, Z(s) = X(s) + Y(s). EXAMPLE 9a Suppose that the experiment consists of ﬂipping a coin 5 times, with the outcome being the resulting sequence of heads and tails. Suppose X is the number of heads in the ﬁrst 3 ﬂips and Y is the number of heads in the ﬁnal 2 ﬂips. Let Z = X + Y. Then, for instance, for the outcome s = (h, t, h, t, h), X(s) = 2 Y(s) = 1 Z(s) = X(s) + Y(s) = 3 meaning that the outcome (h, t, h, t, h) results in 2 heads in the ﬁrst three ﬂips, 1 head in the ﬁnal two ﬂips, and a total of 3 heads in the ﬁve ﬂips. . Let p(s) = P({s}) be the probability that s is the outcome of the experiment. Because we can write any event A as the ﬁnite or countably inﬁnite union of the mutually exclusive events {s}, s ∈ A, it follows by the axioms of probability that P(A) = p(s) s∈A

When A = S, the preceding equation gives p(s) 1= s∈S

Section 4.9

Expected Value of Sums of Random Variables

165

Now, let X be a random variable, and consider E[X]. Because X(s) is the value of X when s is the outcome of the experiment, it seems intuitive that E[X]—the weighted average of the possible values of X, with each value weighted by the probability that X assumes that value—should equal a weighted average of the values X(s), s ∈ S, with X(s) weighted by the probability that s is the outcome of the experiment. We now prove this intuition. Proposition 9.1. E[X] =

X(s) p(s)

s∈S

Proof. Suppose that the distinct values of X are xi , i Ú 1. For each i, let Si be the event that X is equal to xi . That is, Si = {s : X(s) = xi }. Then, E[X] = xi P{X = xi } i

=

xi P(Si )

i

=

i

= = =

xi

s∈Si

i

s∈Si

i

s∈Si

p(s)

xi p(s) X(s)p(s)

X(s)p(s)

s∈S

where the ﬁnal equality follows because S1 , S2 , . . . are mutually exclusive events whose union is S. EXAMPLE 9b Suppose that two independent ﬂips of a coin that comes up heads with probability p are made, and let X denote the number of heads obtained. Because P(X = 0) = P(t, t) = (1 − p)2 , P(X = 1) = P(h, t) + P(t, h) = 2p(1 − p) P(X = 2) = P(h, h) = p2 it follows from the deﬁnition of expected value that E[X] = 0 · (1 − p)2 + 1 · 2p(1 − p) + 2 · p2 = 2p which agrees with E[X] = X(h, h)p2 + X(h, t)p(1 − p) + X(t, h)(1 − p)p + X(t, t)(1 − p)2 = 2p2 + p(1 − p) + (1 − p)p = 2p

.

We now prove the important and useful result that the expected value of a sum of random variables is equal to the sum of their expectations.

166

Chapter 4

Random Variables

Corollary 9.2. For random variables X1 , X2 , . . . , Xn , ⎡ ⎤ n n ⎣ ⎦ E Xi = E[Xi ] i=1

Proof. Let Z =

n

E[Z] =

i=1 Xi .

i=1

Then, by Proposition 9.1,

Z(s)p(s)

s∈S

=

X1 (s) + X2 (s) + . . . + Xn (s) p(s) s∈S

=

X1 (s)p(s) +

s∈S

X2 (s)p(s) + . . . +

s∈S

Xn (s)p(s)

s∈S

= E[X1 ] + E[X2 ] + . . . + E[Xn ]

.

EXAMPLE 9c Find the expected value of the sum obtained when n fair dice are rolled. Solution. Let X be the sum. We will compute E[X] by using the representation X=

n

Xi

i=1

where Xi is the upturned value on die i. Because Xi is equally likely to be any of the values from 1 to 6, it follows that E[Xi ] =

6

i(1/6) = 21/6 = 7/2

i=1

which yields the result ⎤ ⎡ n n Xi ⎦ = E[Xi ] = 3.5 n E[X] = E ⎣ i=1

i=1

.

EXAMPLE 9d Find the expected total number of successes that result from n trials when trial i is a success with probability pi , i = 1, . . . , n. Solution. Letting % Xi =

1, 0,

if trial i is a success if trial i is a failure

we have the representation X=

n i=1

Xi

Section 4.9

Expected Value of Sums of Random Variables

167

Consequently, E[X] =

n

E[Xi ] =

i=1

n

pi

i=1

Note that this result does not require that the trials be independent. It includes as a special case the expected value of a binomial random variable, which assumes independent trials and all pi = p, and thus has mean np. It also gives the expected value of a hypergeometric random variable representing the number of white balls selected when n balls are randomly selected, without replacement, from an urn of N balls of which m are white. We can interpret the hypergeometric as representing the number of successes in n trials, where trial i is said to be a success if the ith ball selected is white. Because the ith ball selected is equally likely to be any of the N balls and thus has probability m/N of being white, it follows that the hypergeometric is the number of successes in n trials in which each trial is a success with probability p = m/N. Hence, even though these hypergeometric trials are dependent, it follows from the result of Example 9d that the expected value of the hypergeometric is np = nm/N. . EXAMPLE 9e Derive an expression for the variance of the number of successful trials in Example 9d, and apply it to obtain the variance of a binomial random variable with parameters n and p, and of a hypergeometric random variable equal to the number of white balls chosen when n balls are randomly chosen from an urn containing N balls of which m are white. Solution. Letting X be the number of successful trials, and using the same represen tation for X—namely, X = ni=1 Xi —as in the previous example, we have ⎡ ⎞⎤ ⎞⎛ ⎛ n n ⎟⎥ ⎢ ⎠⎜ ⎝ E[X 2 ] = E ⎢ X Xj ⎠⎥ ⎝ i ⎦ ⎣ i=1

j=1

⎡ ⎛ ⎞⎤ n ⎢ ⎝ ⎥ = E⎣ Xi Xi + Xj ⎠⎦ jZi

i=1

⎤ ⎡ n n Xi2 + Xi Xj ⎦ = E⎣ i=1 jZi

i=1

=

n

E[Xi2 ] +

i

E[Xi Xj ]

i=1 jZi

i=1

=

n

pi +

n

E[Xi Xj ]

(9.1)

i=1 jZi

where the ﬁnal equation used that Xi2 = Xi . However, because the possible values of both Xi and Xj are 0 or 1, it follows that % 1, if Xi = 1, Xj = 1 Xi Xj = 0, otherwise

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Chapter 4

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Hence,

E[Xi Xj ] = P{Xi = 1, Xj = 1} = P(trials i and j are successes)

Now, on the one hand, if X is binomial, then, for i Z j, the results of trial i and trial j are independent, with each being a success with probability p. Therefore, E[Xi Xj ] = p2 ,

i Z j

Together with Equation (9.1), the preceding equation shows that, for a binomial random variable X, E[X 2 ] = np + n(n − 1)p2 implying that Var(X) = E[X 2 ] − (E[X])2 = np + n(n − 1)p2 − n2 p2 = np(1 − p) On the other hand, if X is hypergeometric, then, given that a white ball is chosen in trial i, each of the other N − 1 balls, of which m − 1 are white, is equally likely to be the jth ball chosen, for j Z i. Consequently, for j Z i, P{Xi = 1, Xj = 1} = P{Xi = 1}P{Xj = 1|Xi = 1} =

mm − 1 N N − 1

Using pi = m/N, we now obtain, from Equation (9.1), E[X 2 ] =

mm − 1 nm + n(n − 1) N N N − 1

Consequently, mm − 1 nm + n(n − 1) − Var(X) = N N N − 1

nm N

2

which, as shown in Example 8j, can be simpliﬁed to yield n − 1 Var(X) = np(1 − p) 1 − N − 1 where p = m/N. 4.10

.

PROPERTIES OF THE CUMULATIVE DISTRIBUTION FUNCTION Recall that, for the distribution function F of X, F(b) denotes the probability that the random variable X takes on a value that is less than or equal to b. Following are some properties of the cumulative distribution function (c.d.f.) F: 1. F is a nondecreasing function; that is, if a < b, then F(a) … F(b). 2. lim F(b) = 1. b→q 3. lim F(b) = 0. b→−q 4. F is right continuous. That is, for any b and any decreasing sequence bn , n Ú 1, that converges to b, lim F(bn ) = F(b). n→q Property 1 follows, as was noted in Section 4.1, because, for a < b, the event {X … a} is contained in the event {X … b} and so cannot have a larger probability. Properties 2, 3, and 4 all follow from the continuity property of probabilities

Section 4.10

Properties of the Cumulative Distribution Function

169

(Section 2.6). For instance, to prove property 2, we note that if bn increases to q, then the events {X … bn }, n Ú 1, are increasing events whose union is the event {X < q}. Hence, by the continuity property of probabilities, lim P{X … bn } = P{X < q} = 1

n→q

which proves property 2. The proof of property 3 is similar and is left as an exercise. To prove property 4, we note that if bn decreases to b, then {X … bn }, n Ú 1, are decreasing events whose intersection is {X … b}. The continuity property then, yields lim P{X … bn } = P{X … b} n

which veriﬁes property 4. All probability questions about X can be answered in terms of the c.d.f., F. For example, P{a < X … b} = F(b) − F(a) for all a < b (8.1) This equation can best be seen to hold if we write the event {X … b} as the union of the mutually exclusive events {X … a} and {a < X … b}. That is, {X … b} = {X … a} ∪ {a < X … b} so

P{X … b} = P{X … a} + P{a < X … b}

which establishes Equation (9.1). If we want to compute the probability that X is strictly less than b, we can again apply the continuity property to obtain / % 1 P{X < b} = P lim X … b − n→q n 1 = lim P X … b − n→q n 1 = lim F b − n→q n Note that P{X < b} does not necessarily equal F(b), since F(b) also includes the probability that X equals b. EXAMPLE 10a The distribution function of the random variable X is given by ⎧ ⎪ 0 x < 0 ⎪ ⎪ ⎪ x ⎪ ⎪ ⎪ 0 … x < 1 ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎨2 1 … x < 2 F(x) = 3 ⎪ ⎪ ⎪ ⎪ 11 ⎪ ⎪ ⎪ 2 … x < 3 ⎪ ⎪ 12 ⎪ ⎪ ⎪ ⎩1 3 … x

170

Chapter 4

Random Variables F(x)

1 11 — 12 2– 3 1– 2

1

2

3

x

FIGURE 4.8: Graph of F(x).

A graph of F(x) is presented in Figure 4.8. Compute (a) P{X < 3}, (b) P{X = 1}, (c) P{X > 12 }, and (d) P{2 < X … 4}. / % 11 1 1 = lim F 3 − = Solution. (a) P{X < 3} = lim P X … 3 − n n n n 12 (b) P{X = 1} = P{X … 1} − P{X < 1} 2 1 1 1 = F(1) − lim F 1 − = − = n n 3 2 6 (c) / % / % 1 1 =1 − P X … P X > 2 2 1 3 =1 − F = 2 4 (d) P{2 < X … 4} = F(4) − F(2) 1 = 12

.

SUMMARY A real-valued function deﬁned on the outcome of a probability experiment is called a random variable. If X is a random variable, then the function F(x) deﬁned by F(x) = P{X … x} is called the distribution function of X. All probabilities concerning X can be stated in terms of F.

Summary

171

A random variable whose set of possible values is either ﬁnite or countably inﬁnite is called discrete. If X is a discrete random variable, then the function p(x) = P{X = x} is called the probability mass function of X. Also, the quantity E[X] deﬁned by E[X] = xp(x) x:p(x)>0

is called the expected value of X. E[X] is also commonly called the mean or the expectation of X. A useful identity states that, for a function g, E[g(X)] = g(x)p(x) x:p(x)>0

The variance of a random variable X, denoted by Var(X), is deﬁned by Var(X) = E[(X − E[X])2 ] The variance, which is equal to the expected square of the difference between X and its expected value, is a measure of the spread of the possible values of X. A useful identity is Var(X) = E[X 2 ] − (E[X])2 √ The quantity Var(X) is called the standard deviation of X. We now note some common types of discrete random variables. The random variable X whose probability mass function is given by n pi (1 − p)n−i i = 0, . . . , n p(i) = i is said to be a binomial random variable with parameters n and p. Such a random variable can be interpreted as being the number of successes that occur when n independent trials, each of which results in a success with probability p, are performed. Its mean and variance are given by E[X] = np

Var(X) = np(1 − p)

The random variable X whose probability mass function is given by p(i) =

e−λ λi i!

i Ú 0

is said to be a Poisson random variable with parameter λ. If a large number of (approximately) independent trials are performed, each having a small probability of being successful, then the number of successful trials that result will have a distribution which is approximately that of a Poisson random variable. The mean and variance of a Poisson random variable are both equal to its parameter λ. That is, E[X] = Var(X) = λ The random variable X whose probability mass function is given by p(i) = p(1 − p)i−1

i = 1, 2, . . .

172

Chapter 4

Random Variables

is said to be a geometric random variable with parameter p. Such a random variable represents the trial number of the ﬁrst success when each trial is independently a success with probability p. Its mean and variance are given by E[X] =

1 p

Var(X) =

1 − p p2

The random variable X whose probability mass function is given by i − 1 pr (1 − p)i−r i Ú r p(i) = r − 1 is said to be a negative binomial random variable with parameters r and p. Such a random variable represents the trial number of the rth success when each trial is independently a success with probability p. Its mean and variance are given by E[X] =

r p

Var(X) =

r(1 − p) p2

A hypergeometric random variable X with parameters n, N, and m represents the number of white balls selected when n balls are randomly chosen from an urn that contains N balls of which m are white. The probability mass function of this random variable is given by m N − m i n − i i = 0, . . . , m p(i) = N n With p = m/N, its mean and variance are E[X] = np Var(X) =

N − n np(1 − p) N − 1

An important property of the expected value is that the expected value of a sum of random variables is equal to the sum of their expected values. That is, ⎡ ⎤ n n E⎣ Xi ⎦ = E[Xi ] i=1

i=1

PROBLEMS 4.1. Two balls are chosen randomly from an urn containing 8 white, 4 black, and 2 orange balls. Suppose that we win $2 for each black ball selected and we lose $1 for each white ball selected. Let X denote our winnings. What are the possible values of X, and what are the probabilities associated with each value? 4.2. Two fair dice are rolled. Let X equal the product of the 2 dice. Compute P{X = i} for i = 1, . . . , 36.

4.3. Three dice are rolled. By assuming that each of the 63 = 216 possible outcomes is equally likely, ﬁnd the probabilities attached to the possible values that X can take on, where X is the sum of the 3 dice. 4.4. Five men and 5 women are ranked according to their scores on an examination. Assume that no two scores are alike and all 10! possible rankings are equally likely. Let X denote the highest ranking achieved by a woman. (For instance, X = 1

Problems

4.5.

4.6.

4.7.

4.8.

4.9. 4.10.

4.11.

if the top-ranked person is female.) Find P{X = i}, i = 1, 2, 3, . . . , 8, 9, 10. Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed n times. What are the possible values of X? In Problem 5, for n = 3, if the coin is assumed fair, what are the probabilities associated with the values that X can take on? Suppose that a die is rolled twice. What are the possible values that the following random variables can take on: (a) the maximum value to appear in the two rolls; (b) the minimum value to appear in the two rolls; (c) the sum of the two rolls; (d) the value of the ﬁrst roll minus the value of the second roll? If the die in Problem 7 is assumed fair, calculate the probabilities associated with the random variables in parts (a) through (d). Repeat Example 1b when the balls are selected with replacement. In Example 1d, compute the conditional probability that we win i dollars, given that we win something; compute it for i = 1, 2, 3. (a) An integer N is to be selected at random from {1, 2, . . . , (10)3 } in the sense that each integer has the same probability of being selected. What is the probability that N will be divisible by 3? by 5? by 7? by 15? by 105? How would your answer change if (10)3 is replaced by (10)k as k became larger and larger? (b) An important function in number theory— one whose properties can be shown to be related to what is probably the most important unsolved problem of mathematics, the ¨ Riemann hypothesis—is the Mobius function μ(n), deﬁned for all positive integral values n as follows: Factor n into its prime factors. If there is a repeated prime factor, as in 12 = 2 · 2 · 3 or 49 = 7 · 7, then μ(n) is deﬁned to equal 0. Now let N be chosen at random from {1, 2, . . . (10)k }, where k is large. Determine P{μ(N) = 0} as k→q. Hint: To compute P{μ(N) Z 0}, use the identity q Pi2 − 1 i=1

Pi2

=

3 8 24 48 6 ··· = 2 4 9 25 49 π

where Pi is the ith-smallest prime. (The number 1 is not a prime.) 4.12. In the game of Two-Finger Morra, 2 players show 1 or 2 ﬁngers and simultaneously guess the number of ﬁngers their opponent will show. If only one of the players guesses correctly, he wins an amount

173

(in dollars) equal to the sum of the ﬁngers shown by him and his opponent. If both players guess correctly or if neither guesses correctly, then no money is exchanged. Consider a speciﬁed player, and denote by X the amount of money he wins in a single game of Two-Finger Morra. (a) If each player acts independently of the other, and if each player makes his choice of the number of ﬁngers he will hold up and the number he will guess that his opponent will hold up in such a way that each of the 4 possibilities is equally likely, what are the possible values of X and what are their associated probabilities? (b) Suppose that each player acts independently of the other. If each player decides to hold up the same number of ﬁngers that he guesses his opponent will hold up, and if each player is equally likely to hold up 1 or 2 ﬁngers, what are the possible values of X and their associated probabilities? 4.13. A salesman has scheduled two appointments to sell encyclopedias. His ﬁrst appointment will lead to a sale with probability .3, and his second will lead independently to a sale with probability .6. Any sale made is equally likely to be either for the deluxe model, which costs $1000, or the standard model, which costs $500. Determine the probability mass function of X, the total dollar value of all sales. 4.14. Five distinct numbers are randomly distributed to players numbered 1 through 5. Whenever two players compare their numbers, the one with the higher one is declared the winner. Initially, players 1 and 2 compare their numbers; the winner then compares her number with that of player 3, and so on. Let X denote the number of times player 1 is a winner. Find P{X = i}, i = 0, 1, 2, 3, 4. 4.15. The National Basketball Association (NBA) draft lottery involves the 11 teams that had the worst won–lost records during the year. A total of 66 balls are placed in an urn. Each of these balls is inscribed with the name of a team: Eleven have the name of the team with the worst record, 10 have the name of the team with the second-worst record, 9 have the name of the team with the thirdworst record, and so on (with 1 ball having the name of the team with the 11th-worst record). A ball is then chosen at random, and the team whose name is on the ball is given the ﬁrst pick in the draft of players about to enter the league. Another ball is then chosen, and if it “belongs” to a team different from the one that received the ﬁrst draft pick, then the team to which it belongs receives the second draft pick. (If the ball belongs

174

Chapter 4

Random Variables

to the team receiving the ﬁrst pick, then it is discarded and another one is chosen; this continues until the ball of another team is chosen.) Finally, another ball is chosen, and the team named on the ball (provided that it is different from the previous two teams) receives the third draft pick. The remaining draft picks 4 through 11 are then awarded to the 8 teams that did not “win the lottery,” in inverse order of their won–lost records. For instance, if the team with the worst record did not receive any of the 3 lottery picks, then that team would receive the fourth draft pick. Let X denote the draft pick of the team with the worst record. Find the probability mass function of X. 4.16. In Problem 15, let team number 1 be the team with the worst record, let team number 2 be the team with the second-worst record, and so on. Let Yi denote the team that gets draft pick number i. (Thus, Y1 = 3 if the ﬁrst ball chosen belongs to team number 3.) Find the probability mass function of (a) Y1 , (b) Y2 , and (c) Y3 . 4.17. Suppose that the distribution function of X is given by ⎧ ⎪ 0 b < 0 ⎪ ⎪ ⎪ ⎪ ⎪ b ⎪ ⎪ 0 … b < 1 ⎪ ⎪ 4 ⎪ ⎪ ⎪ ⎨1 b − 1 + 1 … b < 2 F(b) = ⎪2 4 ⎪ ⎪ ⎪ ⎪ 11 ⎪ ⎪ 2 … b < 3 ⎪ ⎪ ⎪ 12 ⎪ ⎪ ⎪ ⎩1 3 … b (a) Find P{X = i}, i = 1, 2, 3. (b) Find P{ 12 < X < 32 }. 4.18. Four independent ﬂips of a fair coin are made. Let X denote the number of heads obtained. Plot the probability mass function of the random variable X − 2. 4.19. If the distribution function of X is given by ⎧ ⎪ 0 b < 0 ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ 0 … b < 1 ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ ⎪3 ⎪ ⎪ 1 … b < 2 ⎨5 F(b) = ⎪ 4 ⎪ ⎪ 2 … b < 3 ⎪ ⎪ ⎪5 ⎪ ⎪ ⎪ ⎪9 ⎪ ⎪ 3 … b < 3.5 ⎪ ⎪ 10 ⎪ ⎪ ⎪ ⎩1 b Ú 3.5 calculate the probability mass function of X.

4.20. A gambling book recommends the following “winning strategy” for the game of roulette: Bet$1 on red. If red appears which has probability 18 38 , then take the $1 proﬁt andquit. If red does not appear 20 and you lose this bet which has probability 38 of occurring , make additional $1 bets on red on each of the next two spins of the roulette wheel and then quit. Let X denote your winnings when you quit. (a) Find P{X > 0}. (b) Are you convinced that the strategy is indeed a “winning” strategy? Explain your answer! (c) Find E[X]. 4.21. Four buses carrying 148 students from the same school arrive at a football stadium. The buses carry, respectively, 40, 33, 25, and 50 students. One of the students is randomly selected. Let X denote the number of students that were on the bus carrying the randomly selected student. One of the 4 bus drivers is also randomly selected. Let Y denote the number of students on her bus. (a) Which of E[X] or E[Y] do you think is larger? Why? (b) Compute E[X] and E[Y]. 4.22. Suppose that two teams play a series of games that ends when one of them has won i games. Suppose that each game played is, independently, won by team A with probability p. Find the expected number of games that are played when (a) i = 2 and (b) i = 3. Also, show in both cases that this number is maximized when p = 12 . 4.23. You have $1000, and a certain commodity presently sells for $2 per ounce. Suppose that after one week the commodity will sell for either $1 or $4 an ounce, with these two possibilities being equally likely. (a) If your objective is to maximize the expected amount of money that you possess at the end of the week, what strategy should you employ? (b) If your objective is to maximize the expected amount of the commodity that you possess at the end of the week, what strategy should you employ? 4.24. A and B play the following game: A writes down either number 1 or number 2, and B must guess which one. If the number that A has written down is i and B has guessed correctly, B receives i units from A. If B makes a wrong guess, B pays 34 unit to A. If B randomizes his decision by guessing 1 with probability p and 2 with probability 1 − p, determine his expected gain if (a) A has written down number 1 and (b) A has written down number 2. What value of p maximizes the minimum possible value of B’s expected gain, and what is this maximin value? (Note that B’s expected

Problems

4.25.

4.26.

4.27.

4.28.

4.29.

gain depends not only on p, but also on what A does.) Consider now player A. Suppose that she also randomizes her decision, writing down number 1 with probability q. What is A’s expected loss if (c) B chooses number 1 and (d) B chooses number 2? What value of q minimizes A’s maximum expected loss? Show that the minimum of A’s maximum expected loss is equal to the maximum of B’s minimum expected gain. This result, known as the minimax theorem, was ﬁrst established in generality by the mathematician John von Neumann and is the fundamental result in the mathematical discipline known as the theory of games. The common value is called the value of the game to player B. Two coins are to be ﬂipped. The ﬁrst coin will land on heads with probability .6, the second with probability .7. Assume that the results of the ﬂips are independent, and let X equal the total number of heads that result. (a) Find P{X = 1}. (b) Determine E[X]. One of the numbers 1 through 10 is randomly chosen. You are to try to guess the number chosen by asking questions with “yes–no” answers. Compute the expected number of questions you will need to ask in each of the following two cases: (a) Your ith question is to be “Is it i?” i = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. (b) With each question you try to eliminate onehalf of the remaining numbers, as nearly as possible. An insurance company writes a policy to the effect that an amount of money A must be paid if some event E occurs within a year. If the company estimates that E will occur within a year with probability p, what should it charge the customer in order that its expected proﬁt will be 10 percent of A? A sample of 3 items is selected at random from a box containing 20 items of which 4 are defective. Find the expected number of defective items in the sample. There are two possible causes for a breakdown of a machine. To check the ﬁrst possibility would cost C1 dollars, and, if that were the cause of the breakdown, the trouble could be repaired at a cost of R1 dollars. Similarly, there are costs C2 and R2 associated with the second possibility. Let p and 1 − p denote, respectively, the probabilities that the breakdown is caused by the ﬁrst and second possibilities. Under what conditions on p, Ci , Ri , i = 1, 2, should we check the ﬁrst possible cause of breakdown and then the second, as opposed to reversing the checking order, so as to minimize the expected cost involved in returning the machine to working order?

175

Note: If the ﬁrst check is negative, we must still check the other possibility. 4.30. A person tosses a fair coin until a tail appears for the ﬁrst time. If the tail appears on the nth ﬂip, the person wins 2n dollars. Let X denote the player’s winnings. Show that E[X] = +q. This problem is known as the St. Petersburg paradox. (a) Would you be willing to pay $1 million to play this game once? (b) Would you be willing to pay $1 million for each game if you could play for as long as you liked and only had to settle up when you stopped playing? 4.31. Each night different meteorologists give us the probability that it will rain the next day. To judge how well these people predict, we will score each of them as follows: If a meteorologist says that it will rain with probability p, then he or she will receive a score of 1 − (1 − p)2 1 − p2

if it does rain if it does not rain

We will then keep track of scores over a certain time span and conclude that the meteorologist with the highest average score is the best predictor of weather. Suppose now that a given meteorologist is aware of our scoring mechanism and wants to maximize his or her expected score. If this person truly believes that it will rain tomorrow with probability p∗ , what value of p should he or she assert so as to maximize the expected score? 4.32. To determine whether they have a certain disease, 100 people are to have their blood tested. However, rather than testing each individual separately, it has been decided ﬁrst to place the people into groups of 10. The blood samples of the 10 people in each group will be pooled and analyzed together. If the test is negative, one test will sufﬁce for the 10 people, whereas if the test is positive, each of the 10 people will also be individually tested and, in all, 11 tests will be made on this group. Assume that the probability that a person has the disease is .1 for all people, independently of each other, and compute the expected number of tests necessary for each group. (Note that we are assuming that the pooled test will be positive if at least one person in the pool has the disease.) 4.33. A newsboy purchases papers at 10 cents and sells them at 15 cents. However, he is not allowed to return unsold papers. If his daily demand is a binomial random variable with n = 10, p = 13 , approximately how many papers should he purchase so as to maximize his expected proﬁt? 4.34. In Example 4b, suppose that the department store incurs an additional cost of c for each unit of unmet

176

4.35.

4.36.

4.37. 4.38.

4.39.

4.40.

4.41.

4.42.

4.43.

4.44.

Chapter 4

Random Variables

demand. (This type of cost is often referred to as a goodwill cost because the store loses the goodwill of those customers whose demands it cannot meet.) Compute the expected proﬁt when the store stocks s units, and determine the value of s that maximizes the expected proﬁt. A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $1.10; if they are different colors, then you win −$1.00. (That is, you lose $1.00.) Calculate (a) the expected value of the amount you win; (b) the variance of the amount you win. Consider Problem 22 with i = 2. Find the variance of the number of games played, and show that this number is maximized when p = 12 . Find Var(X) and Var(Y) for X and Y as given in Problem 21. If E[X] = 1 and Var(X) = 5, ﬁnd (a) E[(2 + X)2 ]; (b) Var(4 + 3X). A ball is drawn from an urn containing 3 white and 3 black balls. After the ball is drawn, it is replaced and another ball is drawn. This process goes on indeﬁnitely. What is the probability that, of the ﬁrst 4 balls drawn, exactly 2 are white? On a multiple-choice exam with 3 possible answers for each of the 5 questions, what is the probability that a student will get 4 or more correct answers just by guessing? A man claims to have extrasensory perception. As a test, a fair coin is ﬂipped 10 times and the man is asked to predict the outcome in advance. He gets 7 out of 10 correct. What is the probability that he would have done at least this well if he had no ESP? Suppose that, in ﬂight, airplane engines will fail with probability 1 − p, independently from engine to engine. If an airplane needs a majority of its engines operative to complete a successful ﬂight, for what values of p is a 5-engine plane preferable to a 3-engine plane? A communications channel transmits the digits 0 and 1. However, due to static, the digit transmitted is incorrectly received with probability .2. Suppose that we want to transmit an important message consisting of one binary digit. To reduce the chance of error, we transmit 00000 instead of 0 and 11111 instead of 1. If the receiver of the message uses “majority” decoding, what is the probability that the message will be wrong when decoded? What independence assumptions are you making? A satellite system consists of n components and functions on any given day if at least k of the n components function on that day. On a rainy day

each of the components independently functions with probability p1 , whereas on a dry day they each independently function with probability p2 . If the probability of rain tomorrow is α, what is the probability that the satellite system will function? 4.45. A student is getting ready to take an important oral examination and is concerned about the possibility of having an “on” day or an “off” day. He ﬁgures that if he has an on day, then each of his examiners will pass him, independently of each other, with probability .8, whereas if he has an off day, this probability will be reduced to .4. Suppose that the student will pass the examination if a majority of the examiners pass him. If the student feels that he is twice as likely to have an off day as he is to have an on day, should he request an examination with 3 examiners or with 5 examiners? 4.46. Suppose that it takes at least 9 votes from a 12member jury to convict a defendant. Suppose also that the probability that a juror votes a guilty person innocent is .2, whereas the probability that the juror votes an innocent person guilty is .1. If each juror acts independently and if 65 percent of the defendants are guilty, ﬁnd the probability that the jury renders a correct decision. What percentage of defendants is convicted? 4.47. In some military courts, 9 judges are appointed. However, both the prosecution and the defense attorneys are entitled to a peremptory challenge of any judge, in which case that judge is removed from the case and is not replaced. A defendant is declared guilty if the majority of judges cast votes of guilty, and he or she is declared innocent otherwise. Suppose that when the defendant is, in fact, guilty, each judge will (independently) vote guilty with probability .7, whereas when the defendant is, in fact, innocent, this probability drops to .3. (a) What is the probability that a guilty defendant is declared guilty when there are (i) 9, (ii) 8, and (iii) 7 judges? (b) Repeat part (a) for an innocent defendant. (c) If the prosecution attorney does not exercise the right to a peremptory challenge of a judge, and if the defense is limited to at most two such challenges, how many challenges should the defense attorney make if he or she is 60 percent certain that the client is guilty? 4.48. It is known that diskettes produced by a certain company will be defective with probability .01, independently of each other. The company sells the diskettes in packages of size 10 and offers a money-back guarantee that at most 1 of the 10 diskettes in the package will be defective. The guarantee is that the customer can return the entire package of diskettes if he or she ﬁnds more

Problems than one defective diskette in it. If someone buys 3 packages, what is the probability that he or she will return exactly 1 of them? 4.49. When coin 1 is ﬂipped, it lands on heads with probability .4; when coin 2 is ﬂipped, it lands on heads with probability .7. One of these coins is randomly chosen and ﬂipped 10 times. (a) What is the probability that the coin lands on heads on exactly 7 of the 10 ﬂips? (b) Given that the ﬁrst of these ten ﬂips lands heads, what is the conditional probability that exactly 7 of the 10 ﬂips land on heads? 4.50. Suppose that a biased coin that lands on heads with probability p is ﬂipped 10 times. Given that a total of 6 heads results, ﬁnd the conditional probability that the ﬁrst 3 outcomes are (a) h, t, t (meaning that the ﬁrst ﬂip results in heads, the second in tails, and the third in tails); (b) t, h, t. 4.51. The expected number of typographical errors on a page of a certain magazine is .2. What is the probability that the next page you read contains (a) 0 and (b) 2 or more typographical errors? Explain your reasoning! 4.52. The monthly worldwide average number of airplane crashes of commercial airlines is 3.5. What is the probability that there will be (a) at least 2 such accidents in the next month; (b) at most 1 accident in the next month? Explain your reasoning! 4.53. Approximately 80,000 marriages took place in the state of New York last year. Estimate the probability that, for at least one of these couples, (a) both partners were born on April 30; (b) both partners celebrated their birthday on the same day of the year. State your assumptions. 4.54. Suppose that the average number of cars abandoned weekly on a certain highway is 2.2. Approximate the probability that there will be (a) no abandoned cars in the next week; (b) at least 2 abandoned cars in the next week. 4.55. A certain typing agency employs 2 typists. The average number of errors per article is 3 when typed by the ﬁrst typist and 4.2 when typed by the second. If your article is equally likely to be typed by either typist, approximate the probability that it will have no errors. 4.56. How many people are needed so that the probability that at least one of them has the same birthday as you is greater than 12 ?

177

4.57. Suppose that the number of accidents occurring on a highway each day is a Poisson random variable with parameter λ = 3. (a) Find the probability that 3 or more accidents occur today. (b) Repeat part (a) under the assumption that at least 1 accident occurs today. 4.58. Compare the Poisson approximation with the correct binomial probability for the following cases: (a) P{X = 2} when n = 8, p = .1; (b) P{X = 9} when n = 10, p = .95; (c) P{X = 0} when n = 10, p = .1; (d) P{X = 4} when n = 9, p = .2. 4.59. If you buy a lottery ticket in 50 lotteries, in each of 1 which your chance of winning a prize is 100 , what is the (approximate) probability that you will win a prize (a) at least once? (b) exactly once? (c) at least twice? 4.60. The number of times that a person contracts a cold in a given year is a Poisson random variable with parameter λ = 5. Suppose that a new wonder drug (based on large quantities of vitamin C) has just been marketed that reduces the Poisson parameter to λ = 3 for 75 percent of the population. For the other 25 percent of the population, the drug has no appreciable effect on colds. If an individual tries the drug for a year and has 2 colds in that time, how likely is it that the drug is beneﬁcial for him or her? 4.61. The probability of being dealt a full house in a hand of poker is approximately .0014. Find an approximation for the probability that, in 1000 hands of poker, you will be dealt at least 2 full houses. 4.62. Consider n independent trials, each of which results in one of the outcomes 1, . . . , k with respec k tive probabilities p1 , . . . , pk , i=1 pi = 1. Show that if all the pi are small, then the probability that no trial outcome occurs more than once is approximately equal to exp(−n(n − 1) i p2i /2). 4.63. People enter a gambling casino at a rate of 1 every 2 minutes. (a) What is the probability that no one enters between 12:00 and 12:05? (b) What is the probability that at least 4 people enter the casino during that time? 4.64. The suicide rate in a certain state is 1 suicide per 100,000 inhabitants per month. (a) Find the probability that, in a city of 400,000 inhabitants within this state, there will be 8 or more suicides in a given month.

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(b) What is the probability that there will be at least 2 months during the year that will have 8 or more suicides? (c) Counting the present month as month number 1, what is the probability that the ﬁrst month to have 8 or more suicides will be month number i, i Ú 1? What assumptions are you making? 4.65. Each of 500 soldiers in an army company independently has a certain disease with probability 1/103 . This disease will show up in a blood test, and to facilitate matters, blood samples from all 500 soldiers are pooled and tested. (a) What is the (approximate) probability that the blood test will be positive (that is, at least one person has the disease)? Suppose now that the blood test yields a positive result. (b) What is the probability, under this circumstance, that more than one person has the disease? One of the 500 people is Jones, who knows that he has the disease. (c) What does Jones think is the probability that more than one person has the disease? Because the pooled test was positive, the authorities have decided to test each individual separately. The ﬁrst i − 1 of these tests were negative, and the ith one—which was on Jones—was positive. (d) Given the preceding, scenario, what is the probability, as a function of i, that any of the remaining people have the disease? 4.66. A total of 2n people, consisting of n married couples, are randomly seated (all possible orderings being equally likely) at a round table. Let Ci denote the event that the members of couple i are seated next to each other, i = 1, . . . , n. (a) Find P(Ci ). (b) For j Z i, ﬁnd P(Cj |Ci ). (c) Approximate the probability, for n large, that there are no married couples who are seated next to each other. 4.67. Repeat the preceding problem when the seating is random but subject to the constraint that the men and women alternate. 4.68. In response to an attack of 10 missiles, 500 antiballistic missiles are launched. The missile targets of the antiballistic missiles are independent, and each antiballstic missile is equally likely to go towards any of the target missiles. If each antiballistic missile independently hits its target with probability .1, use the Poisson paradigm to approximate the probability that all missiles are hit.

4.69. A fair coin is ﬂipped 10 times. Find the probability that there is a string of 4 consecutive heads by (a) using the formula derived in the text; (b) using the recursive equations derived in the text. (c) Compare your answer with that given by the Poisson approximation. 4.70. At time 0, a coin that comes up heads with probability p is ﬂipped and falls to the ground. Suppose it lands on heads. At times chosen according to a Poisson process with rate λ, the coin is picked up and ﬂipped. (Between these times the coin remains on the ground.) What is the probability that the coin is on its head side at time t? Hint What would be the conditional probability if there were no additional ﬂips by time t, and what would it be if there were additional ﬂips by time t? 4.71. Consider a roulette wheel consisting of 38 numbers 1 through 36, 0, and double 0. If Smith always bets that the outcome will be one of the numbers 1 through 12, what is the probability that (a) Smith will lose his ﬁrst 5 bets; (b) his ﬁrst win will occur on his fourth bet? 4.72. Two athletic teams play a series of games; the ﬁrst team to win 4 games is declared the overall winner. Suppose that one of the teams is stronger than the other and wins each game with probability .6, independently of the outcomes of the other games. Find the probability, for i = 4, 5, 6, 7, that the stronger team wins the series in exactly i games. Compare the probability that the stronger team wins with the probability that it would win a 2-outof-3 series. 4.73. Suppose in Problem 72 that the two teams are evenly matched and each has probability 12 of winning each game. Find the expected number of games played. 4.74. An interviewer is given a list of people she can interview. If the interviewer needs to interview 5 people, and if each person (independently) agrees to be interviewed with probability 23 , what is the probability that her list of people will enable her to obtain her necessary number of interviews if the list consists of (a) 5 people and (b) 8 people? For part (b), what is the probability that the interviewer will speak to exactly (c) 6 people and (d) 7 people on the list? 4.75. A fair coin is continually ﬂipped until heads appears for the 10th time. Let X denote the number of tails that occur. Compute the probability mass function of X. 4.76. Solve the Banach match problem (Example 8e) when the left-hand matchbox originally contained

Theoretical Exercises

4.77.

4.78.

4.79.

4.80.

N1 matches and the right-hand box contained N2 matches. In the Banach matchbox problem, ﬁnd the probability that, at the moment when the ﬁrst box is emptied (as opposed to being found empty), the other box contains exactly k matches. An urn contains 4 white and 4 black balls. We randomly choose 4 balls. If 2 of them are white and 2 are black, we stop. If not, we replace the balls in the urn and again randomly select 4 balls. This continues until exactly 2 of the 4 chosen are white. What is the probability that we shall make exactly n selections? Suppose that a batch of 100 items contains 6 that are defective and 94 that are not defective. If X is the number of defective items in a randomly drawn sample of 10 items from the batch, ﬁnd (a) P{X = 0} and (b) P{X > 2}. A game popular in Nevada gambling casinos is Keno, which is played as follows: Twenty numbers are selected at random by the casino from the set of numbers 1 through 80. A player can select from 1 to 15 numbers; a win occurs if some fraction of the player’s chosen subset matches any of the 20 numbers drawn by the house. The payoff is a function of the number of elements in the player’s selection and the number of matches. For instance, if the player selects only 1 number, then he or she wins if this number is among the set of 20, and the payoff is $2.2 won for every dollar bet. (As the player’s probability of winning in this case is 1 4 , it is clear that the “fair” payoff should be $3 won for every $1 bet.) When the player selects 2 numbers, a payoff (of odds) of $12 won for every $1 bet is made when both numbers are among the 20, (a) What would be the fair payoff in this case? Let Pn, k denote the probability that exactly k of the n numbers chosen by the player are among the 20 selected by the house. (b) Compute Pn, k (c) The most typical wager at Keno consists of selecting 10 numbers. For such a bet the casino pays off as shown in the following table. Compute the expected payoff:

179

Keno Payoffs in 10 Number Bets Number of matches Dollars won for each $1 bet 0–4 5 6 7 8 9 10

-1 1 17 179 1, 299 2, 599 24, 999

4.81. In Example 8i, what percentage of i defective lots does the purchaser reject? Find it for i = 1, 4. Given that a lot is rejected, what is the conditional probability that it contained 4 defective components? 4.82. A purchaser of transistors buys them in lots of 20. It is his policy to randomly inspect 4 components from a lot and to accept the lot only if all 4 are nondefective. If each component in a lot is, independently, defective with probability .1, what proportion of lots is rejected? 4.83. There are three highways in the county. The number of daily accidents that occur on these highways are Poisson random variables with respective parameters .3, .5, and .7. Find the expected number of accidents that will happen on any of these highways today. 4.84. Suppose that 10 balls are put into 5 boxes, with each ball independently being put in box i with probability pi , 5i=1 pi = 1. (a) Find the expected number of boxes that do not have any balls. (b) Find the expected number of boxes that have exactly 1 ball. 4.85. There are k types of coupons. Independently of the types of previously collected coupons, each new coupon collected is of type i with probability k pi , i=1 pi = 1. If n coupons are collected, ﬁnd the expected number of distinct types that appear in this set. (That is, ﬁnd the expected number of types of coupons that appear at least once in the set of n coupons.)

THEORETICAL EXERCISES 4.1. There are N distinct types of coupons, and each time one is obtained it will, independently of past choices, be of type i with probability Pi , i = 1, . . . , N. Let T denote the number one need select

to obtain at least one of each type. Compute P{T = n}. Hint: Use an argument similar to the one used in Example 1e.

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Random Variables

4.2. If X has distribution function F, what is the distribution function of eX ? 4.3. If X has distribution function F, what is the distribution function of the random variable αX + β, where α and β are constants, α Z 0? 4.4. For a nonnegative integer-valued random variable N, show that E[N] =

q

P{N Ú i}

i=1

Hint:

q

P{N Ú i} =

i=1

q q

P{N = k}. Now inter-

i=1 k=i

change the order of summation. 4.5. For a nonnegative integer-valued random variable N, show that q

iP{N > i} =

i=0

Hint:

q

iP{N > i} =

i=0

1 (E[N 2 ] − E[N]) 2 q

i

q

P{N = k}. Now

i=0 k=i+1

interchange the order of summation. 4.6. Let X be such that P{X = 1} = p = 1 − P{X = −1} Find c Z 1 such that E[cX ] = 1. 4.7. Let X be a random variable having expected value μ and variance σ 2 . Find the expected value and variance of X − μ Y= σ 4.8. Find Var(X) if P(X = a) = p = 1 − P(X = b) 4.9. Show how the derivation of the binomial probabilities n i P{X = i} = p (1 − p)n−i , i = 0, . . . , n i leads to a proof of the binomial theorem (x + y)n =

n n i=0

i

xi yn−i

when x and y are nonnegative. x Hint: Let p = x+y . 4.10. Let X be a binomial random variable with parameters n and p. Show that 1 − (1 − p)n+1 1 = E X + 1 (n + 1)p

4.11. Consider n independent sequential trials, each of which is successful with probability p. If there is a total of k successes, show that each of the n!/[k!(n − k)!] possible arrangements of the k successes and n − k failures is equally likely. 4.12. There are n components lined up in a linear arrangement. Suppose that each component independently functions with probability p. What is the probability that no 2 neighboring components are both nonfunctional? Hint: Condition on the number of defective components and use the results of Example 4c of Chapter 1. 4.13. Let X be a binomial random variable with parameters (n, p). What value of p maximizes P{X = k}, k = 0, 1, . . . , n? This is an example of a statistical method used to estimate p when a binomial (n, p) random variable is observed to equal k. If we assume that n is known, then we estimate p by choosing that value of p which maximizes P{X = k}. This is known as the method of maximum likelihood estimation. 4.14. A family has n children with probability αpn , n Ú 1, where α … (1 − p)/p. (a) What proportion of families has no children? (b) If each child is equally likely to be a boy or a girl (independently of each other), what proportion of families consists of k boys (and any number of girls)? 4.15. Suppose that n independent tosses of a coin having probability p of coming up heads are made. Show that the probability that an even number of heads results is 12 [1 + (q − p)n ], where q = 1 − p. Do this by proving and then utilizing the identity [n/2]

i=0

n 2i

p2i qn−2i =

3 12 (p + q)n + (q − p)n 2

where [n/2] is the largest integer less than or equal to n/2. Compare this exercise with Theoretical Exercise 3.5 of Chapter 3. 4.16. Let X be a Poisson random variable with parameter λ. Show that P{X = i} increases monotonically and then decreases monotonically as i increases, reaching its maximum when i is the largest integer not exceeding λ. Hint: Consider P{X = i}/P{X = i − 1}. 4.17. Let X be a Poisson random variable with parameter λ. (a) Show that P{X is even} =

5 14 1 + e−2λ 2

Theoretical Exercises by using the result of Theoretical Exercise 15 and the relationship between Poisson and binomial random variables. (b) Verify the formula in part (a) directly by making use of the expansion of e−λ + eλ . 4.18. Let X be a Poisson random variable with parameter λ. What value of λ maximizes P{X = k}, k Ú 0? 4.19. Show that X is a Poisson random variable with parameter λ, then E[X n ] = λE[(X + 1)n−1 ] Now use this result to compute E[X 3 ]. 4.20. Consider n coins, each of which independently comes up heads with probability p. Suppose that n is large and p is small, and let λ = np. Suppose that all n coins are tossed; if at least one comes up heads, the experiment ends; if not, we again toss all n coins, and so on. That is, we stop the ﬁrst time that at least one of the n coins come up heads. Let X denote the total number of heads that appear. Which of the following reasonings concerned with approximating P{X = 1} is correct (in all cases, Y is a Poisson random variable with parameter λ)? (a) Because the total number of heads that occur when all n coins are rolled is approximately a Poisson random variable with parameter λ, P{X = 1} L P{Y = 1} = λe−λ (b) Because the total number of heads that occur when all n coins are rolled is approximately a Poisson random variable with parameter λ, and because we stop only when this number is positive, P{X = 1} L P{Y = 1|Y > 0} =

λe−λ 1 − e−λ

(c) Because at least one coin comes up heads, X will equal 1 if none of the other n − 1 coins come up heads. Because the number of heads resulting from these n − 1 coins is approximately Poisson with mean (n − 1)p L λ, P{X = 1} L P{Y = 0} = e−λ 4.21. From a set of n randomly chosen people, let Eij denote the event that persons i and j have the same birthday. Assume that each person is equally likely to have any of the 365 days of the year as his or her birthday. Find (a) P(E3,4 |E1,2 ); (b) P(E1,3 |E1,2 ); (c) P(E2,3 |E1,2 ∩ E1,3 ).

181

What can you conclude from your answers to n parts (a)–(c) about the independence of the 2 events Eij ? 4.22. An urn contains 2n balls, of which 2 are numbered 1, 2 are numbered 2, . . . , and 2 are numbered n. Balls are successively withdrawn 2 at a time without replacement. Let T denote the ﬁrst selection in which the balls withdrawn have the same number (and let it equal inﬁnity if none of the pairs withdrawn has the same number). We want to show that, for 0 < α < 1, lim P{T > αn} = e−α/2 n

To verify the preceding formula, let Mk denote the number of pairs withdrawn in the ﬁrst k selections, k = 1, . . . , n. (a) Argue that when n is large, Mk can be regarded as the number of successes in k (approximately) independent trials. (b) Approximate P{Mk = 0} when n is large. (c) Write the event {T > αn} in terms of the value of one of the variables Mk . (d) Verify the limiting probability given for P{T > αn}. 4.23. Consider a random collection of n individuals. In approximating the probability that no 3 of these individuals share the same birthday, a better Poisson approximation than that obtained in the text (at least for values of n between 80 and 90) is obtained by letting Ei be the event that there are at least 3 birthdays on day i, i = 1, . . . , 365. (a) Find P(Ei ). (b) Give an approximation for the probability that no 3 individuals share the same birthday. (c) Evaluate the preceding when n = 88 (which can be shown to be the smallest value of n for which the probability exceeds .5). 4.24. Here is another way to obtain a set of recursive equations for determining Pn , the probability that there is a string of k consecutive heads in a sequence of n ﬂips of a fair coin that comes up heads with probability p: (a) Argue that, for k < n, there will be a string of k consecutive heads if either 1. there is a string of k consecutive heads within the ﬁrst n − 1 ﬂips, or 2. there is no string of k consecutive heads within the ﬁrst n − k − 1 ﬂips, ﬂip n − k is a tail, and ﬂips n − k + 1, . . . , n are all heads. (b) Using the preceding, relate Pn to Pn−1 . Starting with Pk = pk , the recursion can be used to obtain Pk+1 , then Pk+1 , and so on, up to Pn .

182

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Random Variables

4.25. Suppose that the number of events that occur in a speciﬁed time is a Poisson random variable with parameter λ. If each event is counted with probability p, independently of every other event, show that the number of events that are counted is a Poisson random variable with parameter λp. Also, give an intuitive argument as to why this should be so. As an application of the preceding result, suppose that the number of distinct uranium deposits in a given area is a Poisson random variable with parameter λ = 10. If, in a ﬁxed period of time, each deposit is discovered inde1 pendently with probability 50 , ﬁnd the probability that (a) exactly 1, (b) at least 1, and (c) at most 1 deposit is discovered during that time. 4.26. Prove * n λi 1 q −x n e−λ = e x dx i! n! λ

4.31.

4.32.

i=0

Hint: Use integration by parts. 4.27. If X is a geometric random variable, show analytically that P{X = n + k|X > n} = P{X = k} Using the interpretation of a geometric random variable, give a verbal argument as to why the preceding equation is true. 4.28. Let X be a negative binomial random variable with parameters r and p, and let Y be a binomial random variable with parameters n and p. Show that P{X > n} = P{Y < r} Hint: Either one could attempt an analytical proof of the preceding equation, which is equivalent to proving the identity q r−1 i − 1 n pr (1 − p)i−r = r − 1 i

i=n+1

i=0

* pi (1 − p)n−i or one could attempt a proof that uses the probabilistic interpretation of these random variables. That is, in the latter case, start by considering a sequence of independent trials having a common probability p of success. Then try to express the events {X > n} and {Y < r} in terms of the outcomes of this sequence. 4.29. For a hypergeometric random variable, determine P{X = k + 1}/P{X = k} 4.30. Balls numbered 1 through N are in an urn. Suppose that n, n … N, of them are randomly selected

4.33. 4.34.

without replacement. Let Y denote the largest number selected. (a) Find the probability mass function of Y. (b) Derive an expression for E[Y] and then use Fermat’s combinatorial identity (see Theoretical Exercise 11 of Chapter 1) to simplify the expression. A jar contains m + n chips, numbered 1, 2, . . . , n + m. A set of size n is drawn. If we let X denote the number of chips drawn having numbers that exceed each of the numbers of those remaining, compute the probability mass function of X. A jar contains n chips. Suppose that a boy successively draws a chip from the jar, each time replacing the one drawn before drawing another. The process continues until the boy draws a chip that he has previously drawn. Let X denote the number of draws, and compute its probability mass function. Show that Equation (8.6) follows from Equation (8.5). From a set of n elements, a nonempty subset is chosen at random in the sense that all of the nonempty subsets are equally likely to be selected. Let X denote the number of elements in the chosen subset. Using the identities given in Theoretical Exercise 12 of Chapter 1, show that E[X] = 2 − Var(X) =

n n−1 1 2

n · 22n−2 − n(n + 1)2n−2 (2n − 1)2

Show also that, for n large, Var(X)

n 4

in the sense that the ratio Var(X) to n/4 approaches 1 as n approaches q. Compare this formula with the limiting form of Var(Y) when P{Y = i} = 1/n, i = 1, . . . , n. 4.35. An urn initially contains one red and one blue ball. At each stage, a ball is randomly chosen and then replaced along with another of the same color. Let X denote the selection number of the ﬁrst chosen ball that is blue. For instance, if the ﬁrst selection is red and the second blue, then X is equal to 2. (a) Find P{X > i}, i Ú 1. (b) Show that, with probability 1, a blue ball is eventually chosen. (That is, show that P{X < q} = 1.) (c) Find E[X].

Self-Test Problems and Exercises 4.36. Suppose the possible values of X are {xi }, the possible values of Y are {yj }, and the possible values of X + Y are {zk }. Let Ak denote the set of all pairs of indices (i, j) such that xi + yj = zk ; that is, Ak = {(i, j) : xi + yj = zk }. (a) Argue that P{X + Y = zk } =

183

(c) Using the formula from part (b), argue that E[X + Y] = (xi + yj )P{X = xi , i

j

Y = yj } (d) Show that

P{X = xi , Y = yj }

(i,j)∈Ak

P(X = xi ) =

P(X = xi , Y = yj ),

j

P(Y = yj ) =

(b) Show that E[X + Y] =

P{X = xi , Y = yj }

i

(xi + yj )P{X = xi ,

k (i,j)∈Ak

Y = yj }

(e) Prove that E[X + Y] = E[X] + E[Y]

SELF-TEST PROBLEMS AND EXERCISES 4.1. Suppose that the random variable X is equal to the number of hits obtained by a certain baseball player in his next 3 at bats. If P{X = 1} = .3, P{X = 2} = .2, and P{X = 0} = 3P{X = 3}, ﬁnd E[X]. 4.2. Suppose that X takes on one of the values 0, 1, and 2. If for some constant c, P{X = i} = cP{X = i − 1}, i = 1, 2, ﬁnd E[X]. 4.3. A coin that, when ﬂipped, comes up heads with probability p is ﬂipped until either heads or tails has occurred twice. Find the expected number of ﬂips. 4.4. A certain community is composed of m families, r ni of which have i children, ni = m. If one of i=1

the families is randomly chosen, let X denote the number of children in that family. If one of the r ini children is randomly chosen, let Y denote i=1

the total number of children in the family of that child. Show that E[Y] Ú E[X]. 4.5. Suppose that P{X = 0} = 1 − P{X = 1}. If E[X] = 3Var(X), ﬁnd P{X = 0}. 4.6. There are 2 coins in a bin. When one of them is ﬂipped, it lands on heads with probability .6, and when the other is ﬂipped, it lands on heads with probability .3. One of these coins is to be randomly chosen and then ﬂipped. Without knowing which coin is chosen, you can bet any amount up to 10 dollars, and you then either win that amount if the coin comes up heads or lose it if it comes up tails. Suppose, however, that an insider is willing to sell you, for an amount C, the information as to which coin was selected. What is your expected payoff if you buy this information? Note that if you buy

it and then bet x, you will end up either winning x − C or −x − C (that is, losing x + C in the latter case). Also, for what values of C does it pay to purchase the information? 4.7. A philanthropist writes a positive number x on a piece of red paper, shows the paper to an impartial observer, and then turns it face down on the table. The observer then ﬂips a fair coin. If it shows heads, she writes the value 2x and, if tails, the value x/2, on a piece of blue paper, which she then turns face down on the table. Without knowing either the value x or the result of the coin ﬂip, you have the option of turning over either the red or the blue piece of paper. After doing so and observing the number written on that paper, you may elect to receive as a reward either that amount or the (unknown) amount written on the other piece of paper. For instance, if you elect to turn over the blue paper and observe the value 100, then you can elect either to accept 100 as your reward or to take the amount (either 200 or 50) on the red paper. Suppose that you would like your expected reward to be large. (a) Argue that there is no reason to turn over the red paper ﬁrst, because if you do so, then no matter what value you observe, it is always better to switch to the blue paper. (b) Let y be a ﬁxed nonnegative value, and consider the following strategy: Turn over the blue paper, and if its value is at least y, then accept that amount. If it is less than y, then switch to the red paper. Let Ry (x) denote the reward obtained if the philanthropist writes the amount x and you employ this strategy. Find E[Ry (x)]. Note that E[R0 (x)] is the

184

Chapter 4

Random Variables

expected reward if the philanthropist writes the amount x when you employ the strategy of always choosing the blue paper. 4.8. Let B(n, p) represent a binomial random variable with parameters n and p. Argue that P{B(n, p) … i} = 1 − P{B(n, 1 − p) … n − i − 1}

4.9. 4.10.

4.11.

4.12.

4.13.

4.14.

4.15.

Hint: The number of successes less than or equal to i is equivalent to what statement about the number of failures? If X is a binomial random variable with expected value 6 and variance 2.4, ﬁnd P{X = 5}. An urn contains n balls numbered 1 through n. If you withdraw m balls randomly in sequence, each time replacing the ball selected previously, ﬁnd P{X = k}, k = 1, . . . , m, where X is the maximum of the m chosen numbers. Hint: First ﬁnd P{X … k}. Teams A and B play a series of games, with the ﬁrst team to win 3 games being declared the winner of the series. Suppose that team A independently wins each game with probability p. Find the conditional probability that team A wins (a) the series given that it wins the ﬁrst game; (b) the ﬁrst game given that it wins the series. A local soccer team has 5 more games left to play. If it wins its game this weekend, then it will play its ﬁnal 4 games in the upper bracket of its league, and if it loses, then it will play its ﬁnal games in the lower bracket. If it plays in the upper bracket, then it will independently win each of its games in this bracket with probability .4, and if it plays in the lower bracket, then it will independently win each of its games with probability .7. If the probability that the team wins its game this weekend is .5, what is the probability that it wins at least 3 of its ﬁnal 4 games? Each of the members of a 7-judge panel independently makes a correct decision with probability .7. If the panel’s decision is made by majority rule, what is the probability that the panel makes the correct decision? Given that 4 of the judges agreed, what is the probability that the panel made the correct decision? On average, 5.2 hurricanes hit a certain region in a year. What is the probability that there will be 3 or fewer hurricanes hitting this year? The number of eggs laid on a tree leaf by an insect of a certain type is a Poisson random variable with parameter λ. However, such a random variable can be observed only if it is positive, since if it is 0 then we cannot know that such an insect was on the leaf. If we let Y denote the observed number of eggs, then

P{Y = i} = P{X = i|X > 0} where X is Poisson with parameter λ. Find E[Y]. 4.16. Each of n boys and n girls, independently and randomly, chooses a member of the other sex. If a boy and girl choose each other, they become a couple. Number the girls, and let Gi be the event that girl number i is part of a couple. Let P0 = 1 − P(∪ni=1 Gi ) be the probability that no couples are formed. (a) What is P(Gi )? (b) What is P(Gi |Gj )? (c) When n is large, approximate P0 . (d) When n is large, approximate Pk , the probability that exactly k couples are formed. (e) Use the inclusion–exclusion identity to evaluate P0 . 4.17. A total of 2n people, consisting of n married couples, are randomly divided into n pairs. Arbitrarily number the women, and let Wi denote the event that woman i is paired with her husband. (a) Find P(Wi ). (b) For i Z j, ﬁnd P(Wi |Wj ). (c) When n is large, approximate the probability that no wife is paired with her husband. (d) If each pairing must consist of a man and a woman, what does the problem reduce to? 4.18. A casino patron will continue to make $5 bets on red in roulette until she has won 4 of these bets. (a) What is the probability that she places a total of 9 bets? (b) What is her expected winnings when she stops? Remark: On each bet, she will either win $5 with 20 probability 18 38 or lose $5 with probability 38 . 4.19. When three friends go for coffee, they decide who will pay the check by each ﬂipping a coin and then letting the “odd person” pay. If all three ﬂips produce the same result (so that there is no odd person), then they make a second round of ﬂips, and they continue to do so until there is an odd person. What is the probability that (a) exactly 3 rounds of ﬂips are made? (b) more than 4 rounds are needed? 4.20. Show that if X is a geometric random variable with parameter p, then E[1/X] =

−p log(p) 1 − p

Hint: You will need to evaluate an expression of q -a ai /i. To do so, write ai /i = 0 xi−1 dx, the form i=1

and then interchange the sum and the integral. 4.21. Suppose that P{X = a} = p,

P{X = b} = 1 − p

Self-Test Problems and Exercises (a) Show that X−b a−b is a Bernoulli random variable. (b) Find Var(X). 4.22. Each game you play is a win with probability p. You plan to play 5 games, but if you win the ﬁfth game, then you will keep on playing until you lose. (a) Find the expected number of games that you play. (b) Find the expected number of games that you lose. 4.23. Balls are randomly withdrawn, one at a time without replacement, from an urn that initially has N white and M black balls. Find the probability that n white balls are drawn before m black balls, n … N, m … M. 4.24. Ten balls are to be distributed among 5 urns, with each ball going into urn i with probabil 5 ity pi , i=1 pi = 1. Let Xi denote the number of

185

balls that go into urn i. Assume that events corresponding to the locations of different balls are independent. (a) What type of random variable is Xi ? Be as speciﬁc as possible. (b) For i Z j, what type of random variable is Xi + Xj ? (c) Find P{X1 + X2 + X3 = 7}. 4.25. For the match problem (Example 5m Chapter 2), ﬁnd (a) the expected number of matches. (b) the variance of the number of matches.

in

4.26. Let α be the probability that a geometric random variable X with parameter p is an even number. q (a) Find α by using the identity α = i=1 P{X = 2i}. (b) Find α by conditioning on whether X = 1 or X > 1.

C H A P T E R

5

Continuous Random Variables 5.1 5.2 5.3 5.4 5.5 5.6 5.7

INTRODUCTION EXPECTATION AND VARIANCE OF CONTINUOUS RANDOM VARIABLES THE UNIFORM RANDOM VARIABLE NORMAL RANDOM VARIABLES EXPONENTIAL RANDOM VARIABLES OTHER CONTINUOUS DISTRIBUTIONS THE DISTRIBUTION OF A FUNCTION OF A RANDOM VARIABLE

5.1

INTRODUCTION In Chapter 4, we considered discrete random variables—that is, random variables whose set of possible values is either ﬁnite or countably inﬁnite. However, there also exist random variables whose set of possible values is uncountable. Two examples are the time that a train arrives at a speciﬁed stop and the lifetime of a transistor. Let X be such a random variable. We say that X is a continuous† random variable if there exists a nonnegative function f , deﬁned for all real x ∈ (−q, q), having the property that, for any set B of real numbers,‡ * P{X ∈ B} = f (x) dx (1.1) B

The function f is called the probability density function of the random variable X. (See Figure 5.1.) In words, Equation (1.1) states that the probability that X will be in B may be obtained by integrating the probability density function over the set B. Since X must assume some value, f must satisfy * q f (x) dx 1 = P{X ∈ (−q, q)} = −q

All probability statements about X can be answered in terms of f . For instance, from Equation (1.1), letting B = [a, b], we obtain * P{a … X … b} =

b

f (x) dx

(1.2)

a

† Sometimes called absolutely continuous. ‡ Actually, for technical reasons Equation (1.1) is true only for the measurable sets B, which, fortunately, include all sets of practical interest.

186

Section 5.1

Introduction

187

f

a

x

b

P(a ⱕ X ⱕ b) = area of shaded region FIGURE 5.1: Probability density function f.

If we let a = b in Equation (1.2), we get

*

P{X = a} =

a

f (x) dx = 0

a

In words, this equation states that the probability that a continuous random variable will assume any ﬁxed value is zero. Hence, for a continuous random variable, * a f (x) dx P{X < a} = P{X … a} = F(a) = −q

EXAMPLE 1a Suppose that X is a continuous random variable whose probability density function is given by 0 C(4x − 2x2 ) 0 < x < 2 f (x) = 0 otherwise (a) What is the value of C? (b) Find P{X > 1}. Solution. (a) Since f is a probability density function, we must have implying that * 2 (4x − 2x2 ) dx = 1 C

-q

−q f (x) dx

= 1,

0

or

2x3 C 2x2 − 3

6x=2 6 6 =1 6 6 x=0

or C= Hence, (b) P{X > 1} =

-q 1

f (x) dx =

3 2 8 1 (4x

3 8

− 2x2 ) dx =

1 2

.

188

Chapter 5

Continuous Random Variables

EXAMPLE 1b The amount of time in hours that a computer functions before breaking down is a continuous random variable with probability density function given by 0 f (x) =

λe−x/100 x Ú 0 0 x < 0

What is the probability that (a) a computer will function between 50 and 150 hours before breaking down? (b) it will function for fewer than 100 hours? Solution. (a) Since * 1=

q

−q

* f (x) dx = λ

q

e−x/100 dx

0

we obtain 6q 1 = −λ(100)e−x/100 60 = 100λ or

λ=

1 100

Hence, the probability that a computer will function between 50 and 150 hours before breaking down is given by * P{50 < X < 150} =

150 50

= e−1/2

6150 1 −x/100 e dx = −e−x/100 650 100 − e−3/2 L .384

(b) Similarly, *

100

P{X < 100} = 0

6100 1 −x/100 e dx = −e−x/100 60 = 1 − e−1 L .633 100

In other words, approximately 63.3 percent of the time, a computer will fail before registering 100 hours of use. . EXAMPLE 1c The lifetime in hours of a certain kind of radio tube is a random variable having a probability density function given by ⎧ ⎪ x … 100 ⎨0 f (x) = 100 ⎪ ⎩ 2 x > 100 x What is the probability that exactly 2 of 5 such tubes in a radio set will have to be replaced within the ﬁrst 150 hours of operation? Assume that the events Ei , i = 1, 2, 3, 4, 5, that the ith such tube will have to be replaced within this time are independent.

