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Problem Books in Mathematics

Edited by P.R. Halmos

T. Cacoullos

Exercises in Probability

With 22 Illustrations

Springer-Verlag New York Berlin Heidelberg

London Paris Tokyo

T. Cacoullos Department of Mathematics University of Athens Athens 157 10 Greece

Editor

Paul R. Halmos Department of Mathematics Santa Clara University Santa Clara, CA 95053 U.S.A.

Mathematics Subject Classification (1980): 60-01

Library of Congress Cataloging-in-Publication Data Cacoullos, T. (Theophilos)

Exercises in probability. (Problems in mathematics) Translated from the Greek. Bibliography: p. 1. Probabilities-Problems, exercises, etc. I. Title.

II. Series. QA273.25.C33 1989 519.2'076 88-4932

Printed on acid-free paper.

Original Greek edition published in two volumes, Askiseis Theorias Pithanotiton (Exercises in Probability) and Askiseis-Provlemata Theorias Pithanotition (Exercises-Problems in Probability Theory), in 1971 and 1972, respectively, by Leousis-Mastroyiannis, Athens, Greece.

© 1989 by Springer-Verlag New York Inc.

Softcover reprint of the hardcover 1989

All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer-Verlag, 175 Fifth Avenue, New York, NY 10010, U.S.A.), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use of general descriptive names, trade names, trademarks, etc. in this publication, even if the former are not especially identified, is not to be taken as a sign that such names, as understood by the Trade Marks and Merchandise Marks Act, may accordingly be used freely by anyone.

Typeset by Asco Trade Typesetting Ltd., Hong Kong. Printed and bound by R.R. Donnelley & Sons, Harrisonburg, Virginia. Printed in the United States of America.

9 8 7 6 5 4 3 2 1

ISBN-13: 978-1-4612-8863-3 e-ISBN-13: 978-1-4612-4526-1 DOl: 10.1007/978-1-4612-4526-1

Preface

This problem book can serve as a companion text for an introductory or intermediate-level one- or two-semester probability course, usually given to undergraduate or first-year graduate students in American universities. Those who will benefit most are those with a good grasp of calculus and the inclination and patience required by mathematical thinking; and yet many more, with less formal mathematical background and sophistication, can benefit greatly from the elements of theory and solved problems, especially of Part I. In particular, the important part of discrete probabilities with problems of a combinatorial nature, relating to finite sample spaces and equally likely cases, may be challenging and entertaining to the student and amateur alike, through the spectrum from mathematics to the social sciences.

For some time, since I have been teaching courses in probability theory both to undergraduates at the University of Athens (School of Science) and graduate students in American universities, I have noted the need of students for a systematic collection of elementary and intermediate-level exercises in this area. The relevant international bibliography is rather limited. The aim of the present collection is to fill such a gap.

This book is based on the two volumes of the Greek edition entitled Exercises in Probability by the same author. Of the 430 exercises contained in the original Greek edition (1972), 329 (with solutions) were selected for the present translation. A small number of problems in stochastic processes is not included here, since it is felt that a minimal treatment of this important area requires more theory and a larger number of problems. However, an addendum of over 160 exercises and certain complements of theory and problems (with answers or hints) from the recent (1985) second Greek edition is included here as Supplement I: Miscellaneous Exercises, and as Supplement II: Complements and Problems.

VI Preface

In addition to the Supplements, the present book is divided into three parts. Part I: Elementary Probabilities (Chapters 1-4), Part II: Advanced Topics (Chapter 5-10), and Part III: Solutions (to the exercises of Part I and Part II). In most chapters, the exercises are preceded by some basic theory and formulas.

In Part I (with 170 exercises) emphasis is given to classical probabilities (equally likely cases). It concerns mainly one-dimensional, discrete, and continuous distributions (Chapter 2). The expected value, variance, and moments of a distribution are treated in Chapter 3. Chapter 4, in spite of its elementary nature, contains quite a few challenging problems. In general, problems of a higher level of difficulty are marked with an asterisk throughout the text.

Part II deals with more advanced topics: multivariate distributions (Chapter 5), generating and characteristic functions (Chapter 6), distribution of functions of random variables (Chapter 7), and Laws of Large Numbers and Central Limit Theorems (Chapter 8). Chapter 9 deals with special topics: stochastic inequalities, geometrical probabilities, and applications of difference equations in probability. The last chapter (Chapter to) contains general overview exercises.

Clearly, a collection of problems owes a lot to the relevant bibliography. The exercises come from many sources, some of them are new, a large number are variations of more or less standard exercises and problems of elementary or intermediate level, and a few are based on research papers. Such problems, as well as complements, are also marked with an asterisk. The level and sources are indicated in the select Bibliography.

It is hoped that the present edition will make proportionately as many friends as it did among the ten thousand or so students all over Greece who have used the text since its first edition in 1971-72. Naturally, one benefits from such a text by first really trying to solve a problem, before falling into the "temptation" of looking up the solution. That is why solutions are given separately, in the last part of the book.

Thanks are due to my colleagues Dr. Ch. Charalambides and Dr. Ourania Chrysaphinou for preparing some solutions for the first Greek edition, as well as preparing a first draft of a major part of this translation.

New York June, 1987

T. CACOULLOS

Visiting Professor Columbia University

Contents

Preface v

Part I Elementary Probabilities

CHAPTER I

Basic Probabilities. Discrete Spaces

Basic Definitions and Formulas Exercises

1. Sets. Events: 1-8 2. Combinatorics: 9-16 3. Properties of Binomial Coefficients: 17-25 4. Properties of Probability: 26-34 5. Classical Probabilities. Equally Likely Cases: 35-53 6. Independent Events. Conditional Probability: 54-79

CHAPTER 2

Distributions. Random Variables

Elements of Theory Exercises

1. Discrete Distributions: 80-89 2. Continuous Distributions: 90-100

CHAPTER 3

Expectation. Variance. Moments

Elements of Theory Exercises

1. Theoretical Exercises: 101-113 2. Mean and Variance: 114-125

3

3 7 7 8 8 9

10 13

17

17 20 20 21

24

24 26 26 28

Vlll

CHAPTER 4

General Problems: 126-170

Part II Advanced Topics

CHAPTER 5

Multivariate Distributions

Elements of Theory Exercises: 171-190

CHAPTER 6

Generating Functions. Characteristic Functions


CHAPTER 7

Distribution of Functions of Random Variables


CHAPTER 8

Contents

30

37

39

39 43

47

47 51

55

55 58

Limit Theorems. Laws of Large Numbers. Central Limit Theorems 63

Elements of Theory 63 Exercises: 251-269 67

CHAPTER 9

Special Topics: Inequalities, Geometrical Probabilities, Difference Equations

Elements of Theory A. Inequalities B. Geometrical Probabilities C. Difference Equations

Exercises A. Inequalities: 270-282 B. Geometrical Probabilities: 283-289 C. Difference Equations: 290-300

CHAPTER 10

General Exercises: 301-329

Supplements

SUPPLEMENT I

Miscellaneous Exercises: 1-1-1-56

70

70 70 72 72 74 74 75 76

78

83

85

Contents

SUPPLEMENT II

Complements and Problems

1. Multivariate Distributions: 1.1-1.27 2. Generating Functions: 2.1-2.22 3. Transformation of Random Variables: 3.1-3.15 4. Convergence of Random Variables: 4.1-4.19 5. Miscellaneous Complements and Problems: 5.1-5.29

Part III Solutions

Solutions: 1-329

Bibliography

Index

ix

94

94 101 106 109 115

123

125

243

245

PART I

ELEMENTARY PROBABILITIES

CHAPTER 1

Basic Probabilities. Discrete Spaces

Basic Definitions and Formulas

1. Sample space 0: The totality of possible outcomes of a random (chance) experiment.

The individual outcomes are called elementary or simple events or points (cases in the classical definition of probability by Laplace).