Section 5.1

Introduction

189

Solution. From the statement of the problem, we have * 150 f (x) dx P(Ei ) = 0 * 150 = 100 x−2 dx 100

1 = 3 Hence, from the independence of the events Ei , it follows that the desired probability is 2 3 1 80 2 5 = . 2 3 3 243 The relationship between the cumulative distribution F and the probability density f is expressed by * a f (x) dx F(a) = P{X ∈ (−q, a]} = −q

Differentiating both sides of the preceding equation yields d F(a) = f (a) da That is, the density is the derivative of the cumulative distribution function. A somewhat more intuitive interpretation of the density function may be obtained from Equation (1.2) as follows: / * a+ε/2 % ε ε … X … a + = f (x) dx L εf (a) P a − 2 2 a−ε/2 when ε is small and when f (·) is continuous at x = a. In other words, the probability that X will be contained in an interval of length ε around the point a is approximately εf (a). From this result we see that f (a) is a measure of how likely it is that the random variable will be near a. EXAMPLE 1d If X is continuous with distribution function FX and density function fX , ﬁnd the density function of Y = 2X. Solution. We will determine fY in two ways. The ﬁrst way is to derive, and then differentiate, the distribution function of Y: FY (a) = P{Y … a} = P{2X … a} = P{X … a/2} = FX (a/2) Differentiation gives fY (a) =

1 fX (a/2) 2

190

Chapter 5

Continuous Random Variables

Another way to determine fY is to note that … Y … a + } 2 2 = P{a − … 2X … a + } 2 2 a a = P{ − … X … + } 2 4 2 4 L fX (a/2) 2

fY (a) L P{a −

Dividing through by gives the same result as before. 5.2

.

EXPECTATION AND VARIANCE OF CONTINUOUS RANDOM VARIABLES In Chapter 4, we deﬁned the expected value of a discrete random variable X by xP{X = x} E[X] = x

If X is a continuous random variable having probability density function f (x), then, because f (x) dx L P{x … X … x + dx} for dx small it is easy to see that the analogous deﬁnition is to deﬁne the expected value of X by * q xf (x) dx E[X] = −q

EXAMPLE 2a Find E[X] when the density function of X is % 2x if 0 … x … 1 f (x) = 0 otherwise Solution.

* E[X] =

xf (x) dx *

=

1

2x2 dx

0

2 = 3 EXAMPLE 2b The density function of X is given by % 1 if 0 … x … 1 f (x) = 0 otherwise Find E[eX ].

.

Section 5.2

Expectation and Variance of Continuous Random Variables

191

Solution. Let Y = eX . We start by determining FY , the probability distribution function of Y. Now, for 1 … x … e, FY (x) = P{Y … x} = P{eX … x} = P{X … log(x)} * log(x) f (y) dy = 0

= log(x) By differentiating FY (x), we can conclude that the probability density function of Y is given by 1 1 … x … e fY (x) = x Hence,

* X

q

E[e ] = E[Y] =

xfY (x) dx *−q e dx = 1

=e − 1

.

Although the method employed in Example 2b to compute the expected value of a function of X is always applicable, there is, as in the discrete case, an alternative way of proceeding. The following is a direct analog of Proposition 4.1. of Chapter 4. Proposition 2.1. If X is a continuous random variable with probability density function f (x), then, for any real-valued function g, * q E[g(X)] = g(x)f (x) dx −q

An application of Proposition 2.1 to Example 2b yields * 1 X ex dx sincef (x) = 1, 0 < x < 1 E[e ] = 0

=e − 1 which is in accord with the result obtained in that example. The proof of Proposition 2.1 is more involved than that of its discrete random variable analog. We will present such a proof under the provision that the random variable g(X) is nonnegative. (The general proof, which follows the argument in the case we present, is indicated in Theoretical Exercises 2 and 3.) We will need the following lemma, which is of independent interest. Lemma 2.1 For a nonnegative random variable Y, * q E[Y] = P{Y > y} dy 0

192

Chapter 5

Continuous Random Variables

Proof. We present a proof when Y is a continuous random variable with probability density function fY . We have * q * q* q P{Y > y} dy = fY (x) dx dy 0

0

y

-q where we have used the fact that P{Y > y} = y fY (x) dx. Interchanging the order of integration in the preceding equation yields * q * q * x P{Y > y} dy = dy fY (x) dx 0 0 0 * q = xfY (x) dx 0

= E[Y]

.

From Lemma 2.1, for any function g for which g(x) Ú 0, * q E[g(X)] = P{g(X) > y} dy 0 * q* f (x) dx dy =

Proof of Proposition 2.1.

x:g(x)>y * g(x)

*

0

*

x:g(x)>0 0

= =

dy f (x) dx

g(x)f (x) dx x:g(x)>0

which completes the proof. EXAMPLE 2c A stick of length 1 is split at a point U that is uniformly distributed over (0, 1). Determine the expected length of the piece that contains the point p, 0 … p … 1. Solution. Let Lp (U) denote the length of the substick that contains the point p, and note that % 1 − U U < p Lp (U) = U U > p (See Figure 5.2.) Hence, from Proposition 2.1, * 1 E[Lp (U)] = Lp (u) du 0 * p * = (1 − u)du + 0

1

u du p

(1 − p)2 1 p2 1 − + − 2 2 2 2 1 = + p(1 − p) 2

=

Section 5.2

Expectation and Variance of Continuous Random Variables

193

1–U 0

U

1

p

(a)

U 0

p

1

U

(b)

FIGURE 5.2: Substick containing point p: (a) U < p; (b) U > p.

Since p(1 − p) is maximized when p = 12 , it is interesting to note that the expected length of the substick containing the point p is maximized when p is the midpoint of the original stick. . EXAMPLE 2d Suppose that if you are s minutes early for an appointment, then you incur the cost cs, and if you are s minutes late, then you incur the cost ks. Suppose also that the travel time from where you presently are to the location of your appointment is a continuous random variable having probability density function f . Determine the time at which you should depart if you want to minimize your expected cost. Solution. Let X denote the travel time. If you leave t minutes before your appointment, then your cost—call it Ct (X)—is given by % c(t − X) ifX … t Ct (X) = k(X − t) ifX Ú t Therefore,

*

E[Ct (X)] =

q

Ct (x)f (x) dx

0

*

* q c(t − x)f (x) dx + k(x − t)f (x) dx 0 t * t * q * * t f (x) dx − c xf (x) dx + k xf (x) dx − kt = ct

=

t

0

0

t

q

f (x) dx

t

The value of t that minimizes E[Ct (X)] can now be obtained by calculus. Differentiation yields d E[Ct (X)] = ct f (t) + cF(t) − ct f (t) − kt f (t) + kt f (t) − k[1 − F(t)] dt = (k + c)F(t) − k Equating the rightmost side to zero shows that the minimal expected cost is obtained when you leave t∗ minutes before your appointment, where t∗ satisﬁes k k + c As in Chapter 4, we can use Proposition 2.1 to show the following. F(t∗ ) =

Corollary 2.1. If a and b are constants, then E[aX + b] = aE[X] + b

.

194

Chapter 5

Continuous Random Variables

The proof of Corollary 2.1 for a continuous random variable X is the same as the one given for a discrete random variable. The only modiﬁcation is that the sum is replaced by an integral and the probability mass function by a probability density function. The variance of a continuous random variable is deﬁned exactly as it is for a discrete random variable, namely, if X is a random variable with expected value μ, then the variance of X is deﬁned (for any type of random variable) by Var(X) = E[(X − μ)2 ] The alternative formula, Var(X) = E[X 2 ] − (E[X])2 is established in a manner similar to its counterpart in the discrete case. EXAMPLE 2e Find Var(X) for X as given in Example 2a. Solution. We ﬁrst compute E[X 2 ].

* 2

E[X ] = =

q

−q * 1

x2 f (x) dx

2x3 dx

0

=

1 2

Hence, since E[X] = 23 , we obtain 1 Var(X) = − 2

2 2 1 = 3 18

.

It can be shown that, for constants a and b, Var(aX + b) = a2 Var(X) The proof mimics the one given for discrete random variables. There are several important classes of continuous random variables that appear frequently in applications of probability; the next few sections are devoted to a study of some of them. 5.3

THE UNIFORM RANDOM VARIABLE A random variable is said to be uniformly distributed over the interval (0, 1) if its probability density function is given by % 1 0 < x < 1 (3.1) f (x) = 0 otherwise -q Note that Equation (3.1) is a density function, since f (x) Ú 0 and −q f (x) dx = -1 0 dx = 1. Because f (x) > 0 only when x ∈ (0, 1), it follows that X must assume a value in interval (0, 1). Also, since f (x) is constant for x ∈ (0, 1), X is just as likely to

Section 5.3 f(a)

The Uniform Random Variable

195

F(a) 1

1 ——– –␣

␣

a

␣

(a)

a

(b)

FIGURE 5.3: Graph of (a) f(a) and (b) F(a) for a uniform (α, β) random variable.

be near any value in (0, 1) as it is to be near any other value. To verify this statement, note that, for any 0 < a < b < 1, * b f (x) dx = b − a P{a … X … b} = a

In other words, the probability that X is in any particular subinterval of (0, 1) equals the length of that subinterval. In general, we say that X is a uniform random variable on the interval (α, β) if the probability density function of X is given by ⎧ ⎪ 1 ⎨ if α < x < β (3.2) f (x) = β − α ⎪ ⎩0 otherwise -a Since F(a) = −q f (x) dx, it follows from Equation (3.2) that the distribution function of a uniform random variable on the interval (α, β) is given by ⎧ ⎪ 0 a … α ⎪ ⎪ ⎨ a − α α < a < β F(a) = β − α ⎪ ⎪ ⎪ ⎩1 a Ú β Figure 5.3 presents a graph of f (a) and F(a). EXAMPLE 3a Let X be uniformly distributed over (α, β). Find (a) E[X] and (b) Var(X). Solution. (a)

* E[X] =

q

−q * β

xf (x) dx

x dx β − α α β 2 − α2 = 2(β − α) β + α = 2 =

196

Chapter 5

Continuous Random Variables

In words, the expected value of a random variable that is uniformly distributed over some interval is equal to the midpoint of that interval. (b) To ﬁnd Var(X), we ﬁrst calculate E[X 2 ]. * β 1 E[X 2 ] = x2 dx β − α α β 3 − α3 = 3(β − α) β 2 + αβ + α 2 = 3 Hence, (α + β)2 β 2 + αβ + α 2 − 3 4 (β − α)2 = 12

Var(X) =

Therefore, the variance of a random variable that is uniformly distributed over some interval is the square of the length of that interval divided by 12. . EXAMPLE 3b If X is uniformly distributed over (0, 10), calculate the probability that (a) X < 3, (b) X > 6, and (c) 3 < X < 8. * 3 3 1 dx = Solution. (a) P{X < 3} = 10 0 10 * 10 4 1 dx = (b) P{X > 6} = 10 10 6 * 8 1 1 dx = (c) P{3 < X < 8} = . 2 3 10 EXAMPLE 3c Buses arrive at a speciﬁed stop at 15-minute intervals starting at 7 A . M . That is, they arrive at 7, 7:15, 7:30, 7:45, and so on. If a passenger arrives at the stop at a time that is uniformly distributed between 7 and 7:30, ﬁnd the probability that he waits (a) less than 5 minutes for a bus; (b) more than 10 minutes for a bus. Solution. Let X denote the number of minutes past 7 that the passenger arrives at the stop. Since X is a uniform random variable over the interval (0, 30), it follows that the passenger will have to wait less than 5 minutes if (and only if) he arrives between 7:10 and 7:15 or between 7:25 and 7:30. Hence, the desired probability for part (a) is * 15 * 30 1 1 1 P{10 < X < 15} + P{25 < X < 30} = dx + dx = 3 10 30 25 30 Similarly, he would have to wait more than 10 minutes if he arrives between 7 and 7:05 or between 7:15 and 7:20, so the probability for part (b) is P{0 < X < 5} + P{15 < X < 20} =

1 3

.

Section 5.3

The Uniform Random Variable

197

The next example was ﬁrst considered by the French mathematician Joseph L. F. Bertrand in 1889 and is often referred to as Bertrand’s paradox. It represents our initial introduction to a subject commonly referred to as geometrical probability. EXAMPLE 3d Consider a random chord of a circle. What is the probability that the length of the chord will be greater than the side of the equilateral triangle inscribed in that circle? Solution. As stated, the problem is incapable of solution because it is not clear what is meant by a random chord. To give meaning to this phrase, we shall reformulate the problem in two distinct ways. The ﬁrst formulation is as follows: The position of the chord can be determined by its distance from the center of the circle. This distance can vary between 0 and r, the radius of the circle. Now, the length of the chord will be greater than the side of the equilateral triangle inscribed in the circle if the distance from the chord to the center of the circle is less than r/2. Hence, by assuming that a random chord is a chord whose distance D from the center of the circle is uniformly distributed between 0 and r, we see that the probability that the length of the chord is greater than the side of an inscribed equilateral triangle is % / r r/2 1 P D < = = 2 r 2 For our second formulation of the problem, consider an arbitrary chord of the circle; through one end of the chord, draw a tangent. The angle θ between the chord and the tangent, which can vary from 0◦ to 180◦ , determines the position of the chord. (See Figure 5.4.) Furthermore, the length of the chord will be greater than the side of the inscribed equilateral triangle if the angle θ is between 60◦ and 120◦ . Hence, assuming that a random chord is a chord whose angle θ is uniformly distributed between 0◦ and 180◦ , we see that the desired answer in this formulation is P{60 < θ < 120} =

1 120 − 60 = 180 3

Note that random experiments could be performed in such a way that 12 or 13 would be the correct probability. For instance, if a circular disk of radius r is thrown on a table ruled with parallel lines a distance 2r apart, then one and only one of these lines would cross the disk and form a chord. All distances from this chord to the center of the disk would be equally likely, so that the desired probability that the chord’s length will be greater than the side of an inscribed equilateral triangle is 12 . In contrast, if the

A

FIGURE 5.4

198

Chapter 5

Continuous Random Variables

experiment consisted of rotating a needle freely about a point A on the edge (see . Figure 5.4) of the circle, the desired answer would be 13 . 5.4

NORMAL RANDOM VARIABLES We say that X is a normal random variable, or simply that X is normally distributed, with parameters μ and σ 2 if the density of X is given by 1 2 2 f (x) = √ e−(x−μ) /2σ 2π σ

− q < x < q

This density function is a bell-shaped curve that is symmetric about μ. (See Figure 5.5.) .399

–3

–2

–1

0 (a)

1

2

3

.399 ——–

–

2

–

+

+ 2

(b) FIGURE 5.5: Normal density function: (a) μ = 0, σ = 1; (b) arbitrary μ, σ 2 .

The normal distribution was introduced by the French mathematician Abraham DeMoivre in 1733, who used it to approximate probabilities associated with binomial random variables when the binomial parameter n is large. This result was later extended by Laplace and others and is now encompassed in a probability theorem known as the central limit theorem, which is discussed in Chapter 8. The central limit theorem, one of the two most important results in probability theory,† gives a theoretical base to the often noted empirical observation that, in practice, many random phenomena obey, at least approximately, a normal probability distribution. Some examples of random phenomena obeying this behavior are the height of a man, the velocity in any direction of a molecule in gas, and the error made in measuring a physical quantity. To prove that f (x) is indeed a probability density function, we need to show that * q 1 2 2 e−(x−μ) /2σ dx = 1 √ 2πσ −q † The other is the strong law of large numbers.

Section 5.4

Normal Random Variables

199

Making the substitution y = (x − μ)/σ , we see that * q * q 1 1 2 −(x−μ)2 /2σ 2 e dx = √ e−y /2 dy √ 2π σ −q 2π −q Hence, we must show that

*

q −q

Toward this end, let I =

e−y

-q

−y2 /2 dy. −q e

*

2

I = =

q

2 /2

√ 2π

Then

−y2 /2

e *−q q *

dy =

q

−q −q

* dy

e−(y

q −q

e−x

2 +x2 )/2

2 /2

dx

dy dx

We now evaluate the double integral by means of a change of variables to polar coordinates. (That is, let x = r cos θ, y = r sin θ, and dy dx = r dθ dr.) Thus, * q * 2π 2 2 e−r /2 r dθ dr I = 0 0 * q 2 re−r /2 dr = 2π 0

= −2πe−r

2 /2

6q 6 0

= 2π √ Hence, I = 2π, and the result is proved. An important fact about normal random variables is that if X is normally distributed with parameters μ and σ 2 , then Y = aX + b is normally distributed with parameters aμ + b and a2 σ 2 . To prove this statement, suppose that a > 0. (The proof when a < 0 is similar.) Let FY denote the cumulative distribution function of Y. Then FY (x) = P{Y … x} = P{aX + b … x} x − b } = P{X … a x − b = FX ( ) a where FX is the cumulative distribution function of X. By differentiation, the density function of Y is then 1 x − b fX ( ) a a 1 x − b − μ)2 /2σ 2 } =√ exp{−( a 2π aσ 1 =√ exp{−(x − b − aμ)2 /2(aσ )2 } 2π aσ

fY (x) =

which shows that Y is normal with parameters aμ + b and a2 σ 2 .

200

Chapter 5

Continuous Random Variables

An important implication of the preceding result is that if X is normally distributed with parameters μ and σ 2 , then Z = (X − μ)/σ is normally distributed with parameters 0 and 1. Such a random variable is said to be a standard, or a unit, normal random variable. We now show that the parameters μ and σ 2 of a normal random variable represent, respectively, its expected value and variance. EXAMPLE 4a Find E[X] and Var(X) when X is a normal random variable with parameters μ and σ 2 . Solution. Let us start by ﬁnding the mean and variance of the standard normal random variable Z = (X − μ)/σ . We have * q E[Z] = xfZ (x) dx −q * q 1 2 =√ xe−x /2 dx 2π −q 1 2 = − √ e−x /2 |q −q 2π =0 Thus, Var(Z) = E[Z2 ] * q 1 2 x2 e−x /2 dx =√ 2π −q Integration by parts (with u = x and dv = xe−x

2 /2

1 2 Var(Z) = √ (−xe−x /2 |q −q 2π * q 1 2 =√ e−x /2 dx 2π −q =1

) now gives * q 2 + e−x /2 dx) −q

Because X = μ + σ Z, the preceding yields the results E[X] = μ + σ E[Z] = μ and

Var(X) = σ 2 Var(Z) = σ 2

.

It is customary to denote the cumulative distribution function of a standard normal random variable by (x). That is, * x 1 2

(x) = √ e−y /2 dy 2π −q The values of (x) for nonnegative x are given in Table 5.1. For negative values of x,

(x) can be obtained from the relationship

(−x) = 1 − (x)

− q < x < q

(4.1)

Section 5.4

Normal Random Variables

201

TABLE 5.1: AREA (x) UNDER THE STANDARD NORMAL CURVE TO THE LEFT OF X

X

.00

.01

.02

.03

.04

.05

.06

.07

.08

.09

.0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.3 3.4

.5000 .5398 .5793 .6179 .6554 .6915 .7257 .7580 .7881 .8159 .8413 .8643 .8849 .9032 .9192 .9332 .9452 .9554 .9641 .9713 .9772 .9821 .9861 .9893 .9918 .9938 .9953 .9965 .9974 .9981 .9987 .9990 .9993 .9995 .9997

.5040 .5438 .5832 .6217 .6591 .6950 .7291 .7611 .7910 .8186 .8438 .8665 .8869 .9049 .9207 .9345 .9463 .9564 .9649 .9719 .9778 .9826 .9864 .9896 .9920 .9940 .9955 .9966 .9975 .9982 .9987 .9991 .9993 .9995 .9997

.5080 .5478 .5871 .6255 .6628 .6985 .7324 .7642 .7939 .8212 .8461 .8686 .8888 .9066 .9222 .9357 .9474 .9573 .9656 .9726 .9783 .9830 .9868 .9898 .9922 .9941 .9956 .9967 .9976 .9982 .9987 .9991 .9994 .9995 .9997

.5120 .5517 .5910 .6293 .6664 .7019 .7357 .7673 .7967 .8238 .8485 .8708 .8907 .9082 .9236 .9370 .9484 .9582 .9664 .9732 .9788 .9834 .9871 .9901 .9925 .9943 .9957 .9968 .9977 .9983 .9988 .9991 .9994 .9996 .9997

.5160 .5557 .5948 .6331 .6700 .7054 .7389 .7704 .7995 .8264 .8508 .8729 .8925 .9099 .9251 .9382 .9495 .9591 .9671 .9738 .9793 .9838 .9875 .9904 .9927 .9945 .9959 .9969 .9977 .9984 .9988 .9992 .9994 .9996 .9997

.5199 .5596 .5987 .6368 .6736 .7088 .7422 .7734 .8023 .8289 .8531 .8749 .8944 .9115 .9265 .9394 .9505 .9599 .9678 .9744 .9798 .9842 .9878 .9906 .9929 .9946 .9960 .9970 .9978 .9984 .9989 .9992 .9994 .9996 .9997

.5239 .5636 .6026 .6406 .6772 .7123 .7454 .7764 .8051 .8315 .8554 .8770 .8962 .9131 .9279 .9406 .9515 .9608 .9686 .9750 .9803 .9846 .9881 .9909 .9931 .9948 .9961 .9971 .9979 .9985 .9989 .9992 .9994 .9996 .9997

.5279 .5675 .6064 .6443 .6808 .7157 .7486 .7794 .8078 .8340 .8577 .8790 .8980 .9147 .9292 .9418 .9525 .9616 .9693 .9756 .9808 .9850 .9884 .9911 .9932 .9949 .9962 .9972 .9979 .9985 .9989 .9992 .9995 .9996 .9997

.5319 .5714 .6103 .6480 .6844 .7190 .7517 .7823 .8106 .8365 .8599 .8810 .8997 .9162 .9306 .9429 .9535 .9625 .9699 .9761 .9812 .9854 .9887 .9913 .9934 .9951 .9963 .9973 .9980 .9986 .9990 .9993 .9995 .9996 .9997

.5359 .5753 .6141 .6517 .6879 .7224 .7549 .7852 .8133 .8389 .8621 .8830 .9015 .9177 .9319 .9441 .9545 .9633 .9706 .9767 .9817 .9857 .9890 .9916 .9936 .9952 .9964 .9974 .9981 .9986 .9990 .9993 .9995 .9997 .9998

The proof of Equation (4.1), which follows from the symmetry of the standard normal density, is left as an exercise. This equation states that if Z is a standard normal random variable, then P{Z … −x} = P{Z > x}

− q < x < q

Since Z = (X − μ)/σ is a standard normal random variable whenever X is normally distributed with parameters μ and σ 2 , it follows that the distribution function of X can be expressed as a − μ a − μ X − μ … = FX (a) = P{X … a} = P σ σ σ

202

Chapter 5

Continuous Random Variables

EXAMPLE 4b If X is a normal random variable with parameters μ = 3 and σ 2 = 9, ﬁnd (a) P{2 < X < 5}; (b) P{X > 0}; (c) P{|X − 3| > 6}. Solution. (a) %

2 − 3 P{2 < X < 5} = P 3 % 1 =P − < 3 2 = − 3 2 − = 3

/ X − 3 5 − 3 < < 3 3 / 2 Z < 3 1

− 3 1 1 − L .3779 3

(b) % P{X > 0} = P

X − 3 0 − 3 > 3 3

/ = P{Z > −1} = 1 − (−1) = (1) L .8413

(c) P{|X − 3| > 6} = P{X > 9} + P{X < −3} / % / % 9 − 3 X − 3 −3 − 3 X − 3 > < + P =P 3 3 3 3 = P{Z > 2} + P{Z < −2} = 1 − (2) + (−2) = 2[1 − (2)] L .0456

.

EXAMPLE 4c An examination is frequently regarded as being good (in the sense of determining a valid grade spread for those taking it) if the test scores of those taking the examination can be approximated by a normal density function. (In other words, a graph of the frequency of grade scores should have approximately the bell-shaped form of the normal density.) The instructor often uses the test scores to estimate the normal parameters μ and σ 2 and then assigns the letter grade A to those whose test score is greater than μ + σ , B to those whose score is between μ and μ + σ , C to those whose score is between μ − σ and μ, D to those whose score is between μ − 2σ and μ − σ , and F to those getting a score below μ − 2σ . (This strategy is sometimes referred to as grading “on the curve.”) Since

Section 5.4

Normal Random Variables

203

/ X − μ > 1 = 1 − (1) L .1587 P{X > μ + σ } = P σ % / X − μ P{μ < X < μ + σ } = P 0 < < 1 = (1) − (0) L .3413 σ % / X − μ P{μ − σ < X < μ} = P −1 < < 0 σ = (0) − (−1) L .3413 % / X − μ P{μ − 2σ < X < μ − σ } = P −2 < < −1 σ = (2) − (1) L .1359 % / X − μ P{X < μ − 2σ } = P < −2 = (−2) L .0228 σ %

it follows that approximately 16 percent of the class will receive an A grade on the examination, 34 percent a B grade, 34 percent a C grade, and 14 percent a D grade; 2 percent will fail. . EXAMPLE 4d An expert witness in a paternity suit testiﬁes that the length (in days) of human gestation is approximately normally distributed with parameters μ = 270 and σ 2 = 100. The defendant in the suit is able to prove that he was out of the country during a period that began 290 days before the birth of the child and ended 240 days before the birth. If the defendant was, in fact, the father of the child, what is the probability that the mother could have had the very long or very short gestation indicated by the testimony? Solution. Let X denote the length of the gestation, and assume that the defendant is the father. Then the probability that the birth could occur within the indicated period is P{X > 290orX < 240} = P{X > 290} + P{X < 240} / % / % X − 270 X − 270 > 2 + P < −3 =P 10 10 = 1 − (2) + 1 − (3) L .0241

.

EXAMPLE 4e Suppose that a binary message—either 0 or 1—must be transmitted by wire from location A to location B. However, the data sent over the wire are subject to a channel noise disturbance, so, to reduce the possibility of error, the value 2 is sent over the wire when the message is 1 and the value −2 is sent when the message is 0. If x, x = ;2, is the value sent at location A, then R, the value received at location B, is given by R = x + N, where N is the channel noise disturbance. When the message is received at location B, the receiver decodes it according to the following rule: If R Ú .5, then 1 is concluded. If R < .5, then 0 is concluded.

204

Chapter 5

Continuous Random Variables

Because the channel noise is often normally distributed, we will determine the error probabilities when N is a standard normal random variable. Two types of errors can occur: One is that the message 1 can be incorrectly determined to be 0, and the other is that 0 can be incorrectly determined to be 1. The ﬁrst type of error will occur if the message is 1 and 2 + N < .5, whereas the second will occur if the message is 0 and −2 + N Ú .5. Hence, P{error|message is 1} = P{N < −1.5} = 1 − (1.5) L .0668 and P{error|message is 0} = P{N Ú 2.5} = 1 − (2.5) L .0062 5.4.1

.

The Normal Approximation to the Binomial Distribution

An important result in probability theory known as the DeMoivre–Laplace limit theorem states that when n is large, a binomial random variable with parameters n and p will have approximately the same distribution as a normal random variable with the same mean and variance as the binomial. This result was proved originally for the special case of p = 12 by DeMoivre in 1733 and was then extended to general p by Laplace in 1812. It formally states that if we “standardize” the binomial. by ﬁrst subtracting its mean np and then dividing the result by its standard deviation np(1 − p), then the distribution function of this standardized random variable (which has mean 0 and variance 1) will converge to the standard normal distribution function as n→q. The DeMoivre–Laplace limit theorem If Sn denotes the number of successes that occur when n independent trials, each resulting in a success with probability p, are performed, then, for any a < b, 7 0 Sn − np … b → (b) − (a) P a … . np(1 − p) as n→q. Because the preceding theorem is only a special case of the central limit theorem, which is presented in Chapter 8, we shall not present a proof. Note that we now have two possible approximations to binomial probabilities: the Poisson approximation, which is good when n is large and p is small, and the normal approximation, which can be shown to be quite good when np(1 − p) is large. (See Figure 5.6.) [The normal approximation will, in general, be quite good for values of n satisfying np(1 − p) Ú 10.] EXAMPLE 4f Let X be the number of times that a fair coin that is ﬂipped 40 times lands on heads. Find the probability that X = 20. Use the normal approximation and then compare it with the exact solution.

Section 5.4 (10, 0.7)

0.30

0.15

0.20

0.10

0.15 0.10

0.05

0.05 0

2

4

6

8

10

0.0

0

5

10 x

x 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0.0

205

(20, 0.7)

0.20

0.25

0.0

Normal Random Variables

(30, 0.7)

0

5

10

15 x

20

25

30

0.14 0.12 0.10 0.08 0.06 0.04 0.02 0.0

15

20

(50, 0.7)

0

10

20

30

40

50

x

FIGURE 5.6: The probability mass function of a binomial (n, p) random variable becomes more and more “normal” as n becomes larger and larger.

Solution. To employ the normal approximation, note that because the binomial is a discrete integer-valued random variable, whereas the normal is a continuous random variable, it is best to write P{X = i} as P{i − 1/2 < X < i + 1/2} before applying the normal approximation (this is called the continuity correction). Doing so gives P{X = 20} = P{19.5 … X < 20.5} 0 7 19.5 − 20 X − 20 20.5 − 20 =P < √ < √ √ 10 10 10 0 7 X − 20 L P −.16 < √ < .16 10 L (.16) − (−.16) L .1272 The exact result is

P{X = 20} =

40 20

40 1 L .1254 2

.

EXAMPLE 4g The ideal size of a ﬁrst-year class at a particular college is 150 students. The college, knowing from past experience that, on the average, only 30 percent of those accepted for admission will actually attend, uses a policy of approving the applications of 450 students. Compute the probability that more than 150 ﬁrst-year students attend this college. Solution. If X denotes the number of students that attend, then X is a binomial random variable with parameters n = 450 and p = .3. Using the continuity correction,

206

Chapter 5

Continuous Random Variables

we see that the normal approximation yields 0 7 X − (450)(.3) 150.5 − (450)(.3) P{X Ú 150.5} = P √ Ú √ 450(.3)(.7) 450(.3)(.7) L 1 − (1.59) L .0559 Hence, less than 6 percent of the time do more than 150 of the ﬁrst 450 accepted actually attend. (What independence assumptions have we made?) . EXAMPLE 4h To determine the effectiveness of a certain diet in reducing the amount of cholesterol in the bloodstream, 100 people are put on the diet. After they have been on the diet for a sufﬁcient length of time, their cholesterol count will be taken. The nutritionist running this experiment has decided to endorse the diet if at least 65 percent of the people have a lower cholesterol count after going on the diet. What is the probability that the nutritionist endorses the new diet if, in fact, it has no effect on the cholesterol level? Solution. Let us assume that if the diet has no effect on the cholesterol count, then, strictly by chance, each person’s count will be lower than it was before the diet with probability 12 . Hence, if X is the number of people whose count is lowered, then the probability that the nutritionist will endorse the diet when it actually has no effect on the cholesterol count is 100 100 1 100 = P{X Ú 64.5} i 2 i=65 ⎫ ⎧ ⎪ ⎪ 1 ⎬ ⎨ X − (100)( ) 2 8 Ú 2.9 =P ⎪ ⎪ ⎭ ⎩ 100( 1 )( 1 ) 2 2 L 1 − (2.9) L .0019

.

EXAMPLE 4i Fifty-two percent of the residents of New York City are in favor of outlawing cigarette smoking in publicly owned areas. Approximate the probability that more than 50 percent of a random sample of n people from New York are in favor of this prohibition when (a) n = 11 (b) n = 101 (c) n = 1001 How large would n have to be to make this probability exceed .95? Solution. Let N denote the number of residents of New York City. To answer the preceding question, we must ﬁrst understand that a random sample of size n is a N sample such that the n people were chosen in such a manner that each of the n subsets of n people had the same chance of being the chosen subset. Consequently,

Section 5.4

Normal Random Variables

207

Sn , the number of people in the sample who are in favor of the smoking prohibition, is a hypergeometric random variable. That is, Sn has the same distribution as the number of white balls obtained when n balls are chosen from an urn of N balls, of which .52N are white. But because N and .52N are both large in comparison with the sample size n, it follows from the binomial approximation to the hypergeometric (see Section 4.8.3) that the distribution of Sn is closely approximated by a binomial distribution with parameters n and p = .52. The normal approximation to the binomial distribution then shows that 7 0 .5n − .52n Sn − .52n P{Sn > .5n} = P √ > √ n(.52)(.48) n(.52)(.48) 7 0 √ Sn − .52n =P √ > −.04 n n(.52)(.48) √ L (.04 n) ⎧ ⎨ (.1328) = .5528, if n = 11

(.4020) = .6562, if n = 101 P{Sn > .5n} L ⎩ (1.2665) = .8973, if n = 1001 √ In order for this probability to be at least .95, we would need (.04 n) > .95. Because

(x) is an increasing function and (1.645) = .95, this means that √ .04 n > 1.645 Thus,

or

n Ú 1691.266

That is, the sample size would have to be at least 1692.

.

Historical Notes Concerning the Normal Distribution The normal distribution was introduced by the French mathematician Abraham DeMoivre in 1733. DeMoivre, who used this distribution to approximate probabilities connected with coin tossing, called it the exponential bell-shaped curve. Its usefulness, however, became truly apparent only in 1809, when the famous German mathematician Karl Friedrich Gauss used it as an integral part of his approach to predicting the location of astronomical entities. As a result, it became common after this time to call it the Gaussian distribution. During the mid- to late 19th century, however, most statisticians started to believe that the majority of data sets would have histograms conforming to the Gaussian bell-shaped form. Indeed, it came to be accepted that it was “normal” for any well-behaved data set to follow this curve. As a result, following the lead of the British statistician Karl Pearson, people began referring to the Gaussian curve by calling it simply the normal curve. (A partial explanation as to why so many data sets conform to the normal curve is provided by the central limit theorem, which is presented in Chapter 8.) Abraham DeMoivre (1667–1754) Today there is no shortage of statistical consultants, many of whom ply their trade in the most elegant of settings. However, the ﬁrst of their breed worked, in the early

208

Chapter 5

Continuous Random Variables

years of the 18th century, out of a dark, grubby betting shop in Long Acres, London, known as Slaughter’s Coffee House. He was Abraham DeMoivre, a Protestant refugee from Catholic France, and, for a price, he would compute the probability of gambling bets in all types of games of chance. Although DeMoivre, the discoverer of the normal curve, made his living at the coffee shop, he was a mathematician of recognized abilities. Indeed, he was a member of the Royal Society and was reported to be an intimate of Isaac Newton. Listen to Karl Pearson imagining DeMoivre at work at Slaughter’s Coffee House: “I picture DeMoivre working at a dirty table in the coffee house with a brokendown gambler beside him and Isaac Newton walking through the crowd to his corner to fetch out his friend. It would make a great picture for an inspired artist.” Karl Friedrich Gauss Karl Friedrich Gauss (1777–1855), one of the earliest users of the normal curve, was one of the greatest mathematicians of all time. Listen to the words of the well-known mathematical historian E. T. Bell, as expressed in his 1954 book Men of Mathematics: In a chapter entitled “The Prince of Mathematicians,” he writes, “Archimedes, Newton, and Gauss; these three are in a class by themselves among the great mathematicians, and it is not for ordinary mortals to attempt to rank them in order of merit. All three started tidal waves in both pure and applied mathematics. Archimedes esteemed his pure mathematics more highly than its applications; Newton appears to have found the chief justiﬁcation for his mathematical inventions in the scientiﬁc uses to which he put them; while Gauss declared it was all one to him whether he worked on the pure or on the applied side.”‡ 5.5

EXPONENTIAL RANDOM VARIABLES A continuous random variable whose probability density function is given, for some λ > 0, by % −λx λe if x Ú 0 f (x) = 0 if x < 0 is said to be an exponential random variable (or, more simply, is said to be exponentially distributed) with parameter λ. The cumulative distribution function F(a) of an exponential random variable is given by F(a) = P{X … a} * a = λe−λx dx 0 6a = −e−λx 60 = 1 − e−λa

a Ú 0

-q Note that F(q) = 0 λe−λx dx = 1, as, of course, it must. The parameter λ will now be shown to equal the reciprocal of the expected value. EXAMPLE 5a Let X be an exponential random variable with parameter λ. Calculate (a) E[X] and (b) Var(X).

Section 5.5

Exponential Random Variables

209

Solution. (a) Since the density function is given by % −λx λe x Ú 0 f (x) = 0 x < 0 we obtain, for n > 0,

*

q

n

E[X ] =

xn λe−λx dx

0

Integrating by parts (with λe−λx = dv and u = xn ) yields * q n n −λx q E[X ] = −x e |0 + e−λx nxn−1 dx 0 * n q −λx n−1 =0 + λe x dx λ 0 n = E[X n−1 ] λ Letting n = 1 and then n = 2 gives E[X] = E[X 2 ] =

2 2 E[X] = 2 λ λ

(b) Hence, Var(X) =

1 λ

2 − λ2

2 1 1 = 2 λ λ

Thus, the mean of the exponential is the reciprocal of its parameter λ, and the variance is the mean squared. . In practice, the exponential distribution often arises as the distribution of the amount of time until some speciﬁc event occurs. For instance, the amount of time (starting from now) until an earthquake occurs, or until a new war breaks out, or until a telephone call you receive turns out to be a wrong number are all random variables that tend in practice to have exponential distributions. (For a theoretical explanation of this phenomenon, see Section 4.7.) EXAMPLE 5b Suppose that the length of a phone call in minutes is an exponential random variable 1 . If someone arrives immediately ahead of you at a public with parameter λ = 10 telephone booth, ﬁnd the probability that you will have to wait (a) more than 10 minutes; (b) between 10 and 20 minutes. Solution. Let X denote the length of the call made by the person in the booth. Then the desired probabilities are (a) P{X > 10} = 1 − F(10) = e−1 L .368

210

Chapter 5

Continuous Random Variables

(b) P{10 < X < 20} = F(20) − F(10) = e−1 − e−2 L .233

.

We say that a nonnegative random variable X is memoryless if P{X > s + t | X > t} = P{X > s} for all s, t Ú 0

(5.1)

If we think of X as being the lifetime of some instrument, Equation (5.1) states that the probability that the instrument survives for at least s + t hours, given that it has survived t hours, is the same as the initial probability that it survives for at least s hours. In other words, if the instrument is alive at age t, the distribution of the remaining amount of time that it survives is the same as the original lifetime distribution. (That is, it is as if the instrument does not “remember” that it has already been in use for a time t.) Equation (5.1) is equivalent to P{X > s + t, X > t} = P{X > s} P{X > t} or

P{X > s + t} = P{X > s}P{X > t}

(5.2)

Since Equation (5.2) is satisﬁed when X is exponentially distributed (for e−λ(s+t) = e−λs e−λt ), it follows that exponentially distributed random variables are memoryless. EXAMPLE 5c Consider a post ofﬁce that is staffed by two clerks. Suppose that when Mr. Smith enters the system, he discovers that Ms. Jones is being served by one of the clerks and Mr. Brown by the other. Suppose also that Mr. Smith is told that his service will begin as soon as either Ms. Jones or Mr. Brown leaves. If the amount of time that a clerk spends with a customer is exponentially distributed with parameter λ, what is the probability that, of the three customers, Mr. Smith is the last to leave the post ofﬁce? Solution. The answer is obtained by reasoning as follows: Consider the time at which Mr. Smith ﬁrst ﬁnds a free clerk. At this point, either Ms. Jones or Mr. Brown would have just left, and the other one would still be in service. However, because the exponential is memoryless, it follows that the additional amount of time that this other person (either Ms. Jones or Mr. Brown) would still have to spend in the post ofﬁce is exponentially distributed with parameter λ. That is, it is the same as if service for that person were just starting at this point. Hence, by symmetry, the probability that the remaining person ﬁnishes before Smith leaves must equal 12 . . It turns out that not only is the exponential distribution memoryless, but it is also the unique distribution possessing this property. To see this, suppose that X is memoryless and let F(x) = P{X > x}. Then, by Equation (5.2), F(s + t) = F(s)F(t) That is, F(·) satisﬁes the functional equation g(s + t) = g(s)g(t)

Section 5.5

Exponential Random Variables

211

However, it turns out that the only right continuous solution of this functional equation is† g(x) = e−λx (5.3) and, since a distribution function is always right continuous, we must have F(x) = e−λx

or F(x) = P{X … x} = 1 − e−λx

which shows that X is exponentially distributed. EXAMPLE 5d Suppose that the number of miles that a car can run before its battery wears out is exponentially distributed with an average value of 10,000 miles. If a person desires to take a 5000-mile trip, what is the probability that he or she will be able to complete the trip without having to replace the car battery? What can be said when the distribution is not exponential? Solution. It follows by the memoryless property of the exponential distribution that the remaining lifetime (in thousands of miles) of the battery is exponential with 1 . Hence, the desired probability is parameter λ = 10 P{remaining lifetime > 5} = 1 − F(5) = e−5λ = e−1/2 L .604 However, if the lifetime distribution F is not exponential, then the relevant probability is 1 − F(t + 5) P{lifetime > t + 5|lifetime > t} = 1 − F(t) where t is the number of miles that the battery had been in use prior to the start of the trip. Therefore, if the distribution is not exponential, additional information is needed (namely, the value of t) before the desired probability can be calculated. . A variation of the exponential distribution is the distribution of a random variable that is equally likely to be either positive or negative and whose absolute value is exponentially distributed with parameter λ, λ# Ú 0. Such a random variable is said to have a Laplace distribution,‡ and its density is given by f (x) =

1 −λ|x| λe 2

− q < x < q

† One can prove Equation (5.3) as follows: If g(s + t) = g(s)g(t), then

g

1 1 1 2 =g + = g2 n n n n

and repeating this yields g(m/n) = gm (1/n). Also, g(1) = g

1 1 1 + + ··· + n n n

= gn

1 n

or

g

1 = (g(1))1/n n

Hence, g(m/n) = (g(1))m/n , which, since g is right continuous, implies that g(x) = (g(1))x . 2 Because g(1) = g 12 Ú 0, we obtain g(x) = e−λx , where λ = − log(g(1)). ‡ It also is sometimes called the double exponential random variable.

212

Chapter 5

Continuous Random Variables

Its distribution function is given by ⎧ * x ⎪ ⎪ λx 1 ⎪ ⎪ ⎨ 2 −q λe dx F(x) = * * 0 ⎪ 1 x −λx ⎪ λx 1 ⎪ λe dx + λe dx ⎪ ⎩2 2 0 −q ⎧ ⎨ 1 eλx x < 0 = 2 ⎩ 1 − 1 e−λx x > 0

x < 0 x > 0

2

EXAMPLE 5e Consider again Example 4e, which supposes that a binary message is to be transmitted from A to B, with the value 2 being sent when the message is 1 and −2 when it is 0. However, suppose now that, rather than being a standard normal random variable, the channel noise N is a Laplacian random variable with parameter λ = 1. Suppose again that if R is the value received at location B, then the message is decoded as follows: If R Ú .5, then 1 is concluded. If R < .5, then 0 is concluded. In this case, where the noise is Laplacian with parameter λ = 1, the two types of errors will have probabilities given by P{error|message 1 is sent} = P{N < −1.5} 1 = e−1.5 2 L .1116 P{error|message 0 is sent} = P{N Ú 2.5} 1 = e−2.5 2 L .041 On comparing this with the results of Example 4e, we see that the error probabilities are higher when the noise is Laplacian with λ = 1 than when it is a standard normal variable. 5.5.1

Hazard Rate Functions

Consider a positive continuous random variable X that we interpret as being the lifetime of some item. Let X have distribution function F and density f . The hazard rate (sometimes called the failure rate) function λ(t) of F is deﬁned by λ(t) =

f (t) F(t)

,

where F = 1 − F

To interpret λ(t), suppose that the item has survived for a time t and we desire the probability that it will not survive for an additional time dt. That is, consider P{X ∈ (t, t + dt)|X > t}. Now,

Section 5.5

Exponential Random Variables

213

P{X ∈ (t, t + dt), X > t} P{X > t} P{X ∈ (t, t + dt)} = P{X > t} f (t) dt L F(t)

P{X ∈ (t, t + dt)|X > t} =

Thus, λ(t) represents the conditional probability intensity that a t-unit-old item will fail. Suppose now that the lifetime distribution is exponential. Then, by the memoryless property, it follows that the distribution of remaining life for a t-year-old item is the same as that for a new item. Hence, λ(t) should be constant. In fact, this checks out, since λ(t) =

f (t)

F(t) λe−λt = −λt e =λ

Thus, the failure rate function for the exponential distribution is constant. The parameter λ is often referred to as the rate of the distribution. It turns out that the failure rate function λ(t) uniquely determines the distribution F. To prove this, note that, by deﬁnition, λ(t) =

d dt F(t)

1 − F(t)

Integrating both sides yields * log(1 − F(t)) = −

t

λ(t) dt + k

0

or

0 * 7 t 1 − F(t) = ek exp − λ(t) dt 0

Letting t = 0 shows that k = 0; thus,

0 * 7 t F(t) = 1 − exp − λ(t) dt

(5.4)

0

Hence, a distribution function of a positive continuous random variable can be speciﬁed by giving its hazard rate function. For instance, if a random variable has a linear hazard rate function—that is, if λ(t) = a + bt then its distribution function is given by F(t) = 1 − e−at−bt and differentiation yields its density, namely,

2 /2

214

Chapter 5

Continuous Random Variables

f (t) = (a + bt)e−(at+bt

2 /2)

t Ú 0

When a = 0, the preceding equation is known as the Rayleigh density function. EXAMPLE 5f One often hears that the death rate of a person who smokes is, at each age, twice that of a nonsmoker. What does this mean? Does it mean that a nonsmoker has twice the probability of surviving a given number of years as does a smoker of the same age? Solution. If λs (t) denotes the hazard rate of a smoker of age t and λn (t) that of a nonsmoker of age t, then the statement at issue is equivalent to the statement that λs (t) = 2λn (t) The probability that an A-year-old nonsmoker will survive until age B, A < B, is P{A-year-old nonsmoker reaches age B} = P{nonsmoker’s lifetime > B|nonsmoker’s lifetime > A} 1 − Fnon (B) = 1 − Fnon (A) 7 0 * B λn (t) dt exp − =

0 * exp −

0 * = exp −

0 A

7 λn (t) dt

0 B

from (5.4)

7 λn (t) dt

A

whereas the corresponding probability for a smoker is, by the same reasoning, 0 * P{A-year-old smoker reaches age B} = exp − 0

B

λs (t) dt

A

*

B

= exp −2 ⎡

7 7 λn (t) dt

A

0 * = ⎣exp −

B

7⎤2 λn (t) dt ⎦

A

In other words, for two people of the same age, one of whom is a smoker and the other a nonsmoker, the probability that the smoker survives to any given age is the square (not one-half) of the corresponding probability for a nonsmoker. For 1 , 50 … t … 60, then the probability that a 50-year-old nonsmoker instance, if λn (t) = 30 −1/3 reaches age 60 is e L .7165, whereas the corresponding probability for a smoker is e−2/3 L .5134. .

Section 5.6

5.6

Other Continuous Distributions

215

OTHER CONTINUOUS DISTRIBUTIONS 5.6.1

The Gamma Distribution

A random variable is said to have a gamma distribution with parameters (α, λ), λ > 0, α > 0, if its density function is given by ⎧ ⎪ ⎨ λe−λx (λx)α−1 x Ú 0 f (x) = (α) ⎪ ⎩0 x < 0 where (α), called the gamma function, is deﬁned as * q e−y yα−1 dy (α) = 0

Integration of (α) by parts yields * q 6q 6 (α) = −e−y yα−1 6 + e−y (α − 1)yα−2 dy 0 0 * q −y α−2 e y dy = (α − 1)

(6.1)

0

= (α − 1)(α − 1) For integral values of α, say, α = n, we obtain, by applying Equation (6.1) repeatedly, (n) = (n − 1)(n − 1) = (n − 1)(n − 2)(n − 2) = ··· = (n − 1)(n − 2) · · · 3 · 2(1) Since (1) =

-q 0

e−x dx = 1, it follows that, for integral values of n, (n) = (n − 1)!

When α is a positive integer, say, α = n, the gamma distribution with parameters (α, λ) often arises, in practice as the distribution of the amount of time one has to wait until a total of n events has occurred. More speciﬁcally, if events are occurring randomly and in accordance with the three axioms of Section 4.7, then it turns out that the amount of time one has to wait until a total of n events has occurred will be a gamma random variable with parameters (n, λ). To prove this, let Tn denote the time at which the nth event occurs, and note that Tn is less than or equal to t if and only if the number of events that have occurred by time t is at least n. That is, with N(t) equal to the number of events in [0, t], P{Tn … t} = P{N(t) Ú n} q = P{N(t) = j} =

j=n q −λt e (λt)j j=n

j!

216

Chapter 5

Continuous Random Variables

where the ﬁnal identity follows because the number of events in [0, t] has a Poisson distribution with parameter λt. Differentiation of the preceding now yields the density function of Tn : f (t) = =

q −λt e j(λt)j−1 λ j=n q j=n

=

j!

−

λe−λt (λt)j−1 − (j − 1)!

q λe−λt (λt)j j=n q j=n

j! λe−λt (λt)j j!

λe−λt (λt)n−1 (n − 1)!

Hence, Tn has the gamma distribution with parameters (n, λ). (This distribution is often referred to in the literature as the n-Erlang distribution.) Note that when n = 1, this distribution reduces to the exponential distribution. The gamma distribution with λ = 12 and α = n/2, n a positive integer, is called the χn2 (read “chi-squared”) distribution with n degrees of freedom. The chi-squared distribution often arises in practice as the distribution of the error involved in attempting to hit a target in n-dimensional space when each coordinate error is normally distributed. This distribution will be studied in Chapter 6, where its relation to the normal distribution is detailed. EXAMPLE 6a Let X be a gamma random variable with parameters α and λ. Calculate (a) E[X] and (b) Var(X). Solution. (a)

* q 1 λxe−λx (λx)α−1 dx (α) 0 * q 1 = λe−λx (λx)α dx λ(α) 0 (α + 1) = λ(α) α = by Equation (6.1) λ

E[X] =

(b) By ﬁrst calculating E[X 2 ], we can show that Var(X) = The details are left as an exercise. 5.6.2

α λ2 .

The Weibull Distribution

The Weibull distribution is widely used in engineering practice due to its versatility. It was originally proposed for the interpretation of fatigue data, but now its use has been extended to many other engineering problems. In particular, it is widely used in the ﬁeld of life phenomena as the distribution of the lifetime of some object, especially when the “weakest link” model is appropriate for the object. That is, consider an

Section 5.6

Other Continuous Distributions

217

object consisting of many parts, and suppose that the object experiences death (failure) when any of its parts fail. It has been shown (both theoretically and empirically) that under these conditions a Weibull distribution provides a close approximation to the distribution of the lifetime of the item. The Weibull distribution function has the form ⎧ 0 x … ν ⎪ ⎪ ⎨ 7 0 (6.2) F(x) = x − ν β ⎪ x > ν 1 − exp − ⎪ ⎩ α A random variable whose cumulative distribution function is given by Equation (6.2) is said to be a Weibull random variable with parameters ν, α, and β. Differentiation yields the density: ⎧ 0 x … ν ⎪ ⎪ ⎨ 0 β−1 β 7 f (x) = β x − ν x − ν ⎪ x > ν exp − ⎪ ⎩ α α α 5.6.3

The Cauchy Distribution

A random variable is said to have a Cauchy distribution with parameter θ, −q < θ < q, if its density is given by f (x) =

1 1 π 1 + (x − θ)2

− q < x < q

EXAMPLE 6b Suppose that a narrow-beam ﬂashlight is spun around its center, which is located a unit distance from the x-axis. (See Figure 5.7.) Consider the point X at which the beam intersects the x-axis when the ﬂashlight has stopped spinning. (If the beam is not pointing toward the x-axis, repeat the experiment.) 1

0

X

x-axis

FIGURE 5.7

As indicated in Figure 5.7, the point X is determined by the angle θ between the ﬂashlight and the y-axis, which, from the physical situation, appears to be uniformly distributed between −π/2 and π/2. The distribution function of X is thus given by F(x) = P{X … x} = P{tan θ … x} = P{θ … tan−1 x} 1 1 = + tan−1 x 2 π

218

Chapter 5

Continuous Random Variables

where the last equality follows since θ, being uniform over (−π/2, π/2), has distribution P{θ … a} =

1 a a − (−π/2) = + π 2 π

−

π π < a < 2 2

Hence, the density function of X is given by f (x) =

d 1 F(x) = dx π(1 + x2 )

− q < x < q .

and we see that X has the Cauchy distribution.†

5.6.4 The Beta Distribution A random variable is said to have a beta distribution if its density is given by ⎧ 1 ⎪ ⎨ xa−1 (1 − x)b−1 0 < x < 1 f (x) = B(a, b) ⎪ ⎩0 otherwise where

1

B(a, b) =

xa−1 (1 − x)b−1 dx

0

The beta distribution can be used to model a random phenomenon whose set of possible values is some ﬁnite interval [c, d]—which, by letting c denote the origin and taking d − c as a unit measurement, can be transformed into the interval [0, 1]. When a = b, the beta density is symmetric about 12 , giving more and more weight to regions about 12 as the common value a increases. (See Figure 5.8.) When b > a, the density is skewed to the left (in the sense that smaller values become more likely); and it is skewed to the right when a > b. (See Figure 5.9.) The relationship (a)(b) (6.3) B(a, b) = (a + b) can be shown to exist between

1

B(a, b) =

xa−1 (1 − x)b−1 dx

0

and the gamma function. † That d (tan−1 x) = 1/(1 + x2 ) can be seen as follows: If y = tan−1 x, then tan y = x, so dx

1= or

d d dy d (tan y) = (tan y) = dx dy dx dy

sin y cos y

dy = dx

cos2 y + sin2 y cos2 y

cos2 y 1 dy 1 = = = 2 2 2 2 dx tan y + 1 x + 1 sin y + cos y

dy dx

Section 5.7

The Distribution of a Function of a Random Variable

219

f(x) a = 10

a=

a=3

1– 4

a=1

0

1

1– 2

x

FIGURE 5.8: Beta densities with parameters (a, b) when a = b.

f(x)

a=6

a=

3– 2

x FIGURE 5.9: Beta densities with parameters (a, b) when a/(a + b) = 1/20.

Upon using Equation (6.1) along with the identity (6.3), it is an easy matter to show that if X is a beta random variable with parameters a and b, then E[X] = Var(X) = Remark. 5.7

a a + b ab (a + b)2 (a + b + 1)

A veriﬁcation of Equation (6.3) appears in Example 7c of Chapter 6. .

THE DISTRIBUTION OF A FUNCTION OF A RANDOM VARIABLE Often, we know the probability distribution of a random variable and are interested in determining the distribution of some function of it. For instance, suppose that we know the distribution of X and want to ﬁnd the distribution of g(X). To do so, it is

220

Chapter 5

Continuous Random Variables

necessary to express the event that g(X) … y in terms of X being in some set. We illustrate with the following examples. EXAMPLE 7a Let X be uniformly distributed over (0, 1). We obtain the distribution of the random variable Y, deﬁned by Y = X n , as follows: For 0 … y … 1, FY (y) = P{Y … y} = P{X n … y} = P{X … y1/n } = FX (y1/n ) = y1/n For instance, the density function of Y is given by ⎧ ⎪ ⎨ 1 y1/n−1 0 … y … 1 fY (y) = n ⎪ ⎩0 otherwise

.

EXAMPLE 7b If X is a continuous random variable with probability density fX , then the distribution of Y = X 2 is obtained as follows: For y Ú 0, FY (y) = P{Y … y} = P{X 2 … y} √ √ = P{− y … X … y} √ √ = FX ( y) − FX (− y) Differentiation yields 1 √ √ fY (y) = √ [fX ( y) + fX (− y)] 2 y

.

EXAMPLE 7c If X has a probability density fX , then Y = |X| has a density function that is obtained as follows: For y Ú 0, FY (y) = P{Y … y} = P{|X| … y} = P{−y … X … y} = FX (y) − FX (−y) Hence, on differentiation, we obtain fY (y) = fX (y) + fX (−y)

y Ú 0 .

Section 5.7

The Distribution of a Function of a Random Variable

221

The method employed in Examples 7a through 7c can be used to prove Theorem 7.1. Theorem 7.1. Let X be a continuous random variable having probability density function fX . Suppose that g(x) is a strictly monotonic (increasing or decreasing), differentiable (and thus continuous) function of x. Then the random variable Y deﬁned by Y = g(X) has a probability density function given by ⎧ 6 6 ⎪ ⎨ f [g−1 (y)] 66 d g−1 (y)66 if y = g(x) for some x X 6 dy 6 fY (y) = ⎪ ⎩0 if y Z g(x) for all x where g−1 (y) is deﬁned to equal that value of x such that g(x) = y. We shall prove Theorem 7.1 when g(x) is an increasing function. Proof. Suppose that y = g(x) for some x. Then, with Y = g(X), FY (y) = P{g(X) … y} = P{X … g−1 (y)} = FX (g−1 (y)) Differentiation gives fY (y) = fX (g−1 (y))

d −1 g (y) dy

.

which agrees with Theorem 7.1, since g−1 (y) is nondecreasing, so its derivative is nonnegative. When y Z g(x) for any x, then FY (y) is either 0 or 1, and in either case fY (y) = 0. EXAMPLE 7d Let X be a continuous nonnegative random variable with density function f , and let Y = X n . Find fY , the probability density function of Y. Solution. If g(x) = xn , then

g−1 (y) = y1/n

and d −1 1 {g (y)} = y1/n−1 dy n Hence, from Theorem 7.1, we obtain fY (y) = For n = 2, this gives

1 1/n−1 1/n y f (y ) n

1 √ fY (y) = √ f ( y) 2 y

which (since X Ú 0) is in agreement with the result of Example 7b.

.

222

Chapter 5

Continuous Random Variables

SUMMARY A random variable X is continuous if there is a nonnegative function f , called the probability density function of X, such that, for any set B, * P{X ∈ B} = f (x) dx B

If X is continuous, then its distribution function F will be differentiable and d F(x) = f (x) dx The expected value of a continuous random variable X is deﬁned by * q xf (x) dx E[X] = −q

A useful identity is that, for any function g, * q E[g(X)] = g(x)f (x) dx −q

As in the case of a discrete random variable, the variance of X is deﬁned by Var(X) = E[(X − E[X])2 ] A random variable X is said to be uniform over the interval (a, b) if its probability density function is given by ⎧ ⎪ ⎨ 1 a … x … b f (x) = b − a ⎪ ⎩0 otherwise Its expected value and variance are E[X] =

a + b 2

Var(X) =

(b − a)2 12

A random variable X is said to be normal with parameters μ and σ 2 if its probability density function is given by 1 2 2 f (x) = √ e−(x−μ) /2σ 2π σ It can be shown that

μ = E[X]

− q < x < q

σ 2 = Var(X)

If X is normal with mean μ and variance σ 2 , then Z, deﬁned by Z=

X − μ σ

is normal with mean 0 and variance 1. Such a random variable is said to be a standard normal random variable. Probabilities about X can be expressed in terms of probabilities about the standard normal variable Z, whose probability distribution function can be obtained either from Table 5.1 or from a website.

Summary

223

When n is large, the probability distribution function of a binomial random variable with parameters n and p can be approximated by that of a normal random variable having mean np and variance np(1 − p). A random variable whose probability density function is of the form % −λx x Ú 0 λe f (x) = 0 otherwise is said to be an exponential random variable with parameter λ. Its expected value and variance are, respectively, E[X] =

1 λ

Var(X) =

1 λ2

A key property possessed only by exponential random variables is that they are memoryless, in the sense that, for positive s and t, P{X > s + t|X > t} = P{X > s} If X represents the life of an item, then the memoryless property states that, for any t, the remaining life of a t-year-old item has the same probability distribution as the life of a new item. Thus, one need not remember the age of an item to know its distribution of remaining life. Let X be a nonnegative continuous random variable with distribution function F and density function f . The function λ(t) =

f (t) 1 − F(t)

t Ú 0

is called the hazard rate, or failure rate, function of F. If we interpret X as being the life of an item, then, for small values of dt, λ(t) dt is approximately the probability that a t-unit-old item will fail within an additional time dt. If F is the exponential distribution with parameter λ, then λ(t) = λ t Ú 0 In addition, the exponential is the unique distribution having a constant failure rate. A random variable is said to have a gamma distribution with parameters α and λ if its probability density function is equal to f (x) =

λe−λx (λx)α−1 (α)

x Ú 0

and is 0 otherwise. The quantity (α) is called the gamma function and is deﬁned by * q (α) = e−x xα−1 dx 0

The expected value and variance of a gamma random variable are, respectively, E[X] =

α λ

Var(X) =

α λ2

224

Chapter 5

Continuous Random Variables

A random variable is said to have a beta distribution with parameters (a, b) if its probability density function is equal to f (x) =

1 xa−1 (1 − x)b−1 B(a, b)

0 … x … 1

and is equal to 0 otherwise. The constant B(a, b) is given by * 1 B(a, b) = xa−1 (1 − x)b−1 dx 0

The mean and variance of such a random variable are, respectively, E[X] =

a a + b

Var(X) =

(a +

ab + b + 1)

b)2 (a

PROBLEMS 5.1. Let X be a random variable with probability density function 0 c(1 − x2 ) −1 < x < 1 f (x) = 0 otherwise (a) What is the value of c? (b) What is the cumulative distribution function of X? 5.2. A system consisting of one original unit plus a spare can function for a random amount of time X. If the density of X is given (in units of months) by 0 Cxe−x/2 x > 0 f (x) = 0 x … 0 what is the probability that the system functions for at least 5 months? 5.3. Consider the function 0 C(2x − x3 ) f (x) = 0

0 < x < 52 otherwise

Could f be a probability density function? If so, determine C. Repeat if f (x) were given by 0 C(2x − x2 ) 0 < x < 52 f (x) = 0 otherwise 5.4. The probability density function of X, the lifetime of a certain type of electronic device (measured in hours), is given by

⎧ ⎪ ⎨ 10 f (x) = x2 ⎪ ⎩0

x > 10 x … 10

(a) Find P{X > 20}. (b) What is the cumulative distribution function of X? (c) What is the probability that, of 6 such types of devices, at least 3 will function for at least 15 hours? What assumptions are you making? 5.5. A ﬁlling station is supplied with gasoline once a week. If its weekly volume of sales in thousands of gallons is a random variable with probability density function 0 5(1 − x)4 0 < x < 1 f (x) = 0 otherwise what must the capacity of the tank be so that the probability of the supply’s being exhausted in a given week is .01? 5.6. Compute E[X] ⎧ if X has a density function given by ⎪ ⎨ 1 xe−x/2 x > 0 ; (a) f (x) = 4 ⎪ ⎩0 otherwise 0 c(1 − x2 ) −1 < x < 1 (b) f (x) = ; 0 otherwise ⎧ ⎪ ⎨5 x > 5 (c) f (x) = x2 . ⎪ ⎩0 x … 5

Problems 5.7. The density function of X is given by 0 a + bx2 0 … x … 1 f (x) = 0 otherwise If E[X] = 35 , ﬁnd a and b. 5.8. The lifetime in hours of an electronic tube is a random variable having a probability density function given by x Ú 0 f (x) = xe−x Compute the expected lifetime of such a tube. 5.9. Consider Example 4b of Chapter 4, but now suppose that the seasonal demand is a continuous random variable having probability density function f . Show that the optimal amount to stock is the value s∗ that satisﬁes F(s∗ ) =

5.10.

5.11.

5.12.

5.13.

b b +

where b is net proﬁt per unit sale, is the net loss per unit unsold, and F is the cumulative distribution function of the seasonal demand. Trains headed for destination A arrive at the train station at 15-minute intervals starting at 7 A . M ., whereas trains headed for destination B arrive at 15-minute intervals starting at 7:05 A . M . (a) If a certain passenger arrives at the station at a time uniformly distributed between 7 and 8 A . M . and then gets on the ﬁrst train that arrives, what proportion of time does he or she go to destination A? (b) What if the passenger arrives at a time uniformly distributed between 7:10 and 8:10 A . M .? A point is chosen at random on a line segment of length L. Interpret this statement, and ﬁnd the probability that the ratio of the shorter to the longer segment is less than 14 . A bus travels between the two cities A and B, which are 100 miles apart. If the bus has a breakdown, the distance from the breakdown to city A has a uniform distribution over (0, 100). There is a bus service station in city A, in B, and in the center of the route between A and B. It is suggested that it would be more efﬁcient to have the three stations located 25, 50, and 75 miles, respectively, from A. Do you agree? Why? You arrive at a bus stop at 10 o’clock, knowing that the bus will arrive at some time uniformly distributed between 10 and 10:30. (a) What is the probability that you will have to wait longer than 10 minutes?

225

(b) If, at 10:15, the bus has not yet arrived, what is the probability that you will have to wait at least an additional 10 minutes? 5.14. Let X be a uniform (0, 1) random variable. Compute E[X n ] by using Proposition 2.1, and then check the result by using the deﬁnition of expectation. 5.15. If X is a normal random variable with parameters μ = 10 and σ 2 = 36, compute (a) P{X > 5}; (b) P{4 < X < 16}; (c) P{X < 8}; (d) P{X < 20}; (e) P{X > 16}. 5.16. The annual rainfall (in inches) in a certain region is normally distributed with μ = 40 and σ = 4. What is the probability that, starting with this year, it will take over 10 years before a year occurs having a rainfall of over 50 inches? What assumptions are you making? 5.17. A man aiming at a target receives 10 points if his shot is within 1 inch of the target, 5 points if it is between 1 and 3 inches of the target, and 3 points if it is between 3 and 5 inches of the target. Find the expected number of points scored if the distance from the shot to the target is uniformly distributed between 0 and 10. 5.18. Suppose that X is a normal random variable with mean 5. If P{X > 9} = .2, approximately what is Var(X)? 5.19. Let X be a normal random variable with mean 12 and variance 4. Find the value of c such that P{X > c} = .10. 5.20. If 65 percent of the population of a large community is in favor of a proposed rise in school taxes, approximate the probability that a random sample of 100 people will contain (a) at least 50 who are in favor of the proposition; (b) between 60 and 70 inclusive who are in favor; (c) fewer than 75 in favor. 5.21. Suppose that the height, in inches, of a 25-year-old man is a normal random variable with parameters μ = 71 and σ 2 = 6.25. What percentage of 25year-old men are over 6 feet, 2 inches tall? What percentage of men in the 6-footer club are over 6 feet, 5 inches? 5.22. The width of a slot of a duralumin forging is (in inches) normally distributed with μ = .9000 and σ = .0030. The speciﬁcation limits were given as .9000 ; .0050. (a) What percentage of forgings will be defective?

226

5.23.

5.24.

5.25.

5.26.

5.27.

5.28.

5.29.

5.30.

Chapter 5

Continuous Random Variables

(b) What is the maximum allowable value of σ that will permit no more than 1 in 100 defectives when the widths are normally distributed with μ = .9000 and σ ? One thousand independent rolls of a fair die will be made. Compute an approximation to the probability that the number 6 will appear between 150 and 200 times inclusively. If the number 6 appears exactly 200 times, ﬁnd the probability that the number 5 will appear less than 150 times. The lifetimes of interactive computer chips produced by a certain semiconductor manufacturer are normally distributed with parameters μ = 1.4 * 106 hours and σ = 3 * 105 hours. What is the approximate probability that a batch of 100 chips will contain at least 20 whose lifetimes are less than 1.8 * 106 ? Each item produced by a certain manufacturer is, independently, of acceptable quality with probability .95. Approximate the probability that at most 10 of the next 150 items produced are unacceptable. Two types of coins are produced at a factory: a fair coin and a biased one that comes up heads 55 percent of the time. We have one of these coins, but do not know whether it is a fair coin or a biased one. In order to ascertain which type of coin we have, we shall perform the following statistical test: We shall toss the coin 1000 times. If the coin lands on heads 525 or more times, then we shall conclude that it is a biased coin, whereas if it lands on heads less than 525 times, then we shall conclude that it is a fair coin. If the coin is actually fair, what is the probability that we shall reach a false conclusion? What would it be if the coin were biased? In 10,000 independent tosses of a coin, the coin landed on heads 5800 times. Is it reasonable to assume that the coin is not fair? Explain. Twelve percent of the population is left handed. Approximate the probability that there are at least 20 left-handers in a school of 200 students. State your assumptions. A model for the movement of a stock supposes that if the present price of the stock is s, then, after one period, it will be either us with probability p or ds with probability 1 − p. Assuming that successive movements are independent, approximate the probability that the stock’s price will be up at least 30 percent after the next 1000 periods if u = 1.012, d = 0.990, and p = .52. An image is partitioned into two regions, one white and the other black. A reading taken from a randomly chosen point in the white section will give a reading that is normally distributed with μ = 4 and σ 2 = 4, whereas one taken from a randomly

chosen point in the black region will have a normally distributed reading with parameters (6, 9). A point is randomly chosen on the image and has a reading of 5. If the fraction of the image that is black is α, for what value of α would the probability of making an error be the same, regardless of whether one concluded that the point was in the black region or in the white region? 5.31. (a) A ﬁre station is to be located along a road of length A, A < q. If ﬁres occur at points uniformly chosen on (0, A), where should the station be located so as to minimize the expected distance from the ﬁre? That is, choose a so as to minimize E[|X − a|]

5.32.

5.33.

5.34.

5.35.

when X is uniformly distributed over (0, A). (b) Now suppose that the road is of inﬁnite length—stretching from point 0 outward to q. If the distance of a ﬁre from point 0 is exponentially distributed with rate λ, where should the ﬁre station now be located? That is, we want to minimize E[|X − a|], where X is now exponential with rate λ. The time (in hours) required to repair a machine is an exponentially distributed random variable with parameter λ = 12 . What is (a) the probability that a repair time exceeds 2 hours? (b) the conditional probability that a repair takes at least 10 hours, given that its duration exceeds 9 hours? The number of years a radio functions is exponentially distributed with parameter λ = 18 . If Jones buys a used radio, what is the probability that it will be working after an additional 8 years? Jones ﬁgures that the total number of thousands of miles that an auto can be driven before it would need to be junked is an exponential random vari1 able with parameter 20 . Smith has a used car that he claims has been driven only 10,000 miles. If Jones purchases the car, what is the probability that she would get at least 20,000 additional miles out of it? Repeat under the assumption that the lifetime mileage of the car is not exponentially distributed, but rather is (in thousands of miles) uniformly distributed over (0, 40). The lung cancer hazard rate λ(t) of a t-year-old male smoker is such that λ(t) = .027 + .00025(t − 40)2

t Ú 40

Assuming that a 40-year-old male smoker survives all other hazards, what is the probability that he survives to (a) age 50 and (b) age 60 without contracting lung cancer?

Theoretical Exercises 5.36. Suppose that the life distribution of an item has the hazard rate function λ(t) = t3 , t > 0. What is the probability that (a) the item survives to age 2? (b) the item’s lifetime is between .4 and 1.4? (c) a 1-year-old item will survive to age 2? 5.37. If X is uniformly distributed over (−1, 1), ﬁnd (a) P{|X| > 12 }; (b) the density function of the random variable |X|. 5.38. If Y is uniformly distributed over (0, 5), what is the probability that the roots of the equation 4x2 + 4xY + Y + 2 = 0 are both real?

227

5.39. If X is an exponential random variable with parameter λ = 1, compute the probability density function of the random variable Y deﬁned by Y = log X. 5.40. If X is uniformly distributed over (0, 1), ﬁnd the density function of Y = eX . 5.41. Find the distribution of R = A sin θ , where A is a ﬁxed constant and θ is uniformly distributed on (−π/2, π/2). Such a random variable R arises in the theory of ballistics. If a projectile is ﬁred from the origin at an angle α from the earth with a speed ν, then the point R at which it returns to the earth can be expressed as R = (v2 /g) sin 2α, where g is the gravitational constant, equal to 980 centimeters per second squared.

THEORETICAL EXERCISES 5.1. The speed of a molecule in a uniform gas at equilibrium is a random variable whose probability density function is given by 0 2 −bx2 x Ú 0 f (x) = ax e 0 x < 0 where b = m/2kT and k, T, and m denote, respectively, Boltzmann’s constant, the absolute temperature of the gas, and the mass of the molecule. Evaluate a in terms of b. 5.2. Show that * q * q P{Y > y} dy − P{Y < −y} dy E[Y] = 0

0

Hint: Show that * 0 * q P{Y < −y} dy = − xfY (x) dx 0 −q * q * q P{Y > y} dy = xfY (x) dx 0

0

5.3. Show that if X has density function f , then * q g(x)f (x) dx E[g(X)] =

5.5. Use the result that, for a nonnegative random variable Y, * q

E[Y] =

P{Y > t} dt

0

to show that, for a nonnegative random variable X, * q nxn−1 P{X > x} dx E[X n ] = 0

Hint: Start with

*

E[X n ] =

q

P{X n > t} dt

0

and make the change of variables t = xn . 5.6. Deﬁne a collection of events Ea , 0 < a < 1, having the property that P(Ea ) = 1 for all a but Ea = 0. P a

Hint: Let X be uniform over (0, 1) and deﬁne each Ea in terms of X. 5.7. The standard deviation of X, denoted SD(X), is given by . SD(X) = Var(X)

−q

Find SD(aX + b) if X has variance σ 2 . 5.8. Let X be a random variable that takes on values Hint: Using Theoretical Exercise 2, start with between 0 and c. That is, P{0 … X … c} = 1. * q * q Show that E[g(X)] = P{g(X) > y} dy − P{g(X) < −y} dy c2 0 0 Var(X) … 4 and then proceed as in the proof given in the text Hint: One approach is to ﬁrst argue that when g(X) Ú 0.

5.4. Prove Corollary 2.1.

E[X 2 ] … cE[X]

228

Chapter 5

Continuous Random Variables

and then use this inequality to show that Var(X) … c2 [α(1 − α)]

where α =

E[X] c

5.9. Show that Z is a standard normal random variable, then, for x > 0, (a) P{Z > x} = P{Z < −x}; (b) P{|Z| > x} = 2P{Z > x}; (c) P{|Z| < x} = 2P{Z < x} − 1. 5.10. Let f (x) denote the probability density function of a normal random variable with mean μ and variance σ 2 . Show that μ − σ and μ + σ are points of inﬂection of this function. That is, show that f (x) = 0 when x = μ − σ or x = μ + σ . 5.11. Let Z be a standard normal random variable Z, and let g be a differentiable function with derivative g . (a) Show that E[g (Z)] = E[Zg(Z)] (b) Show that E[Zn+1 ] = nE[Zn−1 ] (c) Find E[Z4 ]. 5.12. Use the identity of Theoretical Exercise 5 to derive E[X 2 ] when X is an exponential random variable with parameter λ. 5.13. The median of a continuous random variable having distribution function F is that value m such that F(m) = 12 . That is, a random variable is just as likely to be larger than its median as it is to be smaller. Find the median of X if X is (a) uniformly distributed over (a, b); (b) normal with parameters μ, σ 2 ; (c) exponential with rate λ. 5.14. The mode of a continuous random variable having density f is the value of x for which f (x) attains its maximum. Compute the mode of X in cases (a), (b), and (c) of Theoretical Exercise 5.13. 5.15. If X is an exponential random variable with parameter λ, and c > 0, show that cX is exponential with parameter λ/c. 5.16. Compute the hazard rate function of X when X is uniformly distributed over (0, a). 5.17. If X has hazard rate function λX (t), compute the hazard rate function of aX where a is a positive constant. 5.18. Verify that the gamma density function integrates to 1. 5.19. If X is an exponential random variable with mean 1/λ, show that E[X k ] =

k! λk

k = 1, 2, . . .

Hint: Make use of the gamma density function to evaluate the preceding. 5.20. Verify that α Var(X) = 2 λ

5.21.

5.22.

5.23.

5.24.

5.25.

when X is a gamma random variable with parameters α and λ. √ Show that 12 = π. -q Hint: 12 = 0 e−x x−1/2 dx. Make the change √ of variables y = 2x and then relate the resulting expression to the normal distribution. Compute the hazard rate function of a gamma random variable with parameters (α, λ) and show it is increasing when α Ú 1 and decreasing when α … 1. Compute the hazard rate function of a Weibull random variable and show it is increasing when β Ú 1 and decreasing when β … 1. Show that a plot of log(log(1 − F(x))−1 ) against log x will be a straight line with slope β when F(·) is a Weibull distribution function. Show also that approximately 63.2 percent of all observations from such a distribution will be less than α. Assume that v = 0. Let X − ν β Y= α

Show that if X is a Weibull random variable with parameters ν, α, and β, then Y is an exponential random variable with parameter λ = 1 and vice versa. 5.26. If X is a beta random variable with parameters a and b, show that E[X] = Var(X) =

a a + b ab (a + b)2 (a + b + 1)

5.27. If X is uniformly distributed over (a, b), what random variable, having a linear relation with X, is uniformly distributed over (0, 1)? 5.28. Consider the beta distribution with parameters (a, b). Show that (a) when a > 1 and b > 1, the density is unimodal (that is, it has a unique mode) with mode equal to (a − 1)/(a + b − 2); (b) when a … 1, b … 1, and a + b < 2, the density is either unimodal with mode at 0 or 1 or U-shaped with modes at both 0 and 1; (c) when a = 1 = b, all points in [0, 1] are modes. 5.29. Let X be a continuous random variable having cumulative distribution function F. Deﬁne the random variable Y by Y = F(X). Show that Y is uniformly distributed over (0, 1). 5.30. Let X have probability density fX . Find the probability density function of the random variable Y deﬁned by Y = aX + b.

Self-Test Problems and Exercises 5.31. Find the probability density function of Y = eX when X is normally distributed with parameters μ and σ 2 . The random variable Y is said to have a lognormal distribution (since log Y has a normal distribution) with parameters μ and σ 2 . 5.32. Let X and Y be independent random variables that are both equally likely to be either 1, 2, . . . , (10)N , where N is very large. Let D denote the greatest common divisor of X and Y, and let Qk = P{D = k}. (a) Give a heuristic argument that Qk = k12 Q1 . Hint: Note that in order for D to equal k, k must divide both X and Y and also X/k, and Y/k must be relatively prime. (That is, X/k, and Y/k must have a greatest common divisor equal to 1.) (b) Use part (a) to show that

1/k2 =

1

π 2 /6, so Q1 = 6/π 2 . (In number theory, this is known as the Legendre theorem.) (c) Now argue that q Pi2 − 1 Q1 = Pi2 i=1 where Pi is the ith-smallest prime greater than 1. Hint: X and Y will be relatively prime if they have no common prime factors. Hence, from part (b), we see that q Pi2 − 1 6 = 2 2 π Pi which was noted without explanation in Problem 11 of Chapter 4. (The relationship between this problem and Problem 11 of Chapter 4 is that X and Y are relatively prime if XY has no multiple prime factors.) 5.33. Prove Theorem 7.1 when g(x) is a decreasing function.

1 q

q

i=1

Q1 = P{X and Y are relatively prime} =

It is a well-known identity that

229

1/k2

k=1

SELF-TEST PROBLEMS AND EXERCISES 5.1. The number of minutes of playing time of a certain high school basketball player in a randomly chosen game is a random variable whose probability density function is given in the following ﬁgure: .050 .025 10

20

30

40

Find the probability that the player plays (a) over 15 minutes; (b) between 20 and 35 minutes; (c) less than 30 minutes; (d) more than 36 minutes. 5.2. For some constant c, the random variable X has the probability density function % n cx 0 < x < 1 f (x) = 0 otherwise Find (a) c and (b) P{X > x}, 0 < x < 1.

5.3. For some constant c, the random variable X has the probability density function 0 cx4 0 < x < 2 f (x) = 0 otherwise Find (a) E[X] and (b) Var(X). 5.4. The random variable X has the probability density function 0 ax + bx2 0 < x < 1 f (x) = 0 otherwise If E[X] = .6, ﬁnd (a) P{X < 12 } and (b) Var(X). 5.5. The random variable X is said to be a discrete uniform random variable on the integers 1, 2, . . . , n if P{X = i} =

1 n

i = 1, 2, . . . , n

For any nonnegative real number x, let Int(x) (sometimes written as [x]) be the largest integer that is less than or equal to x. Show that U is a uniform random variable on (0, 1), fain

230

5.6.

5.7.

5.8.

5.9.

5.10.

5.11.

Chapter 5

Continuous Random Variables

X = Int(nU) + 1 is a discrete uniform random variable on 1, . . . , n. Your company must make a sealed bid for a construction project. If you succeed in winning the contract (by having the lowest bid), then you plan to pay another ﬁrm 100 thousand dollars to do the work. If you believe that the minimum bid (in thousands of dollars) of the other participating companies can be modeled as the value of a random variable that is uniformly distributed on (70, 140), how much should you bid to maximize your expected proﬁt? To be a winner in a certain game, you must be successful in three successive rounds. The game depends on the value of U, a uniform random variable on (0, 1). If U > .1, then you are successful in round 1; if U > .2, then you are successful in round 2; and if U > .3, then you are successful in round 3. (a) Find the probability that you are successful in round 1. (b) Find the conditional probability that you are successful in round 2 given that you were successful in round 1. (c) Find the conditional probability that you are successful in round 3 given that you were successful in rounds 1 and 2. (d) Find the probability that you are a winner. A randomly chosen IQ test taker obtains a score that is approximately a normal random variable with mean 100 and standard deviation 15. What is the probability that the score of such a person is (a) above 125; (b) between 90 and 110? Suppose that the travel time from your home to your ofﬁce is normally distributed with mean 40 minutes and standard deviation 7 minutes. If you want to be 95 percent certain that you will not be late for an ofﬁce appointment at 1 P . M ., what is the latest time that you should leave home? The life of a certain type of automobile tire is normally distributed with mean 34,000 miles and standard deviation 4000 miles. (a) What is the probability that such a tire lasts over 40,000 miles? (b) What is the probability that it lasts between 30,000 and 35,000 miles? (c) Given that it has survived 30,000 miles, what is the conditional probability that the tire survives another 10,000 miles? The annual rainfall in Cleveland, Ohio is approximately a normal random variable with mean 40.2 inches and standard deviation 8.4 inches. What is the probability that (a) next year’s rainfall will exceed 44 inches? (b) the yearly rainfalls in exactly 3 of the next 7 years will exceed 44 inches?

Assume that if Ai is the event that the rainfall exceeds 44 inches in year i (from now), then the events Ai , i Ú 1, are independent. 5.12. The following table uses 1992 data concerning the percentages of male and female full-time workers whose annual salaries fall into different ranges: Earnings range

Percentage of females

…9999 10,000–19,999 20,000–24,999 25,000–49,999 Ú50,000

Percentage of males

8.6 38.0 19.4 29.2 4.8

4.4 21.1 15.8 41.5 17.2

Suppose that random samples of 200 male and 200 female full-time workers are chosen. Approximate the probability that (a) at least 70 of the women earn $25,000 or more; (b) at most 60 percent of the men earn $25,000 or more; (c) at least three-fourths of the men and at least half the women earn $20,000 or more. 5.13. At a certain bank, the amount of time that a customer spends being served by a teller is an exponential random variable with mean 5 minutes. If there is a customer in service when you enter the bank, what is the probability that he or she will still be with the teller after an additional 4 minutes? 5.14. Suppose that the cumulative distribution function of the random variable X is given by F(x) = 1 − e−x

2

x > 0

Evaluate (a) P{X > 2}; (b) P{1 < X < 3}; (c) the hazard rate function of F; (d) E[X]; (e) Var(X). Hint: For parts (d) and (e), you might want to make use of the results of Theoretical Exercise 5. 5.15. The number of years that a washing machine functions is a random variable whose hazard rate function is given by ⎧ 0 < t < 2 ⎨ .2 λ(t) = .2 + .3(t − 2) 2 … t < 5 ⎩ 1.1 t > 5 (a) What is the probability that the machine will still be working 6 years after being purchased? (b) If it is still working 6 years after being purchased, what is the conditional probability that it will fail within the next 2 years? 5.16. A standard Cauchy random variable has density function f (x) =

1 π(1 + x2 )

− q < x < q

Self-Test Problems and Exercises Show that X is a standard Cauchy random variable, then 1/X is also a standard Cauchy random variable. 5.17. A roulette wheel has 38 slots, numbered 0, 00, and 1 through 36. If you bet 1 on a speciﬁed number then you either win 35 if the roulette ball lands on that number or lose 1 if it does not. If you continually make such bets, approximate the probability that (a) you are winning after 34 bets; (b) you are winning after 1000 bets; (c) you are winning after 100,000 bets. Assume that each roll of the roulette ball is equally likely to land on any of the 38 numbers. 5.18. There are two types of batteries in a bin. When in use, type i batteries last (in hours) an exponentially distributed time with rate λi , i = 1, 2. A battery that is randomly chosen from the bin will be a type 2 i battery with probability pi , pi = 1. If a rani=1

domly chosen battery is still operating after t hours of use, what is the probability that it will still be operating after an additional s hours? 5.19. Evidence concerning the guilt or innocence of a defendant in a criminal investigation can be summarized by the value of an exponential random

231

variable X whose mean μ depends on whether the defendant is guilty. If innocent, μ = 1; if guilty, μ = 2. The deciding judge will rule the defendant guilty if X > c for some suitably chosen value of c. (a) If the judge wants to be 95 percent certain that an innocent man will not be convicted, what should be the value of c? (b) Using the value of c found in part (a), what is the probability that a guilty defendant will be convicted? 5.20. For any real number y, deﬁne y+ by y+ =

y, if y Ú 0 0, if y < 0

Let c be a constant. (a) Show that 1 2 E[(Z − c)+ ] = √ e−c /2 − c(1 − (c)) 2π when Z is a standard normal random variable. (b) Find E[(X − c)+ ] when X is normal with mean μ and variance σ 2 .

C H A P T E R

6

Jointly Distributed Random Variables 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8

JOINT DISTRIBUTION FUNCTIONS INDEPENDENT RANDOM VARIABLES SUMS OF INDEPENDENT RANDOM VARIABLES CONDITIONAL DISTRIBUTIONS: DISCRETE CASE CONDITIONAL DISTRIBUTIONS: CONTINUOUS CASE ORDER STATISTICS JOINT PROBABILITY DISTRIBUTION OF FUNCTIONS OF RANDOM VARIABLES EXCHANGEABLE RANDOM VARIABLES

6.1

JOINT DISTRIBUTION FUNCTIONS Thus far, we have concerned ourselves only with probability distributions for single random variables. However, we are often interested in probability statements concerning two or more random variables. In order to deal with such probabilities, we deﬁne, for any two random variables X and Y, the joint cumulative probability distribution function of X and Y by F(a, b) = P{X … a, Y … b}

− q < a, b < q

The distribution of X can be obtained from the joint distribution of X and Y as follows: FX (a) = P{X … a} = P{X … a, Y < q} = P lim {X … a, Y … b} b→q = lim P{X … a, Y … b} b→q = lim F(a, b) b→q K F(a, q) Note that, in the preceding set of equalities, we have once again made use of the fact that probability is a continuous set (that is, event) function. Similarly, the cumulative distribution function of Y is given by FY (b) = P{Y … b} = lim F(a, b) a→q K F(q, b) 232

Section 6.1

Joint Distribution Functions

233

The distribution functions FX and FY are sometimes referred to as the marginal distributions of X and Y. All joint probability statements about X and Y can, in theory, be answered in terms of their joint distribution function. For instance, suppose we wanted to compute the joint probability that X is greater than a and Y is greater than b. This could be done as follows: P{X > a, Y > b} = 1 =1 =1 =1 =1

− − − − −

P({X > a, Y > b}c ) P({X > a}c ∪ {Y > b}c ) P({X … a} ∪ {Y … b}) (1.1) [P{X … a} + P{Y … b} − P{X … a, Y … b}] FX (a) − FY (b) + F(a, b)

Equation (1.1) is a special case of the following equation, whose veriﬁcation is left as an exercise: P{a1 < X … a2 , b1 < Y … b2 } = F(a2 , b2 ) + F(a1 , b1 ) − F(a1 , b2 ) − F(a2 , b1 )

(1.2)

whenever a1 < a2 , b1 < b2 . In the case when X and Y are both discrete random variables, it is convenient to deﬁne the joint probability mass function of X and Y by p(x, y) = P{X = x, Y = y} The probability mass function of X can be obtained from p(x, y) by pX (x) = P{X = x} = p(x, y) y:p(x,y)>0

Similarly, pY (y) =

p(x, y)

x:p(x,y)>0

EXAMPLE 1a Suppose that 3 balls are randomly selected from an urn containing 3 red, 4 white, and 5 blue balls. If we let X and Y denote, respectively, the number of red and white balls chosen, then the joint probability mass function of X and Y, p(i, j) = P{X = i, Y = j}, is given by 10 12 5 = p(0, 0) = 3 3 220 40 12 5 4 = p(0, 1) = 3 2 1 220 30 12 5 4 = p(0, 2) = 3 1 2 220 4 12 4 = p(0, 3) = 3 3 220

234

Chapter 6

Jointly Distributed Random Variables

30 12 p(1, 0) = = 3 220 60 12 5 4 3 = p(1, 1) = 3 1 1 1 220 18 12 4 3 = p(1, 2) = 3 2 1 220 15 12 5 3 = p(2, 0) = 3 1 2 220 12 12 4 3 = p(2, 1) = 3 1 2 220 1 12 3 = p(3, 0) = 3 3 220 3 1

5 2

These probabilities can most easily be expressed in tabular form, as in Table 6.1. The reader should note that the probability mass function of X is obtained by computing the row sums, whereas the probability mass function of Y is obtained by computing the column sums. Because the individual probability mass functions of X and Y thus appear in the margin of such a table, they are often referred to as the marginal probability mass functions of X and Y, respectively. . TABLE 6.1: P{X = i, Y = j}

j

Row sum = P{X = i}

i

0

1

2

3

0

10 220

40 220

30 220

4 220

84 220

1

30 220

60 220

18 220

0

108 220

2

15 220

12 220

0

0

27 220

3

1 220

0

0

0

1 220

56 220

112 220

48 220

4 220

Column sum = P{Y = j}

EXAMPLE 1b Suppose that 15 percent of the families in a certain community have no children, 20 percent have 1 child, 35 percent have 2 children, and 30 percent have 3. Suppose further that in each family each child is equally likely (independently) to be a boy or a girl. If a family is chosen at random from this community, then B, the number of boys, and G, the number of girls, in this family will have the joint probability mass function shown in Table 6.2.

Section 6.1

Joint Distribution Functions

235

TABLE 6.2: P{B = i, G = j}

j i

0

1

2

3

Row sum = P{B = i}

0 1 2 3

.15 .10 .0875 .0375

.10 .175 .1125 0

.0875 .1125 0 0

.0375 0 0 0

.3750 .3875 .2000 .0375

.3750

.3875

.2000

.0375

Columnsum = P{G = j}

The probabilities shown in Table 6.2 are obtained as follows: P{B = 0, G = 0} = P{no children} = .15 P{B = 0, G = 1} = P{1 girl and total of 1 child}

1 = P{1 child}P{1 girl|1 child} = (.20) 2 P{B = 0, G = 2} = P{2 girls and total of 2 children} = P{2 children}P{2 girls|2 children} = (.35)

2 1 2

We leave the veriﬁcation of the remaining probabilities in the table to the reader. . We say that X and Y are jointly continuous if there exists a function f (x, y), deﬁned for all real x and y, having the property that, for every set C of pairs of real numbers (that is, C is a set in the two-dimensional plane), ** P{(X, Y) ∈ C} = f (x, y) dx dy (1.3) (x,y)∈C

The function f (x, y) is called the joint probability density function of X and Y. If A and B are any sets of real numbers, then, by deﬁning C = {(x, y) : x ∈ A, y ∈ B}, we see from Equation (1.3) that * * P{X ∈ A, Y ∈ B} = f (x, y) dx dy (1.4) B A

Because F(a, b) = P{X ∈ (−q, a], Y ∈ (−q, b]} * b * a = f (x, y) dx dy −q −q

it follows, upon differentiation, that f (a, b) =

∂2 F(a, b) ∂a ∂b

wherever the partial derivatives are deﬁned. Another interpretation of the joint density function, obtained from Equation (1.4), is

236

Chapter 6

Jointly Distributed Random Variables

* P{a < X < a + da, b < Y < b + db} =

d+db * a+da

b

f (x, y) dx dy

a

L f (a, b) da db when da and db are small and f (x, y) is continuous at a, b. Hence, f (a, b) is a measure of how likely it is that the random vector (X, Y) will be near (a, b). If X and Y are jointly continuous, they are individually continuous, and their probability density functions can be obtained as follows: P{X ∈ A} = P{X ∈ A, Y ∈ (−q, q)} * * q f (x, y) dy dx = A −q * fX (x) dx = A

*

where fX (x) =

q

f (x, y) dy

−q

is thus the probability density function of X. Similarly, the probability density function of Y is given by * q

fY (y) =

f (x, y) dx

−q

EXAMPLE 1c The joint density function of X and Y is given by 0 0 < x < q, 0 < y < q 2e−x e−2y f (x, y) = 0 otherwise Compute (a) P{X > 1, Y < 1}, (b) P{X < Y}, and (c) P{X < a}. Solution. (a) 1* q

* P{X > 1, Y < 1} = *

0

2e−x e−2y dx dy

1

6q 2e−2y −e−x 61 dy 0 * 1 = e−1 2e−2y dy =

=e (b)

**

−1

(x,y):x

0

(1 − e−2 )

2e−x e−2y dx dy

P{X < Y} = =

1

0

2e−x e−2y dx dy

Section 6.1

*

Joint Distribution Functions

237

q

2e−2y (1 − e−y )dy * q * q −2y 2e dy − 2e−3y dy = =

0

0

0

2 =1 − 3 1 = 3 (c)

a* q

* P{X < a} = =

*0 a

2e−2y e−x dy dx

0 −x

e

dx

0

= 1 − e−a

.

EXAMPLE 1d Consider a circle of radius R, and suppose that a point within the circle is randomly chosen in such a manner that all regions within the circle of equal area are equally likely to contain the point. (In other words, the point is uniformly distributed within the circle.) If we let the center of the circle denote the origin and deﬁne X and Y to be the coordinates of the point chosen (Figure 6.1), then, since (X, Y) is equally likely to be near each point in the circle, it follows that the joint density function of X and Y is given by 0 c ifx2 + y2 … R2 f (x, y) = 0 ifx2 + y2 > R2 for some value of c. (a) Determine c. (b) Find the marginal density functions of X and Y. (c) Compute the probability that D, the distance from the origin of the point selected, is less than or equal to a. (d) Find E [D]. y

R (X, Y) (0, 0)

x

FIGURE 6.1: Joint probability distribution.

238

Chapter 6

Jointly Distributed Random Variables

Solution. (a) Because

*

q

*

q

−q −q

f (x, y) dy dx = 1

**

it follows that

dy dx = 1

c x2 +y2 …R2

--

We can evaluate x2 +y2 …R2 dy dx either by using polar coordinates or, more simply, by noting that it represents the area of the circle and is thus equal to πR2 . Hence, 1 c= πR2 (b)

* fX (x) =

q −q

f (x, y) dy *

1 dy πR2 x2 +y2 …R2 * c . 1 = dy, where c = R2 − x2 πR2 −c 2 . 2 = R − x2 x2 … R2 πR2 =

and it equals 0 when x2 > R2 . By symmetry, the marginal density of Y is given by 8 2 R2 − y2 y2 … R2 fY (y) = πR2 =0 y2 > R2 . (c) The distribution function of D = X 2 + Y 2 , the distance from the origin, is obtained as follows: For 0 … a … R, . FD (a) = P{ X 2 + Y 2 … a} = P{X 2 + Y 2 … a2 } ** = f (x, y) dy dx x2 +y2 …a2

1 = πR2 =

** dy dx

x2 +y2 …a2

πa2

πR2 a2 = 2 R -where we have used the fact that x2 +y2 …a2 dy dx is the area of a circle of radius a and thus is equal to πa2 .

Section 6.1

Joint Distribution Functions

239

(d) From part (c), the density function of D is fD (a) =

2a R2

Hence, 2 E[D] = 2 R

0 … a … R

*

R

a2 da =

0

2R 3

.

EXAMPLE 1e The joint density of X and Y is given by % −(x+y) 0 < x < q, 0 < y < q e f (x, y) = 0 otherwise Find the density function of the random variable X/Y. Solution. We start by computing the distribution function of X/Y. For a > 0, / % X … a FX/Y (a) = P Y ** = e−(x+y) dx dy x/y…a q * ay

* = =

*0 q

e−(x+y) dx dy

0

(1 − e−ay )e−y dy

0

0 =

−e−y

e−(a+1)y + a + 1

766q 6 6 6 6 0

1 =1 − a + 1 Differentiation shows that the density function of X/Y is given by fX/Y (a) = 1/ . (a + 1)2 , 0 < a < q. We can also deﬁne joint probability distributions for n random variables in exactly the same manner as we did for n = 2. For instance, the joint cumulative probability distribution function F(a1 , a2 , . . . , an ) of the n random variables X1 , X2 , . . . , Xn is deﬁned by F(a1 , a2 , . . . , an ) = P{X1 … a1 , X2 … a2 , . . . , Xn … an } Further, the n random variables are said to be jointly continuous if there exists a function f (x1 , x2 , . . . , xn ), called the joint probability density function, such that, for any set C in n-space, ** * · · · f (x1 , . . . , xn )dx1 dx2 · · · dxn P{(X1 , X2 , . . . , Xn ) ∈ C} = (x1 ,...,xn )∈C

240

Chapter 6

Jointly Distributed Random Variables

In particular, for any n sets of real numbers A1 , A2 , . . . , An , P{X1 ∈ A1 , X2 , ∈ A2 , . . . , Xn ∈ An } * * * = ··· f (x1 , . . . , xn ) dx1 dx2 · · · dxn An

An−1

A1

EXAMPLE 1f The multinomial distribution One of the most important joint distributions is the multinomial distribution, which arises when a sequence of n independent and identical experiments is performed. Suppose that each experiment can result in any one of r possible outcomes, with r pi = 1. If we let Xi denote the number of the respective probabilities p1 , p2 , . . . , pr , i=1

n experiments that result in outcome number i, then P{X1 = n1 , X2 = n2 , . . . , Xr = nr } = whenever

r

n! n n p 1 p 2 · · · pnr r n1 !n2 ! · · · nr ! 1 2

(1.5)

ni = n.

i=1

Equation (1.5) is veriﬁed by noting that any sequence of outcomes for the n experiments that leads to outcome i occurring ni times for i = 1, 2, . . . , r will, by the assumed n n independence of experiments, have probability p1 1 p2 2 . . . pnr r of occurring. Because there are n!/(n1 !n2 ! . . . nr !) such sequences of outcomes (there are n!/n1 ! . . . nr ! different permutations of n things of which n1 are alike, n2 are alike, . . . , nr are alike), Equation (1.5) is established. The joint distribution whose joint probability mass function is speciﬁed by Equation (1.5) is called the multinomial distribution. Note that when r = 2, the multinomial reduces to the binomial distribution. Note also that any sum of a ﬁxed set of the Xi s will have a binomial distribution. binomial random variable with That is, if N ( {1, 2, . . . , r}, then i∈N Xi will be a parameters n and p = i∈N pi . This follows because i∈N Xi represents the number of the n experiments whose outcome is in N, and each experiment will independently have such an outcome with probability i∈N pi . As an application of the multinomial distribution, suppose that a fair die is rolled 9 times. The probability that 1 appears three times, 2 and 3 twice each, 4 and 5 once each, and 6 not at all is 3 2 2 1 1 0 9 9! 1 1 9! 1 1 1 1 1 = . 3!2!2!1!1!0! 6 6 6 6 6 6 3!2!2! 6 6.2

INDEPENDENT RANDOM VARIABLES The random variables X and Y are said to be independent if, for any two sets of real numbers A and B, P{X ∈ A, Y ∈ B} = P{X ∈ A}P{Y ∈ B}

(2.1)

In other words, X and Y are independent if, for all A and B, the events EA = {X ∈ A} and FB = {Y ∈ B} are independent. It can be shown by using the three axioms of probability that Equation (2.1) will follow if and only if, for all a, b, P{X … a, Y … b} = P{X … a}P{Y … b}

Section 6.2

Independent Random Variables

241

Hence, in terms of the joint distribution function F of X and Y, X and Y are independent if F(a, b) = FX (a)FY (b) for all a, b When X and Y are discrete random variables, the condition of independence (2.1) is equivalent to p(x, y) = pX (x)pY (y) for all x, y (2.2) The equivalence follows because, if Equation (2.1) is satisﬁed, then we obtain Equation (2.2) by letting A and B be, respectively, the one-point sets A = {x} and B = {y}. Furthermore, if Equation (2.2) is valid, then, for any sets A, B, p(x, y) P{X ∈ A, Y ∈ B} = y∈B x∈A

=

pX (x)pY (y)

y∈B x∈A

=

y∈B

pY (y)

pX (x)

x∈A

= P{Y ∈ B}P{X ∈ A} and Equation (2.1) is established. In the jointly continuous case, the condition of independence is equivalent to f (x, y) = fX (x)fY (y)

for all x, y

Thus, loosely speaking, X and Y are independent if knowing the value of one does not change the distribution of the other. Random variables that are not independent are said to be dependent. EXAMPLE 2a Suppose that n + m independent trials having a common probability of success p are performed. If X is the number of successes in the ﬁrst n trials, and Y is the number of successes in the ﬁnal m trials, then X and Y are independent, since knowing the number of successes in the ﬁrst n trials does not affect the distribution of the number of successes in the ﬁnal m trials (by the assumption of independent trials). In fact, for integral x and y, m 0 … x … n, n x n−x py (1 − p)m−y P{X = x, Y = y} = p (1 − p) y 0 … y … m x = P{X = x}P{Y = y} In contrast, X and Z will be dependent, where Z is the total number of successes in the n + m trials. (Why?) . EXAMPLE 2b Suppose that the number of people who enter a post ofﬁce on a given day is a Poisson random variable with parameter λ. Show that if each person who enters the post ofﬁce is a male with probability p and a female with probability 1 − p, then the number of males and females entering the post ofﬁce are independent Poisson random variables with respective parameters λp and λ(1 − p).

242

Chapter 6

Jointly Distributed Random Variables

Solution. Let X and Y denote, respectively, the number of males and females that enter the post ofﬁce. We shall show the independence of X and Y by establishing Equation (2.2). To obtain an expression for P{X = i, Y = j}, we condition on X + Y as follows: P{X = i, Y = j} = P{X = i, Y = j|X + Y = i + j}P{X + Y = i + j} + P{X = i, Y = j|X + Y Z i + j}P{X + Y Z i + j} [Note that this equation is merely a special case of the formula P(E) = P(E|F)P(F) + P(E|F c )P(F c ).] Since P{X = i, Y = j|X + Y Z i + j} is clearly 0, we obtain P{X = i, Y = j} = P{X = i, Y = j|X + Y = i + j}P{X + Y = i + j}

(2.3)

Now, because X + Y is the total number of people who enter the post ofﬁce, it follows, by assumption, that P{X + Y = i + j} = e−λ

λi+j (i + j)!

(2.4)

Furthermore, given that i + j people do enter the post ofﬁce, since each person entering will be male with probability p, it follows that the probability that exactly iof them will be male (and thus j of them female) is just the binomial probability i + j pi (1 − p)j . That is, i P{X = i, Y = j|X + Y = i + j} =

i + j i

pi (1 − p)j

(2.5)

Substituting Equations (2.4) and (2.5) into Equation (2.3) yields P{X = i, Y = j} =

i + j i

pi (1 − p)j e−λ

λi+j (i + j)!

(λp)i [λ(1 − p)]j i!j! e−λp (λp)i −λ(1−p) [λ(1 − p)]j e = i! j! = e−λ

(2.6)

Hence, P{X = i} = e−λp

(λp)i −λ(1−p) [λ(1 − p)]j (λp)i = e−λp e i! j! i!

(2.7)

j

and similarly, P{Y = j} = e−λ(1−p)

[λ(1 − p)]j j!

Equations (2.6), (2.7), and (2.8) establish the desired result.

(2.8) .

Section 6.2

Independent Random Variables

243

EXAMPLE 2c A man and a woman decide to meet at a certain location. If each of them independently arrives at a time uniformly distributed between 12 noon and 1 P . M ., ﬁnd the probability that the ﬁrst to arrive has to wait longer than 10 minutes. Solution. If we let X and Y denote, respectively, the time past 12 that the man and the woman arrive, then X and Y are independent random variables, each of which is uniformly distributed over (0, 60). The desired probability, P{X + 10 < Y} + P{Y + 10 < X}, which, by symmetry, equals 2P{X + 10 < Y}, is obtained as follows: ** f (x, y) dx dy 2P{X + 10 < Y} = 2 x+10

**

=2

fX (x)fY (y) dx dy x+10

* =2

60 * y−10

10

2 = (60)2 25 = 36

*

0 60

1 60

2 dx dy

(y − 10) dy

10

.

Our next example presents the oldest problem dealing with geometrical probabilities. It was ﬁrst considered and solved by Buffon, a French naturalist of the 18th century, and is usually referred to as Buffon’s needle problem. EXAMPLE 2d Buffon’s needle problem A table is ruled with equidistant parallel lines a distance D apart. A needle of length L, where L … D, is randomly thrown on the table. What is the probability that the needle will intersect one of the lines (the other possibility being that the needle will be completely contained in the strip between two lines)? Solution. Let us determine the position of the needle by specifying (1) the distance X from the middle point of the needle to the nearest parallel line and (2) the angle θ between the needle and the projected line of length X. (See Figure 6.2.) The needle will intersect a line if the hypotenuse of the right triangle in Figure 6.2 is less than L/2—that is, if L L X < or X < cos θ cos θ 2 2

X

FIGURE 6.2

244

Chapter 6

Jointly Distributed Random Variables

As X varies between 0 and D/2 and θ between 0 and π/2, it is reasonable to assume that they are independent, uniformly distributed random variables over these respective ranges. Hence, / % ** L cos θ = fX (x)fθ (y) dx dy P X < 2 x

* π/2 * L/2 cos y 4 dx dy πD 0 0 * π/2 L 4 cos y dy = πD 0 2 2L = πD

=

∗ EXAMPLE

.

2e Characterization of the normal distribution

Let X and Y denote the horizontal and vertical miss distances when a bullet is ﬁred at a target, and assume that 1. X and Y are independent continuous random variables having differentiable density functions. 2. The joint density f (x, y) = fX (x)fY (y) of X and Y depends on (x, y) only through x2 + y2 . Loosely put, assumption 2 states that the probability of the bullet landing on any point of the x–y plane depends only on the distance of the point from the target and not on its angle of orientation. An equivalent way of phrasing this assumption is to say that the joint density function is rotation invariant. It is a rather interesting fact that assumptions 1 and 2 imply that X and Y are normally distributed random variables. To prove this, note ﬁrst that the assumptions yield the relation (2.9) f (x, y) = fX (x)fY (y) = g(x2 + y2 ) for some function g. Differentiating Equation (2.9) with respect to x yields fX (x)fY (y) = 2xg (x2 + y2 )

(2.10)

Dividing Equation (2.10) by Equation (2.9) gives fX (x) 2xg (x2 + y2 ) = fX (x) g(x2 + y2 ) or

fX (x) g (x2 + y2 ) = 2xfX (x) g(x2 + y2 )

(2.11)

Because the value of the left-hand side of Equation (2.11) depends only on x, whereas the value of the right-hand side depends on x2 + y2 , it follows that the left-hand side must be the same for all x. To see this, consider any x1 , x2 and let y1 , y2 be such that x21 + y21 = x22 + y22 . Then, from Equation (2.11), we obtain g (x21 + y21 ) g (x22 + y22 ) fX (x2 ) fX (x1 ) = = = 2x1 fX (x1 ) 2x2 fX (x2 ) g(x21 + y21 ) g(x22 + y22 )

Section 6.2

Hence,

fX (x) = c or xfX (x)

Independent Random Variables

245

d (log fX (x)) = cx dx

which implies, upon integration of both sides, that log fX (x) = a +

cx2 2

or

fX (x) = kecx

2 /2

-q Since −q fX (x) dx = 1, it follows that c is necessarily negative, and we may write c = −1/σ 2 . Thus, 2 2 fX (x) = ke−x /2σ That is, X is a normal random variable with parameters μ = 0 and σ 2 . A similar argument can be applied to fY (y) to show that fY (y) = √

1 2π σ

e−y

2 /2σ 2

Furthermore, it follows from assumption 2 that σ 2 = σ 2 and that X and Y are thus independent, identically distributed normal random variables with parameters μ = 0 and σ 2 . . A necessary and sufﬁcient condition for the random variables X and Y to be independent is for their joint probability density function (or joint probability mass function in the discrete case) f (x, y) to factor into two terms, one depending only on x and the other depending only on y. Proposition 2.1. The continuous (discrete) random variables X and Y are independent if and only if their joint probability density (mass) function can be expressed as fX,Y (x, y) = h(x)g(y)

− q < x < q, −q < y < q

Proof. Let us give the proof in the continuous case. First, note that independence implies that the joint density is the product of the marginal densities of X and Y, so the preceding factorization will hold when the random variables are independent. Now, suppose that fX,Y (x, y) = h(x)g(y) Then

* 1= =

q

*−q q −q

*

q

fX,Y (x, y) dx dy * q h(x) dx g(y) dy −q

−q

= C1 C2 -q -q where C1 = −q h(x) dx and C2 = −q g(y) dy. Also, * q fX,Y (x, y) dy = C2 h(x) fX (x) = *−q q fY (y) = fX,Y (x, y) dx = C1 g(y) −q

246

Chapter 6

Jointly Distributed Random Variables

Since C1 C2 = 1, it follows that fX,Y (x, y) = fX (x)fY (y) and the proof is complete. EXAMPLE 2f If the joint density function of X and Y is f (x, y) = 6e−2x e−3y

0 < x < q, 0 < y < q

and is equal to 0 outside this region, are the random variables independent? What if the joint density function is f (x, y) = 24xy

0 < x < 1, 0 < y < 1, 0 < x + y < 1

and is equal to 0 otherwise? Solution. In the ﬁrst instance, the joint density function factors, and thus the random variables, are independent (with one being exponential with rate 2 and the other exponential with rate 3). In the second instance, because the region in which the joint density is nonzero cannot be expressed in the form x ∈ A, y ∈ B, the joint density does not factor, so the random variables are not independent. This can be seen clearly by letting % 1 if 0 < x < 1, 0 < y < 1, 0 < x + y < 1 I(x, y) = 0 otherwise and writing f (x, y) = 24xy I(x, y) which clearly does not factor into a part depending only on x and another depending only on y. . The concept of independence may, of course, be deﬁned for more than two random variables. In general, the n random variables X1 , X2 , . . . , Xn are said to be independent if, for all sets of real numbers A1 , A2 , . . . , An , P{X1 ∈ A1 , X2 ∈ A2 , . . . , Xn ∈ An } =

n

P{Xi ∈ Ai }

i=1

As before, it can be shown that this condition is equivalent to P{X1 … a1 , X2 … a2 , . . . , Xn … an } n = P{Xi … ai } for all a1 , a2 , . . . , an i=1

Finally, we say that an inﬁnite collection of random variables is independent if every ﬁnite subcollection of them is independent. EXAMPLE 2g How can a computer choose a random subset? Most computers are able to generate the value of, or simulate, a uniform (0, 1) random variable by means of a built-in subroutine that (to a high degree of approximation)

Section 6.2

Independent Random Variables

247

produces such “random numbers.” As a result, it is quite easy for a computer to simulate an indicator (that is, a Bernoulli) random variable. Suppose I is an indicator variable such that P{I = 1} = p = 1 − P{I = 0} The computer can simulate I by choosing a uniform (0, 1) random number U and then letting I=

1 if U < p 0 if U Ú p

Suppose that we are interested in having the computer select k, k … n, of the num n bers 1, 2, . . . , n in such a way that each of the subsets of size k is equally likely k to be chosen. We now present a method that will enable the computer to solve this task. To generate such a subset, we will ﬁrst simulate, in sequence, n indicator variables I1 , I2 , . . . , In , of which exactly k will equal 1. Those i for which Ii = 1 will then constitute the desired subset. To generate the random variables I1 , . . . , In , start by simulating n independent uniform (0, 1) random variables U1 , U2 , . . . , Un . Now deﬁne ⎧ ⎪ k ⎨1 if U1 < I1 = n ⎪ ⎩0 otherwise and then, once I1 , . . . , Ii are determined, recursively set ⎧ ⎪ k − (I1 + · · · + Ii ) ⎨1 if Ui+1 < Ii+1 = n − i ⎪ ⎩0 otherwise In words, at the (i + 1)th stage we set Ii+1 equal to 1 (and thus put i + 1 into the desired equal to the remaining number of places in the sub⎛ subset) with a probability ⎞ i Ij ⎠, divided by the remaining number of possibilities (namely, set ⎝namely, k − j=1

n − i). Hence, the joint distribution of I1 , I2 , . . . , In is determined from P{I1 = 1} =

k n k −

P{Ii+1 = 1|I1 , . . . , Ii } =

i j=1

n − i

Ij 1 < i < n

The proof that the preceding formula results in all subsets of size k being equally likely to be chosen is by induction on k + n. It is immediate when k + n = 2 (that is, when k = 1, n = 1), so assume it to be true whenever k + n … l. Now, suppose that k + n = l + 1, and consider any subset of size k—say, i1 … i2 … · · · … ik —and consider the following two cases.

248

Chapter 6

Jointly Distributed Random Variables

Case 1:

i1 = 1 P{I1 = Ii2 = · · · = Iik = 1, Ij = 0 otherwise} = P{I1 = 1}P{Ii2 = · · · = Iik = 1, Ij = 0 otherwise|I1 = 1}

Now given that I1 = 1, the remaining elements of the subset are chosen as if a subset of size k − 1 were to be chosen from the n − 1 elements 2, 3, . . . , n. Hence, by the induction hypothesis, the conditional that this will result in a given probability n − 1 . Hence, subset of size k − 1 being selected is 1/ k − 1 P{I1 = Ii2 = · · · = Iik = 1, Ij = 0 otherwise} 1 1 k = = n n − 1 n k − 1 k Case 2:

i1 Z 1 P{Ii1 = Ii2 = · · · = Iik = 1, Ij = 0 otherwise} = P{Ii1 = · · · = Iik = 1, Ij = 0 otherwise|I1 = 0}P{I1 = 0} k 1 1 1 − = = n n − 1 n k k

where the induction hypothesis was used to evaluate the preceding conditional probability. Thus, in all the probability that a given subset of size k will be the subset cases, < n . . chosen is 1 k Remark. The foregoing method for generating a random subset has a very low memory requirement. A faster algorithm that requires somewhat more memory is presented in Section 10.1. (The latter algorithm uses the last k elements of a random permutation of 1, 2, . . . , n.) . EXAMPLE 2h Let X, Y, Z be independent and uniformly distributed over (0, 1). Compute P{X Ú YZ}. Solution. Since fX,Y,Z (x, y, z) = fX (x)fY (y)fZ (z) = 1 we have

0 … x … 1, 0 … y … 1, 0 … z … 1

*** fX, Y, Z (x, y, z) dx dy dz

P{X Ú YZ} = xÚyz

* =

1* 1* 1

dx dy dz 0

0

yz

Section 6.2

*

Independent Random Variables

249

1* 1

=

(1 − yz) dy dz * 1 z = dz 1 − 2 0 3 = 4 0

0

.

EXAMPLE 2i Probabilistic interpretation of half-life Let N(t) denote the number of nuclei contained in a radioactive mass of material at time t. The concept of half-life is often deﬁned in a deterministic fashion by stating this it is an empirical fact that, for some value h, called the half-life, N(t) = 2−t/h N(0)

t > 0

[Note that N(h) = N(0)/2.] Since the preceding implies that, for any nonnegative s and t, N(t + s) = 2−(s+t)/h N(0) = 2−t/h N(s) it follows that no matter how much time s has already elapsed, in an additional time t the number of existing nuclei will decrease by the factor 2−t/h . Because the deterministic relationship just given results from observations of radioactive masses containing huge numbers of nuclei, it would seem that it might be consistent with a probabilistic interpretation. The clue to deriving the appropriate probability model for half-life resides in the empirical observation that the proportion of decay in any time interval depends neither on the total number of nuclei at the beginning at the interval nor on the location of this interval (since N(t + s)/N(s) depends neither on N(s) nor on s). Thus, it appears that the individual nuclei act independently and with a memoryless life distribution. Consequently, since the unique life distribution that is memoryless is the exponential distribution, and since exactly onehalf of a given amount of mass decays every h time units, we propose the following probabilistic model for radioactive decay. Probabilistic interpretation of the half-life h: The lifetimes of the individual nuclei are independent random variables having a life distribution that is exponential with median equal to h. That is, if L represents the lifetime of a given nucleus, then P{L < t} = 1 − 2−t/h (Because P{L < h} =

1 2

and the preceding can be written as / % log 2 P{L < t} = 1 − exp −t h

it can be seen that L indeed has an exponential distribution with median h.) Note that, under the probabilistic interpretation of half-life just given, if one starts with N(0) nuclei at time 0, then N(t), the number of nuclei that remain at time t, will have a binomial distribution with parameters n = N(0) and p = 2−t/h . Results of Chapter 8 will show that this interpretation of half-life is consistent with the deterministic model when considering the proportion of a large number of nuclei that decay over a given time frame. However, the difference between the deterministic and probabilistic interpretation becomes apparent when one considers the actual number of

250

Chapter 6

Jointly Distributed Random Variables

decayed nuclei. We will now indicate this with regard to the question of whether protons decay. There is some controversy over whether or not protons decay. Indeed, one theory predicts that protons should decay with a half-life of about h = 1030 years. To check this prediction empirically, it has been suggested that one follow a large number of protons for, say, one or two years and determine whether any of them decay within that period. (Clearly, it would not be feasible to follow a mass of protons for 1030 years to see whether one-half of it decays.) Let us suppose that we are able to keep track of N(0) = 1030 protons for c years. The number of decays predicted by the deterministic model would then be given by N(0) − N(c) = h(1 − 2−c/h ) 1 − 2−c/h 1/h 1 − 2−cx L lim x x→ 0 = lim (c2−cx log 2) x→0 = c log 2 L .6931c =

1 = 10−30 L 0 h ˆ by L’Hopital’s rule since

For instance, the deterministic model predicts that in 2 years there should be 1.3863 decays, and it would thus appear to be a serious blow to the hypothesis that protons decay with a half-life of 1030 years if no decays are observed over those 2 years. Let us now contrast the conclusions just drawn with those obtained from the probabilistic model. Again, let us consider the hypothesis that the half-life of protons is h = 1030 years, and suppose that we follow h protons for c years. Since there is a huge number of independent protons, each of which will have a very small probability of decaying within this time period, it follows that the number of protons which decay will have (to a very strong approximation) a Poisson distribution with parameter equal to h(1 − 2−c/h ) L c log 2. Thus, P{0 decays} = e−c log 2 = e− log(2 ) = c

1 2c

and, in general, P{n decays} =

2−c [c log 2]n n!

n Ú 0

Thus we see that even though the average number of decays over 2 years is (as predicted by the deterministic model) 1.3863, there is 1 chance in 4 that there will not be any decays, thereby indicating that such a result in no way invalidates the original hypothesis of proton decay. . Remark. Independence is a symmetric relation. The random variables X and Y are independent if their joint density function (or mass function in the discrete case) is the product of their individual density (or mass) functions. Therefore, to say that X is independent of Y is equivalent to saying that Y is independent of X—or just that X and Y are independent. As a result, in considering whether X is independent of Y in situations where it is not at all intuitive that knowing the value of Y will not change the probabilities concerning X, it can be beneﬁcial to interchange the roles of

Section 6.2

Independent Random Variables

251

X and Y and ask instead whether Y is independent of X. The next example illustrates this point. . EXAMPLE 2j If the initial throw of the dice in the game of craps results in the sum of the dice equaling 4, then the player will continue to throw the dice until the sum is either 4 or 7. If this sum is 4, then the player wins, and if it is 7, then the player loses. Let N denote the number of throws needed until either 4 or 7 appears, and let X denote the value (either 4 or 7) of the ﬁnal throw. Is N independent of X? That is, does knowing which of 4 or 7 occurs ﬁrst affect the distribution of the number of throws needed until that number appears? Most people do not ﬁnd the answer to this question to be intuitively obvious. However, suppose that we turn it around and ask whether X is independent of N. That is, does knowing how many throws it takes to obtain a sum of either 4 or 7 affect the probability that that sum is equal to 4? For instance, suppose we know that it takes n throws of the dice to obtain a sum of either 4 or 7. Does this affect the probability distribution of the ﬁnal sum? Clearly not, since all that is important is that its value is either 4 or 7, and the fact that none of the ﬁrst n − 1 throws were either 4 or 7 does not change the probabilities for the nth throw. Thus, we can conclude that X is independent of N, or equivalently, that N is independent of X. As another example, let X1 , X2 , . . . be a sequence of independent and identically distributed continuous random variables, and suppose that we observe these random variables in sequence. If Xn > Xi for each i = 1, . . . , n − 1, then we say that Xn is a record value. That is, each random variable that is larger than all those preceding it is called a record value. Let An denote the event that Xn is a record value. Is An+1 independent of An ? That is, does knowing that the nth random variable is the largest of the ﬁrst n change the probability that the (n + 1)st random variable is the largest of the ﬁrst n + 1? While it is true that An+1 is independent of An , this may not be intuitively obvious. However, if we turn the question around and ask whether An is independent of An+1 , then the result is more easily understood. For knowing that the (n + 1)st value is larger than X1 , . . . , Xn clearly gives us no information about the relative size of Xn among the ﬁrst n random variables. Indeed, by symmetry, it is clear that each of these n random variables is equally likely to be the largest of this set, so P(An |An+1 ) = P(An ) = 1/n. Hence, we can conclude that An and An+1 are independent events. . Remark.

It follows from the identity

P{X1 … a1 , . . . , Xn … an } = P{X1 … a1 }P{X2 … a2 |X1 … a1 } · · · P{Xn … an |X1 … a1 , . . . , Xn−1 … an−1 } that the independence of X1 , . . . , Xn can be established sequentially. That is, we can show that these random variables are independent by showing that X2 is independent of X1 X3 is independent of X1 , X2 X4 is independent of X1 , X2 , X3

Xn

# # # is independent of X1 , . . . , Xn−1

252

Chapter 6

Jointly Distributed Random Variables

6.3

SUMS OF INDEPENDENT RANDOM VARIABLES It is often important to be able to calculate the distribution of X + Y from the distributions of X and Y when X and Y are independent. Suppose that X and Y are independent, continuous random variables having probability density functions fX and fY . The cumulative distribution function of X + Y is obtained as follows: FX+Y (a) = P{X + Y … a} ** = fX (x)fY (y) dx dy x+y…a q * a−y

* = = =

−q −q * q * a−y −q *−q q −q

fX (x)fY (y) dx dy

(3.1)

fX (x) dxfY (y) dy

FX (a − y)fY (y) dy

The cumulative distribution function FX+Y is called the convolution of the distributions FX and FY (the cumulative distribution functions of X and Y, respectively). By differentiating Equation (3.1), we ﬁnd that the probability density function fX+Y of X + Y is given by * q d FX (a − y)fY (y) dy fX+Y (a) = da * q −q d = FX (a − y)fY (y) dy (3.2) da *−q q = fX (a − y)fY (y) dy −q

6.3.1

Identically Distributed Uniform Random Variables

It is not difﬁcult to determine the density function of the sum of two independent uniform (0, 1) random variables. EXAMPLE 3a Sum of two independent uniform random variables If X and Y are independent random variables, both uniformly distributed on (0, 1), calculate the probability density of X + Y. Solution. From Equation (3.2), since % fX (a) = fY (a) = we obtain

* fX+Y (a) =

1

1 0 < a < 1 0 otherwise

fX (a − y) dy

0

For 0 … a … 1, this yields

*

a

fX+Y (a) = 0

dy = a

Section 6.3

Sums of Independent Random Variables

253

f(x)

1

0

1

2

x

FIGURE 6.3: Triangular density function.

For 1 < a < 2, we get

* fX+Y (a) =

1

dy = 2 − a

a−1

Hence,

⎧ ⎨a fX+Y (a) = 2 − a ⎩0

0 … a … 1 1 < a < 2 otherwise

Because of the shape of its density function (see Figure 6.3), the random variable X + Y is said to have a triangular distribution. . Now, suppose that X1 , X2 , . . . , Xn are independent uniform (0, 1) random variables, and let Fn (x) = P{X1 + . . . + Xn … x} Whereas a general formula for Fn (x) is messy, it has a particularly nice form when x … 1. Indeed, we now use mathematical induction to prove that Fn (x) = xn /n! ,

0 … x … 1

Because the proceeding equation is true for n = 1, assume that Fn−1 (x) = xn−1 /(n − 1)! ,

0 … x … 1

Now, writing n i=1

Xi =

n−1

Xi + Xn

i=1

and using the fact that the Xi are all nonnegative, we see from Equation 3.1 that, for 0 … x … 1, * 1 Fn−1 (x − y)fXn (y)dy Fn (x) = 0 * x 1 (x − y)n−1 dy by the induction hypothesis = (n − 1)! 0 = xn /n! which completes the proof.

254

Chapter 6

Jointly Distributed Random Variables

For an interesting application of the preceding formula, let us use it to determine the expected number of independent uniform (0, 1) random variables that need to be summed to exceed 1. That is, with X1 , X2 , . . . being independent uniform (0, 1) random variables, we want to determine E[N], where N = min{n : X1 + . . . + Xn > 1} Noting that N is greater than n > 0 if and only if X1 + . . . + Xn … 1, we see that P{N > n} = Fn (1) = 1/n! , Because

n > 0

P{N > 0} = 1 = 1/0!

we see that, for n > 0, P{N = n} = P{N > n − 1} − P{N > n} =

1 n − 1 1 − = (n − 1)! n! n!

Therefore, E[N] =

q n(n − 1) n! n=1

=

q n=2

1 (n − 2)!

=e

That is, the mean number of independent uniform (0, 1) random variables that must be summed for the sum to exceed 1 is equal to e. 6.3.2

Gamma Random Variables

Recall that a gamma random variable has a density of the form f (y) =

λe−λy (λy)t−1 (t)

0 < y < q

An important property of this family of distributions is that, for a ﬁxed value of λ, it is closed under convolutions. Proposition 3.1. If X and Y are independent gamma random variables with respective parameters (s, λ) and (t, λ), then X + Y is a gamma random variable with parameters (s + t, λ). Proof. Using Equation (3.2), we obtain * a 1 λe−λ(a−y) [λ(a − y)]s−1 λe−λy (λy)t−1 dy fX+Y (a) = (s)(t) 0 * a = Ke−λa (a − y)s−1 yt−1 dy 0 * 1 y −λa s+t−1 = Ke a (1 − x)s−1 xt−1 dx by letting x = a 0 −λa s+t−1 = Ce a

Section 6.3

Sums of Independent Random Variables

255

where C is a constant that does not depend on a. But, as the preceding is a density function and thus must integrate to 1, the value of C is determined, and we have fX+Y (a) =

λe−λa (λa)s+t−1 (s + t)

Hence, the result is proved. It is now a simple matter to establish, by using Proposition 3.1 and induction, that if parameters Xi , i = 1, . . . , n are independent gamma random variables with respective n n Xi is gamma with parameters ti , λ . We leave the (ti , λ), i = 1, . . . , n, then i=1

i=1

proof of this statement as an exercise. EXAMPLE 3b Let X1 , X2 , . . . , Xn be n independent exponential random variables, each having parameter λ. Then, since an exponential random variable with parameter λ is the same as a gamma random variable with parameters (1, λ), it follows from Proposition 3.1 that . X1 + X2 + · · · + Xn is a gamma random variable with parameters (n, λ). n

If Z1 , Z2 , . . . , Zn are independent standard normal random variables, then Y K

i=1

Zi2 is said to have the chi-squared (sometimes seen as χ 2 ) distribution with n

degrees of freedom. Let us compute the density function of Y. When n = 1, Y = Z12 , and from Example 7b of Chapter 5, we see that its probability density function is given by 1 √ √ fZ2 (y) = √ [fZ ( y) + fZ (− y)] 2 y 1 2 = √ √ e−y/2 2 y 2π =

1 −y/2 (y/2)1/2−1 2e

√ π

But we recognize the preceding as the gamma distribution with parameters 12 , 12 . √ [A by-product of this analysis is that 12 = π .] But since each Zi2 is gamma 1 1 2 , 2 2 , it follows from Proposition 3.1 that the χ distribution with n degrees of freedom is just the gamma distribution with parameters n/2, 12 and hence has a probability density function given by 1 −y/2 y n/2−1 e 2 2 y > 0 fχ 2 (y) = n 2 −y/2 n/2−1 e y = y > 0 n n/2 2 2

256

Chapter 6

Jointly Distributed Random Variables

When n is an even integer, (n/2) = [(n/2) − 1]!, whereas when n is odd, (n/2) can be obtained from iterating the relationship = (t − 1)(t − 1) and then using (t) √ 1 5 the previously obtained result that 2 = π . [For instance, 2 = 32 32 = 31 1 3√ 2 2 2 = 4 π.] In practice, the chi-squared distribution often arises as the distribution of the square of the error involved when one attempts to hit a target in n-dimensional space when the coordinate errors are taken to be independent standard normal random variables. It is also important in statistical analysis. 6.3.3

Normal Random Variables

We can also use Equation (3.2) to prove the following important result about normal random variables. Proposition 3.2. If Xi , i = 1, . . . , n, are independent random variables that are norn mally distributed with respective parameters μi , σi2 , i = 1, . . . , n, then Xi is normally n

distributed with parameters

i=1

μi and

n i=1

i=1

σi2 .

Proof of Proposition 3.2: To begin, let X and Y be independent normal random variables with X having mean 0 and variance σ 2 and Y having mean 0 and variance 1. We will determine the density function of X + Y by utilizing Equation (3.2). Now, with 1 1 + σ2 1 = + c= 2 2σ 2 2σ 2 we have

7 7 0 1 (a − y)2 y2 fX (a − y)fY (y) = √ exp − √ exp − 2 2σ 2 2π σ 2π 0 7 0 7 a2 a 1 2 exp − 2 exp −c y − 2y = 2πσ 2σ 1 + σ2 1

0

Hence, from Equation (3.2),

0 7 0 7 a2 1 a2 fX+Y (a) = exp − 2 exp 2πσ 2σ 2σ 2 (1 + σ 2 ) 0 2 7 * q a dy exp −c y − * 1 + σ2 −q 0 7* q 1 a2 exp − = exp{−cx2 } dx 2πσ 2(1 + σ 2 ) −q 0 7 a2 = C exp − 2(1 + σ 2 )

where C does not depend on a. But this implies that X + Y is normal with mean 0 and variance 1 + σ 2 .