2. Discrete sample space 0: 0 is at most a denumerable set of points.

3. Events: Subsets* of O.

4. Union of n events AI"'" An' denoted by Al U A z ... u An, is the realization of at least one of the events AI' ... , An"

5. Realization or occurrence of an event A means the appearance of an element (point) in A.

6. Intersection or product of two events A and B, denoted by A n B or AB, is the simultaneous realization of both events A and B. Similarly for n events.

7. Complement of A, denoted by A C or A', means A does not occur.

8. The difference of two events A and B is defined by A - B = ABC, i.e., A occurs but not B. Thus AC = 0 - A.

9. Sure or certain event: The sample space O.

10. Impossible event: The complement of the sure event, i.e., the empty set.

* Rigorously, measurable subsets of n, i.e., members of a so-called Borel or a-field B(n) on n: a family of subsets of n containing n itself and closed under the set operations of complementation and countable union.

4 1. Basic Probabilities. Discrete Spaces

11. Probability. A probability (function) P is a set function on 0 (strictly on the Borel field B(O) on 0) which satisfies the following three axioms of Kolmogorov:

(I) P(O) = 1. (II) For every event A, P(A) ~ o.

(III) For every sequence of mutually exclusive events

i #j,

12. (a) Equally likely cases. Let 0 be a finite sample space with N elements (points) WI' W z, ... , W N; if

1 P[{w;}] = N' i = 1, ... , N,

then the Wi are called equally likely elementary events (cases). (b) Laplace definition of probability. Let A c 0, where the points of 0 are

assumed equally likely. Then

peA] = number ofpo~nts ~n A number of POllltS III 0

number of favorable cases to A

number of cases in 0

13. Basic formulas for probabilities. (i) P(A C ) = 1 - P(A).

(ii) P(A - B) = P(A) - P(AB). (iii) Addition theorem of probabilities: P(A u B) = P(A) + P(B) - P(AB).

Poincare's theorem. For n events A I' ... , A.,

P(A I U Az U··· u A.) = SI - S2 + ... + (_1)n- IS., (1.1)

where, for each k = 1,2, ... , n, Sk is defined by

Sk = L P(A i1 A i2 ••• Ai.)' it <i2 <'" <ik

i.e., the summation is over all combinations of the n events taken k at a time (k=I, ... ,n).

14. Conditional probability of B given A:

P(BIA) = P(AB) P(A) ,

P(A) > o.

15. Multiplication formula for probabilities:

P(AB) = P(BIA)P(A) = P(AIB)P(B).

Basic Definitions and Formulas

In general,

16. Independent events (statistically or stochastically): (a) Two events A, B: if P(AB) = P(A)P(B).

5

(b) n events A 1, ... , An: if P(Ait ... Ai.} = P(Ait )··· P(Ai.} for 1 ::;; i1 < i2 < ... < ik ::;; n, 2 ::;; k ::;; n.

17. Independent experiments. Let E 1, E2, ... , En be n chance experiments and 0 1 , ... , On the corresponding sample spaces. The experiments are called stochastically or statistically independent or simply independent if for every Ai c ili (i = 1,2, ... , n),

Remark. Physically independent experiments are assumed statistically independent, e.g., successive throws of a coin, a die, etc.

18. Total probability formula. If Bi n Bj = 0, i =f. j (i,j = 1, ... , n) and A c

(B1 U B2 U ... u Bn), then

(1.3)

19. Bayes's theorem (formula). As in Formula 18,

P B A _ P(AIBi)P(B;) ( d ) - P(AIBdP(Bd + ... + P(AIBn)P(Bn)'

i = 1, ... , n. (1.4)

20. Combinatorics Binomial coefficient. For every real x and any positive integer k we define

( x) = x(x - 1)···(x - k + 1). k k!

In particular, when x is a positive integer, then

(x) x! k = k! (x - k)!

equals the number of combinations of x distinct objects taken k at a time. Pascal's triangle is based on the recursive relation

21. Let S be a set (population) of N distinct elements. (a) A subset of k elements (a k-tuple) is called a sample of size k (a combina

tion). (b) An ordered k-tuple of elements of S is called an ordered sample of size

6 I. Basic Probabilities. Discrete Spaces

k (a permutation). There are

(N) N' k k! = (N _. k)! == (Nh = N(N - 1) .. ' (N - k + 1)

permutations of N objects taken k at a time. (c) S can be divided into k subsets, the first subset (subpopulation) con

taining r l elements, the second r2' etc., the kth rk objects in

ways; (1.5) is called a multinomial coefficient.

22. Placing balls into cells. (a) s distinguishable balls can be placed into n cells in nS ways. (b) s nondistinguishable balls can be placed into n cells in

(1.5)

(1.6)

ways; that is, also the number of ordered n-tuples (rl' r2, ... , rn) of integers rj Z ° which are solutions of the equation

r l + r2 + ... + rn = s.

The numbers rj are referred to as the (cell) occupancy numbers (see Exercise 44).

23. Sampling without replacement. From an urn containing W white balls and B black balls, n balls are drawn (one after another or at once) at random. The probability Pr that r white balls are drawn is given by

N= W+B, (1.7)

max(O, n - B) :.::: r :.::: min(n, W).

The probabilities Pr define the so-called hypergeometric distribution.

24. Multiplication principle. If an "operation" A I can be performed in n I ways, another "operation" A2 in n2 ways, etc., the kth "operation" in nk ways, then the k "operations" can be performed, one after another, in n l n2 .. • nk

ways. This is equivalent to the proposition: If Aj consists of nj points (i = i, ... , k),

then the size (number of points) of the cartesian product

Al x A2 x .. · x Ak={(xI,,,,,xk):xjEAj,i= 1, ... ,k}

is equal to n l n2 ••• nk •

Exercises 7

25. The principle of mathematical induction. For every integer n let Pn be a proposition which is either true or false, i.e., Pn may be true for some nand false for other values of n; if, (i) PI is true, (ii) for every n, Pn true => Pn+1 true, then Pn is true for every n.

Exercises

1. Sets. Events

1. Let n denote the totality of students at the University of Athens and AI' A 2 , A 3, A4 , the sets offreshmen, sophomores,juniors, and seniors, respectively. Moreover, let F denote the set of female students and C the set of Cypriot students. Express in words each of the following sets:

(a) (AI U A 2 ) (\ F; (b) FC'; (c) Al FC; (d) A3FC'; (e) (AI U A 2 )CF.

2. Give the simplified forms of the sets: (a) (A u B) (\ (A u C); (b) (A u B) (\ (A' u B); (c) (A u B) (\ (A' u B) (\

(A' u B').

3. Express each of the following events in terms of the events A, B, and C, and the operations of complementation, union, and intersection:

(a) at least one of the events A, B, C occurs; (b) at most one of the events A, B, C occurs; (c) none of the events A, B, C occurs; (d) all three events occur; (e) exactly one of the events A, B, C occurs; (f) A and B occur but not C; (g) A occurs, if not then B does not occur either.

Give the Venn diagram for each of the above.

4. Let n denote the sample space corresponding to the chance experiment of tossing a coin three times. Let A be the event that heads appear exactly twice, let B be the event that at least two heads appear, and let C the event that heads appear when tails have appeared at least once.

(a) Give the elements of A, B, C; (b) Describe the events: (i) A'B; (ii) A'B'; (iii) AC.

In each of the Exercises 5-8, give the number of points in the sample space n and the points of the events defined therein.

5. A family has 4 children. Let the events: A: boys and girls alternate, B: the first and fourth child are boys, C: as many boys as girls, and D: three successive children of the same sex.

6. A salesman is arranging his schedule for visiting each of three towns a, b, c, twice. A: the first and last visit are in a.


7. Certain diseases cause cell disorders which can be distinguished into four categories. Blood tests are conducted on each of 4 patients and in each case the category is noted. Let A: the patients belong to the same category, B: two patients belong to the same category.

8. An elevator carries two persons and stops at three floors. Let A: they get off at different floors, B: one gets off at the first floor.

2. Combinatorics

9. In how many ways can a lady having 10 dresses, 5 pairs of shoes, and 2 hats be dressed?

10. In how many ways can we place in a bookcase two works each of three volumes and two works each offour volumes, so that the volumes ofthe same work are not separated?

11. In how many ways can r objects be distributed to n persons if there is no restriction on the number of objects that a person may receive?

12. In how many ways can 5 boys and 5 girls be seated around a table so that no 2 boys sit next to each other?

13. Eight points are chosen on the circumference of a circle. How many chords can be drawn by joining these points in all possible ways? If the 8 points are considered vertices of a polygon, how many triangles and how many hexagons can be formed?

14. In how many ways can n persons be seated around a table if 2 arrangements are regarded as the same when each person has the same right and left neighbors? At a dinner at which the n persons are seated randomly, what is the probability that a specific husband is seated next to his wife?

15. In how many ways can 20 recruits be distributed into 4 groups each consisting of 5 recruits? In how many ways can they be distributed into 4 camps, each camp receiving 5 recruits?

16. Using 7 consonants and 5 vowels, how many words consisting of 4 consonants and 3 vowels can we form?

3. Properties of Binomial Coefficients

17. Show that for every x and for every positive integer n:

(i) C ~ 1) + G) = (x: I} (ii) (x) = (x - 1) + (x - 2.) + .. , + (n - 1).

n n-l n-l n-1

Exercises 9

18. If n, m, r, are positive integers prove that

(~)(n ~ m) + (7)(: = 7) + ... + (7)(n ~ m) = (;).

19. For every integer n ~ 2, show the following:

(i) 1 - G) + (;) + ... + (-It(:) = 0;

(ii) G) + 2(;) + 3(;) + ... = n2"-I;

(iii) 2.1.(;) + 3.2-(;) + 4.3.(:) + ... = n(n _1)2"-2.

20. Using Pascal's triangle, show that for every a and positive integers r, n:

(i) i (a - k) = (a + 1) _ (a - n); k=O r r + 1 r + 1

(ii) i (_If,(a)=(_I)"(a-l). k=O k n

21. Using the result of Exercise 18, show that

22. Using the result of Exercise 21, prove that

" (2n)! (2n)2 Jo (v!)2[(n - V)!]2 = n .

23. Using the binomial theorem, show by induction that

i (n)( _1)k-1 = i !. k=1 k k k=1 k

24. As in Exercise 23, show that for positive integers rand n

i (k + r - 1) = (n + r). k=O r - 1 r

Show that this is a special case of Problem 20(ii).

25. In the expansion of (1 + x)" where x> 0 and n is a positive integer, let ak be the term containing Xk. Find the values of k for which ak becomes maximum. What is the maximum term of (1 + e)IOO?

4. Properties of Probability

26. Given peA) = 1/3, PCB) = 1/4, P(AB) = 1/6, find the following probabilities:

peA'), peA' u B), peA u B'), P(AB'), peA' u B').


27. Given P(A) = 3/4 and P(B) = 3/8, show that: (a) P(A u B) ;;:: 3/4, (b). 1/8 ~ P(AB) ~ 3/8. Give inequalities analogous to (a) and (b) for

P(A) = 1/3 and P(B) = 1/4.

28. For any two events A and B show that:

P(AB) - P(A)P(B) = P(A')P(B) - P(A' B) = P(A)P(B') - P(AB').

29. By induction show that for arbitrary events Ai' Az,"" Am the following inequality (due to Bonferoni) holds:

n n

P(A 1 A z··· An) ;;:: L P(A;) - (n - 1) = 1 - L P(Af). ;=1 ;=1

30. For any three events A, B, and C, show that (cf. (1.1»

PEA u B u C] = P(A) + P(B) + P(C) - P(AB) - P(AC)

- P(BC) + P(ABC).

Application: The percentages of students who passed courses A, B, and C are as follows: A: 50%, B: 40%, C: 30%, A and B: 35%, Band C: 20%, and 15% passed all three courses. What is the percentage of students who succeeded in at least one of the three courses?

31. A box contains balls numbered 1,2, ... , n. A ball is drawn at random: (a) What is the probability that its label number is divisible by 3 or 4? (b) Examine the case in (a) as n -+ 00.

32. In terms of P(A), P(B), P(C), P(AB), P(AC), P(BC), and P(ABC) express, for k = 0, 1,2,3, the probabilities that:

(i) exactly k of the events A, B, and C, occur; (ii) at least k of the events A, B, C occur.

33. By induction prove Poincare's theorem, i.e., formula (1.1).

34. * If d(A, B) = P(A 6. B), show that d has all the properties of a distance function; the symmetric difference A 6. B between two sets A and B is defined by A 6.B = AB' u A'B.

5. Classical Probabilities. Equally Likely Cases

35. Three winning tickets are drawn from an urn of 100 tickets. What is the probability of winning for a person who buys: (a) 4 tickets? (b) only one ticket?

36. A bakery makes 80 loaves of bread daily. Ten of them are underweight. An inspector weighs 5 loaves at random. What is the probability that an underweight loaf will be discovered?

Exercises 11

37. Find the probability that among seven persons: (a) no two were born on the same day of the week (Sunday, Monday, etc.); (b) at least two were born on the same day; (c) two were born on a Sunday and two on a Tuesday.

38. A group of 2N boys and 2N girls is randomly divided into two equal groups. What is the probability that each group has the same number of boys and girls?

39. The coefficients a, b, c of the quadradic equation ax2 + bx + c = 0 are determined by throwing a die three times. Find the probabilities that: (a) the roots are real; (b) the roots are complex.

40. In the game of poker a "hand of cards" means 5 cards randomly selected (without replacement) from a deck of 52 cards. What is the probability that a hand of cards:

(a) consists of an ace, a queen, a jack, a king, and a ten of the same suit? (b) contains 4 cards of the same denomination (aces, etc.)? (c) consists of cards with consecutive values, except aces, jacks, kings, and

queens?