Section 6.3

Sums of Independent Random Variables

257

Now, suppose that X1 and X2 are independent normal random variables with Xi having mean μi and variance σi2 , i = 1, 2. Then X1 + X2 = σ2

X2 − μ2 X1 − μ1 + σ2 σ2

+ μ1 + μ2

But since (X1 − μ1 )/σ2 is normal with mean 0 and variance σ12 /σ22 , and (X2 − μ2 )/σ2 is normal with mean 0 and variance 1, it follows from our previous result that (X1 − μ1 )/σ2 + (X2 − μ2 )/σ2 is normal with mean 0 and variance 1 + σ12 /σ22 , implying that X1 + X2 is normal with mean μ1 + μ2 and variance σ22 (1 + σ12 /σ22 ) = σ12 + σ22 . Thus, Proposition 3.2 is established when n = 2. The general case now follows by induction. That is, assume that Proposition 3.2 is true when there are n − 1 random variables. Now consider the case of n, and write n

Xi =

i=1

By the induction hypothesis,

n−1

Therefore, by the result for n = 2,

i=1

Xi + Xn

i=1

Xi is normal with mean

i=1

n

n−1

n i=1

n−1

μi and variance

i=1

Xi is normal with mean

n

n−1 i=1

σi2 .

μi and variance

i=1

σi2 .

EXAMPLE 3c A basketball team will play a 44-game season. Twenty-six of these games are against class A teams and 18 are against class B teams. Suppose that the team will win each game against a class A team with probability .4 and will win each game against a class B team with probability .7. Suppose also that the results of the different games are independent. Approximate the probability that (a) the team wins 25 games or more; (b) the team wins more games against class A teams than it does against class B teams. Solution. (a) Let XA and XB respectively denote the number of games the team wins against class A and against class B teams. Note that XA and XB are independent binomial random variables and E[XA ] = 26(.4) = 10.4 Var(XA ) = 26(.4)(.6) = 6.24 E[XB ] = 18(.7) = 12.6 Var(XB ) = 18(.7)(.3) = 3.78 By the normal approximation to the binomial, XA and XB will have approximately the same distribution as would independent normal random variables with the preceding expected values and variances. Hence, by Proposition 3.2, XA + XB will have

258

Chapter 6

Jointly Distributed Random Variables

approximately a normal distribution with mean 23 and variance 10.02. Therefore, letting Z denote a standard normal random variable, we have P{XA + XB Ú 25} = P{XA + XB Ú 24.5} 0 7 XA + XB − 23 24.5 − 23 =P Ú √ √ 10.02 10.02 0 7 1.5 L P Z Ú √ 10.02 L 1 − P{Z < .4739} L .3178 (b) We note that XA − XB will have approximately a normal distribution with mean −2.2 and variance 10.02. Hence, P{XA − XB Ú 1} = P{XA − XB Ú .5} 0 7 XA − XB + 2.2 .5 + 2.2 =P Ú √ √ 10.02 10.02 0 7 2.7 L P Z Ú √ 10.02 L 1 − P{Z < .8530} L .1968 Therefore, there is approximately a 31.78 percent chance that the team will win at least 25 games and approximately a 19.68 percent chance that it will win more games against class A teams than against class B teams. . The random variable Y is said to be a lognormal random variable with parameters μ and σ if log (Y) is a normal random variable with mean μ and variance σ 2 . That is, Y is lognormal if it can be expressed as Y = eX where X is a normal random variable. EXAMPLE 3d Starting at some ﬁxed time, let S(n) denote the price of a certain security at the end of n additional weeks, n Ú 1. A popular model for the evolution of these prices assumes that the price ratios S(n)/S(n − 1), n Ú 1, are independent and identically distributed lognormal random variables. Assuming this model, with parameters μ = .0165, σ = .0730, what is the probability that (a) the price of the security increases over each of the next two weeks? (b) the price at the end of two weeks is higher than it is today? Solution. Let Z be a standard normal random variable. To solve part (a), we use the fact that log(x) increases in x to conclude that x > 1 if and only if log(x) > log(1) = 0. As a result, we have

Section 6.3

Sums of Independent Random Variables

259

0 7 / S(1) S(1) > 1 = P log > 0 P S(0) S(0) % / −.0165 =P Z > .0730 %

= P{Z < .2260} = .5894 In other words, the probability that the price is up after one week is .5894. Since the successive price ratios are independent, the probability that the price increases over each of the next two weeks is (.5894)2 = .3474. To solve part (b), we reason as follows: / % / % S(2) S(1) S(2) > 1 =P > 1 P S(0) S(1) S(0) 7 0 S(1) S(2) + log > 0 = P log S(1) S(0) S(1) However, log S(2) + log S(0) , being the sum of two independent normal random S(1) variables with a common mean .0165 and a common standard deviation .0730, is a normal random variable with mean .0330 and variance 2(.0730)2 . Consequently, 0 7 / % −.0330 S(2) > 1 =P Z > P √ S(0) .0730 2 = P{Z < .31965} = .6254 6.3.4

.

Poisson and Binomial Random Variables

Rather than attempt to derive a general expression for the distribution of X + Y in the discrete case, we shall consider some examples. EXAMPLE 3e Sums of independent Poisson random variables If X and Y are independent Poisson random variables with respective parameters λ1 and λ2 , compute the distribution of X + Y. Solution. Because the event {X + Y = n} may be written as the union of the disjoint events {X = k, Y = n − k}, 0 … k … n, we have P{X + Y = n} = = =

n k=0 n k=0 n k=0

P{X = k, Y = n − k} P{X = k}P{Y = n − k} e−λ1

λk1 −λ2 λn−k 2 e k! (n − k)!

260

Chapter 6

Jointly Distributed Random Variables −(λ1 +λ2 )

=e =

e−(λ1 +λ2 ) n!

n k=0 n k=0

λk1 λn−k 2 k!(n − k)! n! λk λn−k k!(n − k)! 1 2

e−(λ1 +λ2 ) (λ1 + λ2 )n = n! Thus, X1 + X2 has a Poisson distribution with parameter λ1 + λ2 .

.

EXAMPLE 3f Sums of independent binomial random variables Let X and Y be independent binomial random variables with respective parameters (n, p) and (m, p). Calculate the distribution of X + Y. Solution. Recalling the interpretation of a binomial random variable, and without any computation at all, we can immediately conclude that X + Y is binomial with parameters (n + m, p). This follows because X represents the number of successes in n independent trials, each of which results in a success with probability p; similarly, Y represents the number of successes in m independent trials, each of which results in a success with probability p. Hence, given that X and Y are assumed independent, it follows that X + Y represents the number of successes in n + m independent trials when each trial has a probability p of resulting in a success. Therefore, X + Y is a binomial random variable with parameters (n + m, p). To check this conclusion analytically, note that P{X + Y = k} = = =

n i=0 n

P{X = i, Y = k − i} P{X = i}P{Y = k − i}

i=0 n i=0

n i

i n−i

pq

m k − i

pk−i qm−k+i

r = 0 when j < 0. Thus, where q = 1 − p and where j k n+m−k

P{X + Y = k} = p q

n n i=0

i

m k − i

and the conclusion follows upon application of the combinatorial identity n n m n + m = i k − i k i=0

6.3.5

.

Geometric Random Variables

Let X1 , . . . , Xn be independent geometric random variables, with Xi having parameter pi for i = 1, . . . , n. We are interested in computing the probability mass function

Section 6.3

Sums of Independent Random Variables

261

of their sum Sn = ni=1 Xi . For an application, consider n coins, with coin i having probability pi of coming up heads when ﬂipped, i = 1, . . . , n. Suppose that coin 1 is ﬂipped until heads appears, at which point coin 2 is ﬂipped until it shows heads, and then coin 3 is ﬂipped until it shows heads, and so on. If we let Xi denote the number of ﬂips made with coin i, then X1 , X2 , . . . , Xn will be independent geometric random variables with respective parameters p1 , p2 , . . . , pn , and Sn = ni=1 Xi will represent the total number of ﬂips. If all the pi are equal—say, all pi = p—then Sn has the same distribution as the number of ﬂips of a coin having probability p of coming up heads that are needed to obtain a total of n heads, and so Sn is a negative binomial random variable with probability mass function k − 1 n P{Sn = k} = p (1 − p)k−n , n − 1

k Ú n

As a prelude to determining the probability mass function of Sn when the pi are all distinct, let us ﬁrst consider the case n = 2. Letting qj = 1 − pj , j = 1, 2, we obtain P(S2 = k) =

k−1

P{X1 = j, X2 = k − j}

j=1

=

k−1

P{X1 = j} P{X2 = k − j} (by independence)

j=1

=

k−1

j−1

k−j−1

p1 q1 p2 q2

j=1

= p1 p2 qk−2 2

k−1 (q1 /q2 )j−1 j=1

= p1 p2 qk−2 2

1 − (q1 /q2 )k−1 1 − q1 /q2

p1 p2 qk−1 p1 p2 qk−1 2 1 − q2 − q1 q2 − q1 p1 p2 = p2 qk−1 + p1 qk−1 2 1 p1 − p2 p2 − p1 =

If we now let n = 3 and compute P{S3 = k} by starting with the identity P{S3 = k} =

k−1

P{S2 = j, X3 = k − j} =

j=1

k−1

P{S2 = j}P{X3 = k − j}

j=1

and then substituting the derived formula for the mass function of S2 , we would obtain, after some computations, p2 p3 p1 p3 + p2 qk−1 2 p2 − p1 p3 − p1 p1 − p2 p3 − p2 p1 p2 k−1 + p3 q3 p1 − p3 p2 − p3

P{S3 = k} = p1 qk−1 1

262

Chapter 6

Jointly Distributed Random Variables

The mass functions of S2 and S3 lead to the following conjecture for the mass function of Sn . Proposition 3.3. Let X1 , . . . , Xn be independent geometric random variables, with Xi having parameter pi for i = 1, . . . , n. If all the pi are distinct, then, for k Ú n, P{Sn = k} =

n

pi qk−1 i

jZi

i=1

pj pj − pi

Proof of Proposition 3.3: We will prove this proposition by induction on the value of n + k. Because the proposition is true when n = 2, k = 2, take as the induction hypothesis that it is true for any k Ú n for which n + k … r. Now, suppose k Ú n are such that n + k = r + 1. To compute P{Sn = k}, we condition on whether Xn = 1. This gives P{Sn = k} = P{Sn = k|Xn = 1}P{Xn = 1} + P{Sn = k|Xn > 1}P{Xn > 1} = P{Sn = k|Xn = 1}pn + P{Sn = k|Xn > 1}qn Now, P{Sn = k|Xn = 1} = P{Sn−1 = k − 1|Xn = 1} = P{Sn−1 = k − 1} (by independence) =

n−1

pi qk−2 i

i=1

iZj…n−1

pj pj − pi

(by the induction hypothesis)

Now, if X is geometric with parameter p, then the conditional distribution of X given that it is larger than 1 is the same as the distribution of 1 (the ﬁrst failed trial) plus a geometric with parameter p (the number of additional trials after the ﬁrst until a success occurs). Consequently, P{Sn = k|Xn > 1} = P{X1 + . . . + Xn−1 + Xn + 1 = k} = P{Sn = k − 1} n pj pi qk−2 = i pj − pi i=1

iZj…n

where the ﬁnal equality follows from the induction hypothesis. Thus, from the preceding, we obtain P{Sn = k} = pn

n−1

pi qk−2 i

i=1

= pn

n−1

iZj…n−1

pi qk−2 i

i=1

+ qn pn qk−2 n

iZj…n−1

j

=

n−1 i=1

n pj pj + qn pi qk−2 i pj − pi pj − pi iZj…n

i=1

pj + qn pj − pi

n−1 i=1

pi qk−2 i

iZj…n

pj pj − pi

pj pj − pn

pi qk−2 pn (1 + i

qn ) pn − pi

iZj…n−1

pj pj + pn qk−1 n pj − pi pj − pn j

Section 6.4

Conditional Distributions: Discrete Case

263

Now, using that 1 +

qn pn − pi + qn qi = = pn − pi pn − pi pn − pi

the preceding gives P{Sn = k} = =

n−1 i=1 n

pi qk−1 i

iZj…n

pi qk−1 i

jZi

i=1

pj pj + pn qk−1 n pj − pi pj − pn j

pj pj − pi .

and the proof by induction is complete. 6.4

CONDITIONAL DISTRIBUTIONS: DISCRETE CASE Recall that, for any two events E and F, the conditional probability of E given F is deﬁned, provided that P(F) > 0, by P(E|F) =

P(EF) P(F)

Hence, if X and Y are discrete random variables, it is natural to deﬁne the conditional probability mass function of X given that Y = y, by pX|Y (x|y) = P{X = x|Y = y} P{X = x, Y = y} = P{Y = y} p(x, y) = pY (y) for all values of y such that pY (y) > 0. Similarly, the conditional probability distribution function of X given that Y = y is deﬁned, for all y such that pY (y) > 0, by FX|Y (x|y) = P{X … x|Y = y} = pX|Y (a|y) a…x

In other words, the deﬁnitions are exactly the same as in the unconditional case, except that everything is now conditional on the event that Y = y. If X is independent of Y, then the conditional mass function and the distribution function are the same as the respective unconditional ones. This follows because if X is independent of Y, then pX|Y (x|y) = P{X = x|Y = y} P{X = x, Y = y} = P{Y = y} P{X = x}P{Y = y} = P{Y = y} = P{X = x}

264

Chapter 6

Jointly Distributed Random Variables

EXAMPLE 4a Suppose that p(x, y), the joint probability mass function of X and Y, is given by p(0, 0) = .4 p(0, 1) = .2 p(1, 0) = .1 p(1, 1) = .3 Calculate the conditional probability mass function of X given that Y = 1. Solution. We ﬁrst note that pY (1) =

p(x, 1) = p(0, 1) + p(1, 1) = .5

x

Hence, pX|Y (0|1) =

2 p(0, 1) = pY (1) 5

pX|Y (1|1) =

3 p(1, 1) = pY (1) 5

and .

EXAMPLE 4b If X and Y are independent Poisson random variables with respective parameters λ1 and λ2 , calculate the conditional distribution of X given that X + Y = n. Solution. We calculate the conditional probability mass function of X given that X + Y = n as follows: P{X = k, X + Y = n} P{X + Y = n} P{X = k, Y = n − k} = P{X + Y = n} P{X = k}P{Y = n − k} = P{X + Y = n}

P{X = k|X + Y = n} =

where the last equality follows from the assumed independence of X and Y. Recalling (Example 3e) that X + Y has a Poisson distribution with parameter λ1 + λ2 , we see that the preceding equals e−λ1 λk1 e−λ2 λn−k 2 P{X = k|X + Y = n} = k! (n − k)!

e−(λ1 +λ2 ) (λ1 + λ2 )n n!

−1

λk1 λn−k n! 2 (n − k)! k! (λ1 + λ2 )n k n−k λ1 λ2 n = k λ 1 + λ2 λ1 + λ2 =

In other words, the conditional distribution of X given that X + Y = n is the binomial distribution with parameters n and λ1 /(λ1 + λ2 ). . We can also talk about joint conditional distributions, as is indicated in the next two examples.

Section 6.4

Conditional Distributions: Discrete Case

265

EXAMPLE 4c Consider the multinomial distribution with joint probability mass function P{Xi = ni , i = 1, . . . , k} =

n! n n p 1 · · · pkk , n1 ! · · · nk ! 1

ni Ú 0,

k

ni = n

i=1

Such a mass function results when n independent trials are performed, with each trial resulting in outcome i with probability pi , ki=1 pi = 1. The random variables Xi , i = 1, . . . , k, represent, respectively, the number of trials that result in outcome i, i = 1, . . . , k. Suppose we are given that nj of the trials resulted in outcome j, for j = r + 1, . . . , k, where kj=r+1 nj = m … n. Then, because each of the other n − m trials must have resulted in one of the trials 1, . . . , r, it would seem that the conditional distribution of X1 , . . . , Xr is the multinomial distribution on n − m trials with respective trial outcome probabilities P{outcome i|outcome is not any of r + 1, . . . , k} = where Fr = 1, . . . , r.

r

i=1 pi

pi , i = 1, . . . , r Fr

is the probability that a trial results in one of the outcomes

Solution. To verify this intuition, let n1 , . . . , nr , be such that

r

i=1 ni

= n − m. Then

P{X1 = n1 , . . . , Xr = nr |Xr+1 = nr+1 , . . . Xk = nk } =

P{X1 = n1 , . . . , Xk = nk } P{Xr+1 = nr+1 , . . . Xk = nk }

=

nk n1 nr nr+1 n! n1 !···nk ! p1 · · · pr pr+1 · · · pk n n−m nr+1 n! pr+1 · · · pkk (n−m)!nr+1 !···nk ! Fr

where the probability in the denominator was obtained by regarding outcomes 1, . . . , r as a single outcome having probability Fr , thus showing that the probability is a multinomial probability on n trials with outcome probabilities Fr , pr+1 , . . . , pk . Because r n i=1 i = n − m, the preceding can be written as P{X1 = n1 , . . . , Xr = nr |Xr+1 = nr+1 , . . . Xk = nk } = and our intuition is upheld.

pr (n − m)! p1 n1 ( ) · · · ( )nr n1 ! · · · nr ! Fr Fr .

EXAMPLE 4d Consider n independent trials, with each trial being a success with probability p. Given a total of k successes, show that all possible orderings of the k successes and n − k failures are equally likely. Solution. We want to show that, given a total of k successes, each of the nk possible orderings of k successes and n − k failures is equally likely. Let X denote the number of successes, and consider any ordering of k successes and n − k failures, say, o = (s, s, f , f , . . . , f ). Then

266

Chapter 6

Jointly Distributed Random Variables

P(o, X = k) P(X = k) P(o) = P(X = k) pk (1 − p)n−k = n k n−k k p (1 − p) 1 = n

P(o|X = k) =

k

6.5

.

CONDITIONAL DISTRIBUTIONS: CONTINUOUS CASE If X and Y have a joint probability density function f (x, y), then the conditional probability density function of X given that Y = y is deﬁned, for all values of y such that fY (y) > 0, by f (x, y) fX|Y (x|y) = fY (y) To motivate this deﬁnition, multiply the left-hand side by dx and the right-hand side by (dx dy)/dy to obtain f (x, y) dx dy fY (y) dy P{x … X … x + dx, y … Y … y + dy} L P{y … Y … y + dy} = P{x … X … x + dx|y … Y … y + dy}

fX|Y (x|y) dx =

In other words, for small values of dx and dy, fX|Y (x|y)dx represents the conditional probability that X is between x and x + dx given that Y is between y and y + dy. The use of conditional densities allows us to deﬁne conditional probabilities of events associated with one random variable when we are given the value of a second random variable. That is, if X and Y are jointly continuous, then, for any set A, * P{X ∈ A|Y = y} = fX|Y (x|y) dx A

In particular, by letting A = (−q, a], we can deﬁne the conditional cumulative distribution function of X given that Y = y by * a fX|Y (x|y) dx FX|Y (a|y) K P{X … a|Y = y} = −q

The reader should note that, by using the ideas presented in the preceding discussion, we have been able to give workable expressions for conditional probabilities, even though the event on which we are conditioning (namely, the event {Y = y}) has probability 0. EXAMPLE 5a The joint density of X and Y is given by 0 12 5 x(2 − x − y) f (x, y) = 0

0 < x < 1, 0 < y < 1 otherwise

Compute the conditional density of X given that Y = y, where 0 < y < 1.

Section 6.5

Conditional Distributions: Continuous Case

267

Solution. For 0 < x < 1, 0 < y < 1, we have f (x, y) fY (y) f (x, y) = -q −q f (x, y) dx x(2 − x − y) = -1 0 x(2 − x − y) dx x(2 − x − y) = 2 3 − y/2 6x(2 − x − y) = 4 − 3y

fX|Y (x|y) =

.

EXAMPLE 5b Suppose that the joint density of X and Y is given by ⎧ −x/y e−y ⎪ ⎨e 0 < x < q, 0 < y < q f (x, y) = y ⎪ ⎩0 otherwise Find P{X > 1|Y = y}. Solution. We ﬁrst obtain the conditional density of X given that Y = y. f (x, y) fY (y) e−x/y e−y /y = −y - q −x/y dx e 0 (1/y)e 1 = e−x/y y

fX|Y (x|y) =

Hence,

*

q

1 −x/y e dx 1 y 6q 6 = −e−x/y 6

P{X > 1|Y = y} =

−1/y

1

=e

.

If X and Y are independent continuous random variables, the conditional density of X given that Y = y is just the unconditional density of X. This is so because, in the independent case, fX|Y (x|y) =

fX (x)fY (y) f (x, y) = = fX (x) fY (y) fY (y)

We can also talk about conditional distributions when the random variables are neither jointly continuous nor jointly discrete. For example, suppose that X is a continuous random variable having probability density function f and N is a discrete random variable, and consider the conditional distribution of X given that N = n. Then

268

Chapter 6

Jointly Distributed Random Variables

P{x < X < x + dx|N = n} dx P{N = n|x < X < x + dx} P{x < X < x + dx} = P{N = n} dx and letting dx approach 0 gives lim dx→0

P{N = n|X = x} P{x < X < x + dx|N = n} = f (x) dx P{N = n}

thus showing that the conditional density of X given that N = n is given by fX|N (x|n) =

P{N = n|X = x} f (x) P{N = n}

EXAMPLE 5c The Bivariate Normal Distribution One of the most important joint distributions is the bivariate normal distribution. We say that the random variables X, Y have a bivariate normal distribution if, for constants μx , μy , σx > 0, σy > 0, −1 < ρ < 1, their joint density function is given, for all −q < x, y < q, by ⎧ ⎨ 1 1 x − μx 2 . exp − f (x, y) = ⎩ 2(1 − ρ 2 ) σx 2πσx σy 1 − ρ 2 ⎤⎫ 2 ⎪ y − μy (x − μx )(y − μy ) ⎬ ⎦ + − 2ρ ⎪ σy σx σy ⎭ We now determine the conditional density of X given that Y = y. In doing so, we will continually collect all factors that do not depend on x and represent them by the -q constants Ci . The ﬁnal constant will then be found by using that −q fX|Y (x|y) dx = 1. We have f (x, y) fX|Y (x|y) = fY (y) = C1 f (x, y) ⎧ ⎫ 2 ⎨ x(y − μy ) ⎬ 1 x − μx = C2 exp − − 2ρ ⎭ ⎩ 2(1 − ρ 2 ) σx σx σy ⎧ ⎫ ⎡ ⎤⎪ ⎪ ⎨ ⎬ 1 ⎣x2 − 2x μx + ρ σx (y − μy ) ⎦ = C3 exp − 2 2 ⎪ ⎪ σy ⎩ 2σx (1 − ρ ) ⎭ ⎫ ⎧ ⎡ ⎤2 ⎪ ⎪ ⎬ ⎨ 1 ⎣x − μx + ρ σx (y − μy ) ⎦ = C4 exp − 2 2 ⎪ ⎪ σy ⎭ ⎩ 2σx (1 − ρ ) Recognizing the preceding equation as a normal density, we can conclude that, given Y = y, the random variable X is normally distributed with mean μx + ρ σσxy (y − μy )

and variance σx2 (1 − ρ 2 ). Also, because the joint density of Y, X is exactly the same as that of X, Y, except that μx , σx are interchanged with μy , σy , it similarly follows that

Section 6.5

Conditional Distributions: Continuous Case

269

the conditional distribution of Y given X = x is the normal distribution with mean σ μy + ρ σyx (x − μx ) and variance σy2 (1 − ρ 2 ). It follows from these results that the necessary and sufﬁcient condition for the bivariate normal random variables X and Y to be independent is that ρ = 0 (a result that also follows directly from their joint density, because it is only when ρ = 0 that the joint density factors into two terms, one depending only on x and the other only on y). 1√ With C = , the marginal density of X can be obtained from 2 2πσx σy

1−ρ

* fX (x) =

q

f (x, y) dy ⎧ ⎡ 2 ⎪ 2 * q ⎨ y − μy 1 x − μ x ⎣ =C exp − + 2 ⎪ σx σy −q ⎩ 2(1 − ρ ) ⎫ (x − μx )(y − μy ) ⎬ dy −2ρ ⎭ σx σy −q

Making the change of variables w =

y−μy σy

gives

0

7 1 x − μx 2 fX (x) = Cσy exp − σx 2(1 − ρ 2 ) 0 7 * q 1 x − μx 2 exp − w dw * w − 2ρ σx 2(1 − ρ 2 ) −q 0 7 x − μx 2 1 (1 − ρ 2 ) = Cσy exp − σx 2(1 − ρ 2 ) 0 7 * q 1 x − μx 2 * exp − dw w − ρ σx 2(1 − ρ 2 ) −q Because

2 7 1 ρ . dw = 1 exp − (x − μx ) w − σx 2(1 − ρ 2 ) 2π(1 − ρ 2 ) −q 1

we see that

*

q

0

8 2 2 fX (x) = Cσy 2π(1 − ρ 2 ) e−(x−μx ) /2σx 1 2 2 =√ e−(x−μx ) /2σx 2π σx

That is, X is normal with mean μx and variance σx2 . Similarly, Y is normal with mean μy and variance σy2 . . EXAMPLE 5d Consider n + m trials having a common probability of success. Suppose, however, that this success probability is not ﬁxed in advance but is chosen from a uniform (0, 1)

270

Chapter 6

Jointly Distributed Random Variables

population. What is the conditional distribution of the success probability given that the n + m trials result in n successes? Solution. If we let X denote the probability that a given trial is a success, then X is a uniform (0, 1) random variable. Also, given that X = x, the n + m trials are independent with common probability of success x, so N, the number of successes, is a binomial random variable with parameters (n + m, x). Hence, the conditional density of X given that N = n is fX|N (x|n) =

=

P{N = n|X = x}fX (x) P{N = n} n + m xn (1 − x)m n

P{N = n} = cxn (1 − x)m

0 < x < 1

where c does not depend on x. Thus, the conditional density is that of a beta random variable with parameters n + 1, m + 1. The preceding result is quite interesting, for it states that if the original or prior (to the collection of data) distribution of a trial success probability is uniformly distributed over (0, 1) [or, equivalently, is beta with parameters (1, 1)] then the posterior (or conditional) distribution given a total of n successes in n + m trials is beta with parameters (1 + n, 1 + m). This is valuable, for it enhances our intuition as to what it means to assume that a random variable has a beta distribution. . ∗ 6.6

ORDER STATISTICS Let X1 , X2 , . . . , Xn be n independent and identically distributed continuous random variables having a common density f and distribution function F. Deﬁne X(1) = smallest of X1 , X2 , . . . , Xn X(2) = second smallest of X1 , X2 , . . . , Xn

X(j)

# # # = jth smallest of X1 , X2 , . . . , Xn

X(n)

# # # = largest of X1 , X2 , . . . , Xn

The ordered values X(1) … X(2) … · · · … X(n) are known as the order statistics corresponding to the random variables X1 , X2 , . . . , Xn . In other words, X(1) , . . . , X(n) are the ordered values of X1 , . . . , Xn . The joint density function of the order statistics is obtained by noting that the order statistics X(1) , . . . , X(n) will take on the values x1 … x2 … · · · … xn if and only if, for some permutation (i1 , i2 , . . . , in ) of (1, 2, . . . , n), X1 = xi1 , X2 = xi2 , . . . , Xn = xin

Section 6.6

Order Statistics

271

Since, for any permutation (i1 , . . . , in ) of (1, 2, . . . , n), %

ε ε ε ε < X1 < xi1 + , . . . , xin − < Xn < xin + P xi1 − 2 2 2 2 n L ε fX1 , · · · , Xn (xi1 , . . . , xin ) = εn f (xi1 ) · · · f (xin ) = εn f (x1 ) · · · f (xn )

/

it follows that, for x1 < x2 < · · · < xn , / % ε ε ε ε < X(1) < x1 + , . . . , xn − < X(n) < xn + P x1 − 2 2 2 2 n L n! ε f (x1 ) · · · f (xn ) Dividing by εn and letting ε→0 yields fX(1) ,...,X(n) (x1 , x2 , . . . , xn ) = n!f (x1 ) · · · f (xn ) x1 < x2 < · · · < xn

(6.1)

Equation (6.1) is most simply explained by arguing that, in order for the vector

X(1) , . . . , X(n) to equal x1 , . . . , xn , it is necessary and sufﬁcient for X1 , . . . , Xn to equal one of the n! permutations of x1 , . . . , xn . Since the probability (density) that

X1 , . . . , Xn equals any given permutation of x1 , . . . , xn is just f (x1 ) · · · f (xn ), Equation (6.1) follows. EXAMPLE 6a Along a road 1 mile long are 3 people “distributed at random.” Find the probability that no 2 people are less than a distance of d miles apart when d … 12 . Solution. Let us assume that “distributed at random” means that the positions of the 3 people are independent and uniformly distributed over the road. If Xi denotes the position of the ith person, then the desired probability is P{X(i) > X(i−1) + d, i = 2, 3}. Because fX(1) ,X(2) ,X(3) (x1 , x2 , x3 ) = 3!

0 < x1 < x2 < x3 < 1

it follows that * * * P{X(i) > X(i−1) + d, i = 2, 3} = * = 3! * =6

xi >xj−1 +d

fX(1) ,X(2) ,X(3) (x1 , x2 , x3 ) dx1 dx2 dx3

1−2d * 1−d * 1

0 x1 +d 1−2d * 1−d

x2 +d

dx3 dx2 dx1

(1 − d − x2 ) dx2 dx1

0

* =6

x1 +d 1−2d * 1−2d−x1

y2 dy2 dx1 0

0

272

Chapter 6

Jointly Distributed Random Variables

where we have made the change of variables y2 = 1 − d − x2 . Continuing the string of equalities yields *

1−2d

=3

(1 − 2d − x1 )2 dx1

0

*

1−2d

=3 0

y21 dy1

= (1 − 2d)3 Hence, the desired probability that no 2 people are within a distance d of each other when 3 people are uniformly and independently distributed over an interval of size 1 is (1 − 2d)3 when d … 12 . In fact, the same method can be used to prove that when n people are distributed at random over the unit interval, the desired probability is [1 − (n − 1)d]n

when d …

1 n − 1 .

The proof is left as an exercise.

The density function of the jth-order statistic X(j) can be obtained either by integrating the joint density function (6.1) or by direct reasoning as follows: In order for X(j) to equal x, it is necessary for j − 1 of the n values X1 , . . . , Xn to be less than x, n − j of them to be greater than x, and 1 of them to equal x. Now, the probability density that any given set of j − 1 of the Xi ’s are less than x, another given set of n − j are all greater than x, and the remaining value is equal to x equals [F(x)]j−1 [1 − F(x)]n−j f (x) Hence, since there are

n j − 1, n − j, 1

=

n! (n − j)!(j − 1)!

different partitions of the n random variables X1 , . . . , Xn into the preceding three groups, it follows that the density function of X(j) is given by fX(j) (x) =

n! [F(x)]j−1 [1 − F(x)]n−j f (x) (n − j)!(j − 1)!

(6.2)

EXAMPLE 6b When a sample of 2n + 1 random variables (that is, when 2n + 1 independent and identically distributed random variables) is observed, the (n + 1)st smallest is called the sample median. If a sample of size 3 from a uniform distribution over (0, 1) is observed, ﬁnd the probability that the sample median is between 14 and 34 . Solution. From Equation (6.2), the density of X(2) is given by fX(2) (x) =

3! x(1 − x) 1!1!

0 < x < 1

Section 6.6

Order Statistics

273

Hence, %

1 3 P < X(2) < 4 4

/

*

3/4

=6

x(1 − x) dx

1/4

0 =6

766x=3/4 6 11 6 − = 6 2 3 6 16

x2

x3

.

x=1/4

The cumulative distribution function of X(j) can be found by integrating Equation (6.2). That is, n! FX(j) (y) = (n − j)!(j − 1)!

*

y −q

[F(x)]j−1 [1 − F(x)]n−j f (x) dx

(6.3)

However, FX(j) (y) could also have been derived directly by noting that the jth order statistic is less than or equal to y if and only if there are j or more of the Xi ’s that are less than or equal to y. Thus, because the number of Xi ’s that are less than or equal to y is a binomial random variable with parameters n, p = F(y), it follows that FX(j) (y) = P{X(j) … y} = P{j or more of the Xi ’s are … y} n n [F(y)]k [1 − F(y)]n−k = k

(6.4)

k=j

If, in Equations (6.3) and (6.4), we take F to be the uniform (0, 1) distribution [that is, f (x) = 1, 0 < x < 1], then we obtain the interesting analytical identity n n k=j

k

k

y (1 − y)

n−k

n! = (n − j)!(j − 1)!

*

y

xj−1 (1 − x)n−j dx

0 … y … 1 (6.5)

0

By employing the same type of argument that we used in establishing Equation (6.2), we can show that the joint density function of the order statistics X(i) and X(j) when i < j is fX(i) ,X(j) (xi , xj ) =

n! [F(xi )]i−1 (i − 1)!(j − i − 1)!(n − j)!

(6.6)

* [F(xj ) − F(xi )]j−i−1 [1 − F(xj )]n−j f (xi )f (xj ) for all xi < xj . EXAMPLE 6c Distribution of the range of a random sample Suppose that n independent and identically distributed random variables X1 , X2 , . . . , Xn are observed. The random variable R deﬁned by R = X(n) − X(1) is called the range of the observed random variables. If the random variables Xi have distribution function F and density function f , then the distribution of R can be obtained from Equation (6.6) as follows: For a Ú 0,

274

Chapter 6

Jointly Distributed Random Variables

P{R … a} = P{X(n) − X(1) … a} ** = fX(1) ,X(n) (x1 , xn ) dx1 dxn xn −x1 …a q * x1 +a

* =

−q x1

n! [F(xn ) − F(x1 )]n−2 f (x1 )f (xn ) dxn dx1 (n − 2)!

Making the change of variable y = F(xn ) − F(x1 ), dy = f (xn ) dxn yields * F(x1 +a)−F(x1 ) * x1 +a n−2 [F(xn ) − F(x1 )] f (xn ) dxn = yn−2 dy x1

0

= Thus,

* P{R … a} = n

q −q

1 [F(x1 + a) − F(x1 )]n−1 n − 1

[F(x1 + a) − F(x1 )]n−1 f (x1 ) dx1

(6.7)

Equation (6.7) can be evaluated explicitly only in a few special cases. One such case is when the Xi ’s are all uniformly distributed on (0, 1). In this case, we obtain, from Equation (6.7), that for 0 < a < 1, * 1 P{R < a} = n [F(x1 + a) − F(x1 )]n−1 f (x1 ) dx1 0 * 1−a * 1 n−1 =n a dx1 + n (1 − x1 )n−1 dx1 0

1−a n−1

= n(1 − a)a

n

+ a

Differentiation yields the density function of the range: given in this case by 0 0 … a … 1 n(n − 1)an−2 (1 − a) fR (a) = 0 otherwise That is, the range of n independent uniform (0, 1) random variables is a beta random variable with parameters n − 1, 2. .

6.7

JOINT PROBABILITY DISTRIBUTION OF FUNCTIONS OF RANDOM VARIABLES Let X1 and X2 be jointly continuous random variables with joint probability density function fX1 ,X2 . It is sometimes necessary to obtain the joint distribution of the random variables Y1 and Y2 , which arise as functions of X1 and X2 . Speciﬁcally, suppose that Y1 = g1 (X1 , X2 ) and Y2 = g2 (X1 , X2 ) for some functions g1 and g2 . Assume that the functions g1 and g2 satisfy the following conditions: 1. The equations y1 = g1 (x1 , x2 ) and y2 = g2 (x1 , x2 ) can be uniquely solved for x1 and x2 in terms of y1 and y2 , with solutions given by, say, x1 = h1 (y1 , y2 ), x2 = h2 (y1 , y2 ). 2. The functions g1 and g2 have continuous partial derivatives at all points (x1 , x2 ) and are such that the 2 * 2 determinant

Section 6.7

Joint Distribution of Functions of Random Variables

6 6 6 6 6 J(x1 , x2 ) = 6 6 6 6

∂g1 ∂x1 ∂g2 ∂x1

275

6 6 6 6 ∂g1 ∂g2 ∂g1 ∂g2 6 − Z 0 6 K 6 ∂x1 ∂x2 ∂x2 ∂x1 ∂g2 6 ∂x2 6 ∂g1 ∂x2

at all points (x1 , x2 ). Under these two conditions, it can be shown that the random variables Y1 and Y2 are jointly continuous with joint density function given by fY1 Y2 (y1 , y2 ) = fX1 ,X2 (x1 , x2 )|J(x1 , x2 )|−1

(7.1)

where x1 = h1 (y1 , y2 ), x2 = h2 (y1 , y2 ). A proof of Equation (7.1) would proceed along the following lines: * * P{Y1 … y1 , Y2 … y2 } =

fX1 ,X2 (x1 , x2 ) dx1 dx2

(7.2)

(x1 , x2 ) : g1 (x1 , x2 ) … y1 g2 (x1 , x2 ) … y2

The joint density function can now be obtained by differentiating Equation (7.2) with respect to y1 and y2 . That the result of this differentiation will be equal to the righthand side of Equation (7.1) is an exercise in advanced calculus whose proof will not be presented in this book. EXAMPLE 7a Let X1 and X2 be jointly continuous random variables with probability density function fX1 ,X2 . Let Y1 = X1 + X2 , Y2 = X1 − X2 . Find the joint density function of Y1 and Y2 in terms of fX1 ,X2 . Solution. Let g1 (x1 , x2 ) = x1 + x2 and g2 (x1 , x2 ) = x1 − x2 . Then 6 6 61 1 6 6 = −2 J(x1 , x2 ) = 66 1 −1 6 Also, since the equations y1 = x1 + x2 and y2 = x1 − x2 have x1 = (y1 + y2 )/2, x2 = (y1 − y2 )/2 as their solution, it follows from Equation (7.1) that the desired density is 1 fY1 ,Y2 (y1 , y2 ) = fX1 ,X2 2

y1 + y2 y1 − y2 , 2 2

For instance, if X1 and X2 are independent uniform (0, 1) random variables, then 0 fY1 ,Y2 (y1 , y2 ) =

1 2

0

0 … y1 + y2 … 2, 0 … y1 − y2 … 2 otherwise

276

Chapter 6

Jointly Distributed Random Variables

or if X1 and X2 are independent exponential random variables with respective parameters λ1 and λ2 , then fY1 ,Y2 (y1 , y2 ) ⎧ 0 7 ⎪ y y λ + y − y λ ⎪ 2 2 2 1 1 1 ⎨ exp −λ1 − λ2 y1 + y2 Ú 0, y1 − y2 Ú 0 2 2 2 = ⎪ ⎪ ⎩ 0 otherwise

R

Y

⍜ X

FIGURE 6.4: • = Random point. (X, Y) = (R, Θ).

Finally, if X1 and X2 are independent standard normal random variables, then 1 −[(y1 +y2 )2 /8+(y1 −y2 )2 /8] e 4π 1 −(y2 +y2 )/4 e 1 2 = 4π 1 1 2 2 = √ e−y1 /4 √ e−y2 /4 4π 4π

fY1 ,Y2 (y1 , y2 ) =

Thus, not only do we obtain (in agreement with Proposition 3.2) that both X1 + X2 and X1 − X2 are normal with mean 0 and variance 2, but we also conclude that these two random variables are independent. (In fact, it can be shown that if X1 and X2 are independent random variables having a common distribution function F, then X1 + X2 will be independent of X1 − X2 if and only if F is a normal distribution function.) . EXAMPLE 7b Let (X, Y) denote a random point in the plane, and assume that the rectangular coordinates X and Y are independent standard normal random variables. We are interested in the joint distribution of R, Θ, the polar coordinate representation of (x, y). (See Figure 6.4.) Suppose.ﬁrst that X and Y are both positive. For x and y positive, letting r = g1 (x, y) = x2 + y2 and θ = g2 (x, y) = tan−1 y/x, we see that x ∂g1 =. ∂x x2 + y2 y ∂g1 =. 2 ∂y x + y2

Section 6.7

Joint Distribution of Functions of Random Variables

277

−y −y 1 ∂g2 = 2 = 2 2 ∂x 1 + (y/x) x x + y2 x 1 ∂g2 = 2 = 2 ∂y x[1 + (y/x) ] x + y2 Hence, J(x, y) =

x2 y2 1 1 + =. = 2 2 3/2 2 2 3/2 r (x + y ) (x + y ) x2 + y2

Because the conditional joint density function of X, Y given that they are both positive is f (x, y) 2 2 2 = e−(x +y )/2 , x > 0, y > 0 P(X > 0, Y > 0) π . we see that the conditional joint density function of R = X 2 + Y 2 and = tan−1 (Y/X), given that X and Y are both positive, is f (x, y|X > 0, Y > 0) =

f (r, θ|X > 0, Y > 0) =

2 −r2/2 re , π

0 < θ < π/2,

0 < r < q

Similarly, we can show that 2 −r2/2 re , π 2 2/2 f (r, θ|X < 0, Y < 0) = re−r , π 2 2/2 f (r, θ|X > 0, Y < 0) = re−r , π f (r, θ|X < 0, Y > 0) =

π/2 < θ < π, π < θ < 3π/2, 3π/2 < θ < 2π,

0 < r < q 0 < r < q 0 < r < q

As the joint density is an equally weighted average of these 4 conditional joint densities, we obtain that the joint density of R, Θ is given by f (r, θ) =

1 −r2 /2 re 2π

0 < θ < 2π,

0 < r < q

Now, this joint density factors into the marginal densities for R and Θ, so R and Θ are independent random variables, with Θ being uniformly distributed over (0, 2π) and R having the Rayleigh distribution with density f (r) = re−r

2 /2

0 < r < q

(For instance, when one is aiming at a target in the plane, if the horizontal and vertical miss distances are independent standard normals, then the absolute value of the error has the preceding Rayleigh distribution.) This result is quite interesting, for it certainly is not evident a priori that a random vector whose coordinates are independent standard normal random variables will have an angle of orientation that not only is uniformly distributed, but also is independent of the vector’s distance from the origin.

278

Chapter 6

Jointly Distributed Random Variables

If we wanted the joint distribution of R2 and Θ, then, since the transformation d = g1 (x, y) = x2 + y2 and θ = g2 (x, y) = tan−1 y/x has the Jacobian 6 6 6 6 2x 2y 6 6 6 6 −y x J=6 6=2 6 2 6 6 x + y2 x2 + y2 6 it follows that f (d, θ) =

1 −d/2 1 e 2 2π

0 < d < q,

0 < θ < 2π

Therefore, R2 and Θ are independent, with R2 having an exponential distribution with parameter 12 . But because R2 = X 2 + Y 2 , it follows by deﬁnition that R2 has a chi-squared distribution with 2 degrees of freedom. Hence, we have a veriﬁcation of the result that the exponential distribution with parameter 12 is the same as the chi-squared distribution with 2 degrees of freedom. The preceding result can be used to simulate (or generate) normal random variables by making a suitable transformation on uniform random variables. Let U1 and U2 be independent random variables, each uniformly distributed over (0, 1). We will transform U1 , U2 into two independent unit normal random variables X1 and X2 by ﬁrst considering the polar coordinate representation (R, Θ) of the random vector (X1 , X2 ). From the preceding, R2 and Θ will be independent, and, in addition, R2 = X12 + X22 will have an exponential distribution with parameter λ = 12 . But −2 log U1 has such a distribution, since, for x > 0, / % x P{−2 log U1 < x} = P log U1 > − 2 = P{U1 > e−x/2 } = 1 − e−x/2 Also, because 2πU2 is a uniform (0, 2π) random variable, we can use it to generate Θ. That is, if we let R2 = −2 log U1 Θ = 2πU2 then R2 can be taken to be the square of the distance from the origin and θ can be taken to be the angle of orientation of (X1 , X2 ). Now, since X1 = R cos Θ, X2 = R sin Θ, it follows that . X1 = −2 log U1 cos(2πU2 ) . X2 = −2 log U1 sin(2πU2 ) are independent standard normal random variables.

.

EXAMPLE 7c If X and Y are independent gamma random variables with parameters (α, λ) and (β, λ), respectively, compute the joint density of U = X + Y and V = X/(X + Y).

Section 6.7

Joint Distribution of Functions of Random Variables

279

Solution. The joint density of X and Y is given by λe−λx (λx)α−1 λe−λy (λy)β−1 (α) (β) λα+β e−λ(x+y) xα−1 yβ−1 = (α)(β)

fX,Y (x, y) =

Now, if g1 (x, y) = x + y, g2 (x, y) = x/(x + y), then ∂g1 ∂g1 = =1 ∂x ∂y

so

y ∂g2 = ∂x (x + y)2

6 6 1 1 6 6 −x y J(x, y) = 6 6 6 (x + y)2 (x + y)2

x ∂g2 =− ∂y (x + y)2 6 6 6 1 6 6=− 6 x + y 6

Finally, as the equations u = x + y, v = x/(x + y) have as their solutions x = uv, y = u(1 − v), we see that fU,V (u, v) = fX,Y [uv, u(1 − v)]u =

λe−λu (λu)α+β−1 vα−1 (1 − v)β−1 (α + β) (α + β) (α)(β)

Hence, X + Y and X/(X + Y) are independent, with X + Y having a gamma distribution with parameters (α + β, λ) and X/(X + Y) having a beta distribution with parameters (α, β). The preceding reasoning also shows that B(α, β), the normalizing factor in the beta density, is such that *

1

B(α, β) K

vα−1 (1 − v)β−1 dv

0

(α)(β) = (α + β) This entire result is quite interesting. For suppose there are n + m jobs to be performed, each (independently) taking an exponential amount of time with rate λ to be completed and suppose that we have two workers to perform these jobs. Worker I will do jobs 1, 2, . . . , n, and worker II will do the remaining m jobs. If we let X and Y denote the total working times of workers I and II, respectively, then (either from the foregoing result or from Example 3b) X and Y will be independent gamma random variables having parameters (n, λ) and (m, λ), respectively. It then follows that, independently of the working time needed to complete all n + m jobs (that is, of X + Y), the proportion of this work that will be performed by worker I has a beta distribution with parameters (n, m). . When the joint density function of the n random variables X1 , X2 , . . . , Xn is given and we want to compute the joint density function of Y1 , Y2 , . . . , Yn , where Y1 = g1 (X1 , . . . , Xn )

Y2 = g2 (X1 , . . . , Xn ), . . .

Yn = gn (X1 , . . . , Xn )

280

Chapter 6

Jointly Distributed Random Variables

the approach is the same—namely, we assume that the functions gi have continuous partial derivatives and that the Jacobian determinant. 6 6 6 ∂g ∂g 6 ∂ g 1 1 1 6 6 ··· 6 6 ∂xn 6 6 ∂x1 ∂x2 6 6 6 ∂g2 ∂g2 ∂g2 66 6 J(x1 , . . . , xn ) = 6 Z 0 ··· ∂xn 66 6 ∂x1 ∂x2 6 6 ∂gn 6 6 ∂gn ∂gn 6 6 ··· 6 ∂x1 ∂x2 ∂xn 6 at all points (x1 , . . . , xn ). Furthermore, we suppose that the equations y1 = g1 (x1 , . . . , xn ), y2 = g2 (x1 , . . . , xn ), . . . , yn = gn (x1 , . . . , xn ) have a unique solution, say, x1 = h1 (y1 , . . . , yn ), . . . , xn = hn (y1 , . . . , yn ). Under these assumptions, the joint density function of the random variables Yi is given by fY1 ,...,Yn (y1 , . . . , yn ) = fX1 ,...,Xn (x1 , . . . , xn )|J(x1 , . . . , xn )|−1

(7.3)

where xi = hi (y1 , . . . , yn ), i = 1, 2, . . . , n. EXAMPLE 7d Let X1 , X2 , and X3 be independent standard normal random variables. If Y1 = X1 + X2 + X3 , Y2 = X1 − X2 , and Y3 = X1 − X3 , compute the joint density function of Y1 , Y2 , Y3 . Solution. Letting Y1 = X1 + X2 + X3 , Y2 = X1 − X2 , Y3 = X1 − X3 , the Jacobian of these transformations is given by 6 6 61 1 16 6 6 J = 66 1 −1 0 66 = 3 6 1 0 −1 6 As the preceding transformations yield that X1 =

Y1 + Y2 + Y3 3

X2 =

Y1 − 2Y2 + Y3 3

X3 =

Y1 + Y2 − 2Y3 3

we see from Equation (7.3) that fY1 , Y2 , Y3 (y1 , y2 , y3 ) 1 y1 + y2 + y3 y1 − 2y2 + y3 y1 + y2 − 2y3 = fX1 , X2 , X3 , , 3 3 3 3 Hence, as fX1 , X2 , X3 (x1 , x2 , x3 ) =

1 − 3i=1 x2i /2 e (2π)3/2

we see that fY1 , Y2 , Y3 (y1 , y2 , y3 ) =

1 e−Q(y1 , y2 , y3 )/2 3(2π)3/2

Section 6.7

Joint Distribution of Functions of Random Variables

281

where Q(y1 , y2 , y3 ) = =

y1 + y2 + y3 3

2

+

y1 − 2y2 + y3 3

2

+

y1 + y2 − 2y3 3

y21 2 2 2 + y22 + y23 − y2 y3 3 3 3 3

2

.

EXAMPLE 7e Let X1 , X2 , . . . , Xn be independent and identically distributed exponential random variables with rate λ. Let Yi = X1 + · · · + Xi

i = 1, . . . , n

(a) Find the joint density function of Y1 , . . . , Yn . (b) Use the result of part (a) to ﬁnd the density of Yn . Solution. (a) The Jacobian of the transformations Y1 Yn = X1 + · · · + Xn is 6 6 1 0 0 0 ··· 6 6 1 1 0 0 ··· 6 6 1 1 1 0 ··· J = 66 ··· 6 ··· 6 ··· ··· 6 6 1 1 1 1 ···

= X1 , Y2 = X1 + X2 , . . ., 6 0 66 0 66 0 66 6 6 6 6 16

Since only the ﬁrst term of the determinant will be nonzero, we have J = 1. Now, the joint density function of X1 , . . . , Xn is given by fX1 ,...,Xn (x1 , . . . , xn ) =

n

λe−λxi

0 < xi < q, i = 1, . . . , n

i=1

Hence, because the preceding transformations yield X1 = Y1 , X2 = Y2 − Y1 , . . . , Xi = Yi − Yi−1 , . . . , Xn = Yn − Yn−1 it follows from Equation (7.3) that the joint density function of Y1 , . . . , Yn is fY1 ,...,Yn (y1 , y2 , . . . , yn ) = fX1 ,...,Xn (y1 , y2 − y1 , . . . , yi − yi−1 , . . . , yn − yn−1 ) ⎧ ⎡ ⎤⎫ ⎪ ⎪ n ⎨ ⎬ n = λ exp −λ ⎣y1 + (yi − yi−1 )⎦ ⎪ ⎪ ⎩ ⎭ i=2 = λn e−λyn n −λyn

=λ e

0 < y1 , 0 < yi − yi−1 , i = 2, . . . , n 0 < y1 < y2 < · · · < yn

282

Chapter 6

Jointly Distributed Random Variables

(b) To obtain the marginal density of Yn , let us integrate out the other variables one at a time. Doing this gives * y2 λn e−λyn dy1 fY2 ,...,Yn (y2 , . . . , yn ) = 0

= λn y2 e−λyn Continuing, we obtain

*

fY3 ,...,Yn (y3 , . . . , yn ) =

y3

0 < y2 < y3 < · · · < yn

λn y2 e−λyn dy2

0

= λn

y23 −λyn e 2

0 < y3 < y4 < · · · < yn

The next integration yields fY4 ,...,Yn (y4 , . . . , yn ) = λn

y34 −λyn e 3!

0 < y4 < · · · < yn

Continuing in this fashion gives fYn (yn ) = λn

yn−1 n e−λyn (n − 1)!

0 < yn

which, in agreement with the result obtained in Example 3b, shows that X1 + . · · · + Xn is a gamma random variable with parameters n and λ. ∗ 6.8

EXCHANGEABLE RANDOM VARIABLES The random variables X1 , X2 , . . . , Xn are said to be exchangeable if, for every permutation i1 , . . . , in of the integers 1, . . . , n, P{Xi1 … x1 , Xi2 … x2 , . . . , Xin … xn } = P{X1 … x1 , X2 … x2 , . . . , Xn … xn } for all x1 , . . . , xn . That is, the n random variables are exchangeable if their joint distribution is the same no matter in which order the variables are observed. Discrete random variables will be exchangeable if P{Xi1 = x1 , Xi2 = x2 , . . . , Xin = xn } = P{X1 = x1 , X2 = x2 , . . . , Xn = xn } for all permutations i1 , . . . , in , and all values x1 , . . . , xn . This is equivalent to stating that p(x1 , x2 , . . . , xn ) = P{X1 = x1 , . . . , Xn = xn } is a symmetric function of the vector (x1 , . . . , xn ), which means that its value does not change when the values of the vector are permuted. EXAMPLE 8a Suppose that balls are withdrawn one at a time and without replacement from an urn that initially contains n balls, of which k are considered special, in such a manner that each withdrawal is equally likely to be any of the balls that remain in the urn at the time. Let Xi = 1 if the ith ball withdrawn is special and let Xi = 0 otherwise. We will show that the random variables X1 , . . . , Xn are exchangeable. To do so, let (x1 , . . . , xn ) be a vector consisting of k ones and n − k zeros. However, before considering the joint mass function evaluated at (x1 , . . . , xn ), let us try to gain some insight by

Section 6.8

Exchangeable Random Variables

283

considering a ﬁxed such vector—for instance, consider the vector (1, 1, 0, 1, 0, . . . , 0, 1), which is assumed to have k ones and n − k zeros. Then p(1, 1, 0, 1, 0, . . . , 0, 1) =

11 kk − 1n − kk − 2n − k − 1 ··· nn − 1n − 2n − 3 n − 4 21

which follows because the probability that the ﬁrst ball is special is k/n, the conditional probability that the next one is special is (k − 1)/(n − 1), the conditional probability that the next one is not special is (n − k)/(n − 2), and so on. By the same argument, it follows that p(x1 , . . . , xn ) can be expressed as the product of n fractions. The successive denominator terms of these fractions will go from n down to 1. The numerator term at the location where the vector (x1 , . . . , xn ) is 1 for the ith time is k − (i − 1), and where it is 0 for the ith time it is n − k − (i − 1). Hence, since the vector (x1 , . . . , xn ) consists of k ones and n − k zeros, we obtain p(x1 , . . . , xn ) =

k!(n − k)! n!

xi = 0, 1,

n

xi = k

i=1

Since this is a symmetric function of (x1 , . . . , xn ), it follows that the random variables are exchangeable. . Remark. Another way to obtain the preceding formula for the joint probability mass function is to regard all the n balls as distinguishable from each other. Then, since the outcome of the experiment is an ordering of these balls, it follows that there are n! equally likely outcomes. Finally, because the number of outcomes having special and nonspecial balls in speciﬁed places is equal to the number of ways of permuting the special and the nonspecial balls among themselves, namely k!(n − k)!, we obtain the preceding density function. . It is easily seen that if X1 , X2 , . . . , Xn are exchangeable, then each Xi has the same probability distribution. For instance, if X and Y are exchangeable discrete random variables, then P{X = x, Y = y} = P{X = y, Y = x} = P{Y = x} P{X = x} = y

y

For example, it follows from Example 8a that the ith ball withdrawn will be special with probability k/n, which is intuitively clear, since each of the n balls is equally likely to be the ith one selected. EXAMPLE 8b In Example 8a, let Yi denote the selection number of the ﬁrst special ball withdrawn, let Y2 denote the additional number of balls that are then withdrawn until the second special ball appears, and, in general, let Yi denote the additional number of balls withdrawn after the (i − 1)st special ball is selected until the ith is selected, i = 1, . . . , k. For instance, if n = 4, k = 2 and X1 = 1, X2 = 0, X3 = 0, X4 = 1, then Y1 = 1, Y2 = 3. Now, Y1 = i1 , Y2 = i2 , . . . , Yk = ik 3 Xi1 = Xi1 +i2 = · · · = Xi1 +···+ik = 1, Xj = 0, otherwise; thus, from the joint mass function of the Xi , we obtain P{Y1 = i1 , Y2 = i2 , . . . , Yk = ik } =

k!(n − k)! n!

i1 + · · · + ik … n

Hence, the random variables Y1 , . . . , Yk are exchangeable. Note that it follows from this result that the number of cards one must select from a well-shufﬂed deck until

284

Chapter 6

Jointly Distributed Random Variables

an ace appears has the same distribution as the number of additional cards one must select after the ﬁrst ace appears until the next one does, and so on. . EXAMPLE 8c The following is known as Polya’s urn model: Suppose that an urn initially contains n red and m blue balls. At each stage, a ball is randomly chosen, its color is noted, and it is then replaced along with another ball of the same color. Let Xi = 1 if the ith ball selected is red and let it equal 0 if the ith ball is blue, i Ú 1. To obtain a feeling for the joint probabilities of these Xi , note the following special cases: P{X1 = 1, X2 = 1, X3 = 0, X4 = 1, X5 = 0} n + 1 m n + 2 m + 1 n = n + m n + m + 1 n + m + 2 n + m + 3 n + m + 4 n(n + 1)(n + 2)m(m + 1) = (n + m)(n + m + 1)(n + m + 2)(n + m + 3)(n + m + 4) and P{X1 = 0, X2 = 1, X3 = 0, X4 = 1, X5 = 1} n m + 1 n + 1 n + 2 m = n + m n + m + 1 n + m + 2 n + m + 3 n + m + 4 n(n + 1)(n + 2)m(m + 1) = (n + m)(n + m + 1)(n + m + 2)(n + m + 3)(n + m + 4) By the same reasoning, for any sequence x1 , . . . , xk that contains r ones and k − r zeros, we have P{X1 = x1 , . . . , Xk = xk } n(n + 1) · · · (n + r − 1)m(m + 1) · · · (m + k − r − 1) = (n + m) · · · (n + m + k − 1) Therefore, for any value of k, the random variables X1 , . . . , Xk are exchangeable. An interesting corollary of the exchangeability in this model is that the probability that the ith ball selected is red is the same as the probability that the ﬁrst ball selected n . (For an intuitive argument for this initially nonintuitive result, is red, namely, n+m imagine that all the n + m balls initially in the urn are of different types. That is, one is a red ball of type 1, one is a red ball of type 2, . . ., one is a red ball type of n, one is a blue ball of type 1, and so on, down to the blue ball of type m. Suppose that when a ball is selected it is replaced along with another of its type. Then, by symmetry, the ith ball selected is equally likely to be of any of the n + m distinct types. Because n n .) . of these n + m types are red, the probability is n+m Our ﬁnal example deals with continuous random variables that are exchangeable. EXAMPLE 8d Let X1 , X2 , . . . , Xn be independent uniform (0, 1) random variables, and denote their order statistics by X(1) , . . . , X(n) . That is, X(j) is the jth smallest of X1 , X2 , . . . , Xn .

Summary

285

Also, let Y1 = X(1) , Yi = X(i) − X(i−1) ,

i = 2, . . . n

Show that Y1 , . . . , Yn are exchangeable. Solution. The transformations y1 = x1 , . . . , yi = xi − xi−1 yield

xi = y1 + · · · + yi

i = 2, . . . , n

i = 1, . . . , n

As it is easy to see that the Jacobian of the preceding transformations is equal to 1, so, from Equation (7.3), we obtain fY1 ,...,Yn (y1 , y2 , . . . , yn ) = f (y1 , y1 + y2 , . . . , y1 + · · · + yn ) where f is the joint density function of the order statistics. Hence, from Equation (6.1), we obtain that fY1 ,...,Yn (y1 , y2 , . . . , yn ) = n! 0 < y1 < y1 + y2 < · · · < y1 + · · · + yn < 1 or, equivalently, fY1 ,...,Yn (y1 , y2 , . . . , yn ) = n!

0 < yi < 1, i = 1, . . . , n,

y1 + · · · + yn < 1

Because the preceding joint density is a symmetric function of y1 , . . . , yn , we see that the random variables Y1 , . . . , Yn are exchangeable. . SUMMARY The joint cumulative probability distribution function of the pair of random variables X and Y is deﬁned by F(x, y) = P{X … x, Y … y}

− q < x, y < q

All probabilities regarding the pair can be obtained from F. To ﬁnd the individual probability distribution functions of X and Y, use FX (x) = lim F(x, y) y→q

FY (y) = lim F(x, y) x→q

If X and Y are both discrete random variables, then their joint probability mass function is deﬁned by p(i, j) = P{X = i, Y = j} The individual mass functions are p(i, j) P{X = i} =

P{Y = j} =

j

p(i, j)

i

The random variables X and Y are said to be jointly continuous if there is a function f (x, y), called the joint probability density function, such that for any two-dimensional set C, ** f (x, y) dx dy

P{(X, Y) ∈ C} = C

286

Chapter 6

Jointly Distributed Random Variables

It follows from the preceding formula that P{x < X < x + dx, y < Y < y + dy} L f (x, y) dx dy If X and Y are jointly continuous, then they are individually continuous with density functions * q * q f (x, y)dy fY (y) = f (x, y) dx fX (x) = −q

−q

The random variables X and Y are independent if, for all sets A and B, P{X ∈ A, Y ∈ B} = P{X ∈ A}P{Y ∈ B} If the joint distribution function (or the joint probability mass function in the discrete case, or the joint density function in the continuous case) factors into a part depending only on x and a part depending only on y, then X and Y are independent. In general, the random variables X1 , . . . , Xn are independent if, for all sets of real numbers A1 , . . . , An , P{X1 ∈ A1 , . . . , Xn ∈ An } = P{X1 ∈ A1 } · · · P{Xn ∈ An } If X and Y are independent continuous random variables, then the distribution function of their sum can be obtained from the identity * q FX (a − y)fY (y)dy FX+Y (a) = −q

If Xi , i = 1, . . . , n, are independent normal random variables with respective paramn n n eters μi and σi2 , i = 1, . . . , n, then Xi is normal with parameters μi and σi2 . i=1

i=1

i=1

If Xi , i = 1, . . . , n, are independent Poisson random variables with respective paramn n eters λi , i = 1, . . . , n, then Xi is Poisson with parameter λi . i=1

i=1

If X and Y are discrete random variables, then the conditional probability mass function of X given that Y = y is deﬁned by P{X = x|Y = y} =

p(x, y) pY (y)

where p is their joint probability mass function. Also, if X and Y are jointly continuous with joint density function f , then the conditional probability density function of X given that Y = y is given by fX|Y (x|y) =

f (x, y) fY (y)

The ordered values X(1) … X(2) … · · · … X(n) of a set of independent and identically distributed random variables are called the order statistics of that set. If the random variables are continuous and have density function f , then the joint density function of the order statistics is f (x1 , . . . , xn ) = n!f (x1 ) · · · f (xn )

x1 … x2 … · · · … xn

The random variables X1 , . . . , Xn are called exchangeable if the joint distribution of Xi1 , . . . , Xin is the same for every permutation i1 , . . . , in of 1, . . . , n.