41. In bridge, "a hand of cards" consists of 13 cards drawn at random (without replacement) from a deck of 52 cards (the 52 cards are equally distributed to four players). Find the probabilities that "a hand of cards" will contain:

(i) Vi clubs, V2 spades, and V3 diamonds; (ii) V aces (v = 0, 1, ... ,4);

(iii) Vi aces and V2 kings. In a bridge game find the probabilities that:

(a) each player has an ace; (b) some player has all the aces; (c) some player has Vi aces and his partner V2 aces (Vi + V2 :s; 4).

42. Six girls are to enter a dance with 10 boys to form a ring so that every girl is between two boys:

(a) What is the probability that some specified boy remains between 2 boys?

(b) A spectator notices that a certain girl enters next to a certain boy. Is it random?

43. Five letters are selected at random one after another from the 26 letters of the English alphabet; (a) with replacement, (b) without replacement. Find; for each of the cases (a) and (b), the probabilities that the word formed: (i) contains an "a", (ii) consits of vowels, (iii) is the word "woman".

44. * Prove that the number of ways in which r indistinguishable balls can be distributed in n cells (or, equivalently, the number of different solutions of


the equation Xl + X2 + '" + Xn = r where Xi 2 0 (i = 1, 2, ... , n) are integers),

(n + r - 1) (n + r - 1) ~M~ 1 = .

n - r

45. * Consider r indistinguishable balls randomly distributed in n cells (Bose-Einstein statistics). What is the probability that exactly m cells remain empty?

46. From an ordinary deck of 52 cards, cards are drawn successively until an ace appears. What is the probability that the first ace will appear: (a) at the nth draw? (b) after the nth card?

47.* Birthday problem. In a classroom there are v students. (a) What is the probability that at least two students have the same

birthday? (b) What is the minimum value of v which secures probability 1/2 that at

least two have a common birthday?

48. * N letters are placed at random in N envelopes. Show that the prob

ability that each letter will be placed in a wrong envelope is t (_I)k (k\)' k-2 .

49.* An urn contains nr balls numbered 1,2, ... , n in such a way that r balls bear the same number i for each i = 1, 2, ... , n. N balls are drawn at random without replacement. Find the probability that, (a) exactly m of the numbers will appear in the sample, (b) each of the n numbers will appear at least once.

50. * (Continuation). Balls are drawn until each of the numbers 1, 2, ... , n appears at least once. What is the probability that m balls will be needed?

51. N men run out of a men's club after a fire and each takes a coat and a hat. Prove that:

(a) the probability that no one will take his own coat and hat is

N k(N - k)! k~l (-1) N! kl;

(b) the probability that each man takes a wrong coat and a wrong hat is

[ t (-I)k~J2. k-2 k.

52.* Suppose every packet of the detergent TIDE contains a coupon bearing one of the letters ofthe word TIDE. A customer who has all the letters of the word gets a free packet. All the letters have the same possibility of appearing in a packet. Find the probability that a housewife who buys 8 packets will get: (i) one free packet, (ii) two free packets.

53.* An urn contains Nl white and N2 black balls. When two balls are randomly drawn the probability that both be white is 1/2.

Exercises 13

(a) What is the minimum value of Nl? (b) What is the minimum value of Nl when N2 is an even number? (c) What is the minimum value of the total number N = Nl + N2 of balls

in the urn?

6. Independent Events. Conditional Probability

54. If the events A, B are independent, show that the pairs (A, B'), (A', B), (A', B') also consist of independent events.

55. If the events A, B, and C are mutually independent, then the pairs (A, BC), (B, AC), (C, AB) also consist of independent events.

56. Suppose that for the independent events A, B, and C we have peA) = a, peA u B u C) = 1 - b, P(ABC) = 1 - c, and P(A'B'C) = x. Prove that the probability x satisfies the equation

ax2 + [ab - (1 - a)(a - c - I)Jx + b(1 - a)(1 - c) = 0.

Hence conclude that

Moreover, show that

(1 - a)2 + ab c>-----

1 - a

x (1 - c)(x + b) PCB) = ,

ax P(C) = --b'

x+

57. Let the events A l , A 2 , ... , An be independent and peA;) = p (i = 1, 2, ... , n). What is the probability that:

(a) at least one of the events will occur? (b) at least m of the events will occur? (c) exactly m of the events will occur?

58. A die is thrown as long as necessary for an ace or a 6 to turn up. Given that no ace turned up at the first two throws, what is the probability that at least three throws will be necessary?

59. A parent particle can be divided into 0, 1, or 2 particles with probabilities 1/4, 1/2, 1/4, respectively. It disappears after splitting. Beginning with one particle, the progenitor, let us denote by Xi the number of particles in the i generation. Find, (a) P(X2 > 0), (b) the probability that Xl = 2 given that X 2 = 1.

60. An urn contains n balls numbered 1, 2, ... , n. We select at random r balls, (a) with replacement, (b) without replacement. What is the probability that the largest selected number is m?

61. Two athletic teams A and B playa series of independent games until one of them wins 4 games. The probability of each team winning in each game


equals 1/2. Find the probability that the series will end, (a) in at most 6 games, (b) in 6 games given that team A won the first two games.

62. It is suspected that a patient has one of the diseases AI, A 2 , A 3 • Suppose that the population percentages suffering from these illnesses are in the ratio 2: 1 : 1. The patient is given a test which turns out to be positive in 25% of the cases of AI, 50% of A 2 , and 90% of A 3 • Given that out of three tests taken by the patient two were positive, find the probability for each of the three illnesses.

63. The population of Nicosia (Cyprus) is 75% Greek and 25% Turkish. 20% of the Greeks and 10% of the Turks speak English. A visitor to the town meets someone who speaks English. What is the probability that he is a Greek? Interpret your answer in terms of the population of the town.

64. Two absent-minded room mates, mathematicians, forget their umbrellas in some way or another. A always takes his umbrella when he goes out, while B forgets to take his umbrella with probability 1/2. Each of them foregets his umbrella at a shop with probability 1/4. After visiting three shops they return home. Find the probability that:

(a) they have both umbrellas; (b) they have only one umbrella; (c) B has lost his umbrella given that there is only one umbrella after their

return.

65. Consider families of n children and let A be the event that a family has children of both sexes, and let B be the event that there is at most one girl in the family. Show that the only value of n for which the events A and Bare independent is n = 3, assuming that each child has probability 1/2 of being a boy.

66. At the college entrance examination each candidate is admitted or rejected according to whether he has passed or failed the test. Of the candidates who are really capable, 80% pass the test; and of the incapable, 25% pass the test. Given that 40% of the candidates are really capable, find the proportion of capable college students.

67. Huyghens problem. A and B throw alternately a pair of dice in that order. A wins ifhe scores 6 points before B gets 7 points, in which case B wins. If A starts the game what is his probability of winning?

68. Three players PI' P2 , and P3 throw a die in that order and, by the rules of the game, the first one to obtain an ace will be the winner. Find their probabilities of winning.

69. N players AI' A 2 , ••. , AN throw a biased coin whose probability of heads equals p. Al starts (the game), A2 plays second, etc. The first one to throw heads wins. Find the probability that Ak (k = 1,2, ... , N) will be the winner.

Exercises 15

70. Urn A contains WI white balls and bi black balls. Urn B contains Wz white balls and bz black balls. A ball is drawn from A and is placed into B, and then a ball is transferred from B to A. Finally, a ball is selected from A. What is the probability that the ball will be white?