Problems

287

PROBLEMS 6.1. Two fair dice are rolled. Find the joint probability mass function of X and Y when (a) X is the largest value obtained on any die and Y is the sum of the values; (b) X is the value on the ﬁrst die and Y is the larger of the two values; (c) X is the smallest and Y is the largest value obtained on the dice. 6.2. Suppose that 3 balls are chosen without replacement from an urn consisting of 5 white and 8 red balls. Let Xi equal 1 if the ith ball selected is white, and let it equal 0 otherwise. Give the joint probability mass function of (a) X1 , X2 ; (b) X1 , X2 , X3 . 6.3. In Problem 2, suppose that the white balls are numbered, and let Yi equal 1 if the ith white ball is selected and 0 otherwise. Find the joint probability mass function of (a) Y1 , Y2 ; (b) Y1 , Y2 , Y3 . 6.4. Repeat Problem 2 when the ball selected is replaced in the urn before the next selection. 6.5. Repeat Problem 3a when the ball selected is replaced in the urn before the next selection. 6.6. A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defective ones are identiﬁed. Denote by N1 the number of tests made until the ﬁrst defective is identiﬁed and by N2 the number of additional tests until the second defective is identiﬁed. Find the joint probability mass function of N1 and N2 . 6.7. Consider a sequence of independent Bernoulli trials, each of which is a success with probability p. Let X1 be the number of failures preceding the ﬁrst success, and let X2 be the number of failures between the ﬁrst two successes. Find the joint mass function of X1 and X2 . 6.8. The joint probability density function of X and Y is given by f (x, y) = c(y2 − x2 )e−y

− y … x … y, 0 < y < q

(a) Find c. (b) Find the marginal densities of X and Y. (c) Find E[X]. 6.9. The joint probability density function of X and Y is given by xy 6 2 x + 0 < x < 1, 0 < y < 2 f (x, y) = 7 2

(a) Verify that this is indeed a joint density function. (b) Compute the density function of X. (c) Find P{X > Y}. (d) Find P{Y > 12 |X < 12 }. (e) Find E[X]. (f) Find E[Y]. 6.10. The joint probability density function of X and Y is given by f (x, y) = e−(x+y)

0 … x < q, 0 … y < q

Find (a) P{X < Y} and (b) P{X < a}. 6.11. A television store owner ﬁgures that 45 percent of the customers entering his store will purchase an ordinary television set, 15 percent will purchase a plasma television set, and 40 percent will just be browsing. If 5 customers enter his store on a given day, what is the probability that he will sell exactly 2 ordinary sets and 1 plasma set on that day? 6.12. The number of people that enter a drugstore in a given hour is a Poisson random variable with parameter λ = 10. Compute the conditional probability that at most 3 men entered the drugstore, given that 10 women entered in that hour. What assumptions have you made? 6.13. A man and a woman agree to meet at a certain location about 12:30 P . M . If the man arrives at a time uniformly distributed between 12:15 and 12:45, and if the woman independently arrives at a time uniformly distributed between 12:00 and 1 P . M ., ﬁnd the probability that the ﬁrst to arrive waits no longer than 5 minutes. What is the probability that the man arrives ﬁrst? 6.14. An ambulance travels back and forth at a constant speed along a road of length L. At a certain moment of time, an accident occurs at a point uniformly distributed on the road. [That is, the distance of the point from one of the ﬁxed ends of the road is uniformly distributed over (0, L).] Assuming that the ambulance’s location at the moment of the accident is also uniformly distributed, and assuming independence of the variables, compute the distribution of the distance of the ambulance from the accident. 6.15. The random vector (X, Y) is said to be uniformly distributed over a region R in the plane if, for some constant c, its joint density is % c if(x, y) ∈ R f (x, y) = 0 otherwise

288

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(a) Show that 1/c = area of region R. Suppose that (X, Y) is uniformly distributed over the square centered at (0, 0) and with sides of length 2. (b) Show that X and Y are independent, with each being distributed uniformly over (−1, 1). (c) What is the probability that (X, Y) lies in the circle of radius 1 centered at the origin? That is, ﬁnd P{X 2 + Y 2 … 1}. 6.16. Suppose that n points are independently chosen at random on the circumference of a circle, and we want the probability that they all lie in some semicircle. That is, we want the probability that there is a line passing through the center of the circle such that all the points are on one side of that line, as shown in the following diagram:

Let P1 , . . . , Pn denote the n points. Let A denote the event that all the points are contained in some semicircle, and let Ai be the event that all the points lie in the semicircle beginning at the point Pi and going clockwise for 180◦ , i = 1, . . . , n. (a) Express A in terms of the Ai . (b) Are the Ai mutually exclusive? (c) Find P(A). 6.17. Three points X1 , X2 , X3 are selected at random on a line L. What is the probability that X2 lies between X1 and X3 ? 6.18. Two points are selected randomly on a line of length L so as to be on opposite sides of the midpoint of the line. [In other words, the two points X and Y are independent random variables such that X is uniformly distributed over (0, L/2) and Y is uniformly distributed over (L/2, L).] Find the probability that the distance between the two points is greater than L/3. 6.19. Show that f (x, y) = 1/x, 0 < y < x < 1, is a joint density function. Assuming that f is the joint density function of X, Y, ﬁnd (a) the marginal density of Y; (b) the marginal density of X; (c) E[X]; (c) E[Y].

6.20. The joint density of X and Y is given by 0 x > 0, y > 0 xe−(x+y) f (x, y) = 0 otherwise Are X and Y independent? If, instead, f (x, y) were given by % 2 0 < x < y, 0 < y < 1 f (x, y) = 0 otherwise would X and Y be independent? 6.21. Let f (x, y) = 24xy 0 … x … 1, 0 … y … 1, 0 … x + y … 1 and let it equal 0 otherwise. (a) Show that f (x, y) is a joint probability density function. (b) Find E[X]. (c) Find E[Y]. 6.22. The joint density function of X and Y is % x + y 0 < x < 1, 0 < y < 1 f (x, y) = 0 otherwise (a) Are X and Y independent? (b) Find the density function of X. (c) Find P{X + Y < 1}. 6.23. The random variables X and Y have joint density function f (x, y) = 12xy(1 − x) 0 < x < 1, 0 < y < 1 and equal to 0 otherwise. (a) Are X and Y independent? (b) Find E[X]. (c) Find E[Y]. (d) Find Var(X). (e) Find Var(Y). 6.24. Consider independent trials, each of which results in outcome i, i = 0, 1, . . . , k, with probability k pi = 1. Let N denote the number of trials pi , i=0

needed to obtain an outcome that is not equal to 0, and let X be that outcome. (a) Find P{N = n}, n Ú 1. (b) Find P{X = j}, j = 1, . . . , k. (c) Show that P{N = n, X = j} = P{N = n}P{X = j}. (d) Is it intuitive to you that N is independent of X? (e) Is it intuitive to you that X is independent of N? 6.25. Suppose that 106 people arrive at a service station at times that are independent random variables,

Problems

6.26.

6.27.

6.28.

6.29.

6.30.

6.31.

each of which is uniformly distributed over (0, 106 ). Let N denote the number that arrive in the ﬁrst hour. Find an approximation for P{N = i}. Suppose that A, B, C, are independent random variables, each being uniformly distributed over (0, 1). (a) What is the joint cumulative distribution function of A, B, C? (b) What is the probability that all of the roots of the equation Ax2 + Bx + C = 0 are real? If X1 and X2 are independent exponential random variables with respective parameters λ1 and λ2 , ﬁnd the distribution of Z = X1 /X2 . Also compute P{X1 < X2 }. The time that it takes to service a car is an exponential random variable with rate 1. (a) If A. J. brings his car in at time 0 and M. J. brings her car in at time t, what is the probability that M. J.’s car is ready before A. J.’s car? (Assume that service times are independent and service begins upon arrival of the car.) (b) If both cars are brought in at time 0, with work starting on M. J.’s car only when A. J.’s car has been completely serviced, what is the probability that M. J.’s car is ready before time 2? The gross weekly sales at a certain restaurant is a normal random variable with mean $2200 and standard deviation $230. What is the probability that (a) the total gross sales over the next 2 weeks exceeds $5000; (b) weekly sales exceed $2000 in at least 2 of the next 3 weeks? What independence assumptions have you made? Jill’s bowling scores are approximately normally distributed with mean 170 and standard deviation 20, while Jack’s scores are approximately normally distributed with mean 160 and standard deviation 15. If Jack and Jill each bowl one game, then assuming that their scores are independent random variables, approximate the probability that (a) Jack’s score is higher; (b) the total of their scores is above 350. According to the U.S. National Center for Health Statistics, 25.2 percent of males and 23.6 percent of females never eat breakfast. Suppose that random samples of 200 men and 200 women are chosen. Approximate the probability that (a) at least 110 of these 400 people never eat breakfast; (b) the number of the women who never eat breakfast is at least as large as the number of the men who never eat breakfast.

289

6.32. The expected number of typographical errors on a page of a certain magazine is .2. What is the probability that an article of 10 pages contains (a) 0 and (b) 2 or more typographical errors? Explain your reasoning! 6.33. The monthly worldwide average number of airplane crashes of commercial airlines is 2.2. What is the probability that there will be (a) more than 2 such accidents in the next month? (b) more than 4 such accidents in the next 2 months? (c) more than 5 such accidents in the next 3 months? Explain your reasoning! 6.34. Jay has two jobs to do, one after the other. Each attempt at job i takes one hour and is successful with probability pi . If p1 = .3 and p2 = .4, what is the probability that it will take Jay more than 12 hours to be successful on both jobs? 6.35. In Problem 4, calculate the conditional probability mass function of X1 given that (a) X2 = 1; (b) X2 = 0. 6.36. In Problem 3, calculate the conditional probability mass function of Y1 given that (a) Y2 = 1; (b) Y2 = 0. 6.37. In Problem 5, calculate the conditional probability mass function of Y1 given that (a) Y2 = 1; (b) Y2 = 0. 6.38. Choose a number X at random from the set of numbers {1, 2, 3, 4, 5}. Now choose a number at random from the subset no larger than X, that is, from {1, . . . , X}. Call this second number Y. (a) Find the joint mass function of X and Y. (b) Find the conditional mass function of X given that Y = i. Do it for i = 1, 2, 3, 4, 5. (c) Are X and Y independent? Why? 6.39. Two dice are rolled. Let X and Y denote, respectively, the largest and smallest values obtained. Compute the conditional mass function of Y given X = i, for i = 1, 2, . . . , 6. Are X and Y independent? Why? 6.40. The joint probability mass function of X and Y is given by 1 8 1 p(2, 1) = 8 p(1, 1) =

1 4 1 p(2, 2) = 2 p(1, 2) =

(a) Compute the conditional mass function of X given Y = i, i = 1, 2. (b) Are X and Y independent?

290

Chapter 6

Jointly Distributed Random Variables

(c) Compute P{XY … 3}, P{X + Y > 2}, P{X/Y > 1}. 6.41. The joint density function of X and Y is given by f (x, y) = xe−x(y+1)

x > 0, y > 0

(a) Find the conditional density of X, given Y = y, and that of Y, given X = x. (b) Find the density function of Z = XY. 6.42. The joint density of X and Y is 2

2

f (x, y) = c(x − y )e

6.43.

6.44.

6.45.

6.46.

6.47.

−x

6.50. Let Z1 and Z2 be independent standard normal random variables. Show that X, Y has a bivariate normal distribution when X = Z1 , Y = Z1 + Z2 . 6.51. Derive the distribution of the range of a sample of size 2 from a distribution having density function f (x) = 2x, 0 < x < 1. 6.52. Let X and Y denote the coordinates of a point uniformly chosen in the circle of radius 1 centered at the origin. That is, their joint density is f (x, y) =

0 … x < q, −x … y … x

Find the conditional distribution of Y, given X = x. An insurance company supposes that each person has an accident parameter and that the yearly number of accidents of someone whose accident parameter is λ is Poisson distributed with mean λ. They also suppose that the parameter value of a newly insured person can be assumed to be the value of a gamma random variable with parameters s and α. If a newly insured person has n accidents in her ﬁrst year, ﬁnd the conditional density of her accident parameter. Also, determine the expected number of accidents that she will have in the following year. If X1 , X2 , X3 are independent random variables that are uniformly distributed over (0, 1), compute the probability that the largest of the three is greater than the sum of the other two. A complex machine is able to operate effectively as long as at least 3 of its 5 motors are functioning. If each motor independently functions for a random amount of time with density function f (x) = xe−x , x > 0, compute the density function of the length of time that the machine functions. If 3 trucks break down at points randomly distributed on a road of length L, ﬁnd the probability that no 2 of the trucks are within a distance d of each other when d … L/2. Consider a sample of size 5 from a uniform distribution over (0, 1). Compute the probability that the median is in the interval 14 , 34 .

6.48. If X1 , X2 , X3 , X4 , X5 are independent and identically distributed exponential random variables with the parameter λ, compute (a) P{min(X1 , . . . , X5 ) … a}; (b) P{max(X1 , . . . , X5 ) … a}. 6.49. Let X(1) , X(2) , . . . , X(n) be the order statistics of a set of n independent uniform (0, 1) random variables. Find the conditional distribution of X(n) given that X(1) = s1 , X(2) = s2 , . . . , X(n−1) = sn−1 .

1 π

x2 + y 2 … 1

Find the joint density function of the polar coordinates R = (X 2 + Y 2 )1/2 and = tan−1 Y/X. 6.53. If X and Y are independent random variables both uniformly distributed over. (0, 1), ﬁnd the joint density function of R = X 2 + Y 2 , Θ = tan−1 Y/X. 6.54. If U is uniform on (0, 2π) and Z, independent of U, is exponential with rate 1, show directly (without using the results of Example 7b) that X and Y deﬁned by √ X = 2Z cos U √ Y = 2Z sin U are independent standard normal random variables. 6.55. X and Y have joint density function f (x, y) =

6.56.

6.57.

6.58.

6.59.

1 x2 y2

x Ú 1, y Ú 1

(a) Compute the joint density function of U = XY, V = X/Y. (b) What are the marginal densities? If X and Y are independent and identically distributed uniform random variables on (0, 1), compute the joint density of (a) U = X + Y, V = X/Y; (b) U = X, V = X/Y; (c) U = X + Y, V = X/(X + Y). Repeat Problem 6.56 when X and Y are independent exponential random variables, each with parameter λ = 1. If X1 and X2 are independent exponential random variables, each having parameter λ, ﬁnd the joint density function of Y1 = X1 + X2 and Y2 = eX1 . If X, Y, and Z are independent random variables having identical density functions f (x) = e−x , 0 < x < q, derive the joint distribution of U = X + Y, V = X + Z, W = Y + Z.

Theoretical Exercises 6.60. In Example 8b, let Yk+1 = n + 1 −

k

Yi . Show

i=1

that Y1 , . . . , Yk , Yk+1 are exchangeable. Note that Yk+1 is the number of balls one must observe to obtain a special ball if one considers the balls in their reverse order of withdrawal.

291

6.61. Consider an urn containing n balls numbered 1, . . . , n, and suppose that k of them are randomly withdrawn. Let Xi equal 1 if ball number i is removed and let Xi be 0 otherwise. Show that X1 , . . . , Xn are exchangeable.

THEORETICAL EXERCISES 6.1. Verify Equation (1.2). 6.2. Suppose that the number of events occurring in a given time period is a Poisson random variable with parameter λ. If each event is classiﬁed as a type i event with probability pi , i = 1, . . . , n, pi = 1, independently of other events, show that the numbers of type i events that occur, i = 1, . . . , n, are independent Poisson random variables with respective parameters λpi , i = 1, . . . , n. 6.3. Suggest a procedure for using Buffon’s needle problem to estimate π. Surprisingly enough, this was once a common method of evaluating π. 6.4. Solve Buffon’s needle problem when L > D. 2L (1 − sin θ ) + 2θ/π, where cos θ = ANSWER : πD D/L. 6.5. If X and Y are independent continuous positive random variables, express the density function of (a) Z = X/Y and (b) Z = XY in terms of the density functions of X and Y. Evaluate the density functions in the special case where X and Y are both exponential random variables. 6.6. If X and Y are jointly continuous with joint density function fX,Y (x, y), show that X + Y is continuous with density function * q fX,Y (x, t − x) dx fX+Y (t) = −q

6.7. (a) If X has a gamma distribution with parameters (t, λ), what is the distribution of cX, c > 0? (b) Show that 1 2 χ 2λ 2n has a gamma distribution with parameters n, λ 2 is a when n is a positive integer and χ2n chi-squared random variable with 2n degrees of freedom. 6.8. Let X and Y be independent continuous random variables with respective hazard rate functions λX (t) and λY (t), and set W = min(X, Y). (a) Determine the distribution function of W in terms of those of X and Y.

(b) Show that λW (t), the hazard rate function of W, is given by λW (t) = λX (t) + λY (t) 6.9. Let X1 , . . . , Xn be independent exponential random variables having a common parameter λ. Determine the distribution of min(X1 , . . . , Xn ). 6.10. The lifetimes of batteries are independent exponential random variables, each having parameter λ. A ﬂashlight needs 2 batteries to work. If one has a ﬂashlight and a stockpile of n batteries, what is the distribution of time that the ﬂashlight can operate? 6.11. Let X1 , X2 , X3 , X4 , X5 be independent continuous random variables having a common distribution function F and density function f , and set I = P{X1 < X2 < X3 < X4 < X5 } (a) Show that I does not depend on F. Hint: Write I as a ﬁve-dimensional integral and make the change of variables ui = F(xi ), i = 1, . . . , 5. (b) Evaluate I. (c) Give an intuitive explanation for your answer to (b). 6.12. Show that the jointly continuous (discrete) random variables X1 , . . . , Xn are independent if and only if their joint probability density (mass) function f (x1 , . . . , xn ) can be written as f (x1 , . . . , xn ) =

n

gi (xi )

i=1

for nonnegative functions gi (x), i = 1, . . . , n. 6.13. In Example 5c we computed the conditional density of a success probability for a sequence of trials when the ﬁrst n + m trials resulted in n successes. Would the conditional density change if we speciﬁed which n of these trials resulted in successes?

292

Chapter 6

Jointly Distributed Random Variables

6.14. Suppose that X and Y are independent geometric random variables with the same parameter p. (a) Without any computations, what do you think is the value of P{X = i|X + Y = n}?

6.15.

6.16.

6.17.

6.18.

Hint: Imagine that you continually ﬂip a coin having probability p of coming up heads. If the second head occurs on the nth ﬂip, what is the probability mass function of the time of the ﬁrst head? (b) Verify your conjecture in part (a). Consider a sequence of independent trials, with each trial being a success with probability p. Given that the kth success occurs on trial n, show that all possible outcomes of the ﬁrst n − 1 trials that consist of k − 1 successes and n − k failures are equally likely. If X and Y are independent binomial random variables with identical parameters n and p, show analytically that the conditional distribution of X given that X + Y = m is the hypergeometric distribution. Also, give a second argument that yields the same result without any computations. Hint: Suppose that 2n coins are ﬂipped. Let X denote the number of heads in the ﬁrst n ﬂips and Y the number in the second n ﬂips. Argue that given a total of m heads, the number of heads in the ﬁrst n ﬂips has the same distribution as the number of white balls selected when a sample of size m is chosen from n white and n black balls. Suppose that Xi , i = 1, 2, 3 are independent Poisson random variables with respective means λi , i = 1, 2, 3. Let X = X1 + X2 and Y = X2 + X3 . The random vector X, Y is said to have a bivariate Poisson distribution. Find its joint probability mass function. That is, ﬁnd P{X = n, Y = m}. Suppose X and Y are both integer-valued random variables. Let p(i|j) = P(X = i|Y = j) and q(j|i) = P(Y = j|X = i) Show that p(i|j) P(X = i, Y = j) = p(i|j) i q(j|i)

6.19. Let X1 , X2 , X3 be independent and identically distributed continuous random variables. Compute (a) P{X1 > X2 |X1 > X3 }; (b) P{X1 > X2 |X1 < X3 }; (c) P{X1 > X2 |X2 > X3 }; (d) P{X1 > X2 |X2 < X3 }.

6.20. Let U denote a random variable uniformly distributed over (0, 1). Compute the conditional distribution of U given that (a) U > a; (b) U < a; where 0 < a < 1. 6.21. Suppose that W, the amount of moisture in the air on a given day, is a gamma random variable with parameters (t, β). That is, its density is f (w) = βe−βw (βw)t−1 / (t), w > 0. Suppose also that given that W = w, the number of accidents during that day—call it N—has a Poisson distribution with mean w. Show that the conditional distribution of W given that N = n is the gamma distribution with parameters (t + n, β + 1). 6.22. Let W be a gamma random variable with parameters (t, β), and suppose that conditional on W = w, X1 , X2 , . . . , Xn are independent exponential random variables with rate w. Show that the conditional distribution of W given that X1 = x1 , X2 = x2 , . . . , Xn = xn is gamma with paramen ters t + n, β + xi . i=1

6.23. A rectangular array of mn numbers arranged in n rows, each consisting of m columns, is said to contain a saddlepoint if there is a number that is both the minimum of its row and the maximum of its column. For instance, in the array 1 0 .5

3 −2 12

2 6 3

the number 1 in the ﬁrst row, ﬁrst column is a saddlepoint. The existence of a saddlepoint is of signiﬁcance in the theory of games. Consider a rectangular array of numbers as described previously and suppose that there are two individuals— A and B—that are playing the following game: A is to choose one of the numbers 1, 2, . . . , n and B one of the numbers 1, 2, . . . , m. These choices are announced simultaneously, and if A chose i and B chose j, then A wins from B the amount speciﬁed by the number in the ith row, jth column of the array. Now suppose that the array contains a saddlepoint—say the number in the row r and column k—call this number xrk . Now if player A chooses row r, then that player can guarantee herself a win of at least xrk (since xrk is the minimum number in the row r). On the other hand, if player B chooses column k, then he can guarantee that he will lose no more than xrk (since xrk is the maximum number in the column k). Hence, as A has a way of playing that guarantees her a win of xrk and as B has a way of playing that guarantees he will lose no more than xrk , it seems reasonable to take

Self-Test Problems and Exercises

6.24.

6.25.

6.26.

6.27. 6.28.

6.29. 6.30.

6.31.

these two strategies as being optimal and declare that the value of the game to player A is xrk . If the nm numbers in the rectangular array described are independently chosen from an arbitrary continuous distribution, what is the probability that the resulting array will contain a saddle-point? If X is exponential with rate λ, ﬁnd P{[X] = n, X − [X] … x}, where [x] is deﬁned as the largest integer less than or equal to x. Can you conclude that [X] and X − [X] are independent? Suppose that F(x) is a cumulative distribution function. Show that (a) F n (x) and (b) 1 − [1 − F(x)]n are also cumulative distribution functions when n is a positive integer. Hint: Let X1 , . . . , Xn be independent random variables having the common distribution function F. Deﬁne random variables Y and Z in terms of the Xi so that P{Y … x} = F n (x) and P{Z … x} = 1 − [1 − F(x)]n . Show that if n people are distributed at random along a road L miles long, then the probability that no 2 people are less than a distance D miles apart is when D … L/(n − 1), [1 − (n − 1)D/L]n . What if D > L/(n − 1)? Establish Equation (6.2) by differentiating Equation (6.4). Show that the median of a sample of size 2n + 1 from a uniform distribution on (0, 1) has a beta distribution with parameters (n + 1, n + 1). Verify Equation (6.6), which gives the joint density of X(i) and X(j) . Compute the density of the range of a sample of size n from a continuous distribution having density function f . Let X(1) … X(2) … · · · … X(n) be the ordered values of n independent uniform (0, 1) random variables.

293

Prove that for 1 … k … n + 1, P{X(k) − X(k−1) > t} = (1 − t)n where X(0) K 0, X(n+1) K t. 6.32. Let X1 , . . . , Xn be a set of independent and identically distributed continuous random variables having distribution function F, and let X(i) , i = 1, . . . , n denote their ordered values. If X, independent of the Xi , i = 1, . . . , n, also has distribution F, determine (a) P{X > X(n) }; (b) P{X > X(1) }; (c) P{X(i) < X < X(j) }, 1 … i < j … n. 6.33. Let X1 , . . . , Xn be independent and identically distributed random variables having distribution function F and density f . The quantity M K [X(1) + X(n) ]/2, deﬁned to be the average of the smallest and largest values in X1 , . . ., Xn , is called the midrange of the sequence. Show that its distribution function is * m FM (m) = n [F(2m − x) − F(x)]n−1 f (x) dx −q

6.34. Let X1 , . . . , Xn be independent uniform (0, 1) random variables. Let R = X(n) − X(1) denote the range and M = [X(n) + X(1) ]/2 the midrange of X1 , . . ., Xn . Compute the joint density function of R and M. 6.35. If X and Y are independent standard normal random variables, determine the joint density function of X U=X V= Y Then use your result to show that X/Y has a Cauchy distribution.

SELF-TEST PROBLEMS AND EXERCISES 6.1. Each throw of an unfair die lands on each of the odd numbers 1, 3, 5 with probability C and on each of the even numbers with probability 2C. (a) Find C. (b) Suppose that the die is tossed. Let X equal 1 if the result is an even number, and let it be 0 otherwise. Also, let Y equal 1 if the result is a number greater than three and let it be 0 otherwise. Find the joint probability mass function of X and Y. Suppose now that 12 independent tosses of the die are made. (c) Find the probability that each of the six outcomes occurs exactly twice.

(d) Find the probability that 4 of the outcomes are either one or two, 4 are either three or four, and 4 are either ﬁve or six. (e) Find the probability that at least 8 of the tosses land on even numbers. 6.2. The joint probability mass function of the random variables X, Y, Z is p(1, 2, 3) = p(2, 1, 1) = p(2, 2, 1) = p(2, 3, 2) =

1 4

Find (a) E[XYZ], and (b) E[XY + XZ + YZ]. 6.3. The joint density of X and Y is given by f (x, y) = C(y − x)e−y

− y < x < y,

0

294

Chapter 6

Jointly Distributed Random Variables

(a) Find C. (b) Find the density function of X. (c) Find the density function of Y. (d) Find E[X]. (e) Find E[Y]. 6.4. Let r = r1 + . . . + rk , where all ri are positive integers. Argue that if X1 , . . . , Xr has a multinomial distribution, then so does Y1 , . . . , Yk where, with r0 = 0, ri−1 +ri

Yi =

Xj ,

i … k

j=ri−1 +1

That is, Y1 is the sum of the ﬁrst r1 of the X s, Y2 is the sum of the next r2 , and so on. 6.5. Suppose that X, Y, and Z are independent random variables that are each equally likely to be either 1 or 2. Find the probability mass function of (a) XYZ, (b) XY + XZ + YZ, and (c) X 2 + YZ. 6.6. Let X and Y be continuous random variables with joint density function ⎧ ⎨ x + cy 0 < x < 1, 1 < y < 5 f (x, y) = 5 ⎩0 otherwise where c is a constant. (a) What is the value of c? (b) Are X and Y independent? (c) Find P{X + Y > 3}. 6.7. The joint density function of X and Y is % xy 0 < x < 1, 0 < y < 2 f (x, y) = 0 otherwise (a) Are X and Y independent? (b) Find the density function of X. (c) Find the density function of Y. (d) Find the joint distribution function. (e) Find E[Y]. (f) Find P{X + Y < 1}. 6.8. Consider two components and three types of shocks. A type 1 shock causes component 1 to fail, a type 2 shock causes component 2 to fail, and a type 3 shock causes both components 1 and 2 to fail. The times until shocks 1, 2, and 3 occur are independent exponential random variables with respective rates λ1 , λ2 , and λ3 . Let Xi denote the time at which component i fails, i = 1, 2. The random variables X1 , X2 are said to have a joint bivariate exponential distribution. Find P{X1 > s, X2 > t}. 6.9. Consider a directory of classiﬁed advertisements that consists of m pages, where m is very large. Suppose that the number of advertisements per

page varies and that your only method of ﬁnding out how many advertisements there are on a speciﬁed page is to count them. In addition, suppose that there are too many pages for it to be feasible to make a complete count of the total number of advertisements and that your objective is to choose a directory advertisement in such a way that each of them has an equal chance of being selected. (a) If you randomly choose a page and then randomly choose an advertisement from that page, would that satisfy your objective? Why or why not? Let n(i) denote the number of advertisements on page i, i = 1, . . . , m, and suppose that whereas these quantities are unknown, we can assume that they are all less than or equal to some speciﬁed value n. Consider the following algorithm for choosing an advertisement. Step 1. Choose a page at random. Suppose it is page X. Determine n(X) by counting the number of advertisements on page X. Step 2. “Accept” page X with probability n(X)/n. If page X is accepted, go to step 3. Otherwise, return to step 1. Step 3. Randomly choose one of the advertisements on page X.

(b)

(c)

(d)

(e)

(f)

Call each pass of the algorithm through step 1 an iteration. For instance, if the ﬁrst randomly chosen page is rejected and the second accepted, than we would have needed 2 iterations of the algorithm to obtain an advertisement. What is the probability that a single iteration of the algorithm results in the acceptance of an advertisement on page i? What is the probability that a single iteration of the algorithm results in the acceptance of an advertisement? What is the probability that the algorithm goes through k iterations, accepting the jth advertisement on page i on the ﬁnal iteration? What is the probability that the jth advertisement on page i is the advertisement obtained from the algorithm? What is the expected number of iterations taken by the algorithm?

6.10. The “random” parts of the algorithm in Self-Test Problem 8 can be written in terms of the generated values of a sequence of independent uniform (0, 1) random variables, known as random numbers. With [x] deﬁned as the largest integer less than or equal to x, the ﬁrst step can be written as follows:

Self-Test Problems and Exercises Step 1. Generate a uniform (0, 1) random variable U. Let X = [mU] + 1, and determine the value of n(X). (a) Explain why the above is equivalent to step 1 of Problem 8. Hint: What is the probability mass function of X? (b) Write the remaining steps of the algorithm in a similar style. 6.11. Let X1 , X2 , . . . be a sequence of independent uniform (0, 1) random variables. For a ﬁxed constant c, deﬁne the random variable N by N = min{n : Xn > c} Is N independent of XN ? That is, does knowing the value of the ﬁrst random variable that is greater than c affect the probability distribution of when this random variable occurs? Give an intuitive explanation for your answer. 6.12. The accompanying dartboard is a square whose sides are of length 6:

10 20 30

The three circles are all centered at the center of the board and are of radii 1, 2, and 3, respectively. Darts landing within the circle of radius 1 score 30 points, those landing outside this circle, but within the circle of radius 2, are worth 20 points, and those landing outside the circle of radius 2, but within the circle of radius 3, are worth 10 points. Darts that do not land within the circle of radius 3 do not score any points. Assuming that each dart that you throw will, independently of what occurred on your previous throws, land on a point uniformly distributed in the square, ﬁnd the probabilities of the accompanying events: (a) You score 20 on a throw of the dart. (b) You score at least 20 on a throw of the dart. (c) You score 0 on a throw of the dart. (d) The expected value of your score on a throw of the dart. (e) Both of your ﬁrst two throws score at least 10. (f) Your total score after two throws is 30. 6.13. A model proposed for NBA basketball supposes that when two teams with roughly the same record play each other, the number of points scored in a quarter by the home team minus the number

295

scored by the visiting team is approximately a normal random variable with mean 1.5 and variance 6. In addition, the model supposes that the point differentials for the four quarters are independent. Assume that this model is correct. (a) What is the probability that the home team wins? (b) What is the conditional probability that the home team wins, given that it is behind by 5 points at halftime? (c) What is the conditional probability that the home team wins, given that it is ahead by 5 points at the end of the ﬁrst quarter? 6.14. Let N be a geometric random variable with parameter p. Suppose that the conditional distribution of X given that N = n is the gamma distribution with parameters n and λ. Find the conditional probability mass function of N given that X = x. 6.15. Let X and Y be independent uniform (0, 1) random variables. (a) Find the joint density of U = X, V = X + Y. (b) Use the result obtained in part (a) to compute the density function of V. 6.16. You and three other people are to place bids for an object, with the high bid winning. If you win, you plan to sell the object immediately for 10 thousand dollars. How much should you bid to maximize your expected proﬁt if you believe that the bids of the others can be regarded as being independent and uniformly distributed between 7 and 11 thousand dollars? 6.17. Find the probability that X1 , X2 , . . . , Xn is a permutation of 1, 2, . . . , n, when X1 , X2 , . . . , Xn are independent and (a) each is equally likely to be any of the values 1, . . . , n; (b) each has the probability mass function P{Xi = j} = pj , j = 1, . . . , n. 6.18. Let X1 , . . . , Xn and Y1 , . . . , Yn be independent random vectors, with each vector being a random ordering of k ones and n − k zeroes. That is, their joint probability mass functions are P{X1 = i1 , . . . , Xn = in } = P{Y1 = i1 , . . . , Yn = in } n 1 = , ij = 0, 1, ij = k n j=1 k Let N=

n i=1

|Xi − Yi |

296

Chapter 6

Jointly Distributed Random Variables

denote the number of coordinates at which the two vectors have different values. Also, let M denote the number of values of i for which Xi = 1, Yi = 0. (a) Relate N to M. (b) What is the distribution of M? (c) Find E[N]. (d) Find Var(N). ∗ 6.19. Let Z , Z , . . . , Z be independent standard norn 2 1 mal random variables, and let Sj =

j i=1

Zi

(a) What is the conditional distribution of Sn given that Sk = y, for k = 1, . . . , n? (b) Show that, for 1 … k … n, the conditional distribution of Sk given that Sn = x is normal with mean xk/n and variance k(n − k)/n. 6.20. Let X1 , X2 , . . . be a sequence of independent and identically distributed continuous random variables. Find (a) P{X6 > X1 |X1 = max(X1 , . . . , X5 )} (b) P{X6 > X2 |X1 = max(X1 , . . . , X5 )}

C H A P T E R

7

Properties of Expectation 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9

INTRODUCTION EXPECTATION OF SUMS OF RANDOM VARIABLES MOMENTS OF THE NUMBER OF EVENTS THAT OCCUR COVARIANCE, VARIANCE OF SUMS, AND CORRELATIONS CONDITIONAL EXPECTATION CONDITIONAL EXPECTATION AND PREDICTION MOMENT GENERATING FUNCTIONS ADDITIONAL PROPERTIES OF NORMAL RANDOM VARIABLES GENERAL DEFINITION OF EXPECTATION

7.1

INTRODUCTION In this chapter, we develop and exploit additional properties of expected values. To begin, recall that the expected value of the random variable X is deﬁned by xp(x) E[X] = x

where X is a discrete random variable with probability mass function p(x), and by * q E[X] = xf (x) dx −q

when X is a continuous random variable with probability density function f(x). Since E[X] is a weighted average of the possible values of X, it follows that if X must lie between a and b, then so must its expected value. That is, if P{a … X … b} = 1 then a … E[X] … b To verify the preceding statement, suppose that X is a discrete random variable for which P{a … X … b} = 1. Since this implies that p(x) = 0 for all x outside of the interval [a, b], it follows that E[X] = xp(x) x:p(x)>0

Ú

ap(x)

x:p(x)>0

297

298

Chapter 7

Properties of Expectation

=a

p(x)

x:p(x)>0

=a In the same manner, it can be shown that E[X] … b, so the result follows for discrete random variables. As the proof in the continuous case is similar, the result follows. 7.2

EXPECTATION OF SUMS OF RANDOM VARIABLES For a two-dimensional analog of Propositions 4.1 of Chapter 4 and 2.1 of Chapter 5, which give the computational formulas for the expected value of a function of a random variable, suppose that X and Y are random variables and g is a function of two variables. Then we have the following result. Proposition 2.1. If X and Y have a joint probability mass function p(x,y), then E[g(X, Y)] = g(x, y)p(x, y) y

x

If X and Y have a joint probability density function f(x,y), then * q* q g(x, y)f (x, y) dx dy E[g(X, Y)] = −q −q

Let us give a proof of Proposition 2.1 when the random variables X and Y are jointly continuous with joint density function f (x, y) and when g(X, Y) is a nonnegative random variable. Because g(X, Y) Ú 0, we have, by Lemma 2.1 of Chapter 5, that * q P{g(X, Y) > t} dt

E[g(X, Y)] =

0

Writing

* * P{g(X, Y) > t} = *

shows that

q*

E[g(X, Y)] =

(x,y):g(x,y)>t

f (x, y) dy dx

* (x,y):g(x,y)>t

0

f (x, y) dy dx dt

Interchanging the order of integration gives * * * g(x,y) f (x, y) dt dy dx E[g(X, Y) = x y t=0 * * = g(x, y) f (x, y) dy dx x y

Thus, the result is proven when g(X, Y) is a nonnegative random variable. The general case then follows as in the one-dimensional case. (See Theoretical Exercises 2 and 3 of Chapter 5.) EXAMPLE 2a An accident occurs at a point X that is uniformly distributed on a road of length L. At the time of the accident, an ambulance is at a location Y that is also uniformly

Section 7.2

Expectation of Sums of Random Variables

299

distributed on the road. Assuming that X and Y are independent, ﬁnd the expected distance between the ambulance and the point of the accident. Solution. We need to compute E[|X − Y|]. Since the joint density function of X and Y is 1 f (x, y) = 2 , 0 < x < L, 0 < y < L L it follows from Proposition 2.1 that * L* L 1 |x − y| dy dx E[|X − Y|] = 2 L 0 0 Now,

*

L

* |x − y|dy =

0

=

x

0 x2

*

L

(x − y)dy +

(y − x)dy

x

+

L2 x2 − − x(L − x) 2 2

2 L2 + x2 − xL = 2

Therefore, 1 E[|X − Y|] = 2 L =

*

L 0

L2 + x2 − xL dx 2

L 3

.

For an important application of Proposition 2.1, suppose that E[X] and E[Y] are both ﬁnite and let g(X, Y) = X + Y. Then, in the continuous case, * q* q E[X + Y] = (x + y)f (x, y) dx dy −q *−q * q* q * q q = xf (x, y) dy dx + yf (x, y) dx dy −q −q −q * *−q q q xfX (x) dx + yfY (y) dy = −q

= E[X] + E[Y]

−q

The same result holds in general; thus, whenever E[X] and E[Y] are ﬁnite, E[X + Y] = E[X] + E[Y]

(2.1)

EXAMPLE 2b Suppose that, for random variables X and Y, X Ú Y That is, for any outcome of the probability experiment, the value of the random variable X is greater than or equal to the value of the random variable Y. Since x Ú y is equivalent to the inequality X − Y Ú 0, it follows that E[X − Y] Ú 0, or, equivalently, E[X] Ú E[Y]

.

300

Chapter 7

Properties of Expectation

Using Equation (2.1), we may show by a simple induction proof that if E[Xi ] is ﬁnite for all i = 1, . . . , n, then E[X1 + · · · + Xn ] = E[X1 ] + · · · + E[Xn ]

(2.2)

Equation (2.2) is an extremely useful formula whose utility will now be illustrated by a series of examples. EXAMPLE 2c The sample mean Let X1 , . . . , Xn be independent and identically distributed random variables having distribution function F and expected value μ. Such a sequence of random variables is said to constitute a sample from the distribution F. The quantity X=

n Xi i=1

n

is called the sample mean. Compute E[X]. Solution.

⎤ ⎡ n Xi ⎦ E[X] = E ⎣ n i=1 ⎤ ⎡ n 1 = E⎣ Xi ⎦ n i=1

=

1 n

=μ

n

E[Xi ]

i=1

since E[Xi ] K μ

That is, the expected value of the sample mean is μ, the mean of the distribution. When the distribution mean μ is unknown, the sample mean is often used in statistics to estimate it. . EXAMPLE 2d Boole’s inequality Let A1 , . . . , An denote events, and deﬁne the indicator variables Xi , i = 1, . . . , n, by % 1 if Ai occurs Xi = 0 otherwise Let X=

n

Xi

i=1

so X denotes the number of the events Ai that occur. Finally, let % 1 if X Ú 1 Y= 0 otherwise

Section 7.2

Expectation of Sums of Random Variables

301

so Y is equal to 1 if at least one of the Ai occurs and is 0 otherwise. Now, it is immediate that X Ú Y so

E[X] Ú E[Y]

But since E[X] =

n

E[Xi ] =

n

i=1

P(Ai )

i=1

⎛

and

E[Y] = P{at least one of the Ai occur} = P ⎝

n

⎞ Ai ⎠

i=1

we obtain Boole’s inequality, namely, ⎞ ⎛ n n P(Ai ) P ⎝ Ai ⎠ … i=1

.

i=1

The next three examples show how Equation (2.2) can be used to calculate the expected value of binomial, negative binomial, and hypergeometric random variables. These derivations should be compared with those presented in Chapter 4. EXAMPLE 2e Expectation of a binomial random variable Let X be a binomial random variable with parameters n and p. Recalling that such a random variable represents the number of successes in n independent trials when each trial has probability p of being a success, we have that X = X1 + X2 + · · · + Xn where

% 1 Xi = 0

if the ith trial is a success if the ith trial is a failure

Hence, Xi is a Bernoulli random variable having expectation E[Xi ] = 1(p) + 0(1 − p). Thus, E[X] = E[X1 ] + E[X2 ] + · · · + E[Xn ] = np .

EXAMPLE 2f Mean of a negative binomial random variable If independent trials having a constant probability p of being successes are performed, determine the expected number of trials required to amass a total of r successes. Solution. If X denotes the number of trials needed to amass a total of r successes, then X is a negative binomial random variable that can be represented by X = X1 + X2 + · · · + Xr where X1 is the number of trials required to obtain the ﬁrst success, X2 the number of additional trials until the second success is obtained, X3 the number of additional

302

Chapter 7

Properties of Expectation

trials until the third success is obtained, and so on. That is, Xi represents the number of additional trials required after the (i − 1)st success until a total of i successes is amassed. A little thought reveals that each of the random variables Xi is a geometric random variable with parameter p. Hence, from the results of Example 8b of Chapter 4, E[Xi ] = 1/p, i = 1, 2, . . . , r; thus, E[X] = E[X1 ] + · · · + E[Xr ] =

r p

.

EXAMPLE 2g Mean of a hypergeometric random variable If n balls are randomly selected from an urn containing N balls of which m are white, ﬁnd the expected number of white balls selected. Solution. Let X denote the number of white balls selected, and represent X as X = X1 + · · · + Xm where Xi =

% 1 0

if the ith white ball is selected otherwise

Now E[Xi ] = P{Xi = 1} = P{ith white ball is selected} 1 N − 1 1 n − 1 = N n n = N Hence, E[X] = E[X1 ] + · · · + E[Xm ] =

mn N

We could also have obtained the preceding result by using the alternative representation X = Y1 + · · · + Yn where Yi =

% 1 0

if the ith ball selected is white otherwise

Since the ith ball selected is equally likely to be any of the N balls, it follows that E[Yi ] =

m N

so E[X] = E[Y1 ] + · · · + E[Yn ] =

nm N

.

Section 7.2

Expectation of Sums of Random Variables

303

EXAMPLE 2h Expected number of matches Suppose that N people throw their hats into the center of a room. The hats are mixed up, and each person randomly selects one. Find the expected number of people that select their own hat. Solution. Letting X denote the number of matches, we can compute E[X] most easily by writing X = X1 + X2 + · · · + XN where

% 1 Xi = 0

if the ith person selects his own hat otherwise

Since, for each i, the ith person is equally likely to select any of the N hats, E[Xi ] = P{Xi = 1} =

1 N

Thus,

E[X] = E[X1 ] + · · · + E[XN ] =

1 N

N=1

Hence, on the average, exactly one person selects his own hat.

.

EXAMPLE 2i Coupon-collecting problems Suppose that there are N different types of coupons, and each time one obtains a coupon, it is equally likely to be any one of the N types. Find the expected number of coupons one need amass before obtaining a complete set of at least one of each type. Solution. Let X denote the number of coupons collected before a complete set is attained. We compute E[X] by using the same technique we used in computing the mean of a negative binomial random variable (Example 2f). That is, we deﬁne Xi , i = 0, 1, . . . , N − 1 to be the number of additional coupons that need be obtained after i distinct types have been collected in order to obtain another distinct type, and we note that X = X0 + X1 + · · · + XN−1 When i distinct types of coupons have already been collected, a new coupon obtained will be of a distinct type with probability (N − i)/N. Therefore, N − i i k−1 k Ú 1 P{Xi = k} = N N or, in other words, Xi is a geometric random variable with parameter (N − i)/N. Hence, N E[Xi ] = N − i implying that N N N E[X] = 1 + + + ··· + N − 2 1 N − 1 1 1 + = N 1 + ··· + N − 1 N

.

304

Chapter 7

Properties of Expectation

EXAMPLE 2j Ten hunters are waiting for ducks to ﬂy by. When a ﬂock of ducks ﬂies overhead, the hunters ﬁre at the same time, but each chooses his target at random, independently of the others. If each hunter independently hits his target with probability p, compute the expected number of ducks that escape unhurt when a ﬂock of size 10 ﬂies overhead. Solution. Let Xi equal 1 if the ith duck escapes unhurt and 0 otherwise, for i = 1, 2, . . . , 10. The expected number of ducks to escape can be expressed as E[X1 + · · · + X10 ] = E[X1 ] + · · · + E[X10 ] To compute E[Xi ] = P{Xi = 1}, we note that each of the hunters will, independently, hit the ith duck with probability p/10, so p 10 P{Xi = 1} = 1 − 10 Hence,

p E[X] = 10 1 − 10

10 .

EXAMPLE 2k Expected number of runs Suppose that a sequence of n 1’s and m 0’s is randomly permuted so that each of the (n + m)!/(n!m!) possible arrangements is equally likely. Any consecutive string of 1’s is said to constitute a run of 1’s—for instance, if n = 6, m = 4, and the ordering is 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, then there are 3 runs of 1’s—and we are interested in computing the mean number of such runs. To compute this quantity, let % 1 if a run of 1’s starts at the ith position Ii = 0 otherwise Therefore, R(1), the number of runs of 1, can be expressed as R(1) =

n+m

Ii

i=1

and it follows that E[R(1)] =

n+m

E[Ii ]

i=1

Now, E[I1 ] = P{“1” in position 1} n = n + m

Section 7.2

Expectation of Sums of Random Variables

305

and for 1 < i … n + m, E[Ii ] = P{“0” in position i − 1, “1” in position i} m n = n + mn + m − 1 Hence, E[R(1)] =

nm n + (n + m − 1) n + m (n + m)(n + m − 1)

Similarly, E[R(0)], the expected number of runs of 0’s, is E[R(0)] =

nm m + n + m n + m

and the expected number of runs of either type is E[R(1) + R(0)] = 1 +

2nm n + m

.

EXAMPLE 2l A random walk in the plane Consider a particle initially located at a given point in the plane, and suppose that it undergoes a sequence of steps of ﬁxed length, but in a completely random direction. Speciﬁcally, suppose that the new position after each step is one unit of distance from the previous position and at an angle of orientation from the previous position that is uniformly distributed over (0, 2π). (See Figure 7.1.) Compute the expected square of the distance from the origin after n steps. 2

1 2

1 1

0

1

0

= initial position

1

= position after first step

2

= position after second step FIGURE 7.1

Solution. Letting (Xi , Yi ) denote the change in position at the ith step, i = 1, . . . , n, in rectangular coordinates, we have Xi = cos θi Yi = sin θi

306

Chapter 7

Properties of Expectation

where θi , i = 1, . . . , n, are, by assumption, independent uniform (0, 2π) random vari n n ables. Because the position after n steps has rectangular coordinates Xi , Yi , i=1

i=1

it follows that D2 , the square of the distance from the origin, is given by ⎛ ⎛ ⎞2 ⎞2 n n Xi ⎠ + ⎝ Yi ⎠ D2 = ⎝ =

i=1 n

i=1

(Xi2 + Yi2 ) +

i=1

=n +

(Xi Xj + Yi Yj ) iZj

(cos θi cos θj + sin θi sin θj )

iZj

where cos2 θi + sin2 θi = 1. Taking expectations and using the independence of θi and θj when i Z j and the fact that * 2π 2πE[cos θi ] = cos u du = sin 2π − sin 0 = 0 0 * 2π sin u du = cos 0 − cos 2π = 0 2πE[sin θi ] = 0

we arrive at

E[D2 ] = n

.

EXAMPLE 2m Analyzing the quick-sort algorithm Suppose that we are presented with a set of n distinct values x1 , x2 , . . . , xn and that we desire to put them in increasing order, or as it is commonly stated, to sort them. An efﬁcient procedure for accomplishing this task is the quick-sort algorithm, which is deﬁned as follows. When n = 2, the algorithm compares the two values and then puts them in the appropriate order. When n > 2, one of the elements is randomly chosen—say it is xi —and then all of the other values are compared with xi . Those smaller than xi are put in a bracket to the left of xi and those larger than xi are put in a bracket to the right of xi . The algorithm then repeats itself on these brackets and continues until all values have been sorted. For instance, suppose that we desire to sort the following 10 distinct values: 5, 9, 3, 10, 11, 14, 8, 4, 17, 6 1 We start by choosing one of them at random (that is, each value has probability 10 of being chosen). Suppose, for instance, that the value 10 is chosen. We then compare each of the others to this value, putting in a bracket to the left of 10 all those values smaller than 10 and to the right all those larger. This gives

{5, 9, 3, 8, 4, 6}, 10, {11, 14, 17} We now focus on a bracketed set that contains more than a single value—say the one on the left of the preceding—and randomly choose one of its values—say that 6 is chosen. Comparing each of the values in the bracket with 6 and putting the smaller

Section 7.2

Expectation of Sums of Random Variables

307

ones in a new bracket to the left of 6 and the larger ones in a bracket to the right of 6 gives {5, 3, 4}, 6, {9, 8}, 10, {11, 14, 17} If we now consider the leftmost bracket, and randomly choose the value 4 for comparison then the next iteration yields {3}, 4, {5}, 6, {9, 8}, 10, {11, 14, 17} This continues until there is no bracketed set that contains more than a single value. If we let X denote the number of comparisons that it takes the quick-sort algorithm to sort n distinct numbers, then E[X] is a measure of the effectiveness of this algorithm. To compute E[X], we will ﬁrst express X as a sum of other random variables as follows. To begin, give the following names to the values that are to be sorted: Let 1 stand for the smallest, let 2 stand for the next smallest, and so on. Then, for 1 … i < j … n, let I(i, j) equal 1 if i and j are ever directly compared, and let it equal 0 otherwise. With this deﬁnition, it follows that X=

n n−1

I(i, j)

i=1 j=i+1

implying that

⎡ ⎤ n n−1 ⎢ ⎥ E[X] = E ⎣ I(i, j)⎦ i=1 j=i+1

=

n n−1

E[I(i, j)]

i=1 j=i+1

=

n n−1

P{i and j are ever compared}

i=1 j=i+1

To determine the probability that i and j are ever compared, note that the values i, i + 1, . . . , j − 1, j will initially be in the same bracket (since all values are initially in the same bracket) and will remain in the same bracket if the number chosen for the ﬁrst comparison is not between i and j. For instance, if the comparison number is larger than j, then all the values i, i + 1, . . . , j − 1, j will go in a bracket to the left of the comparison number, and if it is smaller than i, then they will all go in a bracket to the right. Thus all the values i, i + 1, . . . , j − 1, j will remain in the same bracket until the ﬁrst time that one of them is chosen as a comparison value. At that point all the other values between i and j will be compared with this comparison value. Now, if this comparison value is neither i nor j, then upon comparison with it, i will go into a left bracket and j into a right bracket, and thus i and j will be in different brackets and so will never be compared. On the other hand, if the comparison value of the set i, i + 1, . . . , j − 1, j is either i or j, then there will be a direct comparison between i and j. Now, given that the comparison value is one of the values between i and j, it follows that it is equally likely to be any of these j − i + 1 values, and thus the probability that it is either i or j is 2/(j − i + 1). Therefore, we can conclude that P{i and j are ever compared} =

2 j − i + 1

308

Chapter 7

Properties of Expectation

and E[X] =

n n−1 i=1 j=i+1

2 j − i + 1

To obtain a rough approximation of the magnitude of E[X] when n is large, we can approximate the sums by integrals. Now n j=i+1

2 L j − i + 1

*

n

2 dx i+1 x − i + 1 6n = 2 log(x − i + 1)6i+1

= 2 log(n − i + 1) − 2 log(2) L 2 log(n − i + 1) Thus E[X] L

n−1

2 log(n − i + 1)

i=1

*

L 2 =2

n−1

* 1n

log(n − x + 1) dx

log(y)dy 2

= 2(y log(y) − y)|n2 L 2n log(n) Thus we see that when n is large, the quick-sort algorithm requires, on average, approximately 2n log(n) comparisons to sort n distinct values. . EXAMPLE 2n The probability of a union of events Let A1 , . . . An denote events, and deﬁne the indicator variables Xi , i = 1, . . . , n, by % 1 Xi = 0

if Ai occurs otherwise

Now, note that 1 −

n

(1 − Xi ) =

i=1

% 1 0

if ∪ Ai occurs otherwise

Hence, ⎡ E ⎣1 −

n i=1

⎤

⎛ ⎞ n (1 − Xi )⎦ = P ⎝ Ai ⎠ i=1

Section 7.2

Expectation of Sums of Random Variables

309

Expanding the left side of the preceding formula yields ⎡ ⎛ ⎞ n n ⎢ Xi − Xi Xj + Xi Xj Xk P ⎝ Ai ⎠ = E ⎣ i=1

i

i=1

i

⎤

− · · · + (−1)n+1 X1 · · · Xn ⎦ However, Xi1 Xi2 · · · Xik = so

0 1 0

(2.3)

if Ai1 Ai2 · · · Aik occurs otherwise

E[Xi1 · · · Xik ] = P(Ai1 · · · Aik )

Thus, Equation (2.3) is just a statement of the well-known formula for the union of events: P(Ai ) − P(Ai Aj ) + P(Ai Aj Ak ) P(∪Ai ) = i

i

− · · · + (−1)

n+1

i

P(A1 · · · An )

.

When one is dealing with an inﬁnite collection of random variables Xi , i Ú 1, each having a ﬁnite expectation, it is not necessarily true that ⎡ ⎤ q q ⎦ ⎣ Xi = E[Xi ] (2.4) E i=1

i=1

n Xi = lim Xi . Thus, n→q i=1 i=1 ⎡ ⎤ ⎡ ⎤ n q E⎣ Xi ⎦ = E ⎣ lim Xi ⎦ n→q i=1 i=1 ⎡ ⎤ n ? = lim E ⎣ Xi ⎦ n→q

To determine when (2.4) is valid, we note that

= lim n→q =

q

q

i=1 n

E[Xi ]

i=1

E[Xi ]

(2.5)

i=1

Hence, Equation (2.4) is valid whenever we are justiﬁed in interchanging the expectation and limit operations in Equation (2.5). Although, in general, this interchange is not justiﬁed, it can be shown to be valid in two important special cases: 1. The Xi are all nonnegative random variables. (That is, P{Xi Ú 0} = 1 for all i.) q E[|Xi |] < q. 2. i=1

310

Chapter 7

Properties of Expectation

EXAMPLE 2o Consider any nonnegative, integer-valued random variable X. If, for each i Ú 1, we deﬁne % 1 if X Ú i Xi = 0 if X < i then q

Xi =

i=1

X i=1

=

X

q

Xi +

Xi

i=X+1

1 +

i=1

q

0

i=X+1

=X Hence, since the Xi are all nonnegative, we obtain E[X] = =

q i=1 q

E(Xi ) P{X Ú i}

(2.6)

i=1

.

a useful identity. EXAMPLE 2p

Suppose that n elements—call them 1, 2, . . ., n—must be stored in a computer in the form of an ordered list. Each unit of time, a request will be made for one of these elements—i being requested, independently of the past, with probability P(i), i Ú 1, P(i) = 1. Assuming that these probabilities are known, what ordering minimizes i

the average position in the line of the element requested? Solution. Suppose that the elements are numbered so that P(1) Ú P(2) Ú · · · Ú P(n). To show that 1, 2, . . ., n is the optimal ordering, let X denote the position of the requested element. Now, under any ordering—say, O = i1 , i2 , . . . , in , PO {X Ú k} = Ú

n

P(ij )

j=k n

P(j)

j=k

= P1,2, . . . , n {X Ú k} Summing over k and using Equation (2.6) yields Eo [X] Ú E1,2, . . . , n [X] thus showing that ordering the elements in decreasing order of the probability that they are requested minimizes the expected position of the element requested. .

Section 7.2 ∗ 7.2.1

Expectation of Sums of Random Variables

311

Obtaining Bounds from Expectations via the Probabilistic Method

The probabilistic method is a technique for analyzing the properties of the elements of a set by introducing probabilities on the set and then studying an element chosen according to those probabilities. The technique was previously seen in Example 4l of Chapter 3, where it was used to show that a set contained an element that satisﬁed a certain property. In this subsection, we show how it can sometimes be used to bound complicated functions. Let f be a function on the elements of a ﬁnite set S, and suppose that we are interested in m = max f (s) s∈S A useful lower bound for m can often be obtained by letting S be a random element of S for which the expected value of f (S) is computable and then noting that m Ú f (S) implies that m Ú E[f (S)] with strict inequality if f (S) is not a constant random variable. That is, E[f (S)] is a lower bound on the maximum value. EXAMPLE 2q The maximum number of Hamiltonian paths in a tournament A tournament of n > 2 contestants is a tournament in which each of the round-robin n pair of contestants play each other exactly once. Suppose that the players are 2 numbered 1, 2, 3, . . . , n. The permutation i1 , i2 , . . . in is said to be a Hamiltonian path if i1 beats i2 , i2 beats i3 , . . ., and in−1 beats in . A problem of some interest is to determine the largest possible number of Hamiltonian paths. As an illustration, suppose that there are 3 players. On the one hand, one of them wins twice, then there is a single Hamiltonian path. (For instance, if 1 wins twice and 2 beats 3, then the only Hamiltonian path is 1, 2, 3.) On the other hand, if each of the players wins once, than there are 3 Hamiltonian paths. (For instance, if 1 beats 2, 2 beats 3, and 3 beats 1, then 1, 2, 3; 2, 3, 1; and 3, 1, 2, are all Hamiltonians). Hence, when n = 3, there is a maximum of 3 Hamiltonian paths. We now show that there is an outcome of the tournament that results in more than the n!/2n−1 Hamiltonian paths. To begin, let the outcome of the tournament specify n 2 n possible result of each of the games played, and let S denote the set of all 2 2 tournament outcomes. Then, with f (s) deﬁned as the number of Hamiltonian paths that result when the outcome is s ∈ S, we are asked to show that max f (s) Ú s

n! 2n−1

To show this,consider the randomly chosen outcome S that is obtained when the n results of the games are independent, with each contestant being equally likely 2 to win each encounter. To determine E[f (S)], the expected number of Hamiltonian paths that result from the outcome S, number the n! permutations, and, for i = 1, . . . , n!, let % 1, if permutation i is a Hamiltonian Xi = 0, otherwise

312

Chapter 7

Properties of Expectation

Since

f (S) =

Xi

i

it follows that

E[f (S)] =

E[Xi ]

i

Because, by the assumed independence of the outcomes of the games, the probability that any speciﬁed permutation is a Hamiltonian is (1/2)n−1 , it follows that E[Xi ] = P{Xi = 1} = (1/2)n−1 Therefore,

E[f (S)] = n!(1/2)n−1

Since f (S) is not a constant random variable, the preceding equation implies that there is an outcome of the tournament having more than n!/2n−1 Hamiltonian paths. . EXAMPLE 2r A grove of 52 trees is arranged in a circular fashion. If 15 chipmunks live in these trees, show that there is a group of 7 consecutive trees that together house at least 3 chipmunks. Solution. Let the neighborhood of a tree consist of that tree along with the next six trees visited by moving in the clockwise direction. We want to show that, for any choice of living accommodations of the 15 chipmunks, there is a tree that has at least 3 chipmunks living in its neighborhood. To show this, choose a tree at random and let X denote the number of chipmunks that live in its neighborhood. To determine E[X], arbitrarily number the 15 chipmunks and for i = 1, . . . , 15, let % 1, if chipmunk i lives in the neighborhood of the randomly chosen tree Xi = 0, otherwise Because X=

15

Xi

i=1

we obtain that E[X] =

15

E[Xi ]

i=1

However, because Xi will equal 1 if the randomly chosen tree is any of the 7 trees consisting of the tree in which chipmunk i lives along with its 6 neighboring trees when moving in the counterclockwise direction, E[Xi ] = P{Xi = 1} =

7 52

Consequently, 105 > 2 52 showing that there exists a tree with more than 2 chipmunks living in its neighborhood. . E[X] =

Section 7.2 ∗ 7.2.2

Expectation of Sums of Random Variables

313

The Maximum–Minimums Identity

We start with an identity relating the maximum of a set of numbers to the minimums of the subsets of these numbers. Proposition 2.2. For arbitrary numbers xi , i = 1, . . . , n, max xi = xi − min(xi , xj ) + min(xi , xj , xk ) i

i

i

+ . . . + (−1)

i

min(x1 , . . . , xn )

Proof. We will give a probabilistic proof of the proposition. To begin, assume that all the xi are in the interval [0, 1]. Let U be a uniform (0, 1) random variable, and deﬁne the events Ai , i = 1, . . . , n, by Ai = {U < xi }. That is, Ai is the event that the uniform random variable is less than xi . Because at least one of these events Ai will occur if U is less than at least one of the values xi , we have that / % ∪i Ai = U < max xi i

Therefore,

% / P(∪i Ai ) = P U < max xi = max xi i

Also,

i

= > P(Ai ) = P U < xi = xi

In addition, because all of the events Ai1 , . . . , Air will occur if U is less than all the values xi1 , . . . , xir , we see that the intersection of these events is 0 7 Ai1 . . . Air = U < min xij j=1,...r

implying that

7

0 P(Ai1 . . . Air ) = P U < min xij j=1,...r

= min xij j=1,...r

Thus, the proposition follows from the inclusion–exclusion formula for the probability of the union of events: P(Ai ) − P(Ai Aj ) + P(Ai Aj Ak ) P(∪i Ai ) = i

i

+ . . . + (−1)

n+1

i

P(A1 . . . An )

When the xi are nonnegative, but not restricted to the unit interval, let c be such that all the xi are less than c. Then the identity holds for the values yi = xi /c, and the desired result follows by multiplying through by c. When the xi can be negative, let b be such that xi + b > 0 for all i. Therefore, by the preceding, (xi + b) − min(xi + b, xj + b) max(xi + b) = i

i

i

+ · · · + (−1)n+1 min(x1 + b, . . . , xn + b)

314

Chapter 7

Properties of Expectation

Letting M=

xi −

i

min(xi , xj ) + · · · + (−1)n+1 min(x1 , . . . , xn )

i

we can rewrite the foregoing identity as max xi + b = M + b n − i

But

n

0 = (1 − 1) = 1 − n +

n 2

n 2

+ · · · + (−1)n+1

+ · · · + (−1)

n

n n

n n

The preceding two equations show that max xi = M i

and the proposition is proven. It follows from Proposition 2.2 that, for any random variables X1 , . . . , Xn , Xi − min(Xi , Xj ) + · · · + (−1)n+1 min(X1 , . . . , Xn ) max Xi = i

i

i

Taking expectations of both sides of this equality yields the following relationship between the expected value of the maximum and those of the partial minimums: E[Xi ] − E[min(Xi , Xj )] E max Xi = i

i

i

+ · · · + (−1)n+1 E[min(X1 , . . . , Xn )]

(2.7)

EXAMPLE 2s Coupon collecting with unequal probabilities Suppose there are n different types of coupons and that each time one collects a coupon, it is, independently of previous coupons collected, a type i coupon with probn pi = 1. Find the expected number of coupons one needs to collect to ability pi , i=1

obtain a complete set of at least one of each type. Solution. If we let Xi denote the number of coupons one needs collect to obtain a type i, then we can express X as X = max Xi i=1,...,n

Because each new coupon obtained is a type i with probability pi , Xi is a geometric random variable with parameter pi . Also, because the minimum of Xi and Xj is the number of coupons needed to obtain either a type i or a type j, it follows that, for i Z j, min (Xi , Xj ) is a geometric random variable with parameter pi + pj . Similarly, min (Xi , Xj , Xk ), the number needed to obtain any of types i, j, and k, is a geometric

Section 7.3

Moments of the Number of Events that Occur

315

random variable with parameter pi + pj + pk , and so on. Therefore, the identity (2.7) yields E[X] =

1 1 1 − + pi pi + pj pi + pj + pk i

i

i

+ · · · + (−1)n+1

1 p1 + · · · + pn

Noting that

*

q

e−px dx =

0

1 p

and using the identity 1 −

n

(1 − e−pi x ) =

e−pi x −

i

i=1

e−(pi +pj )x + · · · + (−1)n+1 e−(p1 + · · · +pn )x

i

shows, upon integrating the identity, that ⎞ ⎛ * q n ⎝1 − E[X] = (1 − e−pi x )⎠ dx 0

i=1

.

is a more useful computational form. 7.3

MOMENTS OF THE NUMBER OF EVENTS THAT OCCUR Many of the examples solved in the previous section were of the following form: For given events A1 , . . . , An , ﬁnd E[X], where X is the number of these events that occur. The solution then involved deﬁning an indicator variable Ii for event Ai such that % 1, if Ai occurs Ii = 0, otherwise Because X=

n

Ii

i=1

we obtained the result

⎤ ⎡ n n n E[X] = E ⎣ Ii ⎦ = E[Ii ] = P(Ai ) i=1

i=1

(3.1)

i=1

Now suppose we are interested in the number of pairs of events that occur. Because Ii Ij will equal 1 if both Ai and A j occur, and will equal 0 otherwise, it follows that the number of pairs is equal to i

316

Chapter 7

Properties of Expectation

where there are

n 2

terms in the summation. Taking expectations yields X E = E[Ii Ij ] = P(Ai Aj ) 2 i

(3.2)

i

X(X − 1) = P(Ai Aj ) E 2

or

i

giving that E[X 2 ] − E[X] = 2

P(Ai Aj )

(3.3)

i

which yields E[X 2 ], and thus Var(X) = E[X 2 ] − (E[X])2 . Moreover, by considering the number of distinct subsets of k events that all occur, we see that X = Ii1 Ii2 · · · Iik k i1

Taking expectations gives the identity X E = E[Ii1 Ii2 · · · Iik ] = k i1

P(Ai1 Ai2 · · · Aik )

(3.4)

i1

EXAMPLE 3a Moments of binomial random variables Consider n independent trials, with each trial being a success with probability p. Let Ai be the event that trial i is a success. When i Z j, P(Ai Aj ) = p2 . Consequently, Equation (3.2) yields n 2 X 2 p = p E = 2 2 i

or

E[X(X − 1)] = n(n − 1)p2

or Now, E[X] =

n

i=1 P(Ai )

E[X 2 ] − E[X] = n(n − 1)p2 = np, so, from the preceding equation

Var(X) = E[X 2 ] − (E[X])2 = n(n − 1)p2 + np − (np)2 = np(1 − p) which is in agreement with the result obtained in Section 4.6.1. In general, because P(Ai1 Ai2 · · · Aik ) = pk , we obtain from Equation (3.4) that X n k k E = p = p k k i1

or, equivalently, E[X(X − 1) · · · (X − k + 1)] = n(n − 1) · · · (n − k + 1)pk

Section 7.3

Moments of the Number of Events that Occur

317

The successive values E[X k ], k Ú 3, can be recursively obtained from this identity. For instance, with k = 3, it yields E[X(X − 1)(X − 2)] = n(n − 1)(n − 2)p3 or E[X 3 − 3X 2 + 2X] = n(n − 1)(n − 2)p3 or E[X 3 ] = 3E[X 2 ] − 2E[X] + n(n − 1)(n − 2)p3 = 3n(n − 1)p2 + np + n(n − 1)(n − 2)p3

.

EXAMPLE 3b Moments of hypergeometric random variables Suppose n balls are randomly selected from an urn containing N balls, of which m are white. Let Ai be the event that the ith ball selected is white. Then X, the number of white balls selected, is equal to the number of the events A1 , . . . , An that occur. Because the ith ball selected is equally likely to be any of the N balls, of which m are white, P(Ai ) = m/N. Consequently, Equation (3.1) gives that E[X] = ni=1 P(Ai ) = nm/N. Also, since mm − 1 P(Ai Aj ) = P(Ai )P(Aj |Ai ) = NN − 1 we obtain, from Equation (3.2), that m(m − 1) n m(m − 1) X = E = 2 N(N − 1) 2 N(N − 1) i

or E[X(X − 1)] = n(n − 1)

m(m − 1) N(N − 1)

showing that E[X 2 ] = n(n − 1)

m(m − 1) + E[X] N(N − 1)

This formula yields the variance of the hypergeometric, namely, Var(X) = E[X 2 ] − (E[X])2 nm n2 m2 m(m − 1) + − = n(n − 1) N(N − 1) N N2 mn mn (n − 1)(m − 1) + 1 − = N N − 1 N which agrees with the result obtained in Example 8j of Chapter 4. Higher moments of X are obtained by using Equation (3.4). Because P(Ai1 Ai2 · · · Aik ) =

m(m − 1) · · · (m − k + 1) N(N − 1) · · · (N − k + 1)

318

Chapter 7

Properties of Expectation

Equation (3.4) yields

X n m(m − 1) · · · (m − k + 1) E = k k N(N − 1) · · · (N − k + 1)

or E[X(X − 1) · · · (X − k + 1)] = n(n − 1) · · · (n − k + 1)

m(m − 1) · · · (m − k + 1) N(N − 1) · · · (N − k + 1)

.

EXAMPLE 3c Moments in the match problem For i = 1, . . . , N, let Ai be the event that person i selects his or her own hat in the match problem. Then P(Ai Aj ) = P(Ai )P(Aj |Ai ) =

1 1 NN − 1

which follows because, conditional on person i selecting her own hat, the hat selected by person j is equally likely to be any of the other N − 1 hats, of which one is his own. Consequently, with X equal to the number of people who select their own hat, it follows from Equation (3.2) that X N 1 1 E = = 2 N(N − 1 2 N(N − 1) i

thus showing that

E[X(X − 1)] = 1 Therefore, E[X 2 ] = 1 + E[X]. Because E[X] = N i=1 P(Ai ) = 1, we obtain that Var(X) = E[X 2 ] − (E[X])2 = 1.

Hence, both the mean and variance of the number of matches is 1. For higher moments, 1 , we use Equation (3.4), along with the fact that P(Ai1 Ai2 · · · Aik ) = N(N−1)···(N−k+1) to obtain X N 1 E = k k N(N − 1) · · · (N − k + 1) or

E[X(X − 1) · · · (X − k + 1)] = 1

.

EXAMPLE 3d Another coupon-collecting problem Suppose that there are N distinct types of coupons and that, independently of past types collected, each new one obtained is type j with probability pj , N j=1 pj = 1. Find the expected value and variance of the number of different types of coupons that appear among the ﬁrst n collected. Solution. We will ﬁnd it more convenient to work with the number of uncollected types. So, let Y equal the number of different types of coupons collected, and let X = N − Y denote the number of uncollected types. With Ai deﬁned as the event that there are no type i coupons in the collection, X is equal to the number of the

Section 7.3

Moments of the Number of Events that Occur

319

events A1 , . . . , AN that occur. Because the types of the successive coupons collected are independent, and, with probability 1 − pi each new coupon is not type i, we have P(Ai ) = (1 − pi )n Hence, E[X] =

N

i=1 (1

− pi )n , from which it follows that

E[Y] = N − E[X] = N −

N

(1 − pi )n

i=1

Similarly, because each of the n coupons collected is neither a type i nor a type j coupon, with probability 1 − pi − pj , we have P(Ai Aj ) = (1 − pi − pj )n , Thus, E[X(X − 1)] = 2

P(Ai Aj ) = 2

i

or

E[X 2 ] = 2

i Z j (1 − pi − pj )n

i

(1 − pi − pj )n + E[X]

i

Hence, we obtain Var(Y) = Var(X) = E[X 2 ] − (E[X])2 =2

i

(1 − pi − pj )n +

N i=1

⎛ (1 − pi )n − ⎝

N

⎞2 (1 − pi )n ⎠

i=1

In the special case where pi = 1/N, i = 1, . . . , N, the preceding formula gives 1 n E[Y] = N 1 − 1 − N and 1 n 1 2n 2 n 2 + N 1 − − N 1 − Var(Y) = N(N − 1) 1 − N N N

.

EXAMPLE 3e The negative hypergeometric random variables Suppose an urn contains n + m balls, of which n are special and m are ordinary. These items are removed one at a time, with each new removal being equally likely to be any of the balls that remain in the urn. The random variable Y, equal to the number of balls that need be withdrawn until a total of r special balls have been removed, is said to have a negative hypergeometric distribution. The negative hypergeometric distribution bears the same relationship to the hypergeometric distribution as the negative binomial does to the binomial. That is, in both cases, rather than considering a random variable equal to the number of successes in a ﬁxed number of trials (as are the binomial and hypergeometric variables), they refer to the number of trials needed to obtain a ﬁxed number of successes.

320

Chapter 7

Properties of Expectation

To obtain the probability mass function of a negative hypergeometric random variable X, note that X will equal k if both (a) the ﬁrst k − 1 withdrawals consist of r − 1 special and k − r ordinary balls and (b) the kth ball withdrawn is special. Consequently,

P{X = k} =

n m r−1 k−r n+m k−1

n − r + 1 n + m − k + 1

We will not, however, utilize the preceding probability mass function to obtain the mean and variance of Y. Rather, let us number the m ordinary balls as o1 , . . . , om , and then, for each i = 1, . . . , n, let Ai be the event that oi is withdrawn before r special balls have been removed. Then, if X is the number of the events A1 , . . . , Am that occur, it follows that X is the number of ordinary balls that are withdrawn before a total of r special balls have been removed. Consequently, Y=r + X showing that E[Y] = r + E[X] = r +

m

P(Ai )

i=1

To determine P(Ai ), consider the n + 1 balls consisting of oi along with the n special balls. Of these n + 1 balls, oi is equally likely to be the ﬁrst one withdrawn, or the second one withdrawn, . . . , or the ﬁnal one withdrawn. Hence, the probability that it is among the ﬁrst r of these to be selected (and so is removed before a total or r r special balls have been withdrawn) is n+1 . Consequently, P(Ai ) = and E[Y] = r + m

r n + 1

r(n + m + 1) r = n + 1 n + 1

Thus, for instance, the expected number of cards of a well-shufﬂed deck that would need to be turned over until a spade appears is 1 + 39 14 = 3.786, and the expected number of cards that would need to be turned over until an ace appears is 1 + 48 5 = 10.6. To determine Var(Y) = Var(X), we use the identity E[X(X − 1)] = 2 P(Ai Aj ) i

Now, P(Ai Aj ) is the probability that both oi and oj are removed before there have been a total of r special balls removed. So consider the n + 2 balls consisting of oi , oj , and the n special balls. Because all withdrawal orderings of these balls are equally likely, the probability that oi and oj are both among the ﬁrst r + 1 of them to be removed (and so are both removed before r special balls have been withdrawn) is 2 n r(r + 1) r−1 P(Ai Aj ) = 2n+2 = (n + 1)(n + 2) r+1

Section 7.3

Consequently,

Moments of the Number of Events that Occur

321

m r(r + 1) E[X(X − 1)] = 2 2 (n + 1)(n + 2)

so E[X 2 ] = m(m − 1)

r(r + 1) + E[X] (n + 1)(n + 2)

r , this yields Because E[X] = m n+1

r(r + 1) r Var(Y) = Var(X) = m(m − 1) m − (n + 1)(n + 2) n + 1

r m n + 1

2

A little algebra now shows that Var(Y) =

mr(n + 1 − r)(n + m + 1) (n + 1)2 (n + 2)

.

EXAMPLE 3f Singletons in the coupon collector’s problem Suppose that there are n distinct types of coupons and that, independently of past types collected, each new one obtained is equally likely to be any of the n types. Suppose also that one continues to collect coupons until a complete set of at least one of each type has been obtained. Find the expected value and variance of the number of types for which exactly one coupon of that type is collected. Solution. Let X equal the number of types for which exactly one of that type is collected. Also, let Ti denote the ith type of coupon to be collected, and let Ai be the event that there is only a single type Ti coupon in the complete set. Because X is equal to the number of the events A1 , . . . , An that occur, we have E[X] =

n

P(Ai )

i=1

Now, at the moment when the ﬁrst type Ti coupon is collected, there remain n − i types that need to be collected to have a complete set. Because, starting at this moment, each of these n − i + 1 types (the n − i not yet collected and type Ti ) is equally likely to be the last of these types to be collected, it follows that the type Ti will be the 1 . Consequently, last of these types (and so will be a singleton) with probability n−i+1 1 P(Ai ) = n−i+1 , yielding E[X] =

n i=1

1 1 = n − i + 1 i n

i=1

To determine the variance of the number of singletons, let Si, j , for i < j, be the event that the ﬁrst type Ti coupon to be collected is still the only one of its type to have been collected at the moment that the ﬁrst type Tj coupon has been collected. Then P(Ai Aj ) = P(Ai Aj |Si,j )P(Si,j ) Now, P(Si,j ) is the probability that when a type Ti has just been collected, of the n − i + 1 types consisting of type Ti and the n − i as yet uncollected types, a type Ti is not among the ﬁrst j − i of these types to be collected. Because type Ti is equally likely to be the ﬁrst, or second, or . . . , n − i + 1 of these types to be collected, we have P(Si,j ) = 1 −

n + 1 − j j − i = n − i + 1 n + 1 − i

322

Chapter 7

Properties of Expectation

Now, conditional on the event Si,j , both Ai and Aj will occur if, at the time the ﬁrst type Tj coupon is collected, of the n − j + 2 types consisting of types Ti , Tj , and the n − j as yet uncollected types, Ti and Tj are both collected after the other n − j. But this implies that 1 1 P(Ai Aj |Si,j ) = 2 n − j + 2n − j + 1 Therefore, P(Ai Aj ) =

2 , (n + 1 − i)(n + 2 − j)

yielding E[X(X − 1)] = 4

i

i < j

1 (n + 1 − i)(n + 2 − j)

Consequently, using the previous result for E[X], we obtain Var(X) = 4

i

7.4

⎛ ⎞2 n n 1 1 1 ⎠ + − ⎝ (n + 1 − i)(n + 2 − j) i i i=1

.

i=1

COVARIANCE, VARIANCE OF SUMS, AND CORRELATIONS The following proposition shows that the expectation of a product of independent random variables is equal to the product of their expectations. Proposition 4.1. If X and Y are independent, then, for any functions h and g, E[g(X)h(Y)] = E[g(X)]E[h(Y)] Proof. Suppose that X and Y are jointly continuous with joint density f (x, y). Then * q* q E[g(X)h(Y)] = g(x)h(y)f (x, y) dx dy −q *−q * q q = g(x)h(y)fX (x)fY (y) dx dy −q * q *−q q h(y)fY (y)dy g(x)fX (x) dx = −q

= E[h(Y)]E[g(X)]

−q

The proof in the discrete case is similar. Just as the expected value and the variance of a single random variable give us information about that random variable, so does the covariance between two random variables give us information about the relationship between the random variables. Deﬁnition The covariance between X and Y, denoted by Cov (X, Y), is deﬁned by Cov(X, Y) = E[(X − E[X])(Y − E[Y])]

Section 7.4

Covariance, Variance of Sums, and Correlations

323

Upon expanding the right side of the preceding deﬁnition, we see that Cov(X, Y) = E[XY − E[X]Y − XE[Y] + E[Y]E[X]] = E[XY] − E[X]E[Y] − E[X]E[Y] + E[X]E[Y] = E[XY] − E[X]E[Y] Note that if X and Y are independent, then, by Proposition 4.1, Cov(X, Y) = 0. However, the converse is not true. A simple example of two dependent random variables X and Y having zero covariance is obtained by letting X be a random variable such that 1 P{X = 0} = P{X = 1} = P{X = −1} = 3 and deﬁning

% 0 Y= 1

if X Z 0 if X = 0

Now, XY = 0, so E[XY] = 0. Also, E[X] = 0. Thus, Cov(X, Y) = E[XY] − E[X]E[Y] = 0 However, X and Y are clearly not independent. The following proposition lists some of the properties of covariance. Proposition 4.2. (i) Cov(X, Y) = Cov(Y, X) (ii) Cov(X, X) = Var(X) (iii) Cov(aX, Y) = a Cov(X, Y) ⎛ ⎞ m n m n ⎟ ⎜ (iv) Cov ⎝ Xi , Yj ⎠ = Cov(Xi , Yj ) i=1

j=1

i=1 j=1

Proof of Proposition 4.2: Parts (i) and (ii) follow immediately from the deﬁnition of covariance, and part (iii) is left as an exercise for the reader. To prove part (iv), which states that the covariance operation is additive (as is the operation of taking expectations), let μi = E[Xi ] and vj = E[Yj ]. Then ⎤ ⎡ ⎡ ⎤ n n m m ⎥ ⎢ ⎣ ⎦ E Xi = μi , E ⎣ Yj ⎦ = vj i=1

and

i=1

j=1

j=1

⎡ ⎞⎤ ⎛ ⎞⎛ n m n n m m ⎢ ⎟ ⎟⎥ ⎜ ⎝ ⎠⎜ Xi , Yj ⎠ = E ⎢ X − μ Y − vj ⎠⎥ Cov ⎝ ⎝ i i j ⎣ ⎦ ⎞

⎛

i=1

j=1

i=1

i=1

j=1

⎡ ⎤ n m ⎢ ⎥ (Yj − vj )⎦ = E ⎣ (Xi − μi ) i=1

j=1

j=1

324

Chapter 7

Properties of Expectation

⎡ ⎤ m n ⎢ ⎥ = E⎣ (Xi − μi )(Yj − vj )⎦ i=1 j=1

=

m n

E[(Xi − μi )(Yj − vj )]

i=1 j=1

where the last equality follows because the expected value of a sum of random variables is equal to the sum of the expected values. It follows from parts (ii) and (iv) of Proposition 4.2, upon taking Yj = Xj , j = 1, . . . , n, that ⎞ ⎛ ⎛ ⎞ n n n ⎟ ⎜ Var ⎝ Xi ⎠ = Cov ⎝ Xi , Xj ⎠ i=1

i=1

=

n n

j=1

Cov(Xi , Xj )

i=1 j=1

=

n

Var(Xi ) +

Cov(Xi , Xj )

iZj

i=1

Since each pair of indices i, j, i Z j, appears twice in the double summation, the preceding formula is equivalent to ⎛ Var ⎝

n i=1

⎞ Xi ⎠ =

n

Var(Xi ) + 2 Cov(Xi , Xj )

(4.1)

i

i=1

If X1 , . . . , Xn are pairwise independent, in that Xi and Xj are independent for i Z j, then Equation (4.1) reduces to ⎛ ⎞ n n Xi ⎠ = Var(Xi ) Var ⎝ i=1

i=1

The following examples illustrate the use of Equation (4.1). EXAMPLE 4a Let X1 , . . . , Xn be independent and identically distributed random variables having n expected value μ and variance σ 2 , and as in Example 2c, let X = Xi /n be the sami=1

ple mean. The quantities Xi − X, i = 1, . . . , n, are called deviations, as they equal the differences between the individual data and the sample mean. The random variable S2 =

n (Xi − X)2 n − 1 i=1

is called the sample variance. Find (a) Var(X) and (b) E[S2 ].

Section 7.4

Solution.

Covariance, Variance of Sums, and Correlations

325

⎛ ⎞ 2 n 1 (a) Var(X) = Var ⎝ Xi ⎠ n i=1 2 n 1 Var(Xi ) by independence = n i=1

=

σ2 n

(b) We start with the following algebraic identity: n (n − 1)S = (Xi − μ + μ − X)2 2

i=1 n n n 2 2 = (Xi − μ) + (X − μ) − 2(X − μ) (Xi − μ) i=1

i=1

i=1

n (Xi − μ)2 + n(X − μ)2 − 2(X − μ)n(X − μ) = i=1 n = (Xi − μ)2 − n(X − μ)2 i=1

Taking expectations of the preceding yields (n − 1)E[S2 ] =

n

E[(Xi − μ)2 ] − nE[(X − μ)2 ]

i=1

= nσ 2 − nVar(X) = (n − 1)σ 2 where the ﬁnal equality made use of part (a) of this example and the one preceding it made use of the result of Example 2c, namely, that E[X] = μ. Dividing through by n − 1 shows that the expected value of the sample variance is the distribution variance σ 2 . . Our next example presents another method for obtaining the variance of a binomial random variable. EXAMPLE 4b Variance of a binomial random variable Compute the variance of a binomial random variable X with parameters n and p. Solution. Since such a random variable represents the number of successes in n independent trials when each trial has the common probability p of being a success, we may write X = X1 + · · · + Xn where the Xi are independent Bernoulli random variables such that % 1 if the ith trial is a success Xi = 0 otherwise

326

Chapter 7

Properties of Expectation

Hence, from Equation (4.1), we obtain Var(X) = Var(X1 ) + · · · + Var(Xn ) But Var(Xi ) = E[Xi2 ] − (E[Xi ])2 = E[Xi ] − (E[Xi ])2

since Xi2 = Xi

= p − p2 Thus, Var(X) = np(1 − p)

.

EXAMPLE 4c Sampling from a ﬁnite population Consider a set of N people, each of whom has an opinion about a certain subject that is measured by a real number v that represents the person’s “strength of feeling” about the subject. Let vi represent the strength of feeling of person i, i = 1, . . . N. Suppose that the quantities vi , i = 1, . . . , N, are unknown and, to gather information, a group of n of the N people is “randomly chosen” in the sense that all of the N subsets of size n are equally likely to be chosen. These n people are then quesn tioned and their feelings determined. If S denotes the sum of the n sampled values, determine its mean and variance. An important application of the preceding problem is to a forthcoming election in which each person in the population is either for or against a certain candidate or proposition. If we take vi to equal 1 if person i is in favor and 0 if he or she is against, N vi /N represents the proportion of the population that is in favor. To then v = i=1

estimate v, a random sample of n people is chosen, and these people are polled. The proportion of those polled who are in favor—that is, S/n—is often used as an estimate of v. Solution. For each person i, i = 1, . . . , N, deﬁne an indicator variable Ii to indicate whether or not that person is included in the sample. That is, % 1 if person i is in the random sample Ii = 0 otherwise Now, S can be expressed by S=

N

vi Ii

i=1

so E[S] =

N i=1

vi E[Ii ]

Section 7.4

Var(S) =

N

Covariance, Variance of Sums, and Correlations

Var(vi Ii ) + 2

=

Cov(vi Ii , vj Ij )

i

i=1 N

327

v2i Var(Ii ) + 2

vi vj Cov(Ii , Ij )

i

i=1

Because n N n n − 1 E[Ii Ij ] = NN − 1 E[Ii ] =

it follows that Var(Ii ) =

n N

1 −

n N

n(n − 1) Cov(Ii , Ij ) = − N(N − 1) −n(N − n) = 2 N (N − 1)

n N

2

Hence, N vi = nv E[S] = n N i=1

n N

Var(S) =

N N − n 2 2n(N − n) vi − 2 vi vj N N (N − 1) i

i=1

The expression for Var(S) can be simpliﬁed somewhat by using the identity N v2i + 2 vi vj . After some simpliﬁcation, we obtain (v1 + · · · + vN )2 = i=1

i

⎛

N

⎞ v2i

⎜ ⎜ n(N − n) ⎝ i=1 Var(S) = N − 1 N

− v

⎟ ⎟ 2⎠

Consider now the special case in which Np of the v’s are equal to 1 and the remainder equal to 0. Then, in this case, S is a hypergeometric random variable and has mean and variance given, respectively, by E[S] = nv = np and

since v =

Np =p N

n(N − n) Np − p2 N − 1 N n(N − n) p(1 − p) = N − 1

Var(S) =

328

Chapter 7

Properties of Expectation

The quantity S/n, equal to the proportion of those sampled which have values equal to 1, is such that S E =p n S N − n Var = p(1 − p) . n n(N − 1) The correlation of two random variables X and Y, denoted by ρ(X, Y), is deﬁned, as long as Var(X) Var(Y) is positive, by ρ(X, Y) = √ It can be shown that

Cov(X, Y) Var(X)Var(Y)

−1 … ρ(X, Y) … 1

(4.2)

To prove Equation (4.2), suppose that X and Y have variances given by σx2 and σy2 , respectively. Then, on the one hand, Y X + 0 … Var σx σy =

Var(X) Var(Y) 2Cov(X, Y) + + 2 2 σx σy σx σy

= 2[1 + ρ(X, Y)] implying that

−1 … ρ(X, Y)

On the other hand,

X Y 0 … Var − σx σy

Var(X) VarY 2Cov(X, Y) + − σx σy σx2 (−σy )2 = 2[1 − ρ(X, Y)] =

implying that

ρ(X, Y) … 1

which completes the proof of Equation (4.2). In fact, since Var(Z) = 0 implies that Z is constant with probability 1 (this intuitive relationship will be rigorously proven in Chapter 8), it follows from the proof of Equation (4.2) that ρ(X, Y) = 1 implies that Y = a + bX, where b = σy /σx > 0 and ρ(X, Y) = −1 implies that Y = a + bX, where b = −σy /σx < 0. We leave it as an exercise for the reader to show that the reverse is also true: that if Y = a + bX, then ρ(X, Y) is either +1 or −1, depending on the sign of b. The correlation coefﬁcient is a measure of the degree of linearity between X and Y. A value of ρ(X, Y) near +1 or −1 indicates a high degree of linearity between X and Y, whereas a value near 0 indicates that such linearity is absent. A positive

Section 7.4

Covariance, Variance of Sums, and Correlations

329

value of ρ(X, Y) indicates that Y tends to increase when X does, whereas a negative value indicates that Y tends to decrease when X increases. If ρ(X, Y) = 0, then X and Y are said to be uncorrelated. EXAMPLE 4d Let IA and IB be indicator variables for the events A and B. That is, % 1 if A occurs IA = 0 otherwise % 1 if B occurs IB = 0 otherwise Then E[IA ] = P(A) E[IB ] = P(B) E[IA IB ] = P(AB) so Cov(IA , IB ) = P(AB) − P(A)P(B) = P(B)[P(A|B) − P(A)] Thus, we obtain the quite intuitive result that the indicator variables for A and B are either positively correlated, uncorrelated, or negatively correlated, depending on whether P(A|B) is, respectively, greater than, equal to, or less than P(A). . Our next example shows that the sample mean and a deviation from the sample mean are uncorrelated. EXAMPLE 4e Let X1 , . . . , Xn be independent and identically distributed random variables having variance σ 2 . Show that Cov(Xi − X, X) = 0 Solution. We have Cov(Xi − X, X) = Cov(Xi , X) − Cov(X, X) ⎞ ⎛ n 1 ⎟ ⎜ = Cov ⎝Xi , Xj ⎠ − Var(X) n j=1

= =

1 n

n j=1

σ2 n

Cov(Xi , Xj ) −

−

σ2 =0 n

σ2 n

330

Chapter 7

Properties of Expectation

where the next-to-last equality uses the result of Example 4a and the ﬁnal equality follows because 0 0 if j Z i by independence Cov(Xi , Xj ) = σ2 if j = i since Var(Xi ) = σ 2 Although X and the deviation Xi − X are uncorrelated, they are not, in general, independent. However, in the special case where the Xi are normal random variables, it turns out that not only is X independent of a single deviation, but it is independent of the entire sequence of deviations Xj − X, j = 1, . . . , n. This result will be established in Section 7.8, where we will also show that, in this case, the sample mean X and the sample variance S2 are independent, with (n − 1)S2 /σ 2 having a chi-squared distribution with n − 1 degrees of freedom. (See Example 4a for the . deﬁnition of S2 .)

EXAMPLE 4f Consider m independent trials, each of which results in any of r possible outcomes r with probabilities P1 , P2 , . . . , Pr , Pi = 1. If we let Ni , i = 1, . . . , r, denote the num1

ber of the m trials that result in outcome i, then N1 , N2 , . . . , Nr have the multinomial distribution m! n n P1 1 P2 2 · · · Prnr ni = m P{N1 = n1 , N2 = n2 , . . . , Nr = nr } = n1 !n2 ! . . . nr ! r

i=1

For i Z j, it seems likely that when Ni is large, Nj would tend to be small; hence, it is intuitive that they should be negatively correlated. Let us compute their covariance by using Proposition 4.2(iv) and the representation Ni =

m

Ii (k)

and

Nj =

k=1

m

Ij (k)

k=1

where % 1 Ii (k) = 0 % 1 Ij (k) = 0

if trial k results in outcome i otherwise if trial k results in outcome j otherwise

From Proposition 4.2(iv), we have Cov(Ni , Nj ) =

m m

Cov(Ii (k), Ij ())

=1 k=1

Now, on the one hand, when k Z , Cov(Ii (k), Ij ()) = 0

Section 7.5

Conditional Expectation

331

since the outcome of trial k is independent of the outcome of trial . On the other hand, Cov(Ii (), Ij ()) = E[Ii ()Ij ()] − E[Ii ()]E[Ij ()] = 0 − Pi Pj = −Pi Pj where the equation uses the fact that Ii ()Ij () = 0, since trial cannot result in both outcome i and outcome j. Hence, we obtain Cov(Ni , Nj ) = −mPi Pj which is in accord with our intuition that Ni and Nj are negatively correlated. 7.5

.

CONDITIONAL EXPECTATION 7.5.1

Deﬁnitions

Recall that if X and Y are jointly discrete random variables, then the conditional probability mass function of X, given that Y = y, is deﬁned, for all y such that P{Y = y} > 0, by p(x, y) pX|Y (x|y) = P{X = x|Y = y} = pY (y) It is therefore natural to deﬁne, in this case, the conditional expectation of X given that Y = y, for all values of y such that pY (y) > 0, by xP{X = x|Y = y} E[X|Y = y] = x

=

xpX|Y (x|y)

x

EXAMPLE 5a If X and Y are independent binomial random variables with identical parameters n and p, calculate the conditional expected value of X given that X + Y = m. Solution. Let us ﬁrst calculate the conditional probability mass function of X given that X + Y = m. For k … min(n, m), P{X = k|X + Y = m} = = =

=

=

P{X = k, X + Y = m} P{X + Y = m} P{X = k, Y = m − k} P{X + Y = m} P{X = k}P{Y = m − k} P{X + Y = m} n k n n−k p (1 − p) pm−k (1 − p)n−m+k k m − k 2n m p (1 − p)2n−m m n n k m − k 2n m

332

Chapter 7

Properties of Expectation

where we have used the fact (see Example 3f of Chapter 6) that X + Y is a binomial random variable with parameters 2n and p. Hence, the conditional distribution of X, given that X + Y = m, is the hypergeometric distribution, and from Example 2g, we obtain E[X|X + Y = m] =

m 2

.

Similarly, let us recall that if X and Y are jointly continuous with a joint probability density function f (x, y), then the conditional probability density of X, given that Y = y, is deﬁned, for all values of y such that fY (y) > 0, by fX|Y (x|y) =

f (x, y) fY (y)

It is natural, in this case, to deﬁne the conditional expectation of X, given that Y = y, by * E[X|Y = y] =

q

−q

xfX|Y (x|y) dx

provided that fY (y) > 0. EXAMPLE 5b Suppose that the joint density of X and Y is given by f (x, y) =

e−x/y e−y y

0 < x < q, 0 < y < q

Compute E[X|Y = y]. Solution. We start by computing the conditional density f (x, y) fY (y) f (x, y) =* q f (x, y) dx

fX|Y (x|y) =

−q

(1/y)e−x/y e−y

=*

q 0

=*

(1/y)e−x/y e−y dx

(1/y)e−x/y

q

(1/y)e−x/y dx

0

1 = e−x/y y

Section 7.5

Conditional Expectation

333

Hence, the conditional distribution of X, given that Y = y, is just the exponential distribution with mean y. Thus, * q x −x/y E[X|Y = y] = e dx = y . 0 y Remark. Just as conditional probabilities satisfy all of the properties of ordinary probabilities, so do conditional expectations satisfy the properties of ordinary expectations. For instance, such formulas as ⎧ ⎪ g(x)pX|Y (x|y) in the discrete case ⎪ ⎪ ⎨ x E[g(X)|Y = y] = * q ⎪ ⎪ ⎪ g(x)fX|Y (x|y) dx in the continuous case ⎩ −q

and

⎤ ⎡ n n Xi |Y = y⎦ = E[Xi |Y = y] E⎣ i=1

i=1

remain valid. As a matter of fact, conditional expectation given that Y = y can be thought of as being an ordinary expectation on a reduced sample space consisting only of outcomes for which Y = y. . 7.5.2

Computing Expectations by Conditioning

Let us denote by E[X|Y] that function of the random variable Y whose value at Y = y is E[X|Y = y]. Note that E[X|Y] is itself a random variable. An extremely important property of conditional expectations is given by the following proposition. Proposition 5.1. E[X] = E[E[X|Y]] If Y is a discrete random variable, then Equation (5.1) states that E[X|Y = y]P{Y = y} E[X] =

(5.1)

(5.1a)

y

whereas if Y is continuous with density fY (y), then Equation (5.1) states * q E[X] = E[X|Y = y]fY (y) dy −q

(5.1b)

We now give a proof of Equation (5.1) in the case where X and Y are both discrete random variables. Proof of Equation (5.1) when X and Y Are Discrete: We must show that E[X|Y = y]P{Y = y} E[X] = y

(5.2)

334

Chapter 7

Properties of Expectation

Now, the right-hand side of Equation (5.2) can be written as E[X|Y = y]P{Y = y} = xP{X = x|Y = y}P{Y = y} y

y

x

y

x

P{X = x, Y = y} P{Y = y} = x P{Y = y} y x xP{X = x, Y = y} = = x P{X = x, Y = y} x

=

y

xP{X = x}

x

= E[X] and the result is proved. One way to understand Equation (5.2) is to interpret it as follows: To calculate E[X], we may take a weighted average of the conditional expected value of X given that Y = y, each of the terms E[X|Y = y] being weighted by the probability of the event on which it is conditioned. (Of what does this remind you?) This is an extremely useful result that often enables us to compute expectations easily by ﬁrst conditioning on some appropriate random variable. The following examples illustrate its use. EXAMPLE 5c A miner is trapped in a mine containing 3 doors. The ﬁrst door leads to a tunnel that will take him to safety after 3 hours of travel. The second door leads to a tunnel that will return him to the mine after 5 hours of travel. The third door leads to a tunnel that will return him to the mine after 7 hours. If we assume that the miner is at all times equally likely to choose any one of the doors, what is the expected length of time until he reaches safety? Solution. Let X denote the amount of time (in hours) until the miner reaches safety, and let Y denote the door he initially chooses. Now, E[X] = E[X|Y = 1]P{Y = 1} + E[X|Y = 2]P{Y = 2} + E[X|Y = 3]P{Y = 3} 1 = (E[X|Y = 1] + E[X|Y = 2] + E[X|Y = 3]) 3 However, E[X|Y = 1] = 3 E[X|Y = 2] = 5 + E[X] E[X|Y = 3] = 7 + E[X]

(5.3)

To understand why Equation (5.3) is correct, consider, for instance, E[X|Y = 2] and reason as follows: If the miner chooses the second door, he spends 5 hours in the tunnel and then returns to his cell. But once he returns to his cell, the problem is as before; thus his expected additional time until safety is just E[X]. Hence,

Section 7.5

Conditional Expectation

335

E[X|Y = 2] = 5 + E[X]. The argument behind the other equalities in Equation (5.3) is similar. Hence, E[X] =

1 (3 + 5 + E[X] + 7 + E[X]) 3

or

E[X] = 15

.

EXAMPLE 5d Expectation of a sum of a random number of random variables Suppose that the number of people entering a department store on a given day is a random variable with mean 50. Suppose further that the amounts of money spent by these customers are independent random variables having a common mean of $8. Finally, suppose also that the amount of money spent by a customer is also independent of the total number of customers who enter the store. What is the expected amount of money spent in the store on a given day? Solution. If we let N denote the number of customers that enter the store and Xi the amount spent by the ith such customer, then the total amount of money spent can be N Xi . Now, expressed as i=1

⎡ E⎣

N

⎡ ⎡ ⎤⎤ N ⎥ ⎢ Xi ⎦ = E ⎣E ⎣ Xi |N ⎦⎦ ⎤

1

But

1

⎤ ⎡ ⎤ ⎡ N n Xi |N = n⎦ = E ⎣ Xi |N = n⎦ E⎣ 1

1

⎡ ⎤ n = E⎣ Xi ⎦

by the independence of the Xi and N

1

= nE[X] which implies that

⎡ E⎣

where E[X] = E[Xi ] N

⎤ Xi |N ⎦ = NE[X]

1

Thus,

⎡ ⎤ N E⎣ Xi ⎦ = E[NE[X]] = E[N]E[X] i=1

Hence, in our example, the expected amount of money spent in the store is 50 * $8, or $400. . EXAMPLE 5e The game of craps is begun by rolling an ordinary pair of dice. If the sum of the dice is 2, 3, or 12, the player loses. If it is 7 or 11, the player wins. If it is any other number i,

336

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the player continues to roll the dice until the sum is either 7 or i. If it is 7, the player loses; if it is i, the player wins. Let R denote the number of rolls of the dice in a game of craps. Find (a) E[R]; (b) E[R|player wins]; (c) E[R|player loses]. Solution. If we let Pi denote the probability that the sum of the dice is i, then i − 1 , 36

Pi = P14−i =

i = 2, . . . , 7

To compute E[R], we condition on S, the initial sum, giving E[R] =

12

E[R|S = i]Pi

i=2

However, E[R|S = i] =

⎧ ⎨1,

1 , ⎩1 + Pi + P7

if i = 2, 3, 7, 11, 12 otherwise

The preceding equation follows because if the sum is a value i that does not end the game, then the dice will continue to be rolled until the sum is either i or 7, and the number of rolls until this occurs is a geometric random variable with parameter Pi + P7 . Therefore, E[R] = 1 +

6 i=4

10 Pi Pi + Pi + P7 Pi + P7 i=8

= 1 + 2(3/9 + 4/10 + 5/11) = 3.376 To determine E[R|win], let us start by determining p, the probability that the player wins. Conditioning on S yields p=

12

P{win|S = i}Pi

i=2

= P7 + P11 +

6 i=4

10 Pi Pi Pi + Pi Pi + P7 Pi + P7 i=8

= 0.493 where the preceding uses the fact that the probability of obtaining a sum of i before one of 7 is Pi /(Pi + P7 ). Now, let us determine the conditional probability mass function of S, given that the player wins. Letting Qi = P{S = i|win}, we have Q2 = Q3 = Q12 = 0,

Q7 = P7 /p,

Q11 = P11 /p

Section 7.5

Conditional Expectation

337

and, for i = 4, 5, 6, 8, 9, 10, Qi =

P{S = i, win} P{win}

=

Pi P{win|S = i} p

=

Pi2 p(Pi + P7 )

Now, conditioning on the initial sum gives E[R|win, S = i]Qi E[R|win] = i

However, as was noted in Example 2j of Chapter 6, given that the initial sum is i, the number of additional rolls needed and the outcome (whether a win or a loss) are independent. (This is easily seen by ﬁrst noting that, conditional on an initial sum of i, the outcome is independent of the number of additional dice rolls needed and then using the symmetry property of independence, which states that if event A is independent of event B, then event B is independent of event A.) Therefore, E[R|S = i]Qi E[R|win] = i

=1 +

6 i=4

10 Qi Qi + Pi + P7 Pi + P7 i=8

= 2.938 Although we could determine E[R|player loses] exactly as we did E[R|player wins], it is easier to use E[R] = E[R|win]p + E[R|lose](1 − p) implying that E[R|lose] =

E[R] − E[R|win]p = 3.801 1 − p

.

EXAMPLE 5f As deﬁned in Example 5c of Chapter 6, the bivariate normal joint density function of the random variables X and Y is ⎧ ⎡ 2 ⎪ 2 ⎨ y − μy x − μ 1 1 x ⎣ . exp − + f (x, y) = 2 ⎪ σx σy 2πσx σy 1 − ρ 2 ⎩ 2(1 − ρ ) ⎤⎫ ⎪ (x − μx )(y − μy ) ⎬ ⎦ − 2ρ ⎪ σx σy ⎭

338

Chapter 7

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We will now show that ρ is the correlation between X and Y. As shown in Example 5c, μx = E[X], σx2 = Var(X), and μy = E[Y], σy2 = Var(Y). Consequently, Cov(X, Y) σx σy E[XY] − μx μy = σx σy

Corr(X, Y) =

To determine E[XY], we condition on Y. That is, we use the identity 2 3 E[XY] = E E[XY|Y] Recalling from Example 5c that the conditional distribution of X given that Y = y is normal with mean μx + ρ σσxy (y − μy ), we see that E[XY|Y = y] = E[Xy|Y = y] = yE[X|Y = y] σx = y μx + ρ (y − μy ) σy σx 2 = yμx + ρ (y − μy y) σy Consequently, E[XY|Y] = Yμx + ρ implying that

σx 2 (Y − μy Y) σy

σx 2 E[XY] = E Yμx + ρ (Y − μy Y) σy σx = μx E[Y] + ρ E[Y 2 − μy Y] σy σx E[Y 2 ] − μ2y = μx μy + ρ σy σx = μx μy + ρ Var(Y) σy = μx μy + ρσx σy

Therefore, Corr(X, Y) =

ρσx σy =ρ σx σy

.

Sometimes E[X] is easy to compute, and we use the conditioning identity to compute a conditional expected value. This approach is illustrated by our next example. EXAMPLE 5g Consider n independent trials, each of which results in one of the outcomes 1, . . . , k, with respective probabilities p1 , . . . , pk , ki=1 pi = 1. Let Ni denote the number of trials that result in outcome i, i = 1, . . . , k. For i Z j, ﬁnd (a) E[Nj |Ni > 0]

and

(b)

E[Nj |Ni > 1]

Section 7.5

Solution. To solve (a), let

% I=

Then

0, 1,

Conditional Expectation

339

if Ni = 0 if Ni > 0

E[Nj ] = E[Nj |I = 0]P{I = 0} + E[Nj |I = 1]P{I = 1}

or, equivalently, E[Nj ] = E[Nj |Ni = 0]P{Ni = 0} + E[Nj |Ni > 0]P{Ni > 0} Now, the unconditional distribution of Nj is binomial with parameters n, pj . Also, given that Ni = r, each of the n − r trials that do not result in outcome i will, p independently, result in outcome j with probability P(j|not i) = 1−pj i . Consequently, the conditional distribution of Nj , given that Ni = r, is binomial with parameters p n − r, 1−pj i . (For a more detailed argument for this conclusion, see Example 4c of Chapter 6.) Because P{Ni = 0} = (1 − pi )n , the preceding equation yields npj = n

pj (1 − pi )n + E[Nj |Ni > 0](1 − (1 − pi )n 1 − pi

giving the result E[Nj |Ni > 0] = npj

1 − (1 − pi )n−1 1 − (1 − pi )n

We can solve part (b) in a similar manner. Let ⎧ ⎨ 0, if Ni = 0 J = 1, if Ni = 1 ⎩ 2, if N > 1 i Then E[Nj ] = E[Nj |J = 0]P{J = 0} + E[Nj |J = 1]P{J = 1} + E[Nj |J = 2]P{J = 2} or, equivalently, E[Nj ] = E[Nj |Ni = 0]P{Ni = 0} + E[Nj |Ni = 1]P{Ni = 1} + E[Nj |Ni > 1]P{Ni > 1} This equation yields npj = n

pj pj (1 − pi )n + (n − 1) npi (1 − pi )n−1 1 − pi 1 − pi

+ E[Nj |Ni > 1](1 − (1 − pi )n − npi (1 − pi )n−1 ) giving the result E[Nj |Ni > 1] =

npj [1 − (1 − pi )n−1 − (n − 1)pi (1 − pi )n−2 ] 1 − (1 − pi )n − npi (1 − pi )n−1

.

It is also possible to obtain the variance of a random variable by conditioning. We illustrate this approach by the following example.

340

Chapter 7

Properties of Expectation

EXAMPLE 5h Variance of the geometric distribution Independent trials, each resulting in a success with probability p, are successively performed. Let N be the time of the ﬁrst success. Find Var(N). Solution. Let Y = 1 if the ﬁrst trial results in a success and Y = 0 otherwise. Now, Var(N) = E[N 2 ] − (E[N])2 To calculate E[N 2 ], we condition on Y as follows: E[N 2 ] = E[E[N 2 |Y]] However, E[N 2 |Y = 1] = 1 E[N 2 |Y = 0] = E[(1 + N)2 ] These two equations follow because, on the one hand, if the ﬁrst trial results in a success, then, clearly, N = 1; thus, N 2 = 1. On the other hand, if the ﬁrst trial results in a failure, then the total number of trials necessary for the ﬁrst success will have the same distribution as 1 (the ﬁrst trial that results in failure) plus the necessary number of additional trials. Since the latter quantity has the same distribution as N, we obtain E[N 2 |Y = 0] = E[(1 + N)2 ]. Hence, E[N 2 ] = E[N 2 |Y = 1]P{Y = 1} + E[N 2 |Y = 0]P{Y = 0} = p + (1 − p)E[(1 + N)2 ] = 1 + (1 − p)E[2N + N 2 ] However, as was shown in Example 8b of Chapter 4, E[N] = 1/p; therefore, E[N 2 ] = 1 +

2(1 − p) + (1 − p)E[N 2 ] p

or E[N 2 ] =

2 − p p2

Consequently, Var(N) = E[N 2 ] − (E[N])2 2 2 − p 1 = − p p2 1 − p = p2

.

EXAMPLE 5i Consider a gambling situation in which there are r players, with player i initially having ni units, ni > 0, i = 1,..., r. At each stage, two of the players are chosen to play a game, with the winner of the game receiving 1 unit from the loser. Any player whose fortune drops to 0 is eliminated, and this continues until a single player has all n K ri=1 ni units, with that player designated as the victor. Assuming that the results of successive games are independent and that each game is equally likely to be won

Section 7.5

Conditional Expectation

341

by either of its two players, ﬁnd the average number of stages until one of the players has all n units. Solution. To ﬁnd the expected number of stages played, suppose ﬁrst that there are only 2 players, with players 1 and 2 initially having j and n − j units, respectively. Let Xj denote the number of stages that will be played, and let mj = E[Xj ]. Then, for j = 1,..., n − 1, Xj = 1 + Aj where Aj is the additional number of stages needed beyond the ﬁrst stage. Taking expectations gives mj = 1 + E[Aj ] Conditioning on the result of the ﬁrst stage then yields mj = 1 + E[Aj |1 wins ﬁrst stage]1/2 + E[Aj |2 wins ﬁrst stage]1/2 Now, if player 1 wins at the ﬁrst stage, then the situation from that point on is exactly the same as in a problem which supposes that player 1 starts with j + 1 and player 2 with n − ( j + 1) units. Consequently, E[Aj |1 wins ﬁrst stage] = mj+1 and, analogously, E[Aj |2 wins ﬁrst stage] = mj−1 Thus, mj = 1 +

1 1 mj+1 + mj−1 2 2

or, equivalently, mj+1 = 2mj − mj−1 − 2,

j = 1, . . . , n − 1

(5.4)

Using that m0 = 0, the preceding equation yields m2 = 2m1 − 2 m3 = 2m2 − m1 − 2 = 3m1 − 6 = 3(m1 − 2) m4 = 2m3 − m2 − 2 = 4m1 − 12 = 4(m1 − 3) suggesting that mi = i(m1 − i + 1),

i = 1, . . . , n

(5.5)

To prove the preceding equality, we use mathematical induction. Since we’ve already shown the equation to be true for i = 1, 2, we take as the induction hypothesis that it is true whenever i … j < n. Now we must prove that it is true for j + 1. Using Equation (5.4) yields mj+1 = 2mj − mj−1 − 2 = 2j(m1 − j + 1) − (j − 1)(m1 − j + 2) − 2 (by the induction hypothesis) = (j + 1)m1 − 2j2 + 2j + j2 − 3j + 2 − 2 = (j + 1)m1 − j2 − j = (j + 1)(m1 − j)

342

Chapter 7

Properties of Expectation

which completes the induction proof of (5.5). Letting i = n in (5.5), and using that mn = 0, now yields that m1 = n − 1 which, again using (5.5), gives the result mi = i(n − i) Thus, the mean number of games played when there are only 2 players with initial amounts i and n − i is the product of their initial amounts. Because both players play all stages, this is also the mean number of stages involving player 1. Now let us return to the problem involving r players with initial amounts ni , i = r 1,..., r, i=1 ni = n. Let X denote the number of stages needed to obtain a victor, and let Xi denote the number of stages involving player i. Now, from the point of view of player i, starting with ni , he will continue to play stages, independently being equally likely to win or lose each one, until his fortune is either n or 0. Thus, the number of stages he plays is exactly the same as when he has a single opponent with an initial fortune of n − ni . Consequently, by the preceding result it follows that E[Xi ] = ni (n − ni ) so

⎡ E⎣

r

⎤ Xi ⎦ =

i=1

r

2

ni (n − ni ) = n −

i=1

r

n2i

i=1

But because each stage involves two players, 1 Xi 2 r

X=

i=1

Taking expectations now yields

⎞ ⎛ r 1⎝ 2 n2i ⎠ E[X] = n − 2 i=1

It is interesting to note that while our argument shows that the mean number of stages does not depend on the manner in which the teams are selected at each stage, the same is not true for the distribution of the number of stages. To see this, suppose r = 3, n1 = n2 = 1, and n3 = 2. If players 1 and 2 are chosen in the ﬁrst stage, then it will take at least three stages to determine a winner, whereas if player 3 is in the ﬁrst stage, then it is possible for there to be only two stages. . In our next example, we use conditioning to verify a result previously noted in Section 6.3.1: that the expected number of uniform (0, 1) random variables that need to be added for their sum to exceed 1 is equal to e. EXAMPLE 5j Let U1 , U2 , . . . be a sequence of independent uniform (0, 1) random variables. Find E[N] when ⎫ ⎧ n ⎬ ⎨ Ui > 1 N = min n: ⎭ ⎩ i=1

Section 7.5

Conditional Expectation

343

Solution. We will ﬁnd E[N] by obtaining a more general result. For x ∈ [0, 1], let ⎫ ⎧ n ⎬ ⎨ Ui > x N(x) = min n: ⎭ ⎩ i=1

and set m(x) = E[N(x)] That is, N(x) is the number of uniform (0, 1) random variables we must add until their sum exceeds x, and m(x) is its expected value. We will now derive an equation for m(x) by conditioning on U1 . This gives, from Equation (5.1b), *

1

m(x) =

E[N(x)|U1 = y] dy

(5.6)

% 1 1 + m(x − y)

(5.7)

0

Now, E[N(x)|U1 = y] =

if y > x if y … x

The preceding formula is obviously true when y > x. It is also true when y … x, since, if the ﬁrst uniform value is y, then, at that point, the remaining number of uniform random variables needed is the same as if we were just starting and were going to add uniform random variables until their sum exceeded x − y. Substituting Equation (5.7) into Equation (5.6) gives * x m(x − y) dy m(x) = 1 + *0 x by letting =1 + m(u)du u=x − y 0 Differentiating the preceding equation yields m (x) = m(x) or, equivalently, m (x) =1 m(x) Integrating this equation gives log[m(x)] = x + c or m(x) = kex Since m(0) = 1, it follows that k = 1, so we obtain m(x) = ex Therefore, m(1), the expected number of uniform (0, 1) random variables that need to be added until their sum exceeds 1, is equal to e. .

344

Chapter 7

Properties of Expectation

7.5.3

Computing Probabilities by Conditioning

Not only can we obtain expectations by ﬁrst conditioning on an appropriate random variable, but we may also use this approach to compute probabilities. To see this, let E denote an arbitrary event, and deﬁne the indicator random variable X by % 1 if E occurs X= 0 if E does not occur It follows from the deﬁnition of X that E[X] = P(E) E[X|Y = y] = P(E|Y = y)

for any random variable Y

Therefore, from Equations (5.1a) and (5.1b), we obtain P(E|Y = y)P(Y = y) if Y is discrete P(E) = y

* =

q −q

(5.8) P(E|Y = y)fY (y)dy if Y is continuous

Note that if Y is a discrete random variable taking on one of the values y1 , . . . , yn , then, by deﬁning the events Fi , i = 1, . . . , n, by Fi = {Y = yi }, Equation (5.8) reduces to the familiar equation n P(E|Fi )P(Fi ) P(E) = i=1

where F1 , . . . , Fn are mutually exclusive events whose union is the sample space. EXAMPLE 5k The best-prize problem Suppose that we are to be presented with n distinct prizes, in sequence. After being presented with a prize, we must immediately decide whether to accept it or to reject it and consider the next prize. The only information we are given when deciding whether to accept a prize is the relative rank of that prize compared to ones already seen. That is, for instance, when the ﬁfth prize is presented, we learn how it compares with the four prizes we’ve already seen. Suppose that once a prize is rejected, it is lost, and that our objective is to maximize the probability of obtaining the best prize. Assuming that all n! orderings of the prizes are equally likely, how well can we do? Solution. Rather surprisingly, we can do quite well. To see this, ﬁx a value k, 0 … k < n, and consider the strategy that rejects the ﬁrst k prizes and then accepts the ﬁrst one that is better than all of those ﬁrst k. Let Pk (best) denote the probability that the best prize is selected when this strategy is employed. To compute this probability, condition on X, the position of the best prize. This gives Pk (best) =

n

Pk (best|X = i)P(X = i)

i=1

1 Pk (best|X = i) n n

=

i=1

Section 7.5

Conditional Expectation

345

Now, on the one hand, if the overall best prize is among the ﬁrst k, then no prize is ever selected under the strategy considered. That is, Pk (best|X = i) = 0 if i … k On the other hand, if the best prize is in position i, where i > k, then the best prize will be selected if the best of the ﬁrst i − 1 prizes is among the ﬁrst k (for then none of the prizes in positions k + 1, k + 2, . . . , i − 1 would be selected). But, conditional on the best prize being in position i, it is easy to verify that all possible orderings of the other prizes remain equally likely, which implies that each of the ﬁrst i − 1 prizes is equally likely to be the best of that batch. Hence, we have Pk (best|X = i) = P{best of ﬁrst i − 1 is among the ﬁrst k|X = i} k = if i > k i − 1 From the preceding, we obtain n k 1 n i − 1 i=k+1 * n 1 k dx L n k+1 x − 1 n − 1 k = log n k k n L log n k

Pk (best) =

Now, if we consider the function

x n g(x) = log n x

n 1 1 − g (x) = log n x n

then

so

n n g (x) = 0 * log =1*x= x e

Thus, since Pk (best) L g(k), we see that the best strategy of the type considered is to let the ﬁrst n/e prizes go by and then accept the ﬁrst one to appear that is better than all of those. In addition, since g(n/e) = 1/e, the probability that this strategy selects the best prize is approximately 1/e L .36788. Remark. Most people are quite surprised by the size of the probability of obtaining the best prize, thinking that this probability would be close to 0 when n is large. However, even without going through the calculations, a little thought reveals that the probability of obtaining the best prize can be made reasonably large. Consider the strategy of letting half of the prizes go by and then selecting the ﬁrst one to appear that is better than all of those. The probability that a prize is actually selected is the probability that the overall best is among the second half, and this is 12 . In addition, given that a prize is selected, at the time of selection that prize would have been

346

Chapter 7

Properties of Expectation

the best of more than n/2 prizes to have appeared and would thus have probability of at least 12 of being the overall best. Hence, the strategy of letting the ﬁrst half of all prizes go by and then accepting the ﬁrst one that is better than all of those prizes has a probability greater than 14 of obtaining the best prize. . EXAMPLE 5l Let U be a uniform random variable on (0, 1), and suppose that the conditional distribution of X, given that U = p, is binomial with parameters n and p. Find the probability mass function of X. Solution. Conditioning on the value of U gives * 1 P{X = i|U = p}fU (p) dp P{X = i} = 0 * 1 = P{X = i|U = p} dp 0 * 1 n! = pi (1 − p)n−i dp i!(n − i)! 0 Now, it can be shown (a probabilistic proof is given in Section 6.6) that * 1 i!(n − i)! pi (1 − p)n−i dp = (n + 1)! 0 Hence, we obtain P{X = i} =

1 n + 1

i = 0, . . . , n

That is, we obtain the surprising result that if a coin whose probability of coming up heads is uniformly distributed over (0, 1) is ﬂipped n times, then the number of heads occurring is equally likely to be any of the values 0, . . . , n. Because the preceding conditional distribution has such a nice form, it is worth trying to ﬁnd another argument to enhance our intuition as to why such a result is true. To do so, let U, U1 , . . . , Un be n + 1 independent uniform (0, 1) random variables, and let X denote the number of the random variables U1 , . . . , Un that are smaller than U. Since all the random variables U, U1 , . . . , Un have the same distribution, it follows that U is equally likely to be the smallest, or second smallest, or largest of them; so X is equally likely to be any of the values 0, 1, . . . , n. However, given that U = p, the number of the Ui that are less than U is a binomial random variable with parameters n and p, thus establishing our previous result. . EXAMPLE 5m Suppose that X and Y are independent continuous random variables having densities fX and fY , respectively. Compute P{X < Y}. Solution. Conditioning on the value of Y yields * q P{X < Y|Y = y}fY (y) dy P{X < Y} = *−q q = P{X < y|Y = y}fY (y) dy −q

Section 7.5

* = =

q

*−q q −q

Conditional Expectation

347

P{X < y}fY (y} dy by independence FX (y)fY (y) dy

where * FX (y) =

y −q

fX (x) dx

.

EXAMPLE 5n Suppose that X and Y are independent continuous random variables. Find the distribution of X + Y. Solution. By conditioning on the value of Y, we obtain * P{X + Y < a} = = = =

7.5.4

q

*−q q *−q q *−q q −q

P{X + Y < a|Y = y}fY (y) dy P{X + y < a|Y = y}fY (y) dy P{X < a − y}fY (y) dy FX (a − y)fY (y) dy

.

Conditional Variance

Just as we have deﬁned the conditional expectation of X given the value of Y, we can also deﬁne the conditional variance of X given that Y = y: Var(X|Y) K E[(X − E[X|Y])2 |Y] That is, Var(X|Y) is equal to the (conditional) expected square of the difference between X and its (conditional) mean when the value of Y is given. In other words, Var(X|Y) is exactly analogous to the usual deﬁnition of variance, but now all expectations are conditional on the fact that Y is known. There is a very useful relationship between Var(X), the unconditional variance of X, and Var(X|Y), the conditional variance of X given Y, that can often be applied to compute Var(X). To obtain this relationship, note ﬁrst that, by the same reasoning that yields Var(X) = E[X 2 ] − (E[X])2 , we have Var(X|Y) = E[X 2 |Y] − (E[X|Y])2 so E[Var(X|Y)] = E[E[X 2 |Y]] − E[(E[X|Y])2 ] = E[X 2 ] − E[(E[X|Y])2 ]

(5.9)

348

Chapter 7

Properties of Expectation

Also, since E[E[X|Y]] = E[X], we have Var(E[X|Y]) = E[(E[X|Y])2 ] − (E[X])2

(5.10)

Hence, by adding Equations (5.9) and (5.10), we arrive at the following proposition. Proposition 5.2.

The conditional variance formula Var(X) = E[Var(X|Y)] + Var(E[X|Y])

EXAMPLE 5o Suppose that by any time t the number of people that have arrived at a train depot is a Poisson random variable with mean λt. If the initial train arrives at the depot at a time (independent of when the passengers arrive) that is uniformly distributed over (0, T), what are the mean and variance of the number of passengers who enter the train? Solution. For each t Ú 0, let N(t) denote the number of arrivals by t, and let Y denote the time at which the train arrives. The random variable of interest is then N(Y). Conditioning on Y gives E[N(Y)|Y = t] = E[N(t)|Y = t] = E[N(t)] by the independence of Y and N(t) since N(t) is Poisson with mean λt = λt Hence, E[N(Y)|Y] = λY so taking expectations gives E[N(Y)] = λE[Y] =

λT 2

To obtain Var(N(Y)), we use the conditional variance formula: Var(N(Y)|Y = t) = Var(N(t)|Y = t) = Var(N(t)) by independence = λt Thus, Var(N(Y)|Y) = λY E[N(Y)|Y] = λY Hence, from the conditional variance formula, Var(N(Y)) = E[λY] + Var(λY) =λ

T2 T + λ2 2 12

where we have used the fact that Var(Y) = T 2 /12.

.

Section 7.6

Conditional Expectation and Prediction

349

EXAMPLE 5p Variance of a sum of a random number of random variables Let X1 , X2 , . . . be a sequence of independent and identically distributed random variables, and let N be a nonnegative integer-valued random variable that is independent N Xi , we condition on N: of the sequence Xi , i Ú 1. To compute Var i=1

⎤ ⎡ N Xi |N ⎦ = NE[X] E⎣ ⎛ Var ⎝

i=1 N

⎞ Xi |N ⎠ = NVar(X)

i=1

The preceding result follows because, given N, N i=1 Xi is just the sum of a ﬁxed number of independent random variables, so its expectation and variance are just the sums of the individual means and variances, respectively. Hence, from the conditional variance formula, ⎛ ⎞ N Xi ⎠ = E[N]Var(X) + (E[X])2 Var(N) . Var ⎝ i=1

7.6

CONDITIONAL EXPECTATION AND PREDICTION Sometimes a situation arises in which the value of a random variable X is observed and then, on the basis of the observed value, an attempt is made to predict the value of a second random variable Y. Let g(X) denote the predictor; that is, if X is observed to equal x, then g(x) is our prediction for the value of Y. Clearly, we would like to choose g so that g(X) tends to be close to Y. One possible criterion for closeness is to choose g so as to minimize E[(Y − g(X))2 ]. We now show that, under this criterion, the best possible predictor of Y is g(X) = E[Y|X]. Proposition 6.1. E[(Y − g(X))2 ] Ú E[(Y − E[Y|X])2 ] Proof. E[(Y − g(X))2 |X] = E[(Y − E[Y|X] + E[Y|X] − g(X))2 |X] = E[(Y − E[Y|X])2 |X] + E[(E[Y|X] − g(X))2 |X] + 2E[(Y − E[Y|X])(E[Y|X] − g(X))|X]

(6.1)

However, given X, E[Y|X] − g(X), being a function of X, can be treated as a constant. Thus, E[(Y − E[Y|X])(E[Y|X] − g(X))|X] = (E[Y|X] − g(X))E[Y − E[Y|X]|X] = (E[Y|X] − g(X))(E[Y|X] − E[Y|X]) =0

(6.2)

350

Chapter 7

Properties of Expectation

Hence, from Equations (6.1) and (6.2), we obtain E[(Y − g(X))2 |X] Ú E[(Y − E[Y|X])2 |X] and the desired result follows by taking expectations of both sides of the preceding expression. Remark. A second, more intuitive, although less rigorous, argument verifying Proposition 6.1 is as follows. It is straightforward to verify that E[(Y − c)2 ] is minimized at c = E[Y]. (See Theoretical Exercise 1.) Thus, if we want to predict the value of Y when there are no data available to use, the best possible prediction, in the sense of minimizing the mean square error, is to predict that Y will equal its mean. However, if the value of the random variable X is observed to be x, then the prediction problem remains exactly as in the previous (no-data) case, with the exception that all probabilities and expectations are now conditional on the event that X = x. Hence, the best prediction in this situation is to predict that Y will equal its conditional expected value given that X = x, thus establishing Proposition 6.1. . EXAMPLE 6a Suppose that the son of a man of height x (in inches) attains a height that is normally distributed with mean x + 1 and variance 4. What is the best prediction of the height at full growth of the son of a man who is 6 feet tall? Solution. Formally, this model can be written as Y=X + 1 + e where e is a normal random variable, independent of X, having mean 0 and variance 4. The X and Y, of course, represent the heights of the man and his son, respectively. The best prediction E[Y|X = 72] is thus equal to E[Y|X = 72] = E[X + 1 + e|X = 72] = 73 + E[e|X = 72] = 73 + E(e) by independence = 73

.

EXAMPLE 6b Suppose that if a signal value s is sent from location A, then the signal value received at location B is normally distributed with parameters (s, 1). If S, the value of the signal sent at A, is normally distributed with parameters (μ, σ 2 ), what is the best estimate of the signal sent if R, the value received at B, is equal to r? Solution. Let us start by computing the conditional density of S given R. We have fS, R (s, r) fR (r) fS (s)fR|S (r|s) = fR (r)

fS|R (s|r) =

= Ke−(s−μ)

2 /2σ 2

e−(r−s)

2 /2

Section 7.6

Conditional Expectation and Prediction

351

where K does not depend on s. Now, μ 1 1 − + + r s + C1 2 2σ 2 σ2 ⎡ ⎤ 1 + σ2 ⎣ 2 μ + rσ 2 ⎦ s + C1 = s − 2 2σ 2 1 + σ2 2 1 + σ2 (μ + rσ 2 ) = + C2 s − 2σ 2 1 + σ2

(r − s)2 (s − μ)2 = s2 + 2 2σ 2

where C1 and C2 do not depend on s. Hence, ⎧ 2 ⎫ ⎪ ⎪ 2 ⎪ (μ + rσ ) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ − s − ⎪ ⎪ 2 ⎨ ⎬ 1 + σ fS|R (s|r) = C exp ⎪ ⎪ ⎪ ⎪ σ2 ⎪ ⎪ ⎪ ⎪ 2 ⎪ ⎪ 2 ⎪ ⎪ 1 + σ ⎩ ⎭ where C does not depend on s. Thus, we may conclude that the conditional distribution of S, the signal sent, given that r is received, is normal with mean and variance now given by μ + rσ 2 1 + σ2 σ2 Var(S|R = r) = 1 + σ2 E[S|R = r] =

Consequently, from Proposition 6.1, given that the value received is r, the best estimate, in the sense of minimizing the mean square error, for the signal sent is E[S|R = r] =

σ2 1 μ + r 1 + σ2 1 + σ2

Writing the conditional mean as we did previously is informative, for it shows that it equals a weighted average of μ, the a priori expected value of the signal, and r, the value received. The relative weights given to μ and r are in the same proportion to each other as 1 (the conditional variance of the received signal when s is sent) is to σ 2 (the variance of the signal to be sent). . EXAMPLE 6c In digital signal processing, raw continuous analog data X must be quantized, or discretized, in order to obtain a digital representation. In order to quantize the raw data X, an increasing set of numbers ai , i = 0, ;1, ;2, . . ., such that lim ai = q and i→+q lim ai = −q is ﬁxed, and the raw data are then quantized according to the interval i→−q (ai , ai+1 ] in which X lies. Let us denote by yi the discretized value when X ∈ (ai , ai+1 ], and let Y denote the observed discretized value—that is, Y = yi

if ai < X … ai+1

352

Chapter 7

Properties of Expectation

The distribution of Y is given by P{Y = yi } = FX (ai+1 ) − FX (ai ) Suppose now that we want to choose the values yi , i = 0, ;1, ;2, . . . so as to minimize E[(X − Y)2 ], the expected mean square difference between the raw data and their quantized version. (a) Find the optimal values yi , i = 0, ;1, . . .. For the optimal quantizer Y, show that (b) E[Y] = E[X], so the mean square error quantizer preserves the input mean; (c) Var(Y) = Var(X) − E[(X − Y)2 ]. Solution. (a) For any quantizer Y, upon conditioning on the value of Y, we obtain E[(X − Y)2 ] = E[(X − yi )2 |ai < X … ai+1 ]P{ai < X … ai+1 } i

Now, if we let I = i if ai < X … ai+1 then E[(X − yi )2 |ai < X … ai+1 ] = E[(X − yi )2 |I = i] and by Proposition 6.1, this quantity is minimized when yi = E[X|I = i] = E[X|ai < X … ai+1 ] * ai+1 xfX (x) dx = FX (ai+1 ) − FX (ai ) ai Now, since the optimal quantizer is given by Y = E[X|I], it follows that (b) E[Y] = E[X] (c) Var(X) = E[Var(X|I)] + Var(E[X|I]) = E[E[(X − Y)2 |I]] + Var(Y) = E[(X − Y)2 ] + Var(Y)

.