71. Ten percent of a certain population suffer from a serious disease. A person suspected of the disease is given two independent tests. Each test makes a correct diagnosis 90% of the time. Find the probability that the person really has the illness:

(a) given that both tests are positive; (b) given that only one test is positive.

72. A player randomly chooses one of the coins A, B. Coin A has probability of heads 3/4, and coin B has probability of heads 1/4. He tosses the coin twice.

(a) Find the probability that he obtains: (i) two heads, (ii) heads only once. (b) Instead of the above procedure, suppose the player can choose an

unbiased (symmetric) coin, which he tosses twice. Which procedure should he follow in order to maximize the probability of at least one head?

73. Urn A contains 5 black balls and 6 white balls, and urn B contains 8 black balls and 4 white balls. Two balls are transferred from B to A and then a ball is drawn from A.

(a) What is the probability that this ball is white? (b) Given that the ball drawn is white, what is the probability that at least

one white ball was transferred to A?

74. The Pap test makes a correct diagnosis with probability 95%. Given that the test is positive for a lady, what is the probability that she really has the disease? How do you interpret this? Assume that one in every 2,000 women, on average, has the disease.

75. A secretary goes to work following one of three routes A, B, C. Her choice of route is independent of the weather. If it rains, the probabilities of arriving late, following A, B, C, are 0.06, 0.15, 0.12, respectively. The corresponding probabilities, if it does not rain, are 0.05, 0.10, 0.15.

(a) Given that on a sunny day she arrives late, what is the probability that she took route C? Assume that, on average, one in every four days is rainy.

(b) Given that on a day she arrives late, what is the probability that it is a rainy day?

76. At an art exhibition there are 12 paintings of which 10 are original. A visitor selects a painting at random and before he decides to buy, he asks the opinion of an expert about the authenticity of the painting. The expert is right in 9 out of 10 cases on average.

(a) Given that the expert decides that the painting is authentic, what is the probability that this is really the case?

(b) If the expert decides that the painting is a copy, then the visitor returns


it and chooses another one; what is the probablity that his second choice is an original?

77. What is the conditional probability that a hand at poker (see Exercise 40) consists of spades, given that it consists of black cards?

Note: Of the 52 cards of an ordinary deck, the 13 spades and 13 clubs are black whereas the 13 hearts and 13 diamonds are red.

78. The probability Pk that a family has k children is given by Po = Pl = a, Pk = (l - 2a)2-(k-l) (k ~ 2). It is known that a family has two boys. What is the probability that:

(a) the family has only two children? (b) the family has two girls as well?

79. An amplifier may burn one or both of its fuses if one or both of its tubes are defective. Let the events:

Ai: only tube i is defective (i = 1, 2); A3: both tubes are defective; B/ fuse j burns out (j = 1, 2); B3: both fuses burn out. The conditional probabilities of burning the fuses (given the state of the

tubes, i.e., the P[BjIA;]), appear in the following table:

i/j Bl B2 B3

Al 0.7 0.2 0.1 A2 0.3 0.6 0.1 A3 0.2 0.2 0.6

Find the probabilities that: (a) both tubes are defective given that both fuses were burnt out; (b) only tube 2 is defective given that both fuses were burnt out; (c) only tube 2 is defective given that at least one fuse was burnt out. The events A1 , A 2, A3 have probabilities 0.3, 0.2, 0.1, respectively.

CHAPTER 2

Distributions. Random Variables

Elements of Theory

1. A function (strictly measurable) X(w) defined on the sample space n = {w} is called a random or stochastic variable.

A random variable is called discrete ifit takes on (with positive probability) at most a countable set of values, that is, if there exists a sequence Xl' X2' ...

with 00

P[X = x;] = Pi > 0 and I. Pi = 1. (2.1) i=l

The sequence {Pn, n = 1, ... } defines the so-called probability (mass) function or frequency function of X.

A random variable X is called continuous (strictly, absolutely continuous) if for every real c there exists a function f (almost everywhere continuous) such that

P[X :::; c] = foo f(x) dx. (2.2)

The function f is called the probability density function or simply the density of X. It follows that:

2. The distribution function or cumulative distribution function of a random variable X is defined by

F(x) == P[X :::; x]

Thus for a continuous random variable,

F(x) = f~oo f(t) dt

for every real x.

for every x, (2.3)

18 2. Distributions. Random Variables

while for a discrete random variable

F(x) = I P[X = x;J.

From (2.3), it follows that if F is differentiable at x then

dF(x) _ f( dx - x).

Moreover, the probability differential f(x) dx can be interpreted as

f(x) dx = P[x < X < x + dx].

For a continuous random variable X, we have

P[a < X < fJ] = P[a ::; X ::; fJ] = P[a < X ::; fJ] = P[a ::; X < fJ],

i.e., no point carries positive probability and every set A of the real line of zero length carries zero probability.

3. The main discrete distributions (i) The hypergeometric (cf. (1.7)) with parameters N, n, and p;

q = 1 - p. (2.4)

(ii) The binomial with parameters n (number of independent Bernoulli trials) and p (probability of success on each trial):

k = 0, 1, ... , n, q = 1 - p. (2.5)

(iii) The Poisson distribution with parameter 2:

2k P[X = k] == p(kI2) = e-.l. k! ' k = 0,1,2, .... (2.6)

(iv) The geometric with parameter p (the number of Bernoulli trials required before or including the first success):

or

P[X = k] = pqk,

P[X = k] = pqk-l,

k = 0, 1,2, ... ,

k = 1,2,3, .... (2.7)

(v) The Pascal distribution with parameters p and r (number of failures before the rth success in a sequence of Bernoulli trials):

(r + k - 1) (- r) P[X=k]= r-l p'qk= k p'(-q)k, k = 0, 1, .... (2.8)

If r is simply positive (not necessarily an integer) then (2.8) is also called a negative binomial distribution.

Elements of Theory

4. The main continuous distributions (i) Uniform or rectangular in the interval (oc, P) with density

1 f(x) = -p -,

-oc oc < x < p.

(ii) Exponential (or negative exponential) with density

f(x) = ()e- OX , x> O.

19

(iii) Normal or Gaussian or Gauss-Laplace with parameters Ji (mean) and (12 (variance) denoted by N(Ji, (12) with density

1 [1 (x - Ji)2 ] f(x) = .-- exp -- 2 '

(1fo 2 (1 -00 < x < 00.

(iv) Laplace distribution (bilateral or double exponential) with density

f(x) = ie-lxi, - 00 < x < 00. (2.9)

(v) p distribution with parameters p > 0 and q > 0 with density

P(xl ) = r(p + q) P-l(l _ X)q-l p, q r(p)r(q) x ,

_ 1 p-1(1 )q-l ---x -x. B(p, q)

where the r function is defined for every p > 0 by

r(p) = I'" xp-1e-x dx,

0< x < 1,

and the B function is defined for every p > 0, q > 0 by

B(p, q) = II x p- l (1 - X)q-l dx.