It sometimes happens that the joint probability distribution of X and Y is not completely known; or if it is known, it is such that the calculation of E[Y|X = x] is mathematically intractable. If, however, the means and variances of X and Y and the correlation of X and Y are known, then we can at least determine the best linear predictor of Y with respect to X. To obtain the best linear predictor of Y with respect to X, we need to choose a and b so as to minimize E[(Y − (a + bX))2 ]. Now, E[(Y − (a + bX))2 ] = E[Y 2 − 2aY − 2bXY + a2 + 2abX + b2 X 2 ] = E[Y 2 ] − 2aE[Y] − 2bE[XY] + a2 + 2abE[X] + b2 E[X 2 ]

Section 7.6

Conditional Expectation and Prediction

353

Taking partial derivatives, we obtain ∂ E[(Y − a − bX)2 ] = −2E[Y] + 2a + 2bE[X] ∂a ∂ E[(Y − a − bX)2 ] = −2E[XY] + 2aE[X] + 2bE[X 2 ] ∂b

(6.3)

Setting Equations (6.3) to 0 and solving for a and b yields the solutions σy Cov(X, Y) E[XY] − E[X]E[Y] = =ρ 2 2 2 σx E[X ] − (E[X]) σx ρσy E[X] a = E[Y] − bE[X] = E[Y] − σx

b=

(6.4)

where ρ = Correlation(X, Y), σy2 = Var(Y), and σx2 = Var(X). It is easy to verify that the values of a and b from Equation (6.4) minimize E[(Y − a − bX)2 ]; thus, the best (in the sense of mean square error) linear predictor Y with respect to X is μy +

ρσy (X − μx ) σx

where μy = E[Y] and μx = E[X]. The mean square error of this predictor is given by E

σy Y − μy − ρ (X − μx ) σx

2

5 5 4 σy2 4 3 σy 2 = E (Y − μy )2 + ρ 2 2 E (X − μx )2 − 2ρ E (Y − μy )(X − μx ) σx σx = σy2 + ρ 2 σy2 − 2ρ 2 σy2 = σy2 (1 − ρ 2 )

(6.5)

We note from Equation (6.5) that if ρ is near +1 or −1, then the mean square error of the best linear predictor is near zero. . EXAMPLE 6d An example in which the conditional expectation of Y given X is linear in X, and hence in which the best linear predictor of Y with respect to X is the best overall predictor, is when X and Y have a bivariate normal distribution. For, as shown in Example 5c of Chapter 6, in that case, E[Y|X = x] = μy + ρ

σy (x − μx ) σx

.

354

Chapter 7

Properties of Expectation

7.7

MOMENT GENERATING FUNCTIONS The moment generating function M(t) of the random variable X is deﬁned for all real values of t by M(t) = E[etX ] ⎧ ⎪ etx p(x) ⎪ ⎪ ⎨ x = * q ⎪ ⎪ ⎪ etx f (x) dx ⎩ −q

if X is discrete with mass function p(x) if X is continuous with density f (x)

We call M(t) the moment generating function because all of the moments of X can be obtained by successively differentiating M(t) and then evaluating the result at t = 0. For example, d E[etX ] dt d tX =E (e ) dt

M (t) =

(7.1)

= E[XetX ] where we have assumed that the interchange of the differentiation and expectation operators is legitimate. That is, we have assumed that ⎡ ⎤ d d ⎣ tx [etx p(x)] e p(x)⎦ = dt dt x x in the discrete case and d dt

*

* d tx e f (x) dx = [e f (x)] dx dt tx

in the continuous case. This assumption can almost always be justiﬁed and, indeed, is valid for all of the distributions considered in this book. Hence, from Equation (7.1), evaluated at t = 0, we obtain M (0) = E[X] Similarly, d M (t) dt d = E[XetX ] dt d tX (Xe ) =E dt

M (t) =

= E[X 2 etX ] Thus, M (0) = E[X 2 ]

Section 7.7

Moment Generating Functions

355

In general, the nth derivative of M(t) is given by Mn (t) = E[X n etX ] n Ú 1 implying that

Mn (0) = E[X n ]

n Ú 1

We now compute M(t) for some common distributions. EXAMPLE 7a Binomial distribution with parameters n and p If X is a binomial random variable with parameters n and p, then M(t) = E[etX ] n n tk pk (1 − p)n−k e = k k=0 n n = (pet )k (1 − p)n−k k k=0 t

= (pe + 1 − p)n where the last equality follows from the binomial theorem. Differentiation yields M (t) = n(pet + 1 − p)n−1 pet Thus,

E[X] = M (0) = np

Differentiating a second time yields M (t) = n(n − 1)(pet + 1 − p)n−2 (pet )2 + n(pet + 1 − p)n−1 pet so

E[X 2 ] = M (0) = n(n − 1)p2 + np

The variance of X is given by Var(X) = E[X 2 ] − (E[X])2 = n(n − 1)p2 + np − n2 p2 = np(1 − p) .

verifying the result obtained previously. EXAMPLE 7b Poisson distribution with mean λ If X is a Poisson random variable with parameter λ, then M(t) = E[etX ] q tn −λ n e e λ = n! n=0

= e−λ

q (λet )n n=0

n!

356

Chapter 7

Properties of Expectation

= e−λ eλe

t

= exp{λ(et − 1)} Differentiation yields M (t) = λet exp{λ(et − 1)} M (t) = (λet )2 exp{λ(et − 1)} + λet exp{λ(et − 1)} Thus, E[X] = M (0) = λ E[X 2 ] = M (0) = λ2 + λ Var(X) = E[X 2 ] − (E[X])2 =λ Hence, both the mean and the variance of the Poisson random variable equal λ.

.

EXAMPLE 7c Exponential distribution with parameter λ M(t) = E[etX ] * q = etx λe−λx dx 0 * q e−(λ−t)x dx =λ 0

λ = λ − t

for t < λ

We note from this derivation that, for the exponential distribution, M(t) is deﬁned only for values of t less than λ. Differentiation of M(t) yields M (t) =

λ (λ − t)2

M (t) =

2λ (λ − t)3

Hence, E[X] = M (0) =

1 λ

E[X 2 ] = M (0) =

2 λ2

The variance of X is given by Var(X) = E[X 2 ] − (E[X])2 1 = 2 λ

.

EXAMPLE 7d Normal distribution We ﬁrst compute the moment generating function of a unit normal random variable with parameters 0 and 1. Letting Z be such a random variable, we have MZ (t) = E[etZ ] * q 1 2 =√ etx e−x /2 dx 2π −q

Section 7.7

Moment Generating Functions

357

7 (x2 − 2tx) exp − dx 2 −q 7 0 * q t2 (x − t)2 1 + dx exp − =√ 2 2 2π −q * q 1 2 2 e−(x−t) /2 dx = et /2 √ 2π −q 1 =√ 2π

= et

*

0

q

2 /2

Hence, the moment generating function of the unit normal random variable Z is given 2 by MZ (t) = et /2 . To obtain the moment generating function of an arbitrary normal random variable, we recall (see Section 5.4) that X = μ + σ Z will have a normal distribution with parameters μ and σ 2 whenever Z is a unit normal random variable. Hence, the moment generating function of such a random variable is given by MX (t) = E[etX ] = E[et(μ+σ Z) ] = E[etμ etσ Z ] = etμ E[etσ Z ] = etμ MZ (tσ ) = etμ e(tσ ) /2 7 0 σ 2 t2 + μt = exp 2 2

By differentiating, we obtain

7 2 t2 σ + μt (t) = (μ + tσ 2 ) exp MX 2 7 7 0 0 σ 2 t2 σ 2 t2 2 2 2 MX (t) = (μ + tσ ) exp + μt + σ exp + μt 2 2 0

Thus, E[X] = M (0) = μ E[X 2 ] = M (0) = μ2 + σ 2 implying that Var(X) = E[X 2 ] − E([X])2 = σ2

.

Tables 7.1 and 7.2 give the moment generating functions for some common discrete and continuous distributions. An important property of moment generating functions is that the moment generating function of the sum of independent random variables equals the product of the individual moment generating functions. To prove this, suppose that X and Y are

358

Chapter 7

Properties of Expectation TABLE 7.1: DISCRETE PROBABILITY DISTRIBUTION

Binomial with parameters n, p; 0 … p … 1 Poisson with parameter λ > 0 Geometric with parameter 0 … p … 1 Negative binomial with parameters r, p; 0 … p … 1

Probability mass function, p(x)

Moment generating function, M(t)

Mean

Variance

n x p (1 − p)n−x x

(pet + 1 − p)n

np

np(1 − p)

exp{λ(et − 1)}

λ

λ

pet 1 − (1 − p)et

1 p

1 − p p2

r p

r(1 − p) p2

x = 0, 1, . . . , n e−λ

λx x!

x = 0, 1, 2, . . . p(1 − p)x−1 x = 1, 2, . . . n − 1 r p (1 − p)n−r r − 1

pet 1 − (1 − p)et

r

n = r, r + 1, . . .

independent and have moment generating functions MX (t) and MY (t), respectively. Then MX+Y (t), the moment generating function of X + Y, is given by MX+Y (t) = E[et(X+Y) ] = E[etX etY ] = E[etX ]E[etY ] = MX (t)MY (t) where the next-to-last equality follows from Proposition 4.1, since X and Y are independent. Another important result is that the moment generating function uniquely determines the distribution. That is, if MX (t) exists and is ﬁnite in some region about t = 0, then the distribution of X is uniquely determined. For instance, if 10 1 (et + 1)10 , MX (t) = 2 then it follows from Table 7.1 that X is a binomial random variable with parameters 10 and 12 . EXAMPLE 7e Suppose that the moment generating function of a random variable X is given by t M(t) = e3(e −1) . What is P{X = 0}?

TABLE 7.2:

Uniform over (a, b)

Normal with parameters (μ, σ 2 )

⎧ −λx s−1 ⎪ ⎪ ⎨ λe (λx) x Ú 0 (s) f (x) = ⎪ ⎪ ⎩0 x < 0 1

f (x) = √ e 2πσ

−(x−μ)2 /2σ 2

− q < x < q

Mean

Variance

etb − eta t(b − a)

a + b 2

(b − a)2 12

λ λ − t

1 λ

1 λ2

s λ

s λ2

μ

σ2

λ λ − t 0

s

σ 2 t2 exp μt + 2

7

Moment Generating Functions

Gamma with parameters (s, λ), λ > 0

Probability mass function, f (x) ⎧ ⎪ ⎨ 1 a < x < b f (x) = b − a ⎪ ⎩0 otherwise 0 λe−λx x Ú 0 f (x) = 0 x < 0

Moment generating function, M(t)

Section 7.7

Exponential with parameter λ > 0

CONTINUOUS PROBABILITY DISTRIBUTION

359

360

Chapter 7

Properties of Expectation

Solution. We see from Table 7.1 that M(t) = e3(e −1) is the moment generating function of a Poisson random variable with mean 3. Hence, by the one-to-one correspondence between moment generating functions and distribution functions, it follows . that X must be a Poisson random variable with mean 3. Thus, P{X = 0} = e−3 . t

EXAMPLE 7f Sums of independent binomial random variables If X and Y are independent binomial random variables with parameters (n, p) and (m, p), respectively, what is the distribution of X + Y? Solution. The moment generating function of X + Y is given by MX+Y (t) = MX (t)MY (t) = (pet + 1 − p)n (pet + 1 − p)m = (pet + 1 − p)m+n However, (pet + 1 − p)m+n is the moment generating function of a binomial random variable having parameters m + n and p. Thus, this must be the distribution of X + Y. . EXAMPLE 7g Sums of independent Poisson random variables Calculate the distribution of X + Y when X and Y are independent Poisson random variables with means respective λ1 and λ2 . Solution. MX+Y (t) = MX (t)MY (t) = exp{λ1 (et − 1)} exp{λ2 (et − 1)} = exp{(λ1 + λ2 )(et − 1)} Hence, X + Y is Poisson distributed with mean λ1 + λ2 , verifying the result given in Example 3e of Chapter 6. . EXAMPLE 7h Sums of independent normal random variables Show that if X and Y are independent normal random variables with respective parameters (μ1 , σ12 ) and (μ2 , σ22 ), then X + Y is normal with mean μ1 + μ2 and variance σ12 + σ22 . Solution. MX+Y (t) = MX (t)MY (t) 0 7 0 7 σ12 t2 σ22 t2 + μ1 t exp + μ2 t = exp 2 2 0 7 (σ12 + σ22 )t2 + (μ1 + μ2 )t = exp 2 which is the moment generating function of a normal random variable with mean μ1 + μ2 and variance σ12 + σ22 . The desired result then follows because the moment generating function uniquely determines the distribution. .

Section 7.7

Moment Generating Functions

361

EXAMPLE 7i Compute the moment generating function of a chi-squared random variable with n degrees of freedom. Solution. We can represent such a random variable as Z12 + · · · + Zn2 where Z1 , . . . , Zn are independent standard normal random variables. Let M(t) be its moment generating function. Then, by the preceding, 2

M(t) = (E[etZ ])n where Z is a standard normal random variable. Now, * q 1 2 2 tZ2 E[e ] = √ etx e−x /2 dx 2π *−q q 1 2 2 =√ e−x /2σ dx where σ 2 = (1 − 2t)−1 2π −q =σ = (1 − 2t)−1/2 where the next-to-last equality uses the fact that the normal density with mean 0 and variance σ 2 integrates to 1. Therefore, M(t) = (1 − 2t)−n/2

.

EXAMPLE 7j Moment generating function of the sum of a random number of random variables Let X1 , X2 , . . . be a sequence of independent and identically distributed random variables, and let N be a nonnegative, integer-valued random variable that is independent of the sequence X, i Ú 1. We want to compute the moment generating function of Y=

N

Xi

i=1

(In Example 5d, Y was interpreted as the amount of money spent in a store on a given day when both the amount spent by a customer and the number of customers are random variables.) To compute the moment generating function of Y, we ﬁrst condition on N as follows: ⎤ ⎡ ⎤ ⎡ ⎫66 ⎫66 ⎧ ⎧ N n ⎬6 ⎬6 ⎨ ⎨ ⎥ ⎢ ⎥ ⎢ 6 6 ⎥ ⎢ ⎥ X X N = n = E exp t E⎢ exp t 6 i i 6 N = n⎦ ⎦ ⎣ ⎣ ⎭6 ⎭6 ⎩ ⎩ 6 6 1 1 ⎡ ⎫⎤ ⎧ n ⎬ ⎨ ⎥ ⎢ Xi ⎦ = E ⎣exp t ⎭ ⎩ 1

n

= [MX (t)]

362

Chapter 7

Properties of Expectation

where

MX (t) = E[etXi ]

Hence,

E[etY |N] = (MX (t))N

Thus,

MY (t) = E[(MX (t))N ]

The moments of Y can now be obtained upon differentiation, as follows: (t) = E[N(MX (t))N−1 MX (t)] MY

So E[Y] = MY (0) (0)] = E[N(MX (0))N−1 MX = E[NEX] = E[N]E[X]

(7.2)

verifying the result of Example 5d. (In this last set of equalities, we have used the fact that MX (0) = E[e0X ] = 1.) Also, (t) = E[N(N − 1)(MX (t))N−2 (MX (t))2 + N(MX (t))N−1 MX (t)] MY

so (0) E[Y 2 ] = MY

= E[N(N − 1)(E[X])2 + NE[X 2 ]] = (E[X])2 (E[N 2 ] − E[N]) + E[N]E[X 2 ] 2

2

2

(7.3) 2

= E[N](E[X ] − (E[X]) ) + (E[X]) E[N ] = E[N]Var(X) + (E[X])2 E[N 2 ] Hence, from Equations (7.2) and (7.3), we have Var(Y) = E[N]Var(X) + (E[X])2 (E[N 2 ] − (E[N])2 ) = E[N]Var(X) + (E[X])2 Var(N)

.

EXAMPLE 7k Let Y denote a uniform random variable on (0, 1), and suppose that, conditional on Y = p, the random variable X has a binomial distribution with parameters n and p. In Example 5k, we showed that X is equally likely to take on any of the values 0, 1, . . . , n. Establish this result by using moment generating functions. Solution. To compute the moment generating function of X, start by conditioning on the value of Y. Using the formula for the binomial moment generating function gives E[etX |Y = p] = (pet + 1 − p)n

Section 7.7

Moment Generating Functions

363

Now, Y is uniform on (0, 1), so, upon taking expectations, we obtain * 1 E[etX ] = (pet + 1 − p)n dp 0

* et 1 yn dy (by the substitution y = pet + 1 − p) et − 1 1 1 et(n+1) − 1 = n + 1 et − 1 1 (1 + et + e2t + · · · + ent ) = n + 1

=

Because the preceding is the moment generating function of a random variable that is equally likely to be any of the values 0, 1, . . . , n, the desired result follows from the fact that the moment generating function of a random variable uniquely determines its distribution. . 7.7.1

Joint Moment Generating Functions

It is also possible to deﬁne the joint moment generating function of two or more random variables. This is done as follows: For any n random variables X1 , . . . , Xn , the joint moment generating function, M(t1 , . . . , tn ), is deﬁned, for all real values of t1 , . . . , tn , by M(t1 , . . . , tn ) = E[et1 X1 + · · · +tn Xn ] The individual moment generating functions can be obtained from M(t1 , . . . , tn ) by letting all but one of the tj ’s be 0. That is, MXi (t) = E[etXi ] = M(0, . . . , 0, t, 0, . . . , 0) where the t is in the ith place. It can be proven (although the proof is too advanced for this text) that the joint moment generating function M(t1 , . . . , tn ) uniquely determines the joint distribution of X1 , . . . , Xn . This result can then be used to prove that the n random variables X1 , . . . , Xn are independent if and only if M(t1 , . . . , tn ) = MX1 (t1 ) · · · MXn (tn )

(7.4)

For the proof in one direction, if the n random variables are independent, then M(t1 , . . . , tn ) = E[e(t1 X1 + · · · +tn Xn ) ] = E[et1 X1 · · · etn Xn ] = E[et1 X1 ] · · · E[etn Xn ] by independence = MX1 (t1 ) · · · MXn (tn ) For the proof in the other direction, if Equation (7.4) is satisﬁed, then the joint moment generating function M(t1 , . . . , tn ) is the same as the joint moment generating function of n independent random variables, the ith of which has the same distribution as Xi . As the joint moment generating function uniquely determines the joint distribution, this must be the joint distribution; hence, the random variables are independent.

364

Chapter 7

Properties of Expectation

EXAMPLE 7l Let X and Y be independent normal random variables, each with mean μ and variance σ 2 . In Example 7a of Chapter 6, we showed that X + Y and X − Y are independent. Let us now establish this result by computing their joint moment generating function: E[et(X+Y)+s(X−Y) ] = E[e(t+s)X+(t−s)Y ] = E[e(t+s)X ]E[e(t−s)Y ] = eμ(t+s)+σ = e2μt+σ

2 t2

2 (t+s)2 /2

eσ

eμ(t−s)+σ

2 (t−s)2 /2

2 s2

But we recognize the preceding as the joint moment generating function of the sum of a normal random variable with mean 2μ and variance 2σ 2 and an independent normal random variable with mean 0 and variance 2σ 2 . Because the joint moment generating function uniquely determines the joint distribution, it follows that X + Y and X − Y are independent normal random variables. . In the next example, we use the joint moment generating function to verify a result that was established in Example 2b of Chapter 6. EXAMPLE 7m Suppose that the number of events that occur is a Poisson random variable with mean λ and that each event is independently counted with probability p. Show that the number of counted events and the number of uncounted events are independent Poisson random variables with respective means λp and λ(1 − p). Solution. Let X denote the total number of events, and let Xc denote the number of them that are counted. To compute the joint moment generating function of Xc , the number of events that are counted, and X − Xc , the number that are uncounted, start by conditioning on X to obtain E[esXc +t(X−Xc ) |X = n] = etn E[e(s−t)Xc |X = n] = etn (pes−t + 1 − p)n = (pes + (1 − p)et )n which follows because, conditional on X = n, Xc is a binomial random variable with parameters n and p. Hence, E[esXc +t(X−Xc ) |X] = (pes + (1 − p)et )X Taking expectations of both sides of this equation yields E[esXc +t(X−Xc ) ] = E[(pes + (1 − p)et )X ] Now, since X is Poisson with mean λ, it follows that E[etX ] = eλ(e −1) . Therefore, for any positive value a we see (by letting a = et ) that E[aX ] = eλ(a−1) . Thus t

E[esXc +t(X−Xc ) ] = eλ(pe +(1−p)e −1) s

s−1 )

= eλp(e

t

eλ(1−p)(e −1) t

Section 7.8

Additional Properties of Normal Random Variables

365

As the preceding is the joint moment generating function of independent Poisson random variables with respective means λp and λ(1 − p), the result is proven. . 7.8

ADDITIONAL PROPERTIES OF NORMAL RANDOM VARIABLES 7.8.1

The Multivariate Normal Distribution

Let Z1 , . . . , Zn be a set of n independent unit normal random variables. If, for some constants aij , 1 … i … m, 1 … j … n, and μi , 1 … i … m, X1 = a11 Z1 + · · · + a1n Zn + μ1 X2 = a21 Z1 + · · · + a2n Zn + μ2 . . . Xi = ai1 Z1 + · · · + ain Zn + μi . . . Xm = am1 Z1 + · · · + amn Zn + μm then the random variables X1 , . . . , Xm are said to have a multivariate normal distribution. From the fact that the sum of independent normal random variables is itself a normal random variable, it follows that each Xi is a normal random variable with mean and variance given, respectively, by E[Xi ] = μi n Var(Xi ) = a2ij j=1

Let us now consider M(t1 , . . . , tm ) = E[exp{t1 X1 + · · · + tm Xm }] the joint moment generating function of X1 , . . . , Xm . The ﬁrst thing to note is that m since ti Xi is itself a linear combination of the independent normal random varii=1

ables Z1 , . . . , Zn , it is also normally distributed. Its mean and variance are ⎡ ⎤ m m E⎣ ti Xi ⎦ = ti μi i=1

and ⎛ Var ⎝

m

i=1

⎞

⎛

⎞

⎜ ti Xi ⎠ = Cov ⎝

i=1

m i=1

=

m m i=1 j=1

ti Xi ,

m

⎟ tj Xj ⎠

j=1

ti tj Cov(Xi , Xj )

366

Chapter 7

Properties of Expectation

Now, if Y is a normal random variable with mean μ and variance σ 2 , then E[eY ] = MY (t)|t=1 = eμ+σ Thus, M(t1 , . . . , tm ) = exp

⎧ ⎪ m ⎨ ⎪ ⎩ i=1

ti μi +

1 2

2 /2

m m

ti tj Cov(Xi , Xj )

i=1 j=1

⎫ ⎪ ⎬ ⎪ ⎭

which shows that the joint distribution of X1 , . . . , Xm is completely determined from a knowledge of the values of E[Xi ] and Cov(Xi , Xj ), i, j = 1, . . . , m. It can be shown that when m = 2, the multivariate normal distribution reduces to the bivariate normal. EXAMPLE 8a Find P(X < Y) for bivariate normal random variables X and Y having parameters μx = E[X], μy = E[Y], σx2 = Var(X), σy2 = Var(Y), ρ = Corr(X, Y) Solution. Because X − Y is normal with mean E[X − Y] = μx − μy and variance Var(X − Y) = Var(X) + Var(−Y) + 2Cov(X, −Y) = σx2 + σy2 − 2ρσx σy we obtain P{X < Y} = P{X − Y < 0} ⎧ ⎫ ⎪ ⎪ ⎨ X − Y − (μ − μ ) ⎬ −(μx − μy ) x y =P 8 < 8 ⎪ ⎩ σx2 + σy2 − 2ρσx σy ⎭ σx2 + σy2 − 2ρσx σy ⎪ μy − μx = 8 σx2 + σy2 − 2ρσx σy

.

EXAMPLE 8b Suppose that the conditional distribution of X, given that = θ, is normal with mean θ and variance 1. Moreover, suppose that itself is a normal random variable with mean μ and variance σ 2 . Find the conditional distribution of given that X = x. Solution. Rather than using and then simplifying Bayes’s formula, we will solve this problem by ﬁrst showing that X, has a bivariate normal distribution. To do so, note that the joint density function of X, can be written as fX, (x, θ) = fX| (x|θ)f (θ) where fX| (x|θ) is a normal density with mean θ and variance 1. However, if we let Z be a standard normal random variable that is independent of , then the conditional distribution of Z + , given that = θ, is also normal with mean θ and variance 1.

Section 7.8

Additional Properties of Normal Random Variables

367

Consequently, the joint density of Z + , is the same as that of X, . Because the former joint density is clearly bivariate normal (since Z + and are both linear combinations of the independent normal random variables Z and ), it follows that X, has a bivariate normal distribution. Now, E[X] = E[Z + ] = μ Var(X) = Var(Z + ) = 1 + σ 2 and ρ = Corr(X, ) = Corr(Z + , ) Cov(Z + , ) =√ Var(Z + )Var() σ =. 1 + σ2 Because X, has a bivariate normal distribution, the conditional distribution of , given that X = x, is normal with mean ? Var() (x − E[X]) E[|X = x] = E[] + ρ Var(X) =μ +

σ2 (x − μ) 1 + σ2

and variance Var(|X = x) = Var()(1 − ρ 2 ) = 7.8.2

σ2 1 + σ2

.

The Joint Distribution of the Sample Mean and Sample Variance

Let X1 , . . . , Xn be independent normal random variables, each with mean μ and varin ance σ 2 . Let X = Xi /n denote their sample mean. Since the sum of independent i=1

normal random variables is also a normal random variable, it follows that X is a normal random variable with (from Examples 2c and 4a) expected value μ and variance σ 2 /n. Now, recall from Example 4e that Cov(X, Xi − X) = 0,

i = 1, . . . , n

(8.1)

Also, note that since X, X1 − X, X2 − X, . . . , Xn − X are all linear combinations of the independent standard normals (Xi − μ)/σ , i = 1, . . . , n, it follows that X, Xi − X, i = 1, . . . , n has a joint distribution that is multivariate normal. If we let Y be a normal random variable, with mean μ and variance σ 2 /n, that is independent of the Xi , i = 1, . . . , n, then Y, Xi − X, i = 1, . . . , n also has a multivariate normal distribution and, indeed, because of Equation (8.1), has the same expected values and covariances as the random variables X, Xi − X, i = 1, . . . , n. But since a multivariate normal distribution is determined completely by its expected values and covariances,

368

Chapter 7

Properties of Expectation

it follows that Y, Xi − X, i = 1, . . . , n and X, Xi − X, i = 1, . . . , n have the same joint distribution, thus showing that X is independent of the sequence of deviations Xi − X, i = 1, . . . , n. Since X is independent of the sequence of deviations Xi − X, i = 1, . . . , n, it is also n (Xi − X)2 /(n − 1). independent of the sample variance S2 K i=1

Since we already know that X is normal with mean μ and variance σ 2 /n, it remains only to determine the distribution of S2 . To accomplish this, recall, from Example 4a, the algebraic identity n

(n − 1)S2 =

(Xi − X)2

i=1 n

=

(Xi − μ)2 − n(X − μ)2

i=1

Upon dividing the preceding equation by σ 2 , we obtain (n − 1)S2 + σ2

Now,

X − μ √ σ/ n

2

n Xi − μ 2 = σ

(8.2)

i=1

n Xi − μ 2 σ i=1

is the sum of the squares of n independent standard normal random variables and so is a chi-squared random variable with n degrees of freedom. Hence, from Example 7i, its moment generating function is (1 − 2t)−n/2 . Also, because

X − μ √ σ/ n

2

is the square of a standard normal variable, it is a chi-squared random variable with 1 degree of freedom, and so has moment generating function (1 − 2t)−1/2 . Now, we have seen previously that the two random variables on the left side of Equation (8.2) are independent. Hence, as the moment generating function of the sum of independent random variables is equal to the product of their individual moment generating functions, we have 2 /σ 2

E[et(n−1)S

](1 − 2t)−1/2 = (1 − 2t)−n/2

or 2 /σ 2

E[et(n−1)S

] = (1 − 2t)−(n−1)/2

But as (1 − 2t)−(n−1)/2 is the moment generating function of a chi-squared random variable with n − 1 degrees of freedom, we can conclude, since the moment generating function uniquely determines the distribution of the random variable, it follows that that is the distribution of (n − 1)S2 /σ 2 . Summing up, we have shown the following.

Section 7.9

General Deﬁnition of Expectation

369

Proposition 8.1. If X1 , . . . , Xn are independent and identically distributed normal random variables with mean μ and variance σ 2 , then the sample mean X and the sample variance S2 are independent. X is a normal random variable with mean μ and variance σ 2 /n; (n − 1)S2 /σ 2 is a chi-squared random variable with n − 1 degrees of freedom. 7.9

GENERAL DEFINITION OF EXPECTATION Up to this point, we have deﬁned expectations only for discrete and continuous random variables. However, there also exist random variables that are neither discrete nor continuous, and they, too, may possess an expectation. As an example of such a random variable, let X be a Bernoulli random variable with parameter p = 12 , and let Y be a uniformly distributed random variable over the interval [0, 1]. Furthermore, suppose that X and Y are independent, and deﬁne the new random variable W by % X if X = 1 W= Y if X Z 1 Clearly, W is neither a discrete (since its set of possible values, [0, 1], is uncountable) nor a continuous (since P{W = 1} = 12 ) random variable. In order to deﬁne the expectation of an arbitrary random variable, we require the notion of a Stieltjes integral. Before deﬁning this integral, let us recall that, for any -b function g, a g(x) dx is deﬁned by *

b

g(x) dx = lim

a

n

g(xi )(xi − xi−1 )

i=1

where the limit is taken over all a = x0 < x1 < x2 · · · < xn = b as n→q and where max (xi − xi−1 )→0. i=1,...,n

For any distribution function F, we deﬁne the Stieltjes integral of the nonnegative function g over the interval [a, b] by *

b

g(x) dF(x) = lim

a

n

g(xi )[F(xi ) − F(xi−1 )]

i=1

where, as before, the limit is taken over all a = x0 < x1 < · · · < xn = b as n→q and where max (xi − xi−1 )→0. Further, we deﬁne the Stieltjes integral over the whole i=1,...,n

real line by

*

q −q

* g(x) dF(x) =

lim a→ − q b→ + q

b

g(x) dF(x) a

Finally, if g is not a nonnegative function, we deﬁne g+ and g− by % g(x) if g(x) Ú 0 + g (x) = 0 if g(x) < 0 % 0 if g(x) Ú 0 g− (x) = −g(x) if g(x) < 0

370

Chapter 7

Properties of Expectation

Because g(x) = g+ (x) − g− (x) and g+ and g− are both nonnegative functions, it is natural to deﬁne * q * q * q + g(x) dF(x) = g (x) dF(x) − g− (x) dF(x) −q

−q

-q

-q

−q

-q and we say that −q g(x) dF(x) exists as long as −q g+ (x) dF(x) and −q g− (x) dF(x) are not both equal to +q. If X is an arbitrary random variable having cumulative distribution F, we deﬁne the expected value of X by * q x dF(x) (9.1) E[X] = −q

It can be shown that if X is a discrete random variable with mass function p(x), then * q xdF(x) = xp(x) −q

x:p(x)>0

whereas if X is a continuous random variable with density function f (x), then * q * q xdF(x) = xf (x) dx −q

−q

The reader should note that Equation (9.1) yields an intuitive deﬁnition of E[X]; consider the approximating sum n

xi [F(xi ) − F(xi−1 )]

i=1

of E[X]. Because F(xi ) − F(xi−1 ) is just the probability that X will be in the interval (xi−1 , xi ], the approximating sum multiplies the approximate value of X when it is in the interval (xi−1 , xi ] by the probability that it will be in that interval and then sums over all the intervals. Clearly, as these intervals get smaller and smaller in length, we obtain the “expected value” of X. Stieltjes integrals are mainly of theoretical interest because they yield a compact way of deﬁning and dealing with the properties of expectation. For instance, the use of Stieltjes integrals avoids the necessity of having to give separate statements and proofs of theorems for the continuous and the discrete cases. However, their properties are very much the same as those of ordinary integrals, and all of the proofs presented in this chapter can easily be translated into proofs in the general case. SUMMARY If X and Y have a joint probability mass function p(x, y), then g(x, y)p(x, y) E[g(X, Y)] = y

x

whereas if they have a joint density function f (x, y), then * q* q g(x, y)f (x, y) dx dy E[g(X, Y)] = −q −q

A consequence of the preceding equations is that E[X + Y] = E[X] + E[Y]

Summary

371

⎡ ⎤ n n ⎣ ⎦ E Xi = E[Xi ]

which generalizes to

i=1

i=1

The covariance between random variables X and Y is given by Cov(X, Y) = E[(X − E[X])(Y − E[Y])] = E[XY] − E[X]E[Y] A useful identity is

⎞ ⎛ m n m n ⎟ ⎜ Xi , Yj ⎠ = Cov(Xi , Yj ) Cov ⎝ i=1

j=1

i=1 j=1

When n = m and Yi = Xi , i = 1, . . . , n, the preceding formula gives ⎛ ⎞ n n Var ⎝ Xi ⎠ = Var(Xi ) + 2 Cov(Xi , Yj ) i=1

i

i=1

The correlation between X and Y, denoted by ρ(X, Y), is deﬁned by ρ(X, Y) = √

Cov(X, Y) Var(X)Var(Y)

If X and Y are jointly discrete random variables, then the conditional expected value of X, given that Y = y, is deﬁned by E[X|Y = y] = xP{X = x|Y = y] x

If X and Y are jointly continuous random variables, then * q E[X|Y = y] = xfX|Y (x|y) −q

where fX|Y (x|y) =

f (x, y) fY (y)

is the conditional probability density of X given that Y = y. Conditional expectations, which are similar to ordinary expectations except that all probabilities are now computed conditional on the event that Y = y, satisfy all the properties of ordinary expectations. Let E[X|Y] denote that function of Y whose value at Y = y is E[X|Y = y]. A very useful identity is E[X] = E[E[X|Y]] In the case of discrete random variables, this equation reduces to the identity E[X|Y = y]P{Y = y} E[X] = y

and, in the continuous case, to E[X] =

*

q −q

E[X|Y = y]fY (y)

372

Chapter 7

Properties of Expectation

The preceding equations can often be applied to obtain E[X] by ﬁrst “conditioning” on the value of some other random variable Y. In addition, since, for any event A, P(A) = E[IA ], where IA is 1 if A occurs and is 0 otherwise, we can use the same equations to compute probabilities. The conditional variance of X, given that Y = y, is deﬁned by Var(X|Y = y) = E[(X − E[X|Y = y])2 |Y = y] Let Var(X|Y) be that function of Y whose value at Y = y is Var(X|Y = y). The following is known as the conditional variance formula: Var(X) = E[Var(X|Y)] + Var(E[X|Y]) Suppose that the random variable X is to be observed and, on the basis of its value, one must then predict the value of the random variable Y. In such a situation, it turns out that, among all predictors, E[Y|X] has the smallest expectation of the square of the difference between it and Y. The moment generating function of the random variable X is deﬁned by M(t) = E[etX ] The moments of X can be obtained by successively differentiating M(t) and then evaluating the resulting quantity at t = 0. Speciﬁcally, we have 6 6 dn 6 n n = 1, 2, . . . E[X ] = n M(t)6 6 dt t=0

Two useful results concerning moment generating functions are, ﬁrst, that the moment generating function uniquely determines the distribution function of the random variable and, second, that the moment generating function of the sum of independent random variables is equal to the product of their moment generating functions. These results lead to simple proofs that the sum of independent normal (Poisson, gamma) random variables remains a normal (Poisson, gamma) random variable. If X1 , . . . , Xm are all linear combinations of a ﬁnite set of independent standard normal random variables, then they are said to have a multivariate normal distribution. Their joint distribution is speciﬁed by the values of E[Xi ], Cov(Xi , Xj ), i, j = 1, . . . , m. If X1 , . . . , Xn are independent and identically distributed normal random variables, then their sample mean n Xi X= n i=1

and their sample variance n (Xi − X)2 S = n − 1 2

i=1

are independent. The sample mean X is a normal random variable with mean μ and variance σ 2 /n; the random variable (n − 1)S2 /σ 2 is a chi-squared random variable with n − 1 degrees of freedom.

Problems

373

PROBLEMS 7.1. A player throws a fair die and simultaneously ﬂips a fair coin. If the coin lands heads, then she wins twice, and if tails, then one-half of the value that appears on the die. Determine her expected winnings. 7.2. The game of Clue involves 6 suspects, 6 weapons, and 9 rooms. One of each is randomly chosen and the object of the game is to guess the chosen three. (a) How many solutions are possible? In one version of the game, the selection is made and then each of the players is randomly given three of the remaining cards. Let S, W, and R be, respectively, the numbers of suspects, weapons, and rooms in the set of three cards given to a speciﬁed player. Also, let X denote the number of solutions that are possible after that player observes his or her three cards. (b) Express X in terms of S, W, and R. (c) Find E[X]. 7.3. Gambles are independent, and each one results in the player being equally likely to win or lose 1 unit. Let W denote the net winnings of a gambler whose strategy is to stop gambling immediately after his ﬁrst win. Find (a) P{W > 0} (b) P{W < 0} (c) E[W] 7.4. If X and Y have joint density function % 1/y, if 0 < y < 1, 0 < x < y fX,Y (x, y) = 0, otherwise ﬁnd (a) E[XY] (b) E[X] (c) E[Y] 7.5. The county hospital is located at the center of a square whose sides are 3 miles wide. If an accident occurs within this square, then the hospital sends out an ambulance. The road network is rectangular, so the travel distance from the hospital, whose coordinates are (0, 0), to the point (x, y) is |x| + |y|. If an accident occurs at a point that is uniformly distributed in the square, ﬁnd the expected travel distance of the ambulance. 7.6. A fair die is rolled 10 times. Calculate the expected sum of the 10 rolls. 7.7. Suppose that A and B each randomly and independently choose 3 of 10 objects. Find the expected number of objects (a) chosen by both A and B; (b) not chosen by either A or B; (c) chosen by exactly one of A and B.

7.8. N people arrive separately to a professional dinner. Upon arrival, each person looks to see if he or she has any friends among those present. That person then sits either at the table of a friend or at an unoccupied table if none of those present N is a friend. Assuming that each of the 2 pairs of people is, independently, a pair of friends with probability p, ﬁnd the expected number of occupied tables. Hint: Let Xi equal 1 or 0, depending on whether the ith arrival sits at a previously unoccupied table. 7.9. A total of n balls, numbered 1 through n, are put into n urns, also numbered 1 through n in such a way that ball i is equally likely to go into any of the urns 1, 2, . . . , i. Find (a) the expected number of urns that are empty; (b) the probability that none of the urns is empty. 7.10. Consider 3 trials, each having the same probability of success. Let X denote the total number of successes in these trials. If E[X] = 1.8, what is (a) the largest possible value of P{X = 3}? (b) the smallest possible value of P{X = 3}? In both cases, construct a probability scenario that results in P{X = 3} having the stated value. Hint: For part (b), you might start by letting U be a uniform random variable on (0, 1) and then deﬁning the trials in terms of the value of U. 7.11. Consider n independent ﬂips of a coin having probability p of landing on heads. Say that a changeover occurs whenever an outcome differs from the one preceding it. For instance, if n = 5 and the outcome is HHTHT, then there are 3 changeovers. Find the expected number of changeovers. Hint: Express the number of changeovers as the sum of n − 1 Bernoulli random variables. 7.12. A group of n men and n women is lined up at random. (a) Find the expected number of men who have a woman next to them. (b) Repeat part (a), but now assuming that the group is randomly seated at a round table. 7.13. A set of 1000 cards numbered 1 through 1000 is randomly distributed among 1000 people with each receiving one card. Compute the expected number of cards that are given to people whose age matches the number on the card. 7.14. An urn has m black balls. At each stage, a black ball is removed and a new ball that is black with

374

Chapter 7

Properties of Expectation

probability p and white with probability 1 − p is put in its place. Find the expected number of stages needed until there are no more black balls in the urn. NOTE : The preceding has possible applications to understanding the AIDS disease. Part of the body’s immune system consists of a certain class of cells known as T-cells. There are 2 types of Tcells, called CD4 and CD8. Now, while the total number of T-cells in AIDS sufferers is (at least in the early stages of the disease) the same as that in healthy individuals, it has recently been discovered that the mix of CD4 and CD8 T-cells is different. Roughly 60 percent of the T-cells of a healthy person are of the CD4 type, whereas the percentage of the T-cells that are of CD4 type appears to decrease continually in AIDS sufferers. A recent model proposes that the HIV virus (the virus that causes AIDS) attacks CD4 cells and that the body’s mechanism for replacing killed T-cells does not differentiate between whether the killed T-cell was CD4 or CD8. Instead, it just produces a new T-cell that is CD4 with probability .6 and CD8 with probability .4. However, although this would seem to be a very efﬁcient way of replacing killed T-cells when each one killed is equally likely to be any of the body’s T-cells (and thus has probability .6 of being CD4), it has dangerous consequences when facing a virus that targets only the CD4 T-cells. 7.15. In Example 2h, say that i and j, i Z j, form a matched pair if i chooses the hat belonging to j and j chooses the hat belonging to i. Find the expected number of matched pairs. 7.16. Let Z be a standard normal random variable, and, for a ﬁxed x, set % Z if Z > x X= 0 otherwise 1 2 Show that E[X] = √ e−x /2 . 2π 7.17. A deck of n cards numbered 1 through n is thoroughly shufﬂed so that all possible n! orderings can be assumed to be equally likely. Suppose you are to make n guesses sequentially, where the ith one is a guess of the card in position i. Let N denote the number of correct guesses. (a) If you are not given any information about your earlier guesses show that, for any strategy, E[N] = 1. (b) Suppose that after each guess you are shown the card that was in the position in question. What do you think is the best strategy? Show that, under this strategy,

1 1 + + ··· + 1 n n − 1 * n 1 L dx = log n 1 x

E[N] =

(c) Suppose that you are told after each guess whether you are right or wrong. In this case, it can be shown that the strategy which maximizes E[N] is one that keeps on guessing the same card until you are told you are correct and then changes to a new card. For this strategy, show that 1 1 1 + + ··· + E[N] = 1 + 2! 3! n! L e − 1 Hint: For all parts, express N as the sum of indicator (that is, Bernoulli) random variables. 7.18. Cards from an ordinary deck of 52 playing cards are turned face up one at a time. If the 1st card is an ace, or the 2nd a deuce, or the 3rd a three, or . . ., or the 13th a king, or the 14 an ace, and so on, we say that a match occurs. Note that we do not require that the (13n + 1)th card be any particular ace for a match to occur but only that it be an ace. Compute the expected number of matches that occur. 7.19. A certain region is inhabited by r distinct types of a certain species of insect. Each insect caught will, independently of the types of the previous catches, be of type i with probability r Pi , i = 1, . . . , r Pi = 1 1

(a) Compute the mean number of insects that are caught before the ﬁrst type 1 catch. (b) Compute the mean number of types of insects that are caught before the ﬁrst type 1 catch. 7.20. In an urn containing n balls, the ith ball has weight W(i), i = 1, . . . , n. The balls are removed without replacement, one at a time, according to the following rule: At each selection, the probability that a given ball in the urn is chosen is equal to its weight divided by the sum of the weights remaining in the urn. For instance, if at some time i1 , . . . , ir is the set of balls remaining in the urn, then the next selection will be ij with prob r ability W(ij ) W(ik ), j = 1, . . . , r. Compute k=1

the expected number of balls that are withdrawn before ball number 1 is removed. 7.21. For a group of 100 people, compute (a) the expected number of days of the year that are birthdays of exactly 3 people: (b) the expected number of distinct birthdays.

Problems 7.22. How many times would you expect to roll a fair die before all 6 sides appeared at least once? 7.23. Urn 1 contains 5 white and 6 black balls, while urn 2 contains 8 white and 10 black balls. Two balls are randomly selected from urn 1 and are put into urn 2. If 3 balls are then randomly selected from urn 2, compute the expected number of white balls in the trio. Hint: Let Xi = 1 if the ith white ball initially in urn 1 is one of the three selected, and let Xi = 0 otherwise. Similarly, let Yi = 1 if the ith white ball from urn 2 is one of the three selected, and let Yi = 0 otherwise. The number of white balls in the trio 5 8 Yi . can now be written as Xi + 1

1

7.24. A bottle initially contains m large pills and n small pills. Each day, a patient randomly chooses one of the pills. If a small pill is chosen, then that pill is eaten. If a large pill is chosen, then the pill is broken in two; one part is returned to the bottle (and is now considered a small pill) and the other part is then eaten. (a) Let X denote the number of small pills in the bottle after the last large pill has been chosen and its smaller half returned. Find E[X]. Hint: Deﬁne n + m indicator variables, one for each of the small pills initially present and one for each of the m small pills created when a large one is split in two. Now use the argument of Example 2m. (b) Let Y denote the day on which the last large pill is chosen. Find E[Y]. Hint: What is the relationship between X and Y? 7.25. Let X1 , X2 , . . . be a sequence of independent and identically distributed continuous random variables. Let N Ú 2 be such that X1 Ú X2 Ú · · · Ú XN−1 < XN That is, N is the point at which the sequence stops decreasing. Show that E[N] = e. Hint: First ﬁnd P{N Ú n}. 7.26. If X1 , X2 , . . . , Xn are independent and identically distributed random variables having uniform distributions over (0, 1), ﬁnd (a) E[max(X1 , . . . , Xn )]; (b) E[min(X1 , . . . , Xn )]. ∗ 7.27. If 101 items are distributed among 10 boxes, then at least one of the boxes must contain more than 10 items. Use the probabilistic method to prove this result. ∗ 7.28. The k-of-r-out-of-n circular reliability system, k … r … n, consists of n components that are arranged in a circular fashion. Each component is either functional or failed, and the system functions if there is no block of r consecutive components of which

375

at least k are failed. Show that there is no way to arrange 47 components, 8 of which are failed, to make a functional 3-of-12-out-of-47 circular system. ∗ 7.29. There are 4 different types of coupons, the ﬁrst 2 of which compose one group and the second 2 another group. Each new coupon obtained is type i with probability pi , where p1 = p2 = 1/8, p3 = p4 = 3/8. Find the expected number of coupons that one must obtain to have at least one of (a) all 4 types; (b) all the types of the ﬁrst group; (c) all the types of the second group; (d) all the types of either group. 7.30. If X and Y are independent and identically distributed with mean μ and variance σ 2 , ﬁnd E[(X − Y)2 ] 7.31. In Problem 6, calculate the variance of the sum of the rolls. 7.32. In Problem 9, compute the variance of the number of empty urns. 7.33. If E[X] = 1 and Var(X) = 5, ﬁnd (a) E[(2 + X)2 ]; (b) Var(4 + 3X). 7.34. If 10 married couples are randomly seated at a round table, compute (a) the expected number and (b) the variance of the number of wives who are seated next to their husbands. 7.35. Cards from an ordinary deck are turned face up one at a time. Compute the expected number of cards that need to be turned face up in order to obtain (a) 2 aces; (b) 5 spades; (c) all 13 hearts. 7.36. Let X be the number of 1’s and Y the number of 2’s that occur in n rolls of a fair die. Compute Cov(X, Y). 7.37. A die is rolled twice. Let X equal the sum of the outcomes, and let Y equal the ﬁrst outcome minus the second. Compute Cov(X, Y). 7.38. The random variables X and Y have a joint density function given by 0 2e−2x /x 0 … x < q, 0 … y … x f (x, y) = 0 otherwise Compute Cov(X, Y). 7.39. Let X1 , . . . be independent with common mean μ and common variance σ 2 , and set Yn = Xn + Xn+1 + Xn+2 . For j Ú 0, ﬁnd Cov(Yn , Yn+j ).

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7.40. The joint density function of X and Y is given by f (x, y) =

1 −(y+x/y) , e y

x > 0, y > 0

Find E[X], E[Y], and show that Cov(X, Y) = 1. 7.41. A pond contains 100 ﬁsh, of which 30 are carp. If 20 ﬁsh are caught, what are the mean and variance of the number of carp among the 20? What assumptions are you making? 7.42. A group of 20 people consisting of 10 men and 10 women is randomly arranged into 10 pairs of 2 each. Compute the expectation and variance of the number of pairs that consist of a man and a woman. Now suppose the 20 people consist of 10 married couples. Compute the mean and variance of the number of married couples that are paired together. 7.43. Let X1 , X2 , . . . , Xn be independent random variables having an unknown continuous distribution function F, and let Y1 , Y2 , . . . , Ym be independent random variables having an unknown continuous distribution function G. Now order those n + m variables, and let ⎧ ⎨1 if the ith smallest of the n + m variables is from the X sample Ii = ⎩0 otherwise The random variable R =

n+m

7.45. If X1 , X2 , X3 , and X4 are (pairwise) uncorrelated random variables, each having mean 0 and variance 1, compute the correlations of (a) X1 + X2 and X2 + X3 ; (b) X1 + X2 and X3 + X4 . 7.46. Consider the following dice game, as played at a certain gambling casino: Players 1 and 2 roll a pair of dice in turn. The bank then rolls the dice to determine the outcome according to the following rule: Player i, i = 1, 2, wins if his roll is strictly greater than the bank’s. For i = 1, 2, let % 1 if i wins Ii = 0 otherwise

7.47.

7.48.

iIi is the sum of the

i=1

ranks of the X sample and is the basis of a standard statistical procedure (called the Wilcoxon sum-ofranks test) for testing whether F and G are identical distributions. This test accepts the hypothesis that F = G when R is neither too large nor too small. Assuming that the hypothesis of equality is in fact correct, compute the mean and variance of R. Hint: Use the results of Example 3e. 7.44. Between two distinct methods for manufacturing certain goods, the quality of goods produced by method i is a continuous random variable having distribution Fi , i = 1, 2. Suppose that n goods are produced by method 1 and m by method 2. Rank the n + m goods according to quality, and let ⎧ ⎨1 if the jth best was produced from method 1 Xj = ⎩2 otherwise For the vector X1 , X2 , . . . , Xn+m , which consists of n 1’s and m 2’s, let R denote the number of runs of 1. For instance, if n = 5, m = 2, and X = 1, 2, 1, 1, 1, 1, 2, then R = 2. If F1 = F2 (that is, if the two methods produce identically distributed goods), what are the mean and variance of R?

7.49.

7.50.

and show that I1 and I2 are positively correlated. Explain why this result was to be expected. Consider a graph having n vertices labeled 1, 2,. . . , n, and suppose that, between each of the n pairs of distinct vertices, an edge is indepen2 dently present with probability p. The degree of vertex i, designated as Di , is the number of edges that have vertex i as one of their vertices. (a) What is the distribution of Di ? (b) Find ρ(Di , Dj ), the correlation between Di and Dj . A fair die is successively rolled. Let X and Y denote, respectively, the number of rolls necessary to obtain a 6 and a 5. Find (a) E[X]; (b) E[X|Y = 1]; (c) E[X|Y = 5]. There are two misshapen coins in a box; their probabilities for landing on heads when they are ﬂipped are, respectively, .4 and .7. One of the coins is to be randomly chosen and ﬂipped 10 times. Given that two of the ﬁrst three ﬂips landed on heads, what is the conditional expected number of heads in the 10 ﬂips? The joint density of X and Y is given by f (x, y) =

e−x/y e−y , y

0 < x < q,

0 < y < q

Compute E[X 2 |Y = y]. 7.51. The joint density of X and Y is given by f (x, y) =

e−y , y

0 < x < y,

0 < y < q

Compute E[X 3 |Y = y]. 7.52. A population is made up of r disjoint subgroups. Let pi denote the proportion of the population that is in subgroup i, i = 1, . . . , r. If the average weight of the members of subgroup i is wi , i = 1, . . . , r, what is the average weight of the members of the population?

Problems 7.53. A prisoner is trapped in a cell containing 3 doors. The ﬁrst door leads to a tunnel that returns him to his cell after 2 days’ travel. The second leads to a tunnel that returns him to his cell after 4 days’ travel. The third door leads to freedom after 1 day of travel. If it is assumed that the prisoner will always select doors 1, 2, and 3 with respective probabilities .5, .3, and .2, what is the expected number of days until the prisoner reaches freedom? 7.54. Consider the following dice game: A pair of dice is rolled. If the sum is 7, then the game ends and you win 0. If the sum is not 7, then you have the option of either stopping the game and receiving an amount equal to that sum or starting over again. For each value of i, i = 2, . . . , 12, ﬁnd your expected return if you employ the strategy of stopping the ﬁrst time that a value at least as large as i appears. What value of i leads to the largest expected return? Hint: Let Xi denote the return when you use the critical value i. To compute E[Xi ], condition on the initial sum. 7.55. Ten hunters are waiting for ducks to ﬂy by. When a ﬂock of ducks ﬂies overhead, the hunters ﬁre at the same time, but each chooses his target at random, independently of the others. If each hunter independently hits his target with probability .6, compute the expected number of ducks that are hit. Assume that the number of ducks in a ﬂock is a Poisson random variable with mean 6. 7.56. The number of people who enter an elevator on the ground ﬂoor is a Poisson random variable with mean 10. If there are N ﬂoors above the ground ﬂoor, and if each person is equally likely to get off at any one of the N ﬂoors, independently of where the others get off, compute the expected number of stops that the elevator will make before discharging all of its passengers. 7.57. Suppose that the expected number of accidents per week at an industrial plant is 5. Suppose also that the numbers of workers injured in each accident are independent random variables with a common mean of 2.5. If the number of workers injured in each accident is independent of the number of accidents that occur, compute the expected number of workers injured in a week. 7.58. A coin having probability p of coming up heads is continually ﬂipped until both heads and tails have appeared. Find (a) the expected number of ﬂips; (b) the probability that the last ﬂip lands on heads. 7.59. There are n + 1 participants in a game. Each person independently is a winner with probability p. The winners share a total prize of 1 unit.

377

(For instance, if 4 people win, then each of them receives 14 , whereas if there are no winners, then none of the participants receive anything.) Let A denote a speciﬁed one of the players, and let X denote the amount that is received by A. (a) Compute the expected total prize shared by the players. 1 − (1 − p)n+1 (b) Argue that E[X] = . n + 1 (c) Compute E[X] by conditioning on whether A is a winner, and conclude that E[(1 + B)−1 ] =

1 − (1 − p)n+1 (n + 1)p

when B is a binomial random variable with parameters n and p. 7.60. Each of m + 2 players pays 1 unit to a kitty in order to play the following game: A fair coin is to be ﬂipped successively n times, where n is an odd number, and the successive outcomes are noted. Before the n ﬂips, each player writes down a prediction of the outcomes. For instance, if n = 3, then a player might write down (H, H, T), which means that he or she predicts that the ﬁrst ﬂip will land on heads, the second on heads, and the third on tails. After the coins are ﬂipped, the players count their total number of correct predictions. Thus, if the actual outcomes are all heads, then the player who wrote (H, H, T) would have 2 correct predictions. The total kitty of m + 2 is then evenly split up among those players having the largest number of correct predictions. Since each of the coin ﬂips is equally likely to land on either heads or tails, m of the players have decided to make their predictions in a totally random fashion. Speciﬁcally, they will each ﬂip one of their own fair coins n times and then use the result as their prediction. However, the ﬁnal 2 of the players have formed a syndicate and will use the following strategy: One of them will make predictions in the same random fashion as the other m players, but the other one will then predict exactly the opposite of the ﬁrst. That is, when the randomizing member of the syndicate predicts an H, the other member predicts a T. For instance, if the randomizing member of the syndicate predicts (H, H, T), then the other one predicts (T, T, H). (a) Argue that exactly one of the syndicate members will have more than n/2 correct predictions. (Remember, n is odd.) (b) Let X denote the number of the m nonsyndicate players that have more than n/2 correct predictions. What is the distribution of X?

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Properties of Expectation the number of blue balls that are withdrawn. Find Cov(X, Y) (a) by deﬁning appropriate indicator (that is, Bernoulli) random variables

(c) With X as deﬁned in part (b), argue that E[payoff to the syndicate] = (m + 2) 1 *E X + 1

Xi , Yj such that X =

(d) Use part (c) of Problem 59 to conclude that E[payoff to the syndicate] =

2(m + 2) m + 1 1 m+1 * 1− 2

i=1

7.64.

and explicitly compute this number when m = 1, 2, and 3. Because it can be shown that m+1 2(m + 2) 1 1 − > 2 m + 1 2 7.65. it follows that the syndicate’s strategy always gives it a positive expected proﬁt. 7.61. Let X1 , . . . be independent random variables with the common distribution function F, and suppose they are independent of N, a geometric random variable with parameter p. Let M = max(X1 , . . . , XN ). (a) Find P{M … x} by conditioning on N. (b) Find P{M … x|N = 1}. (c) Find P{M … x|N > 1}. (d) Use (b) and (c) to rederive the probability you found in (a). 7.62. Let U1 , U2 , . . . be a sequence of independent uniform (0, 1) random variables. In Example 5i we showed that, for 0 … x … 1, E[N(x)] = ex , where ⎧ ⎫ n ⎨ ⎬ N(x) = min n : Ui > x ⎩ ⎭

7.66. 7.67.

7.68.

i=1

This problem gives another approach to establishing that result. (a) Show by induction on n that, for 0 < x … 1 and all n Ú 0, P{N(x) Ú n + 1} =

xn n!

Hint: First condition on U1 and then use the induction hypothesis. Use part (a) to conclude that

7.69.

7.70. E[N(x)] = ex 7.63. An urn contains 30 balls, of which 10 are red and 8 are blue. From this urn, 12 balls are randomly withdrawn. Let X denote the number of red and Y

10

Xi , Y =

8

Yj

j=1

(b) by conditioning (on either X or Y) to determine E[XY]. Type i light bulbs function for a random amount of time having mean μi and standard deviation σi , i = 1, 2. A light bulb randomly chosen from a bin of bulbs is a type 1 bulb with probability p and a type 2 bulb with probability 1 − p. Let X denote the lifetime of this bulb. Find (a) E[X]; (b) Var(X). The number of winter storms in a good year is a Poisson random variable with mean 3, whereas the number in a bad year is a Poisson random variable with mean 5. If next year will be a good year with probability .4 or a bad year with probability .6, ﬁnd the expected value and variance of the number of storms that will occur. In Example 5c, compute the variance of the length of time until the miner reaches safety. Consider a gambler who, at each gamble, either wins or loses her bet with respective probabilities p and 1 − p. A popular gambling system known as the Kelley strategy is to always bet the fraction 2p − 1 of your current fortune when p > 12 . Compute the expected fortune after n gambles of a gambler who starts with x units and employs the Kelley strategy. The number of accidents that a person has in a given year is a Poisson random variable with mean λ. However, suppose that the value of λ changes from person to person, being equal to 2 for 60 percent of the population and 3 for the other 40 percent. If a person is chosen at random, what is the probability that he will have (a) 0 accidents and (b) exactly 3 accidents in a certain year? What is the conditional probability that he will have 3 accidents in a given year, given that he had no accidents the preceding year? Repeat Problem 68 when the proportion of the population having a value of λ less than x is equal to 1 − e−x . Consider an urn containing a large number of coins, and suppose that each of the coins has some probability p of turning up heads when it is ﬂipped. However, this value of p varies from coin to coin. Suppose that the composition of the urn is such that if a coin is selected at random from it, then

Problems the p-value of the coin can be regarded as being the value of a random variable that is uniformly distributed over [0, 1]. If a coin is selected at random from the urn and ﬂipped twice, compute the probability that (a) the ﬁrst ﬂip results in a head; (b) both ﬂips result in heads. 7.71. In Problem 70, suppose that the coin is tossed n times. Let X denote the number of heads that occur. Show that P{X = i} =

1 n + 1

i = 0, 1, . . . , n

Hint: Make use of the fact that * 1 (a − 1)!(b − 1)! xa−1 (1 − x)b−1 dx = (a + b − 1)! 0 when a and b are positive integers. 7.72. Suppose that in Problem 70 we continue to ﬂip the coin until a head appears. Let N denote the number of ﬂips needed. Find (a) P{N Ú i}, i Ú 0; (b) P{N = i}; (c) E[N]. 7.73. In Example 6b, let S denote the signal sent and R the signal received. (a) Compute E[R]. (b) Compute Var(R). (c) Is R normally distributed? (d) Compute Cov(R, S). 7.74. In Example 6c, suppose that X is uniformly distributed over (0, 1). If the discretized regions are determined by a0 = 0, a1 = 12 , and a2 = 1, calculate the optimal quantizer Y and compute E[(X − Y)2 ]. 7.75. The moment generating function of X is given by MX (t) = exp{2et − 2} and that of Y by MY (t) = ( 34 et + 14 )10 . If X and Y are independent, what are (a) P{X + Y = 2}? (b) P{XY = 0}? (c) E[XY]? 7.76. Let X be the value of the ﬁrst die and Y the sum of the values when two dice are rolled. Compute the joint moment generating function of X and Y. 7.77. The joint density of X and Y is given by 1 2 f (x, y) = √ e−y e−(x−y) /2 2π

0 < y < q, −q < x < q

379

(a) Compute the joint moment generating function of X and Y. (b) Compute the individual moment generating functions. 7.78. Two envelopes, each containing a check, are placed in front of you. You are to choose one of the envelopes, open it, and see the amount of the check. At this point, either you can accept that amount or you can exchange it for the check in the unopened envelope. What should you do? Is it possible to devise a strategy that does better than just accepting the ﬁrst envelope? Let A and B, A < B, denote the (unknown) amounts of the checks, and note that the strategy that randomly selects an envelope and always accepts its check has an expected return of (A + B)/2. Consider the following strategy: Let F(·) be any strictly increasing (that is, continuous) distribution function. Choose an envelope randomly and open it. If the discovered check has the value x, then accept it with probability F(x) and exchange it with probability 1 − F(x). (a) Show that if you employ the latter strategy, then your expected return is greater than (A + B)/2. Hint: Condition on whether the ﬁrst envelope has the value A or B. Now consider the strategy that ﬁxes a value x and then accepts the ﬁrst check if its value is greater than x and exchanges it otherwise. (b) Show that, for any x, the expected return under the x-strategy is always at least (A + B)/2 and that it is strictly larger than (A + B)/2 if x lies between A and B. (c) Let X be a continuous random variable on the whole line, and consider the following strategy: Generate the value of X, and if X = x, then employ the x-strategy of part (b). Show that the expected return under this strategy is greater than (A + B)/2. 7.79. Successive weekly sales, in units of one thousand dollars, have a bivariate normal distribution with common mean 40, common standard deviation 6, and correlation .6. (a) Find the probability that the total of the next 2 weeks’ sales exceeds 90. (b) If the correlation were .2 rather than .6, do you think that this would increase or decrease the answer to (a)? Explain your reasoning. (c) Repeat (a) when the correlation is .2.

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THEORETICAL EXERCISES 7.1. Show that E[(X − a)2 ] is minimized at a = E[X]. 7.2. Suppose that X is a continuous random variable with density function f . Show that E[|X − a|] is minimized when a is equal to the median of F. Hint: Write * E[|X − a|] = |x − a|f (x) dx Now break up the integral into the regions where x < a and where x > a, and differentiate. 7.3. Prove Proposition 2.1 when (a) X and Y have a joint probability mass function; (b) X and Y have a joint probability density function and g(x, y) Ú 0 for all x, y. 7.4. Let X be a random variable having ﬁnite expectation μ and variance σ 2 , and let g(·) be a twice differentiable function. Show that g”(μ) 2 E[g(X)] L g(μ) + σ 2 Hint: Expand g(·) in a Taylor series about μ. Use the ﬁrst three terms and ignore the remainder. 7.5. Let A1 , A2 , . . . , An be arbitrary events, and deﬁne Ck = {at least k of the Ai occur}. Show that n

P(Ck ) =

k=1

n

P(Ak )

k=1

Hint: Let X denote the number of the Ai that occur. Show that both sides of the preceding equation are equal to E[X]. 7.6. In the text, we noted that ⎤ ⎡ q q Xi ⎦ = E[Xi ] E⎣ i=1

i=1

when the Xi are all nonnegative random variables. Since an integral is a limit of sums, one might expect that * q * q E X(t)dt = E[X(t)] dt 0

0

whenever X(t), 0 … t < q, are all nonnegative random variables; and this result is indeed true. Use it to give another proof of the result that, for a nonnegative random variable X, * q E[X) = P{X > t} dt 0

Hint: Deﬁne, for each nonnegative t, the random variable X(t) by % 1 if t < X X(t) = 0 if t Ú X -q Now relate 0 X(t)dt to X. 7.7. We say that X is stochastically larger than Y, written X Úst Y, if, for all t. P{X > t} Ú P{Y > t} Show that if X Úst Y, then E[X] Ú E[Y] when (a) X and Y are nonnegative random variables; (b) X and Y are arbitrary random variables. Hint: Write X as X = X+ − X− where

%

X X = 0 +

if X Ú 0 , if X < 0

−

X =

%

0 if X Ú 0 −X if X < 0

Similarly, represent Y as Y + − Y − . Then make use of part (a). 7.8. Show that X is stochastically larger than Y if and only if E[f (X)] Ú E[f (Y)] for all increasing functions f . Hint: Show that X Úst Y, then E[f (X)] Ú E[f (Y)] by showing that f (X) Úst f (Y) and then using Theoretical Exercise 7.7. To show that if E[f (X)] Ú E[f (Y)] for all increasing functions f , then P{X > t} Ú P{Y > t}, deﬁne an appropriate increasing function f . 7.9. A coin having probability p of landing on heads is ﬂipped n times. Compute the expected number of runs of heads of size 1, of size 2, and of size k, 1 … k … n. 7.10. Let X1 , X2 , . . . , Xn be independent and identically distributed positive random variables. For k … n, ﬁnd ⎡ ⎤ k ⎢ Xi ⎥ ⎢ ⎥ ⎢ i=1 ⎥ ⎥ E⎢ ⎢ ⎥ n ⎢ ⎥ ⎣ Xi ⎦ i=1

7.11. Consider n independent trials, each resulting in any one of r possible outcomes with probabilities P1 , P2 , . . . , Pr . Let X denote the number of outcomes that never occur in any of the trials. Find

Theoretical Exercises E[X] and show that, among all probability vectors P1 , . . . , Pr , E[X] is minimized when Pi = 1/r, i = 1, . . . , r. 7.12. Let X1 , X2 ,... be a sequence of independent random variables having the probability mass function P{Xn = 0} = P{Xn = 2} = 1/2 , n Ú 1 q n The random variable X = n=1 Xn /3 is said to have the Cantor distribution. Find E[X] and Var(X). 7.13. Let X1 , . . . , Xn be independent and identically distributed continuous random variables. We say that a record value occurs at time j, j … n, if Xj Ú Xi for all 1 … i … j. Show that n (a) E[number of record values] = 1/j; j=1 n

(j − 1)/j2 .