(2.10)

(vi) The gamma (r) distribution with parameters it > 0 (scale parameter) and s > 0 (shape parameter) with density

its y(xlit s) = __ e-.<xx s - l

'r(s) , x> O. (2.11 )

When it = 1/2 and 2s = v = integer, the gamma (r) distribution is called a X2

(chi-square) distribution with v degrees of freedom, denoted by X~. (vii) Cauchy distribution with density

-00 < x < 00. (2.12)

(viii) Student's or t distribution with v degrees of freedom. The distribution of the ratio Z/~ where Z is N(O, 1) independent of the X~. It has

density

r(~) 1

fv(t) = (v)( t 2 )(1+V)/2' -00 < t < 00.

for - 1 +-2 v

(ix) F or Snedecor's distribution with m and n degrees of freedom. The distribution of the ratio nx;,/mx~ where the X;' and X~ are independent. It has density

(m+ n) r -2~ (m)m/2 x(m/2)-1

f(x) = (m) (n) ~ ( m )(m+n)/2' r - r - 1 +-x 2 2 n

x> O.

Exercises

1. Discrete Distributions

80. An urn contains 7 white balls numbered 1,2, ... , 7 and 3 black balls numbered 8, 9, 10. Five balls are randomly selected, (a) with replacement, (b) without replacement.

For each of the cases (a) and (b) give the distribution: (I) of the number of white balls in the sample;

(II) of the minimum number in the sample; (III) of the maximum number in the sample; (IV) of the minimum number of balls needed for selecting a white ball.

81. (Continuation). As in Problem 80 with the black balls numbered 1,2, and 3.

82. A machine normally makes items of which 4% are defective. Every hour the producer draws a sample of size 10 for inspection. If the sample contains no defective items he does not stop the machine. What is the probability that the machine will not be stopped when it has started producing items of which 10% are defective.

83. One per thousand of a population is subject to certain kinds of accident each year. Given that an insurance company has insured 5,000 persons from the population, find the probability that at most 2 persons will incur this accident.

84. A certain airline company, having observed that 5% of the persons making reservations on a flight do not show up for the flight, sells 100 seats on a plane that has 95 seats. What is the probability that there will be a seat available for every person who shows up for the flight?

85. Workers in a factory incur accidents at the rate of two accidents per

Exercises 21

week. Calculate the probability that there will be at most two accidents, (i) during 1 week, (ii) during 2 weeks (iii) in each of 2 weeks.

86. Suppose that the suicide rate in a certain state is four suicides per one million inhabitants per month. Find the probability that in a certain town of population 500,000 there will be at most four suicides in a month. Would you find it surprising that during 1 year there were at least 2 months in which more than four suicides occurred?

87. A coin is tossed 30 times. For n = 1,2, ... ,20, what is the conditional probability that exactly 10 + n heads appear given that the first 10 tosses resulted in heads. Show that the conditional probability that 10 + n tosses will result in heads, given that heads appeared at least ten times, equals

C03~ n);n 20 ( 30 ) 1 . k~O 10 + k 2k

88. How many children should a family have so that with probability 0.95 it has at least a boy and at least a girl.

89. Show that the Poisson probabilities p(kIA) satisfy the recurrence relation

A p(kIA) = 'kP(k - l1A),

and hence determine the values of k for which the terms p(kIA) reach their maximum (for given A).

2. Continuous Distributions

90. Verify that each of the following functions f is a probability density function and sketch its graph.

(a) f(x) = 1 - 11 - xl for 0<x<2

(b) 1 13

f(x) = -; 132 + (x - af for -00 < x < 00,

(c) f(x) = ~e-(lx-IlIl/t, 20'

for -00 < x < 00,

(d) f(x) = txe-x/2 for 0< x < 00.

91. The amount of bread (in hundreds of kilos) that a bakery sells in a day is a random variable with density

{ex

f(x) = ~(6 - x)

for 0:::;; x < 3,

for 3:::;; x < 6,

otherwise.


(i) Find the value of c which makes f a probability density function. (ii) What is the probability that the number of kilos of bread that will be

sold in a day is, (a) more than 300 kilos? (b) between 150 and 450 kilos? (iii) Denote by A and B the events in (a) and (b), respectively. Are A and B

independent events?

92. Suppose that the duration in minutes of long-distance telephone conversations follows an exponential density function;

f(x) = te- x /5 for x> O.

Find the probability that the duration of a conversation: (a) will exceed 5 minutes; (b) will be between 5 and 6 minutes; (c) will be less than 3 minutes; (d) will be less than 6 minutes given that it was greater than 3 minutes.

93. A number is randomly chosen from the interval (0, 1). What is the probability that:

(a) its first decimal digit will be a 1; (b) its second decimal digit will be a 5; (c) the first decimal digit of its square root will be a 3?

94. The height of men is normally distributed with mean J1 = 167 cm and standard deviation (J = 3 cm.

(I) What is the percentage of the population of men that have height, (a) greater than 167 cm, (b) greater than 170 cm, (c) between 161 cm and 173 cm?

(II) In a random sample of four men what is the probability that: (i) all will have height greater than 170 cm; (ii) two will have height smaller than the mean (and two bigger than the

mean)?

95. A machine produces bolts the length of which (in centimeters) obeys a normal probability law with mean 5 and standard deviation (J = 0.2. A bolt is called defective if its length falls outside the interval (4.8, 5.2).

(a) What is the proportion of defective bolts that this machine produces? (b) What is the probability that among ten bolts none will be defective?

96. Consider a shop at which customers arrive at random at a rate of twenty persons per hour. What is the probability that the time intervals between succesive arrivals will be:

(a) shorter than 3 minutes; (b) longer than 4 minutes. (c) Suppose that 10% of the customers buy a certain object. Find the

distribution of the number of customers who buy an object in an hour.

97. If X is a continuous random variable with cumulative distribution function F and density function f, show that the random variable Y = X 2 is also continuous and express its cumulative distribution function and density in terms of F and f.

Exercises

98. (Continuation). Find the density of Y = X 2 when X has: (a) the normal distribution N(Il, (j2);

(b) the Laplace distribution (see (2.9)); (c) the Cauchy distribution (see (2.12)).

99. As in Exercises 97 and 98 with Y = IXI.

23

100. The lognormal distribution. If the log X is normally distributed then X is said to have a lognormal distribution. Find its density.

CHAPTER 3

Expectation. Variance. Moments

Elements of Theory

1. The expected value or expectation or mean value or simply the mean of a random variable X, denoted by E(X), is defined by

{~XiP[X = xJ E(X) = t: xf(x) dx

for a discrete X,

for a continuous X,

provided the series or the integral converge absolutely, in which case we say that E(X) exists and write E(X) < 00.

The linearity property of the expectation operation: If E(X) < 00, E(Y) < 00,

then for any constants a and b we have

E[aX + bY] = aE(X) + bE(Y).

The mean value of a function of a random variable: Let

Y = g(X)

be a (measurable) function of X with frequency function fx. Then

{LYJY(YJ,

E(Y) = E[g(X)] = Ii 00

yfy(y) dy, -00

Elements of Theory

= {~g(X;)f(X;) roooo g(x)f(x) dx

25

for X discrete,

for X continuous, (3.1)

where fy denotes the frequency function of Y; (3.1) permits the computation of E(Y) by means of fx without finding first fy.

2. Moments. The moment of order r about the origin of a random variable X is defined by

J1~ == E(X'), r = 1,2, ....

The first moment J1'1 coincides with the mean E(X) == J1. The central moment of order r is defined by

J1, == E(X - J1)' r = 2, 3, ....

The variance of X is the second (order) central moment, i.e.,

E(X - J1)2 = E(X2) - J12. (3.2)

We write V or Var for variance. The positive square root ofVar(X) is called the standard deviation of X, usually denoted by (5.