(b) Var(number of record values) =

j=1

7.14. For Example 2i, show that the variance of the number of coupons needed to amass a full set is equal to N−1 iN i=1

(N − i)2

When N is large, this can be shown to be approximately equal (in the sense that their ratio approaches 1 as N →q) to N 2 π 2 /6. 7.15. Consider n independent trials, the ith of which results in a success with probability Pi . (a) Compute the expected number of successes in the n trials—call it μ. (b) For a ﬁxed value of μ, what choice of P1 , . . . , Pn maximizes the variance of the number of successes? (c) What choice minimizes the variance? ∗ 7.16. Suppose that each of the elements of S = {1, 2, . . . , n} is to be colored either red or blue. Show that if A1 , . . . , Ar are subsets of S, there is a way of doing the coloring so that at most r (1/2)|Ai |−1 of these subsets have all their elei=1

ments the same color (where |A| denotes the number of elements in the set A). 7.17. Suppose that X1 and X2 are independent random variables having a common mean μ. Suppose also that Var(X1 ) = σ12 and Var(X2 ) = σ22 . The value of μ is unknown, and it is proposed that μ be estimated by a weighted average of X1 and X2 . That is, λX1 + (1 − λ)X2 will be used as an estimate of μ for some appropriate value of λ. Which value of λ yields the estimate having the lowest possible

381

variance? Explain why it is desirable to use this value of λ. 7.18. In Example 4f, we showed that the covariance of the multinomial random variables Ni and Nj is equal to −mPi Pj by expressing Ni and Nj as the sum of indicator variables. We could also have obtained that result by using the formula Var(Ni + Nj ) = Var(Ni ) + Var(Nj ) + 2 Cov(Ni , Nj ) (a) What is the distribution of Ni + Nj ? (b) Use the preceding identity to show that Cov(Ni , Nj ) = −mPi Pj . 7.19. Show that X and Y are identically distributed and not necessarily independent, then Cov(X + Y, X − Y) = 0 7.20. The Conditional Covariance Formula. The conditional covariance of X and Y, given Z, is deﬁned by Cov(X, Y|Z) K E[(X − E[X|Z])(Y − E[Y|Z])|Z] (a) Show that Cov(X, Y|Z) = E[XY|Z] − E[X|Z]E[Y|Z] (b) Prove the conditional covariance formula Cov(X, Y) = E[Cov(X, Y|Z)] + Cov(E[X|Z], E[Y|Z]) (c) Set X = Y in part (b) and obtain the conditional variance formula. 7.21. Let X(i) , i = 1, . . . , n, denote the order statistics from a set of n uniform (0, 1) random variables, and note that the density function of X(i) is given by f (x) =

n! xi−1 (1 − x)n−i (i − 1)!(n − i)!

0

(a) Compute Var(X(i) ), i = 1, . . . , n. (b) Which value of i minimizes, and which value maximizes, Var(X(i) )? 7.22. Show that Y = a + bX, then % +1 if b > 0 ρ(X, Y) = −1 if b < 0 7.23. Show that Z is a standard normal random variable and if Y is deﬁned by Y = a + bZ + cZ2 , then ρ(Y, Z) = .

b b2

+ 2c2

382

Chapter 7

Properties of Expectation

7.24. Prove the Cauchy–Schwarz inequality, namely, (E[XY])2 … E[X 2 ]E[Y 2 ] Hint: Unless Y = −tX for some constant, in which case the inequality holds with equality, if follows that, for all t, 0 < E[(tX + Y)2 ] = E[X 2 ]t2 + 2E[XY]t + E[Y 2 ] Hence, the roots of the quadratic equation E[X 2 ]t2 + 2E[XY]t + E[Y 2 ] = 0 must be imaginary, which implies that the discriminant of this quadratic equation must be negative. 7.25. Show that if X and Y are independent, then E[X|Y = y] = E[X]

7.26. 7.27.

7.28. 7.29.

for all y

(a) in the discrete case; (b) in the continuous case. Prove that E[g(X)Y|X] = g(X)E[Y|X]. Prove that if E[Y|X = x] = E[Y] for all x, then X and Y are uncorrelated; give a counterexample to show that the converse is not true. Hint: Prove and use the fact that E[XY] = E[XE[Y|X]]. Show that Cov(X, E[Y|X]) = Cov(X, Y). Let X1 , . . . , Xn be independent and identically distributed random variables. Find E[X1 |X1 + · · · + Xn = x]

7.30. Consider Example 4f, which is concerned with the multinomial distribution. Use conditional expectation to compute E[Ni Nj ], and then use this to verify the formula for Cov(Ni , Nj ) given in Example 4f. 7.31. An urn initially contains b black and w white balls. At each stage, we add r black balls and then withdraw, at random, r balls from the b + w + r balls in the urn. Show that E[number of white balls after stage t] t b + w = w b + w + r 7.32. For an event A, let IA equal 1 if A occurs and let it equal 0 if A does not occur. For a random variable X, show that E[X|A] =

E[XIA ] P(A)

7.33. A coin that lands on heads with probability p is continually ﬂipped. Compute the expected number of ﬂips that are made until a string of r heads in a row is obtained.

Hint: Condition on the time of the ﬁrst occurrence of tails to obtain the equation E[X] = (1 − p)

r

pi−1 (i + E[X])

i=1

+(1 − p)

q

pi−1 r

i=r+1

Simplify and solve for E[X]. 7.34. For another approach to Theoretical Exercise 33, let Tr denote the number of ﬂips required to obtain a run of r consecutive heads. (a) Determine E[Tr |Tr−1 ]. (b) Determine E[Tr ] in terms of E[Tr−1 ]. (c) What is E[T1 ]? (d) What is E[Tr ]? 7.35. The probability generating function of the discrete nonnegative integer valued random variable X having probability mass function pj , j Ú 0, is deﬁned by φ(s) = E[sX ] =

q

pj sj

j=0

Let Y be a geometric random variable with parameter p = 1 − s, where 0 < s < 1. Suppose that Y is independent of X, and show that φ(s) = P{X < Y} 7.36. One ball at a time is randomly selected from an urn containing a white and b black balls until all of the remaining balls are of the same color. Let Ma,b denote the expected number of balls left in the urn when the experiment ends. Compute a recursive formula for Ma,b and solve when a = 3 and b = 5. 7.37. An urn contains a white and b black balls. After a ball is drawn, it is returned to the urn if it is white; but if it is black, it is replaced by a white ball from another urn. Let Mn denote the expected number of white balls in the urn after the foregoing operation has been repeated n times. (a) Derive the recursive equation 1 Mn+1 = 1 − Mn + 1 a + b (b) Use part (a) to prove that Mn = a + b − b 1 −

1 a + b

n

(c) What is the probability that the (n + 1)st ball drawn is white?

Theoretical Exercises 7.38. The best linear predictor of Y with respect to X1 and X2 is equal to a + bX1 + cX2 , where a, b, and c are chosen to minimize E[(Y − (a + bX1 + cX2 ))2 ] Determine a, b, and c. 7.39. The best quadratic predictor of Y with respect to X is a + bX + cX 2 , where a, b, and c are chosen to minimize E[(Y − (a + bX + cX 2 ))2 ]. Determine a, b, and c. 7.40. Use the conditional variance formula to determine the variance of a geometric random variable X having parameter p. 7.41. Let X be a normal random variable with parameters μ = 0 and σ 2 = 1, and let I, independent of X, be such that P{I = 1} = 12 = P{I = 0}. Now deﬁne Y by % X if I = 1 Y= −X if I = 0 In words, Y is equally likely to equal either X or −X. (a) Are X and Y independent? (b) Are I and Y independent? (c) Show that Y is normal with mean 0 and variance 1. (d) Show that Cov(X, Y) = 0. 7.42. It follows from Proposition 6.1 and the fact that the best linear predictor of Y with respect to X is σ μy + ρ σyx (X − μx ) that if E[Y|X] = a + bX then a = μy − ρ

σy μx σx

b=ρ

σy σx

(Why?) Verify this directly. 7.43. Show that, for random variables X and Z,

383

and their number is denoted by X1 . In general, let Xn denote the size of the nth generation. Let q q jPj and σ 2 = (j − μ)2 Pj denote, respecμ= j=0

j=0

tively, the mean and the variance of the number of offspring produced by a single individual. Suppose that X0 = 1—that is, initially there is a single individual in the population. (a) Show that E[Xn ] = μE[Xn−1 ] (b) Use part (a) to conclude that E[Xn ] = μn (c) Show that Var(Xn ) = σ 2 μn−1 + μ2 Var(Xn−1 ) (d) Use part (c) to conclude that ⎧ n ⎪ ⎪ ⎨σ 2 μn−1 μ − 1 μ − 1 Var(Xn ) = ⎪ ⎪ ⎩nσ 2

if μ Z 1 if μ = 1

The model just described is known as a branching process, and an important question for a population that evolves along such lines is the probability that the population will eventually die out. Let π denote this probability when the population starts with a single individual. That is, π = P{population eventually dies out|X0 = 1) (e) Argue that π satisﬁes π=

q

Pj π j

j=0

E[(X − Y)2 ] = E[X 2 ] − E[Y 2 ] where Y = E[X|Z] 7.44. Consider a population consisting of individuals able to produce offspring of the same kind. Suppose that, by the end of its lifetime, each individual will have produced j new offspring with probability Pj , j Ú 0, independently of the number produced by any other individual. The number of individuals initially present, denoted by X0 , is called the size of the zeroth generation. All offspring of the zeroth generation constitute the ﬁrst generation,

Hint: Condition on the number of offspring of the initial member of the population. 7.45. Verify the formula for the moment generating function of a uniform random variable that is given in Table 7.7. Also, differentiate to verify the formulas for the mean and variance. 7.46. For a standard normal random variable Z, let μn = E[Zn ]. Show that ⎧ ⎪ when n is odd ⎨0 μn = (2j)! ⎪ ⎩ 2j j! when n = 2j

384

Chapter 7

Properties of Expectation

Hint: Start by expanding the moment generating function of Z into a Taylor series about 0 to obtain E[etZ ] = et /2 q (t2 /2)j = j! 2

j=0

7.47. Let X be a normal random variable with mean μ and variance σ 2 . Use the results of Theoretical Exercise 46 to show that n n−2j σ 2j (2j)! [n/2] 2j μ n E[X ] = 2j j! j=0

In the preceding equation, [n/2] is the largest integer less than or equal to n/2. Check your answer by letting n = 1 and n = 2. 7.48. If Y = aX + b, where a and b are constants, express the moment generating function of Y in terms of the moment generating function of X. 7.49. The positive random variable X is said to be a lognormal random variable with parameters μ and σ 2 if log(X) is a normal random variable with mean μ and variance σ 2 . Use the normal moment generating function to ﬁnd the mean and variance of a lognormal random variable. 7.50. Let X have moment generating function M(t), and deﬁne (t) = log M(t). Show that (t)|t=0 = Var(X)

7.51. Use Table 7.2 to determine the distribution of n Xi when X1 , . . . , Xn are independent and i=1

identically distributed exponential random variables, each having mean 1/λ. 7.52. Show how to compute Cov(X, Y) from the joint moment generating function of X and Y. 7.53. Suppose that X1 , . . . , Xn have a multivariate normal distribution. Show that X1 , . . . , Xn are independent random variables if and only if Cov(Xi , Xj ) = 0

when i Z j

7.54. If Z is a standard normal random variable, what is Cov(Z, Z2 )? 7.55. Suppose that Y is a normal random variable with mean μ and variance σ 2 , and suppose also that the conditional distribution of X, given that Y = y, is normal with mean y and variance 1. (a) Argue that the joint distribution of X, Y is the same as that of Y + Z, Y when Z is a standard normal random variable that is independent of Y. (b) Use the result of part (a) to argue that X, Y has a bivariate normal distribution. (c) Find E[X], Var(X), and Corr(X, Y). (d) Find E[Y|X = x]. (e) What is the conditional distribution of Y given that X = x?

SELF-TEST PROBLEMS AND EXERCISES 7.1. Consider a list of m names, where the same name may appear more than once on the list. Let n(i), i = 1, . . . , m, denote the number of times that the name in position i appears on the list, and let d denote the number of distinct names on the list. (a) Express d in terms of the variables m, n(i), i = 1, . . . , m. Let U be a uniform (0, 1) random variable, and let X = [mU] + 1. (b) What is the probability mass function of X? (c) Argue that E[m/n(X)] = d. 7.2. An urn has n white and m black balls that are removed one at a time in a randomly chosen order. Find the expected number of instances in which a white ball is immediately followed by a black one. 7.3. Twenty individuals consisting of 10 married couples are to be seated at 5 different tables, with 4 people at each table. (a) If the seating is done “at random,” what is the expected number of married couples that are seated at the same table? (b) If 2 men and 2 women are randomly chosen to be seated at each table, what is the expected

number of married couples that are seated at the same table? 7.4. If a die is to be rolled until all sides have appeared at least once, ﬁnd the expected number of times that outcome 1 appears. 7.5. A deck of 2n cards consists of n red and n black cards. The cards are shufﬂed and then turned over one at a time. Suppose that each time a red card is turned over, we win 1 unit if more red cards than black cards have been turned over by that time. (For instance, if n = 2 and the result is r b r b, then we would win a total of 2 units.) Find the expected amount that we win. 7.6. Let A1 , A2 , . . . , An be events, and let N denote the number of them that occur. Also, let I = 1 if all of these events occur, and let it be 0 otherwise. Prove Bonferroni’s inequality, namely, P(A1 · · · An ) Ú

n

P(Ai ) − (n − 1)

i=1

Hint: Argue ﬁrst that N … n − 1 + I.

Self-Test Problems and Exercises 7.7. Let X be the smallest value obtained when k numbers are randomly chosen from the set 1, . . . , n. Find E[X] by interpreting X as a negative hypergeometric random variable. 7.8. An arriving plane carries r families. A total of nj of these families have checked in a total of j pieces of luggage, nj = r. Suppose that when j the plane lands, the N = jnj pieces of luggage

7.15.

j

come out of the plane in a random order. As soon as a family collects all of its luggage, it immediately departs the airport. If the Sanchez family checked in j pieces of luggage, ﬁnd the expected number of families that depart after they do. ∗ 7.9. Nineteen items on the rim of a circle of radius 1 are to be chosen. Show that, for any choice of these points, there will be an arc of (arc) length 1 that contains at least 4 of them. 7.10. Let X be a Poisson random variable with mean λ. Show that if λ is not too small, then √ Var( X) L .25

7.11.

7.12.

7.13.

7.14.

Hint: Use the result of Theoretical Exercise 4 to √ approximate E[ X]. Suppose in Self-Test Problem 3 that the 20 people are to be seated at seven tables, three of which have 4 seats and four of which have 2 seats. If the people are randomly seated, ﬁnd the expected value of the number of married couples that are seated at the same table. Individuals 1 through n, n > 1, are to be recruited into a ﬁrm in the following manner: Individual 1 starts the ﬁrm and recruits individual 2. Individuals 1 and 2 will then compete to recruit individual 3. Once individual 3 is recruited, individuals 1, 2, and 3 will compete to recruit individual 4, and so on. Suppose that when individuals 1, 2, . . . , i compete to recruit individual i + 1, each of them is equally likely to be the successful recruiter. (a) Find the expected number of the individuals 1, . . . , n who did not recruit anyone else. (b) Derive an expression for the variance of the number of individuals who did not recruit anyone else, and evaluate it for n = 5. The nine players on a basketball team consist of 2 centers, 3 forwards, and 4 backcourt players. If the players are paired up at random into three groups of size 3 each, ﬁnd (a) the expected value and (b) the variance of the number of triplets consisting of one of each type of player. A deck of 52 cards is shufﬂed and a bridge hand of 13 cards is dealt out. Let X and Y denote, respectively, the number of aces and the number of spades in the hand.

7.16.

7.17.

7.18.

7.19.

385

(a) Show that X and Y are uncorrelated. (b) Are they independent? Each coin in a bin has a value attached to it. Each time that a coin with value p is ﬂipped, it lands on heads with probability p. When a coin is randomly chosen from the bin, its value is uniformly distributed on (0, 1). Suppose that after the coin is chosen, but before it is ﬂipped, you must predict whether it will land on heads or on tails. You will win 1 if you are correct and will lose 1 otherwise. (a) What is your expected gain if you are not told the value of the coin? (b) Suppose now that you are allowed to inspect the coin before it is ﬂipped, with the result of your inspection being that you learn the value of the coin. As a function of p, the value of the coin, what prediction should you make? (c) Under the conditions of part (b), what is your expected gain? In Self-Test Problem 1, we showed how to use the value of a uniform (0, 1) random variable (commonly called a random number) to obtain the value of a random variable whose mean is equal to the expected number of distinct names on a list. However, its use required that one choose a random position and then determine the number of times that the name in that position appears on the list. Another approach, which can be more efﬁcient when there is a large amount of replication of names, is as follows: As before, start by choosing the random variable X as in Problem 1. Now identify the name in position X, and then go through the list, starting at the beginning, until that name appears. Let I equal 0 if you encounter that name before getting to position X, and let I equal 1 if your ﬁrst encounter with the name is at position X. Show that E[mI] = d. Hint: Compute E[I] by using conditional expectation. A total of m items are to be sequentially distributed among n cells, with each item independently being put in cell j with probability pj , j = 1, . . . , n. Find the expected number of collisions that occur, where a collision occurs whenever an item is put into a nonempty cell. Let X be the length of the initial run in a random ordering of n ones and m zeroes. That is, if the ﬁrst k values are the same (either all ones or all zeroes), then X Ú k. Find E[X]. There are n items in a box labeled H and m in a box labeled T. A coin that comes up heads with probability p and tails with probability 1 − p is ﬂipped. Each time it comes up heads, an item is removed from the H box, and each time it comes up tails, an item is removed from the T box. (If a box is empty and its outcome occurs, then no items

386

Chapter 7

Properties of Expectation

are removed.) Find the expected number of coin ﬂips needed for both boxes to become empty. Hint: Condition on the number of heads in the ﬁrst n + m ﬂips. 7.20. Let X be a nonnegative random variable having distribution function F. Show that if F(x) = 1 − F(x), then * q E[X n ] = xn−1 F(x) dx 0

Hint: Start with the identity * x xn−1 dx Xn = n 0 * q =n xn−1 IX (x) dx 0

where

% Ix (x) =

1, 0,

if x < X otherwise

∗ 7.21.

Let n a1 , . . . , an , not all equal to 0, be such that i=1 ai = 0. Show that there is a permutation i1 , . . . , in such that nj=1 aij aij+1 < 0. Hint: Use the probabilistic method. (It is interesting that there need not be a permutation whose sum of products of successive pairs is positive. For instance, if n = 3, a1 = a2 = −1, and a3 = 2, there is no such permutation.) 7.22. Suppose that Xi , i = 1, 2, 3, are independent Poisson random variables with respective means λi , i = 1, 2, 3. Let X = X1 + X2 and Y = X2 + X3 . The random vector X, Y is said to have a bivariate Poisson distribution. (a) Find E[X] and E[Y]. (b) Find Cov(X, Y). (c) Find the joint probability mass function P{X = i, Y = j}. 7.23. Let (Xi , Yi ), i = 1, . . . , be a sequence of independent and identically distributed random vectors. That is, X1 , Y1 is independent of, and has the same distribution as X2 , Y2 , and so on. Although Xi and Yi can be dependent, Xi and Yj are independent when i Z j. Let μx = E[Xi ], σy2 = Var(Yi ),

μy = E[Yi ],

σx2 = Var(Xi ),

ρ = Corr(Xi , Yi )

Find Corr( ni=1 Xi , nj=1 Yj ). 7.24. Three cards are randomly chosen without replacement from an ordinary deck of 52 cards. Let X denote the number of aces chosen. (a) Find E[X|the ace of spades is chosen]. (b) Find E[X|at least one ace is chosen].

7.25. Let be the standard normal distribution function, and let X be a normal random variable with mean μ and variance 1. We want to ﬁnd E[ (X)]. To do so, let Z be a standard normal random variable that is independent of X, and let % 1, if Z < X I= 0, if Z Ú X (a) Show that E[I|X = x] = (x). (b) Show that E[ (X)] = P{Z < X}. (c) Show that E[ (X)] = ( √μ ). 2 Hint: What is the distribution of X − Z? The preceding comes up in statistics. Suppose you are about to observe the value of a random variable X that is normally distributed with an unknown mean μ and variance 1, and suppose that you want to test the hypothesis that the mean μ is greater than or equal to 0. Clearly you would want to reject this hypothesis if X is sufﬁciently small. If it results that X = x, then the p-value of the hypothesis that the mean is greater than or equal to 0 is deﬁned to be the probability that X would be as small as x if μ were equal to 0 (its smallest possible value if the hypothesis were true). (A small p-value is taken as an indication that the hypothesis is probably false.) Because X has a standard normal distribution when μ = 0, the p-value that results when X = x is (x). Therefore, the preceding shows that the expected p-value that results when the true mean is μ is ( √μ ). 2

7.26. A coin that comes up heads with probability p is ﬂipped until either a total of n heads or of m tails is amassed. Find the expected number of ﬂips. Hint: Imagine that one continues to ﬂip even after the goal is attained. Let X denote the number of ﬂips needed to obtain n heads, and let Y denote the number of ﬂips needed to obtain m tails. Note that max(X, Y) + min(X, Y) = X + Y. Compute E[max(X, Y)] by conditioning on the number of heads in the ﬁrst n + m − 1 ﬂips. 7.27. A deck of n cards numbered 1 through n, initially in any arbitrary order, is shufﬂed in the following manner: At each stage, we randomly choose one of the cards and move it to the front of the deck, leaving the relative positions of the other cards unchanged. This procedure is continued until all but one of the cards has been chosen. At this point it follows by symmetry that all n! possible orderings are equally likely. Find the expected number of stages that are required. 7.28. Suppose that a sequence of independent trials in which each trial is a success with probability p is

Self-Test Problems and Exercises performed until either a success occurs or a total of n trials has been reached. Find the mean number of trials that are performed. Hint: The computations are simpliﬁed if you use the identity that, for a nonnegative integer valued random variable X, E[X] =

q i=1

P{X Ú i}

387

7.29. Suppose that X and Y are both Bernoulli random variables. Show that X and Y are independent if and only if Cov(X, Y) = 0. 7.30. In the generalized match problem, there are n individuals of whom ni wear hat size i, ri=1 ni = n. There are also n hats, of which hi are of size i, ri=1 hi = n. If each individual randomly chooses a hat (without replacement), ﬁnd the expected number who choose a hat that is their size.

C H A P T E R

8

Limit Theorems 8.1 8.2 8.3 8.4 8.5 8.6

INTRODUCTION CHEBYSHEV’S INEQUALITY AND THE WEAK LAW OF LARGE NUMBERS THE CENTRAL LIMIT THEOREM THE STRONG LAW OF LARGE NUMBERS OTHER INEQUALITIES BOUNDING THE ERROR PROBABILITY WHEN APPROXIMATING A SUM OF INDEPENDENT BERNOULLI RANDOM VARIABLES BY A POISSON RANDOM VARIABLE

8.1

INTRODUCTION The most important theoretical results in probability theory are limit theorems. Of these, the most important are those classiﬁed either under the heading laws of large numbers or under the heading central limit theorems. Usually, theorems are considered to be laws of large numbers if they are concerned with stating conditions under which the average of a sequence of random variables converges (in some sense) to the expected average. By contrast, central limit theorems are concerned with determining conditions under which the sum of a large number of random variables has a probability distribution that is approximately normal.

8.2

CHEBYSHEV’S INEQUALITY AND THE WEAK LAW OF LARGE NUMBERS We start this section by proving a result known as Markov’s inequality. Proposition 2.1. Markov’s inequality If X is a random variable that takes only nonnegative values, then, for any value a > 0, E[X] P{X Ú a} … a Proof. For a > 0, let % 1 if X Ú a I= 0 otherwise and note that, since X Ú 0, I …

X a

Taking expectations of the preceding inequality yields E[I] …

E[X] a

which, because E[I] = P{X Ú a}, proves the result. As a corollary, we obtain Proposition 2.2. 388

Section 8.2

Chebyshev’s Inequality and the Weak Law of Large Numbers

389

Proposition 2.2. Chebyshev’s inequality If X is a random variable with ﬁnite mean μ and variance σ 2 , then, for any value k > 0, σ2 P{|X − μ| Ú k} … 2 k Proof. Since (X − μ)2 is a nonnegative random variable, we can apply Markov’s inequality (with a = k2 ) to obtain P{(X − μ)2 Ú k2 } …

E[(X − μ)2 ] k2

(2.1)

But since (X − μ)2 Ú k2 if and only if |X − μ| Ú k, Equation (2.1) is equivalent to P{|X − μ| Ú k} …

E[(X − μ)2 ] σ2 = k2 k2

and the proof is complete. The importance of Markov’s and Chebyshev’s inequalities is that they enable us to derive bounds on probabilities when only the mean, or both the mean and the variance, of the probability distribution are known. Of course, if the actual distribution were known, then the desired probabilities could be computed exactly and we would not need to resort to bounds. EXAMPLE 2a Suppose that it is known that the number of items produced in a factory during a week is a random variable with mean 50. (a) What can be said about the probability that this week’s production will exceed 75? (b) If the variance of a week’s production is known to equal 25, then what can be said about the probability that this week’s production will be between 40 and 60? Solution. Let X be the number of items that will be produced in a week. (a) By Markov’s inequality, P{X > 75} …

50 2 E[X] = = 75 75 3

(b) By Chebyshev’s inequality, P{|X − 50| Ú 10} …

σ2 1 = 4 102

Hence, P{|X − 50| < 10} Ú 1 −

3 1 = 4 4

so the probability that this week’s production will be between 40 and 60 is at least .75. .

390

Chapter 8

Limit Theorems

As Chebyshev’s inequality is valid for all distributions of the random variable X, we cannot expect the bound on the probability to be very close to the actual probability in most cases. For instance, consider Example 2b. EXAMPLE 2b If X is uniformly distributed over the interval (0, 10), then, since E[X] = 5 and Var(X) = 25 3 , it follows from Chebyshev’s inequality that P{|X − 5| > 4} …

25 L .52 3(16)

whereas the exact result is P{|X − 5| > 4} = .20 Thus, although Chebyshev’s inequality is correct, the upper bound that it provides is not particularly close to the actual probability. Similarly, if X is a normal random variable with mean μ and variance σ 2 , Chebyshev’s inequality states that P{|X − μ| > 2σ } …

1 4

whereas the actual probability is given by 06 7 6 6X − μ6 6 > 2 = 2[1 − (2)] L .0456 P{|X − μ| > 2σ } = P 66 6 σ

.

Chebyshev’s inequality is often used as a theoretical tool in proving results. This use is illustrated ﬁrst by Proposition 2.3 and then, most importantly, by the weak law of large numbers. Proposition 2.3. If Var(X) = 0, then P{X = E[X]} = 1 In other words, the only random variables having variances equal to 0 are those which are constant with probability 1. Proof. By Chebyshev’s inequality, we have, for any n Ú 1, / % 1 =0 P |X − μ| > n Letting n→q and using the continuity property of probability yields 0 / /7 % % 1 1 0 = lim P |X − μ| > = P lim |X − μ| > n→q n→q n n = P{X Z μ} and the result is established.

Section 8.3

The Central Limit Theorem

391

Theorem 2.1 The weak law of large numbers Let X1 , X2 , . . . be a sequence of independent and identically distributed random variables, each having ﬁnite mean E[Xi ] = μ. Then, for any ε > 0, 7 06 6 6 X1 + · · · + Xn 6 P 66 − μ66 Ú ε →0 as n→q n Proof. We shall prove the theorem only under the additional assumption that the random variables have a ﬁnite variance σ 2 . Now, since X1 + · · · + Xn σ2 X1 + · · · + Xn = μ and Var = E n n n it follows from Chebyshev’s inequality that 06 7 6 6 X1 + · · · + Xn 6 σ2 P 66 − μ66 Ú ε … n nε2 and the result is proven. The weak law of large numbers was originally proven by James Bernoulli for the special case where the Xi are 0, 1 (that is, Bernoulli) random variables. His statement and proof of this theorem were presented in his book Ars Conjectandi, which was published in 1713, eight years after his death, by his nephew Nicholas Bernoulli. Note that, because Chebyshev’s inequality was not known in Bernoulli’s time, Bernoulli had to resort to a quite ingenious proof to establish the result. The general form of the weak law of large numbers presented in Theorem 2.1 was proved by the Russian mathematician Khintchine. 8.3

THE CENTRAL LIMIT THEOREM The central limit theorem is one of the most remarkable results in probability theory. Loosely put, it states that the sum of a large number of independent random variables has a distribution that is approximately normal. Hence, it not only provides a simple method for computing approximate probabilities for sums of independent random variables, but also helps explain the remarkable fact that the empirical frequencies of so many natural populations exhibit bell-shaped (that is, normal) curves. In its simplest form the central limit theorem is as follows. Theorem 3.1 The central limit theorem Let X1 , X2 , . . . be a sequence of independent and identically distributed random variables, each having mean μ and variance σ 2 . Then the distribution of X1 + · · · + Xn − nμ √ σ n tends to the standard normal as n→q. That is, for −q < a < q, 7 0 * a 1 X1 + · · · + Xn − nμ 2 e−x /2 dx as n→q … a →√ P √ σ n 2π −q The key to the proof of the central limit theorem is the following lemma, which we state without proof.

392

Chapter 8

Limit Theorems

Lemma 3.1 Let Z1 , Z2 , . . . be a sequence of random variables having distribution functions FZn and moment generating functions MZn , n Ú 1; and let Z be a random variable having distribution function FZ and moment generating function MZ . If MZn (t) → MZ (t) for all t, then FZn (t) → FZ (t) for all t at which FZ (t) is continuous. If we let Z be a standard normal random variable, then, since MZ (t) = et /2 , it 2 follows from Lemma 3.1 that if MZn (t) → et /2 as n → q, then FZn (t) → (t) as n → q. We are now ready to prove the central limit theorem. 2

Proof of the Central Limit Theorem: Let us assume at ﬁrst that μ = 0 and σ 2 = 1. We shall prove the theorem under the assumption that the moment generating function√of the Xi , M(t), exists and is ﬁnite. Now, the moment generating function of Xi / n is given by ⎡ 0 7⎤ t tX i E ⎣exp √ ⎦ = M √ n n Thus, the moment generating function of

n i=1

n √ Xi / n is given by M √tn . Let

L(t) = log M(t) and note that L(0) = 0 M (0) L (0) = M(0) =μ =0 L (0) =

M(0)M (0) − [M (0)]2 [M(0)]2

= E[X 2 ] =1

√ 2 Now, to prove the theorem, we must show that [M(t/ n)]n → et /2 as n → q, or, √ 2 equivalently, that nL(t/ n) → t /2 as n → q. To show this, note that √ √ L(t/ n) −L (t/ n)n−3/2 t ˆ = lim by L’Hopital’s rule lim n→q n→q −2n−2 n−1 √ L (t/ n)t = lim n→q 2n−1/2 √ −L (t/ n)n−3/2 t2 ˆ again by L’Hopital’s rule = lim n→q −2n−3/2 ⎡ ⎤ t t2 ⎦ = lim ⎣L √ n→q n 2 =

t2 2

Section 8.3

The Central Limit Theorem

393

Thus, the central limit theorem is proven when μ = 0 and σ 2 = 1. The result now follows in the general case by considering the standardized random variables Xi∗ = (Xi − μ)/σ and applying the preceding result, since E[Xi∗ ] = 0, Var(Xi∗ ) = 1. Remark.

Although Theorem 3.1 states only that, for each a, 0 7 X1 + · · · + Xn − nμ P … a → (a) √ σ n

it can, in fact, be shown that the convergence is uniform in a. [We say that fn (a) → f (a) uniformly in a if, for each ε > 0, there exists an N such that |fn (a) − f (a)| < ε for all a whenever n Ú N.] . The ﬁrst version of the central limit theorem was proven by DeMoivre around 1733 for the special case where the Xi are Bernoulli random variables with p = 12 . The theorem was subsequently extended by Laplace to the case of arbitrary p. (Since a binomial random variable may be regarded as the sum of n independent and identically distributed Bernoulli random variables, this justiﬁes the normal approximation to the binomial that was presented in Section 5.4.1.) Laplace also discovered the more general form of the central limit theorem given in Theorem 3.1. His proof, however, was not completely rigorous and, in fact, cannot easily be made rigorous. A truly rigorous proof of the central limit theorem was ﬁrst presented by the Russian mathematician Liapounoff in the period 1901–1902. This important theorem is illustrated by the central limit theorem module on the text website. This website yields plots of the density function of the sum of n independent and identically distributed random variables that each take on one of the values 0, 1, 2, 3, 4. When using it, one enters the probability mass function and the desired value of n. Figure 8.1 shows the resulting plots for a speciﬁed probability mass function when (a) n = 5, (b) n = 10, (c) n = 25, and (d) n = 100. EXAMPLE 3a An astronomer is interested in measuring the distance, in light-years, from his observatory to a distant star. Although the astronomer has a measuring technique, he knows that, because of changing atmospheric conditions and normal error, each time a measurement is made it will not yield the exact distance, but merely an estimate. As a result, the astronomer plans to make a series of measurements and then use the average value of these measurements as his estimated value of the actual distance. If the astronomer believes that the values of the measurements are independent and identically distributed random variables having a common mean d (the actual distance) and a common variance of 4 (light-years), how many measurements need he make to be reasonably sure that his estimated distance is accurate to within ;.5 lightyear? Solution. Suppose that the astronomer decides to make n observations. If X1 , X2 , . . . , Xn are the n measurements, then, from the central limit theorem, it follows that n

Zn =

i=1

Xi − nd √ 2 n

394

Chapter 8

Limit Theorems

Central Limit Theorem Enter the probabilities and the number of random variables to be summed. The output gives the mass function of the sum along with its mean and variance.

P0

.25

P1

.15

P2

.1

P3

.2

P4

.3

n =

5

Start

Quit

Mean = 10.75 Variance = 12.6375 0.15 0.10 p(i) 0.05 0.00

0

5

10

15

20

i

FIGURE 8.1(a)

has approximately a standard normal distribution. Hence, ⎫ ⎧ n ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ Xi 0 √ √ 7 n n i=1 ⎪ ⎪ − d … .5⎪ = P −.5 … Zn … .5 P⎪ ⎭ ⎩ − .5 … n 2 2 √ √ √ n n n − φ − = 2 − 1 L 4 4 4 Therefore, if the astronomer wants, for instance, to be 95 percent certain that his estimated value is accurate to within .5 light year, he should make n∗ measurements, where n∗ is such that √ √ n∗ n∗ − 1 = .95 or

= .975 2 4 4 Thus, from Table 5.1 of Chapter 5, √ n∗ = 1.96 or n∗ = (7.84)2 L 61.47 4 As n∗ is not integral valued, he should make 62 observations.

Section 8.3

The Central Limit Theorem

395

Central Limit Theorem Enter the probabilities and the number of random variables to be summed. The output gives the mass function of the sum along with its mean and variance.

P0

.25

P1

.15

P2

.1

P3

.2

P4

.3

n =

10

Start

Quit

Mean = 21.5 Variance = 25.275 0.08 0.06 p(i)

0.04 0.02 0.00

0

10

20

30

40

i

FIGURE 8.1(b)

Note, however, that the preceding analysis has been done under the assumption that the normal approximation will be a good approximation when n = 62. Although this will usually be the case, in general the question of how large n need be before the approximation is “good” depends on the distribution of the Xi . If the astronomer is concerned about this point and wants to take no chances, he can still solve his problem by using Chebyshev’s inequality. Since ⎤ ⎞ ⎡ ⎛ n n X X 4 i⎦ i⎠ E⎣ =d Var ⎝ = n n n i=1

i=1

Chebyshev’s inequality yields ⎧6 ⎫ 6 ⎪ ⎪ 6 n ⎨66 ⎬ 6 Xi 16 4 6 6 P 6 − d6 > .5 … = 2 ⎪ ⎪ n n(.5) 6 ⎩6 i=1 n ⎭ Hence, if he makes n = 16/.05 = 320 observations, he can be 95 percent certain that his estimate will be accurate to within .5 light-year. .

396

Chapter 8

Limit Theorems

Central Limit Theorem Enter the probabilities and the number of random variables to be summed. The output gives the mass function of the sum along with its mean and variance.

P0

.25

P1

.15

P2

.1

P3

.2

P4

.3

n =

25

Start

Quit

Mean = 53.75 Variance = 63.1875 0.05 0.04 0.03 p(i) 0.02 0.01 0.00

0

20

40

60

80

100

i

FIGURE 8.1(c)

EXAMPLE 3b The number of students who enroll in a psychology course is a Poisson random variable with mean 100. The professor in charge of the course has decided that if the number enrolling is 120 or more, he will teach the course in two separate sections, whereas if fewer than 120 students enroll, he will teach all of the students together in a single section. What is the probability that the professor will have to teach two sections? Solution. The exact solution e−100

q (100)i i!

i=120

does not readily yield a numerical answer. However, by recalling that a Poisson random variable with mean 100 is the sum of 100 independent Poisson random variables, each with mean 1, we can make use of the central limit theorem to obtain an approximate solution. If X denotes the number of students that enroll in the course, we have P{X Ú 120} = P{X Ú 119.5}

(the continuity correction)

Section 8.3

The Central Limit Theorem

397

Central Limit Theorem Enter the probabilities and the number of random variables to be summed. The output gives the mass function of the sum along with its mean and variance.

P0

.25

P1

.15

P2

.1

P3

.2

P4

.3

n =

100

Start

Quit

Mean = 215. Variance = 252.75

p(i)

0.030 0.025 0.020 0.015 0.010 0.005 0.000

0

100

200

300

400

i

FIGURE 8.1(d)

0

X − 100 119.5 − 100 =P Ú √ √ 100 100

7

L 1 − (1.95) L .0256 where we have used the fact that the variance of a Poisson random variable is equal to its mean. . EXAMPLE 3c If 10 fair dice are rolled, ﬁnd the approximate probability that the sum obtained is between 30 and 40, inclusive. Solution. Let Xi denote the value of the ith die, i = 1, 2, . . . , 10. Since E(Xi ) =

7 , 2

Var(Xi ) = E[Xi2 ] − (E[Xi ])2 =

35 , 12

398

Chapter 8

Limit Theorems

the central limit theorem yields

⎧ ⎪ ⎪ ⎪ ⎨

⎫ ⎪ ⎪ ⎪ ⎬

X − 35 40.5 − 35 29.5 − 35 … @ … @ @ 350 350 350 ⎪ ⎪ ⎪ ⎭ 12 12 12 L 2 (1.0184) − 1 L .692

P{29.5 … X … 40.5} = P

⎪ ⎪ ⎪ ⎩

.

EXAMPLE 3d Let Xi , i = 1, . . . , 10, be independent random 0 variables,7each uniformly distributed 10 Xi > 6 . over (0, 1). Calculate an approximation to P i=1

Solution. Since E[Xi ] =

Hence,

10

1 and Var(Xi ) = 12 , we have, by the central limit theorem, ⎫ ⎧ 10 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎫ ⎪ ⎧ X − 5 ⎪ ⎪ i ⎪ ⎪ 10 ⎬ ⎨ ⎬ ⎨ 6 − 5 1 > @ Xi > 6 = P P @ ⎭ ⎩ ⎪ 1 1 ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ 10( ) 10( ) ⎪ ⎪ ⎪ ⎪ 12 12 ⎪ ⎭ ⎩ √ L 1 − ( 1.2) L .1367 1 2

Xi will be greater than 6 only 14 percent of the time.

.

i=1

EXAMPLE 3e An instructor has 50 exams that will be graded in sequence. The times required to grade the 50 exams are independent, with a common distribution that has mean 20 minutes and standard deviation 4 minutes. Approximate the probability that the instructor will grade at least 25 of the exams in the ﬁrst 450 minutes of work. Solution. If we let Xi be the time that it takes to grade exam i, then X=

25

Xi

i=1

is the time it takes to grade the ﬁrst 25 exams. Because the instructor will grade at least 25 exams in the ﬁrst 450 minutes of work if the time it takes to grade the ﬁrst 25 exams is less than or equal to 450, we see that the desired probability is P{X … 450}. To approximate this probability, we use the central limit theorem. Now, E[X] =

25

E[Xi ] = 25(20) = 500

i=1

and Var(X) =

25 i=1

Var(Xi ) = 25(16) = 400

Section 8.3

The Central Limit Theorem

399

Consequently, with Z being a standard normal random variable, we have X − 500 450 − 500 P{X … 450} = P{ √ … } √ 400 400 L P{Z … −2.5} = P{Z Ú 2.5} . = 1 − (2.5) = .006 Central limit theorems also exist when the Xi are independent, but not necessarily identically distributed random variables. One version, by no means the most general, is as follows. Theorem 3.2 Central limit theorem for independent random variables Let X1 , X2 , . . . be a sequence of independent random variables having respective means and variances μi = E[Xi ], σi2 = Var(Xi ). If (a) the Xi are uniformly q σi2 = q—then bounded—that is, if for some M, P{|Xi | < M} = 1 for all i, and (b) i=1 ⎫ ⎧ ⎪ ⎪ n ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (Xi − μi ) ⎪ ⎪ ⎪ ⎪ ⎬ ⎨ i=1 A … a → (a) as n → q P B ⎪ ⎪ ⎪ ⎪ B n 2 ⎪ ⎪ ⎪ ⎪ C ⎪ ⎪ σ ⎪ ⎪ i ⎪ ⎪ ⎭ ⎩ i=1 Historical Note Pierre-Simon, Marquis de Laplace The central limit theorem was originally stated and proven by the French mathematician Pierre-Simon, Marquis de Laplace, who came to the theorem from his observations that errors of measurement (which can usually be regarded as being the sum of a large number of tiny forces) tend to be normally distributed. Laplace, who was also a famous astronomer (and indeed was called “the Newton of France”), was one of the great early contributors to both probability and statistics. Laplace was also a popularizer of the uses of probability in everyday life. He strongly believed in its importance, as is indicated by the following quotations of his taken from his published book Analytical Theory of Probability: “We see that the theory of probability is at bottom only common sense reduced to calculation; it makes us appreciate with exactitude what reasonable minds feel by a sort of instinct, often without being able to account for it.. . . It is remarkable that this science, which originated in the consideration of games of chance, should become the most important object of human knowledge.. . . The most important questions of life are, for the most part, really only problems of probability.” The application of the central limit theorem to show that measurement errors are approximately normally distributed is regarded as an important contribution to science. Indeed, in the 17th and 18th centuries the central limit theorem was often called the law of frequency of errors. Listen to the words of Francis Galton (taken from his book Natural Inheritance, published in 1889): “I know of scarcely anything so apt to impress the imagination as the wonderful form of cosmic order expressed by the ‘Law of Frequency of Error.’ The Law would have been personiﬁed by the Greeks and deiﬁed, if they had known of it. It reigns with serenity and in complete self-effacement amidst the wildest confusion. The huger the mob and the greater the apparent anarchy, the more perfect is its sway. It is the supreme law of unreason.”

400

Chapter 8

Limit Theorems

8.4

THE STRONG LAW OF LARGE NUMBERS The strong law of large numbers is probably the best-known result in probability theory. It states that the average of a sequence of independent random variables having a common distribution will, with probability 1, converge to the mean of that distribution. Theorem 4.1 The strong law of large numbers Let X1 , X2 , . . . be a sequence of independent and identically distributed random variables, each having a ﬁnite mean μ = E[Xi ]. Then, with probability 1, X1 + X2 + · · · + Xn →μ as n → q† n As an application of the strong law of large numbers, suppose that a sequence of independent trials of some experiment is performed. Let E be a ﬁxed event of the experiment, and denote by P(E) the probability that E occurs on any particular trial. Letting % 1 if E occurs on the ith trial Xi = 0 if E does not occur on the ith trial we have, by the strong law of large numbers, that with probability 1, X1 + · · · + Xn → E[X] = P(E) n

(4.1)

Since X1 + · · · + Xn represents the number of times that the event E occurs in the ﬁrst n trials, we may interpret Equation (4.1) as stating that, with probability 1, the limiting proportion of time that the event E occurs is just P(E). Although the theorem can be proven without this assumption, our proof of the strong law of large numbers will assume that the random variables Xi have a ﬁnite fourth moment. That is, we will suppose that E[Xi4 ] = K < q. Proof of the Strong Law of Large Numbers: To begin, assume that μ, the mean of n Xi and consider the Xi , is equal to 0. Let Sn = i=1

E[S4n ]

= E[(X1 + · · · + Xn )(X1 + · · · + Xn ) * (X1 + · · · + Xn )(X1 + · · · + Xn )]

Expanding the right side of the preceding equation results in terms of the form Xi4 ,

Xi3 Xj ,

Xi2 Xj2 ,

Xi2 Xj Xk ,

and

Xi Xj Xk Xl

where i, j, k, and l are all different. Because all the Xi have mean 0, it follows by independence that E[Xi3 Xj ] = E[Xi3 ]E[Xj ] = 0 E[Xi2 Xj Xk ] = E[Xi2 ]E[Xj ]E[Xk ] = 0 E[Xi Xj Xk Xl ] = 0 † That is, the strong law of large numbers states that

P{ lim (X1 + · · · + Xn )/n = μ} = 1 n→q

Section 8.4

The Strong Law of Large Numbers

401

4 Now, for a given pair i and j, there will be = 6 terms in the expansion that will 2 equal Xi2 Xj2 . Hence, upon expanding the preceding product and taking expectations term by term, it follows that n E[Xi2 Xj2 ] E[S4n ] = nE[Xi4 ] + 6 2 = nK + 3n(n − 1)E[Xi2 ]E[Xj2 ] where we have once again made use of the independence assumption. Now, since 0 … Var(Xi2 ) = E[Xi4 ] − (E[Xi2 ])2 we have

(E[Xi2 ])2 … E[Xi4 ] = K

Therefore, from the preceding, we obtain E[S4n ] … nK + 3n(n − 1)K

which implies that

S4 E n4 n Therefore,

⎡ E⎣

q S4 n=1

…

⎤

n⎦ n4

=

3K K + 2 3 n n

q n=1

S4 E n4 n

But the preceding implies that, with probability 1,

< q q n=1

S4n /n4 < q. (For if there is a

positive probability that the sum is inﬁnite, then its expected value is inﬁnite.) But the convergence of a series implies that its nth term goes to 0; so we can conclude that, with probability 1, S4 lim n4 = 0 n→q n But if S4n /n4 = (Sn /n)4 goes to 0, then so must Sn /n; hence, we have proven that, with probability 1, Sn →0 as n→q n When μ, the mean of the Xi , is not equal to 0, we can apply the preceding argument to the random variables Xi − μ to obtain that with probability 1, lim n→q

n (Xi − μ) =0 n i=1

or, equivalently, lim n→q which proves the result.

n Xi i=1

n

=μ

402

Chapter 8

Limit Theorems

The strong law is illustrated by two modules on the text website that consider independent and identically distributed random variables which take on one of the values 0, 1, 2, 3, and 4. The modules simulate the values of n such random variables; the proportions of time that each outcome occurs, as well as the resulting sample mean n Xi /n, are then indicated and plotted. When using these modules, which differ only i=1

in the type of graph presented, one enters the probabilities and the desired value of n. Figure 8.2 gives the results of a simulation using a speciﬁed probability mass function and (a) n = 100, (b) n = 1000, and (c) n = 10,000. Many students are initially confused about the difference between the weak and the strong laws of large numbers. The weak law of large numbers states that, for any speciﬁed large value n∗ , (X1 + · · · + Xn∗ )/n∗ is likely to be near μ. However, it does not say that (X1 + · · · + Xn )/n is bound to stay near μ for all values of n larger than n∗ . Thus, it leaves open the possibility that large values of |(X1 + · · · + Xn )/n − μ| can occur inﬁnitely often (though at infrequent intervals). The strong law shows that this cannot occur. In particular, it implies that, with probability 1, for any positive value ε, 6 6 6 6 6 6 n Xi 6 6 − μ 6 6 6 6 1 n will be greater than ε only a ﬁnite number of times. Strong Law Of Large Numbers Enter the probabilities and the number of trials to be simulated. The output gives the total number of times each outcome occurs, and the average of all outcomes.

P0

.1

P1

.2

P2

.3

P3

.35

P4

.05

n =

100

Start

Quit

Theoretical Mean = 2.05 Sample Mean = 1.89

0

1

2

3

4

15

20

30

31

4

FIGURE 8.2(a)

Section 8.5

Other Inequalities

403

Strong Law Of Large Numbers Enter the probabilities and the number of trials to be simulated. The output gives the total number of times each outcome occurs, and the average of all outcomes.

P0

.1

P1

.2

P2

.3

P3

.35

P4

.05

n =

1000

Start

Quit

Theoretical Mean = 2.05 Sample Mean = 2.078

0

1

2

3

4

106

189

285

361

59

FIGURE 8.2(b)

The strong law of large numbers was originally proven, in the special case of Bernoulli random variables, by the French mathematician Borel. The general form of the strong law presented in Theorem 4.1 was proven by the Russian mathematician A. N. Kolmogorov. 8.5

OTHER INEQUALITIES We are sometimes confronted with situations in which we are interested in obtaining an upper bound for a probability of the form P{X − μ Ú a}, where a is some positive value and when only the mean μ = E[X] and variance σ 2 = Var(X) of the distribution of X are known. Of course, since X − μ Ú a > 0 implies that |X − μ| Ú a, it follows from Chebyshev’s inequality that P{X − μ Ú a} … P{|X − μ| Ú a} …

σ2 a2

when

a > 0

However, as the following proposition shows, it turns out that we can do better. Proposition 5.1. One-sided Chebyshev inequality If X is a random variable with mean 0 and ﬁnite variance σ 2 , then, for any a > 0, P{X Ú a} …

σ2

σ2 + a2

404

Chapter 8

Limit Theorems Strong Law Of Large Numbers Enter the probabilities and the number of trials to be simulated. The output gives the total number of times each outcome occurs, and the average of all outcomes.

P0

.1

P1

.2

P2

.3

P3

.35

P4

.05

n =

10000

Start

Quit

Theoretical Mean = 2.05 Sample Mean = 2.0416

0

1

2

3

4

1041

2027

2917

3505

510

FIGURE 8.2(c)

Proof. Let b > 0 and note that X Ú a

is equivalent to X + b Ú a + b

Hence, P{X Ú a} = P{X + b Ú a + b} … P{(X + b)2 Ú (a + b)2 } where the inequality is obtained by noting that since a + b > 0, X + b Ú a + b implies that (X + b)2 Ú (a + b)2 . Upon applying Markov’s inequality, the preceding yields that P{X Ú a} …

E[(X + b)2 ] σ 2 + b2 = (a + b)2 (a + b)2

Letting b = σ 2 /a [which is easily seen to be the value of b that minimizes (σ 2 + b2 )/(a + b)2 ] gives the desired result.

Section 8.5

Other Inequalities

405

EXAMPLE 5a If the number of items produced in a factory during a week is a random variable with mean 100 and variance 400, compute an upper bound on the probability that this week’s production will be at least 120. Solution. It follows from the one-sided Chebyshev inequality that P{X Ú 120} = P{X − 100 Ú 20} …

1 400 = 2 2 400 + (20)

Hence, the probability that this week’s production will be 120 or more is at most 12 . If we attempted to obtain a bound by applying Markov’s inequality, then we would have obtained E(X) 5 = P{X Ú 120} … 120 6 which is a far weaker bound than the preceding one.

.

Suppose now that X has mean μ and variance σ 2 . Since both X − μ and μ − X have mean 0 and variance σ 2 , it follows from the one-sided Chebyshev inequality that, for a > 0, σ2 P{X − μ Ú a} … 2 σ + a2 and P{μ − X Ú a} …

σ2 σ 2 + a2

Thus, we have the following corollary. Corollary 5.1. If E[X] = μ and Var(X) = σ 2 , then, for a > 0, σ2 + a2 σ2 P{X … μ − a} … 2 σ + a2 P{X Ú μ + a} …

σ2

EXAMPLE 5b A set of 200 people consisting of 100 men and 100 women is randomly divided into 100 pairs of 2 each. Give an upper bound to the probability that at most 30 of these pairs will consist of a man and a woman. Solution. Number the men arbitrarily from 1 to 100, and for i = 1, 2, . . . 100, let % 1 if man i is paired with a woman Xi = 0 otherwise Then X, the number of man–woman pairs, can be expressed as X=

100 i=1

Xi

406

Chapter 8

Limit Theorems

Because man i is equally likely to be paired with any of the other 199 people, of which 100 are women, we have 100 E[Xi ] = P{Xi = 1} = 199 Similarly, for i Z j, E[Xi Xj ] = P{Xi = 1, Xj = 1} = P{Xi = 1}P{Xj = 1|Xi = 1} =

100 99 199 197

where P{Xj = 1|Xi = 1} = 99/197, since, given that man i is paired with a woman, man j is equally likely to be paired with any of the remaining 197 people, of which 99 are women. Hence, we obtain E[X] =

100

E[Xi ]

i=1

100 199 L 50.25

= (100)

Var(X) =

100

Var(Xi ) + 2

i

i=1

100 99 = 100 + 2 199 199

100 2

Cov(Xi , Xj )

100 99 − 199 197

100 199

2

L 25.126 The Chebyshev inequality then yields P{X … 30} … P{|X − 50.25| Ú 20.25} …

25.126 L .061 (20.25)2

Thus, there are fewer than 6 chances in a hundred that fewer than 30 men will be paired with women. However, we can improve on this bound by using the one-sided Chebyshev inequality, which yields P{X … 30} = P{X … 50.25 − 20.25} 25.126 … 25.126 + (20.25)2 L .058

.

When the moment generating function of the random variable X is known, we can obtain even more effective bounds on P{X Ú a}. Let M(t) = E[etX ] be the moment generating function of the random variable X. Then, for t > 0, P{X Ú a} = P{etX Ú eta } … E[etX ]e−ta

by Markov’s inequality

Section 8.5

Other Inequalities

407

Similarly, for t < 0, P{X … a} = P{etX Ú eta } … E[etX ]e−ta Thus, we have the following inequalities, known as Chernoff bounds. Proposition 5.2.

Chernoff bounds P{X Ú a} … e−ta M(t) −ta

P{X … a} … e

for all t > 0 for all t < 0

M(t)

Since the Chernoff bounds hold for all t in either the positive or negative quadrant, we obtain the best bound on P{X Ú a} by using the t that minimizes e−ta M(t). EXAMPLE 5c Chernoff bounds for the standard normal random variable If Z is a standard normal random variable, then its moment generating function is 2 M(t) = et /2 , so the Chernoff bound on P{Z Ú a} is given by P{Z Ú a} … e−ta et Now the value of t, t > 0, that minimizes et which is t = a. Thus, for a > 0, we have

2 /2

for all

2 /2−ta

t > 0

is the value that minimizes t2 /2 − ta,

P{Z Ú a} … e−a

2 /2

Similarly, we can show that, for a < 0, P{Z … a} … e−a

2 /2

.

EXAMPLE 5d Chernoff bounds for the Poisson random variable If X is a Poisson random variable with parameter λ, then its moment generating funct tion is M(t) = eλ(e −1) . Hence, the Chernoff bound on P{X Ú i} is P{X Ú i} … eλ(e −1) e−it t

t > 0

Minimizing the right side of the preceding inequality is equivalent to minimizing λ(et − 1) − it, and calculus shows that the minimal value occurs when et = i/λ. Provided that i/λ > 1, this minimizing value of t will be positive. Therefore, assuming that i > λ and letting et = i/λ in the Chernoff bound yields i λ(i/λ−1) λ P{X Ú i} … e i or, equivalently, P{X Ú i} …

e−λ (eλ)i ii

.

EXAMPLE 5e Consider a gambler who is equally likely to either win or lose 1 unit on every play, independently of his past results. That is, if Xi is the gambler’s winnings on the ith play, then the Xi are independent and

408

Chapter 8

Limit Theorems

P{Xi = 1} = P{Xi = −1} = n

Let Sn =

1 2

Xi denote the gambler’s winnings after n plays. We will use the Chernoff

i=1

bound on P{Sn Ú a}. To start, note that the moment generating function of Xi is E[etX ] =

et + e−t 2

Now, using the McLaurin expansions of et and e−t , we see that t

−t

e + e

t3 t2 + + ··· + =1 + t + 2! 3! 7 0 t4 t2 + + ··· =2 1 + 2! 4! =2

t2 t3 1 − t + − + ··· 2! 3!

q t2n (2n)! n=0 q

… 2

n=0

= 2et

(t2 /2)n n!

since (2n)! Ú n!2n

2 /2

Therefore, E[etX ] Ú et

2 /2

Since the moment generating function of the sum of independent random variables is the product of their moment generating functions, we have E[etSn ] = (E[etX ])n … ent

2 /2

Using the preceding result along with the Chernoff bound gives P{Sn Ú a} … e−ta ent

2 /2

t > 0

The value of t that minimizes the right side of the preceding is the value that minimizes nt2 /2 − ta, and this value is t = a/n. Supposing that a > 0 (so that the minimizing t is positive) and letting t = a/n in the preceding inequality yields P{Sn Ú a} … e−a

2 /2n

a > 0

This latter inequality yields, for example, P{S10 Ú 6} … e−36/20 L .1653

Section 8.5

Other Inequalities

409

whereas the exact probability is P{S10 Ú 6} = P{gambler wins at least 8 of the ﬁrst 10 games} 10 10 10 + + 10 9 8 56 L .0547 = = 10 1024 2

.

The next inequality is one having to do with expectations rather than probabilities. Before stating it, we need the following deﬁnition. Deﬁnition A twice-differentiable real-valued function f (x) is said to be convex if f (x) Ú 0 for all x; similarly, it is said to be concave if f (x) … 0. Some examples of convex functions are f (x) = x2 , f (x) = eax , and f (x) = −x1/n for x Ú 0. If f (x) is convex, then g(x) = −f (x) is concave, and vice versa. Proposition 5.3. Jensen’s inequality If f (x) is a convex function, then E[ f (X)] Ú f (E[X]) provided that the expectations exist and are ﬁnite. Proof. Expanding f (x) in a Taylor’s series expansion about μ = E[X] yields f (x) = f (μ) + f (μ)(x − μ) +

f (ξ )(x − μ)2 2

where ξ is some value between x and μ. Since f (ξ ) Ú 0, we obtain f (x) Ú f (μ) + f (μ)(x − μ) Hence,

f (X) Ú f (μ) + f (μ)(X − μ)

Taking expectations yields E[ f (X)] Ú f (μ) + f (μ)E[X − μ] = f (μ) and the inequality is established. EXAMPLE 5f An investor is faced with the following choices: Either she can invest all of her money in a risky proposition that would lead to a random return X that has mean m, or she can put the money into a risk-free venture that will lead to a return of m with probability 1. Suppose that her decision will be made on the basis of maximizing the expected value of u(R), where R is her return and u is her utility function. By Jensen’s inequality, it follows that if u is a concave function, then E[u(X)] … u(m), so the riskfree alternative is preferable, whereas if u is convex, then E[u(X)] Ú u(m), so the risky investment alternative would be preferred. .

410

Chapter 8

Limit Theorems

8.6

BOUNDING THE ERROR PROBABILITY WHEN APPROXIMATING A SUM OF INDEPENDENT BERNOULLI RANDOM VARIABLES BY A POISSON RANDOM VARIABLE In this section, we establish bounds on how closely a sum of independent Bernoulli random variables is approximated by a Poisson random variable with the same mean. Suppose that we want to approximate the sum of independent Bernoulli random variables with respective means p1 , p2 , . . . , pn . Starting with a sequence Y1 , . . . , Yn of independent Poisson random variables, with Yi having mean pi , we will construct a sequence of independent Bernoulli random variables X1 , . . . , Xn with parameters p1 , . . . , pn such that P{Xi Z Yi } … p2i Letting X = clude that

n

Xi and Y =

i=1

n

for each i

Yi , we will use the preceding inequality to con-

i=1

P{X Z Y} …

n

p2i

i=1

Finally, we will show that the preceding inequality implies that, for any set of real numbers A, |P{X ∈ A} − P{Y ∈ A}| …

n

p2i

i=1

Since X is the sum of independent Bernoulli random variables and Y is a Poisson random variable, the latter inequality will yield the desired bound. To show how the task is accomplished, let Yi , i = 1, . . . , n be independent Poisson random variables with respective means pi . Now let U1 , . . . , Un be independent random variables that are also independent of the Yi ’s and which are such that % 0 Ui = 1

with probability (1 − pi )epi with probability 1 − (1 − pi )epi

This deﬁnition implicitly makes use of the inequality e−p Ú 1 − p in assuming that (1 − pi )epi … 1. Next, deﬁne the random variables Xi , i = 1, . . . , n, by Xi =

% 0 1

if Yi = Ui = 0 otherwise

Note that P{Xi = 0} = P{Yi = 0}P{Ui = 0} = e−pi (1 − pi )epi = 1 − pi P{Xi = 1} = 1 − P{Xi = 0} = pi

Section 8.6

Bounding the Error Probability

411

Now, if Xi is equal to 0, then so must Yi equal 0 (by the deﬁnition of Xi ). Therefore, P{Xi Z Yi } = P{Xi = 1, Yi Z 1} = P{Yi = 0, Xi = 1} + P{Yi > 1} = P{Yi = 0, Ui = 1} + P{Yi > 1} = e−pi [1 − (1 − pi )epi ] + 1 − e−pi − pi e−pi = pi − pi e−pi (since 1 − e−p … p)

… p2i Now let X =

n

n

Xi and Y =

i=1

Yi , and note that X is the sum of independent

i=1

Bernoulli random variables and Y is Poisson with the expected value E[Y] = E[X] = n pi . Note also that the inequality X Z Y implies that Xi Z Yi for some i, so i=1

P{X Z Y} … P{Xi Z Yi for some i} n … P{Xi Z Yi } (Boole’s inequality) …

i=1 n

p2i

i=1

For any event B, let IB , the indicator variable for the event B, be deﬁned by % 1 if B occurs IB = 0 otherwise Note that, for any set of real numbers A, I{X∈A} − I{Y∈A} … I{XZY} The preceding inequality follows from the fact that, since an indicator variable is either 0 or 1, the left-hand side equals 1 only when I{X∈A} = 1 and I{Y∈A} = 0. But this would imply that X ∈ A and Y ∈ A, which means that X Z Y, so the right side would also equal 1. Upon taking expectations of the preceding inequality, we obtain P{X ∈ A} − P{Y ∈ A} … P{X Z Y} By reversing X and Y, we obtain, in the same manner, P{Y ∈ A} − P{X ∈ A} … P{X Z Y} Thus, we can conclude that |P{X ∈ A} − P{Y ∈ A}| … P{X Z Y} Therefore, we have proven that with λ =

n i=1

pi ,

6 6 ⎧ ⎫ 6 6 n n 6 ⎬ 6 ⎨ −λ i e λ 6 6 Xi ∈ A − p2i 6 … 6P ⎭ 6 ⎩ i! 6 6 6 i∈A i=1 i=1

412

Chapter 8

Limit Theorems

Remark. When all the pi are equal to p, X is a binomial random variable. Hence, the preceding inequality shows that, for any set of nonnegative integers A, 6 6 6 6 −np (np)i 6 6 e n i n−i 6 … np2 6 − . p (1 − p) 6 6 i i! 6 6 i∈A

i∈A

SUMMARY Two useful probability bounds are provided by the Markov and Chebyshev inequalities. The Markov inequality is concerned with nonnegative random variables and says that, for X of that type, E[X] P{X Ú a} … a for every positive value a. The Chebyshev inequality, which is a simple consequence of the Markov inequality, states that if X has mean μ and variance σ 2 , then, for every positive k, 1 P{|X − μ| Ú kσ } … 2 k The two most important theoretical results in probability are the central limit theorem and the strong law of large numbers. Both are concerned with a sequence of independent and identically distributed random variables. The central limit theorem says that if the random variables have a ﬁnite mean μ and a ﬁnite variance σ 2 , then the distribution of the sum of the ﬁrst n of them is, for large n, approximately that of a normal random variable