The moments of an integer-valued positive random variable, such as the binomial, the Poisson, etc., are more conveniently computed by means of the so-called factorial moments.

The factorial moment Jr, of the random variable X is defined by

Jr, = E[X(X - 1) .. . (X - r + 1)] = E[(X),]. Thus

(3.3)

3. Quantiles and percentiles. For every p (0 < p < 1), the p quantile point or the lOOp percentile point, X p say, of a random variable X (or its distribution) with distribution function F, is defined as a solution of

F(Xp-)::; P ::; F(Xp).

For a continuous and increasing F, Xp is uniquely defined by

F(Xp) = p.

For p = 0.5 we have the median Xo.s. XO. 2S is called the first quartile and XO. 7S the third quartile. The difference XO. 7S - XO. 2S is called interquartile range and can be used as a measure of scatter of the distribution.

4. Probability or factorial moment generating function. This is useful mainly for integer-valued positive random variables and is defined by

00 00

P(t) == E(t X ) = L tkP[X = k] = L Pk tk ; k~O k~O

it exists at least for I t I ::; 1.

26 , 3. Expectation. Variance. Moments

The rth factorial moment ITr is given by

_ [drp(t)] _ <r) ITr - ~ 1=1 - P(1)'

so that, by (3.3),

E(X) = P'(1), Var(X) = P"(I) + P'(I) - [P'(I)Y

The probabilities Pk = P[X = k] are given by

_ ~ [dkP(t)] Pk - k' dt k .

• 1=0

5. Moment generating function of a random variable X defined by

(discrete X),

(continuous X).

If M(t) exists, i.e., there exists a b > 0 such that E(e1x ) < 00 for alII t I < b, then all the moments E(xr) (r = 1, 2, ... ) exist and can be obtained from

E(xr) = [drM(t)] = M<r)(o). dt r 1=0

Exercises

1. Theoretical Exercises

101.* Show that the integral S~~ Ix - mlf(x) dx becomes minimum when m is the median of the distribution with density f.

102. Let f(x) denote the density function of the random variable X. Suppose that X has a symmetric distribution about a, that is, f(x + a) = f(a - x) for every x. Show that the mean E(X) equals a, provided it exists.

103.* Cauchy-Schwarz inequality. Show that E2(XY):::;; E(X2)E(y2) provided the second moments exist.

104. Show that IE(X)I :::;; E(IXI).

105. For every set A of real numbers, we define the indicator function by

{I if x E A,

IA(x) = 0 if x ¢ A.

Exercises

Show that

P(A) = E[IA(X)] = L dFx(x).

106. If E(X) = E(X2) = 0, show that P(X = 0) = 1. Hint: Use Chebyshev's inequality. See (8.5).

107. Show that the mean J.l of a random variable X has the property

min E(X - C)2 = E(X - J.l)2 = V(X).

27

108.* The mean value geometrically. Show that for a continuous random variable X with density function f and cumulative distribution function F

J.l = E(X) = foo [1 - F(x)] dx - J:oo F(x) dx.

Consequently, in the graph below J.l = area(A) - area(B).

y

~~wwwwwwww~ _____________________ x o

109. If the nth moment ofthe random variable X with distribution function F exists, show that

E(X - C)k = k 1'" (x - C)k-l [1 - F(x)] dx

- k foo (x - C)k-l F(x) dx, 1 $; k $; n.

110. Show that the first n moments determine the first n central moments and, conversely, that the first n central moments with the mean determine the first n moments.

111. If X is bounded, i.e., there is a constant k < oosuchthatP[IXI $; k] =

1, then X has moments of every order.

28 3. Expectation. Variance. Moments

112. * If the n-order moment J1~ exists, then show that there exist all J11c (k = 1,2, .... , n - 1).

113. Show that a necessary condition for the mean of a random variable with distribution function F to exist is that

lim xF(x) = lim x[1 - F(x)] = O. x--+-oo x-+oo

2. Mean and Variance

114. * In a lottery, n numbers are selected from the N numbers 1, 2, ... , N. Find the variance of the sum Sn of the selected numbers.

115. An urn contains Nl white balls and N2 black balls; n balls are drawn at random, (a) with replacement, (b) without replacement. What is the expected number of white balls in the sample?

116. A student takes a multiple-choice test consisting of two problems. The first one has 3 possible answers and the second one has 5. The student chooses, at random, one answer as the right one from each of the two problems. Find:

(a) the expected number, E(X), of the right answers X of the student; (b) the Var(X). Generalize.

117. In a lottery that sells 3,000 tickets the first lot wins $1,000, the second $500, and five other lots that come next win $100 each. What is the expected gain of a man who pays 1 dollar to buy a ticket?

118. A die is thrown until the result "ace or even number" appears three times. Find the expected number of throws:

(a) in one performance of the experiment; (b) in ten repetitions.

119. A pays 1 dollar for each participation in the following game: three dice are thrown; if one ace appears he gets 1 dollar, if two aces appear he gets 2 dollars and if three aces appear he gets 8 dollars; otherwise he gets nothing. Is the game fair, i.e., is the expected gain of the player zero? If not, how much should the player receive when three aces appear to make the game fair?

120. A player has 15 dollars. If heads appear on the first toss he receives 1 dollar and he withdraws from the game. If tails appear he bets 2 dollars and if he wins the second toss he withdraws; otherwise he continues playing on a bet of 4 dollars. If he loses, he bets the remaining amount of 8 dollars. What is the expected gain of the player?

121. Of the parcels mailed abroad 10% never reach their destination. Two books may be sent separately or in a single parcel. Each book is worth $2. The postage for each book sent separately is 10¢ and for both books in a single

Exercises 29

parcel 15¢. For each of the two ways of mailing, find: (a) the probability that both books reach their destination; (b) the probability that at least one book reaches its destination; (c) the expected net (after postage) value of the commodities reaching their

destinatination. For each of the three criteria (a), (b), and (c), which way of mailing is preferable?

122. (a) Two indentical coins are thrown once in such a way that if one shows heads so does the other (dependent). Let P[heads] = P and X be the total number of heads that appear. Find the mean and the variance of the random variable X.

(b) Two different coins with probabilities of heads PI and P2 are thrown independently. Find the mean and the variance of the random variable X as defined in (a).

123. Mixed distribution. Let X be a random variable with distribution function

( {I - 0.8e- X

F x) = o for x;:::: 0,

for x < O.

Plot the graph of F(x) and then evaluate E(X). Give an example of a random variable X that has the above distribution.

124. If the random variable X is N(/1, 1), show that the random variable Y = [1 - <l>(X)]lcp(X), where <l> and cp denote the distribution function and the density of N(O, 1), respectively, has mean value 1/11.

125. The truncated Poisson distribution with the zero class missing has probability function

),k

P(X = k) = -(e-:-;'---I-)k!' k = 1,2, ....

Find E(X) and Var(X).

CHAPTER 4

General Problems

126. Three players A, B, and C playa game as follows: At the first stage A and B play against each other while C stays out of the game. The winner plays against C at the second stage. Then the winner of the second game plays against the last loser, and so on. The winner is declared the one who wins two games in succession. Find the elements (simple events) of the sample space n of all possible outcomes of the game. If the probability of each player winning a game is 1/2, determine the probabilities of the corresponding possible outcomes and show that their sum equals one.

127. A symmetrical coin is tossed until the same result appears twice in succession. Describe the sample space in terms of the results H ("heads") and T ("tails"). Find the probabilities of the events:

(a) the experiment terminates earlier or at the seventh trial; (b) an even number of trials is required.

What is the expected number of trials?

128. A pair of dice is tossed six times. What is the probability that all the faces will appear twice?

129. A deck of 52 cards is divided among four players so that each of them gets 13 cards (game of bridge). Find the probability that at least one player will have a complete suit (spades, hearts, diamonds, clubs).

130. Two coins CI and C2 have a probability of falling heads PI and P2' respectively. You win a bet if in three tosses you get at least two heads in succession. You toss the coins alternately starting with either coin. If PI > P2' what coin would you select to start the game?

131. Ten pairs of shoes are in a closet. Four shoes are selected at random. Find the probability that there will be at least one pair among the four shoes selected.

4. General Problems 31

132. Let N cells be numbered 1,2, ... , N. We randomly throw balls into them. The process is continued until a ball falls in the cell bearing the number 1. What is the probability that:

(a) n throws will be necessary? (b) more than n throws will be necessary?

133. (Continuation). The process terminates when anyone of the cells receives two balls.

(a) What is the probability that more than n throws will be necessary? (b) What is the expected number of throws?

134. From a usual deck of cards we draw one card after another. What is the probability that:

(a) the nth card will be the first ace? (b) the first ace will appear among the first n cards?

135. The 52 cards of an ordinary deck of cards are placed successively one after the other and from left to right. Find the probability that the thirteenth spade will appear before the thirteenth diamond.

136. Three balls are drawn at random one after another and without replacement from n balls numbered 1,2, ... , n. Find the probability that the first will bear a number smaller than that of the second ball.

137. An urn contains m kinds of objects all in the same proportion. Objects are drawn with replacement, one after another, until each kind appears at least once. What is the probability Pv that v objects will be required?

138. A newspaper dealer gets n papers every day for sale. The number X of papers sold is a random variable following the Poisson distribution with parameter A.. For each paper sold he earns 1 cent, for each unsold paper he loses b cents. Let Y denote the net gain of the dealer. Find:

(a) E(Y); (b) How many newspapers should he get every day in order to maximize

his profit?

139. Three numbers are selected at random one after another and without replacement from n numbers 1,2, ... , n. What is the probability that the first number drawn will be the smallest and the second number the largest?

140. The events A, B, and C are independent with P(A) = 0.2, P(B) = 0.3, and P( C) = 0.1. Find the probability that at least two will occur among the three events.

141. We throw four coins simultaneously and we repeat it once more. What is the probability that the second throw will result in the same configuration as the first throw when the coins are, (a) distinguishable, (b) indistinguishable.

142. An urn contains w white balls and b black balls. We draw balls successively and without replacement one after another until a white ball appears for the first time. Let X be the required number of draws. Find:

32

(a) the distribution of X; (b) E(X). (c) Using the result of (a) show the identity

4. General Problems

n-w (n-w)(n-w-l) (n-w) ... 2.l n 1 +--+ + ... +-------

n-l (n-l)(n-2) (n-l) ... (w+l)w w

where n = w + b.

143. (Continuation). Let w = b = N and draw balls successively and without replacement one after another until all the 2N balls are drawn. What is the probability that in a stage of the experiment the same number of white and black balls has been drawn?

144. * Let B1 , ••• , Bvo ... , be a partition of the sample space Q and P( C) > O. Show the following theorem of total probability:

00

P[AIC] = L P[BjIC]P[AIBjC]. j~l

145. Each of the three Misses T. C-Rena, Nike, and Galatea-wants to accompany me on a trip. Since I cannot take all of them, I play the following game: I say to them, "I will think of one of the numbers 1, 2, 3. Rena, you guess first. If you guess right you will come with me; otherwise Nike will guess next. If she guesses right she will come with me. Otherwise Galatea will accompany me". Galatea complains that the game is unfair. Is she right? Suppose everyone of them writes down on a piece of paper the number which she guesses I thought of and I follow the above sequence checking Rena's number first, etc. This is repeated until one is chosen to come with me. Galatea and Nike protest this time. Are they justified?

146. A die is thrown v times. Find the probability that each of k (1 ~ k ~ 6) given faces of the die appears at least once.

147. The folded normal distribution has density

I(x) = c~l_e-x2/2, ~

x> O.

Determine the constant c and then evaluate the mean of the distribution.

148. Suppose the lifetime of an electric lamp of a certain type obeys a normal law with mean /1 = 180 hours and standard deviation (J' = 20. In a random sample of four lamps:

(a) What is the probability that all four lamps have a lifetime greater than 200 hours?

(b) The random sample of four lamps is placed in an urn and then we randomly draw one lamp from it; what is the probability that the lamp will have a lifetime greater than 200 hours?

Generalize the conclusion.

149.* The game of "craps" is played as follows. The gambler throws two dice. If at the first throw he gets 7 or 11 he wins, and if he gets 2, 3, or 12 he loses. For each of the other sums the game is continued in two ways:

(a) the gambler continues throwing the two dice until he wins with a 7 or he loses with the results of the outcome of the first throw;

(b) the gambler continues until he loses with 7 or wins with the result of the first throw.

What is the probability of the gambler winning in cases (a) and (b)?

150. Two gamblers A and B agree to playas follows. They throw two dice and if the sum S of the outcomes is < 10 B receives S dollars from A, otherwise B pays A x dollars. Determine x so that the game is fair.

151. An urn contains n white balls and n red balls. We draw two balls at first, and then another two, and so on, until all the balls are drawn. Find the probability that each of the selected pairs consists of a white baH and a red ball.

152. Each of two urns A and B contains n balls numbered 1 to n. We draw one ball from each of the urns. Find the probability that the ball drawn from A bears a number smaller than that drawn from B.

153. The rumor mongers (see Feller, 1957, p. 55). In a town of N + 1 inhabitants, a person tells a rumor to a second person, who in turn repeats it to a third person, and so on. At each step the recipient of the rumor is chosen at random from the N inhabitants available.

(i) Find the probability that the rumor will be told n times without, (a) returning to the originator, (b) being repeated to any person. Do the same problem when at each step the rumor is told to k persons.

(ii) In a large town where the rumor mongers constitute 100p% of the population, what is the probability that the rumor does not return to the originator?

154. * Pascal's problem. In the game of tossing a fair coin, the first one to obtain n successes (heads or tails) wins. Show that the game is fair (i.e., each gambler has a probability of winning equal to 1/2). Suppose that the game was interrupted when the first gambler had won k tosses and the second gambler had won m tosses (0 ::s; k, m < n). Calculate for each gambler the probability of winning if the game is continued. Hence deduce how the stake should be divided after interrupting the game.

155.* Chebyshev's problem. Find the probability that a given fraction min (where m and n are integers) is irreducible.

156. * Let qJ(n) denote Euler's function, i.e., the number of (positive) integers which are primes relative to n and smaller than n. Using a probabilistic argument, show that

qJ(n) = n fI (1 -!), pin p

where the product extends over all prime divisors p of n.

34 4. General Problems

157. * Assume that the number of insect colonies in a certain area follows the Poisson distribution with parameter A, and that the number of insects in a colony has a logarithmic distribution with parameter p. Show that the total number of insects in the area is a negative binomial with parameters q = 1 - P and - A/log(1 - p).

158. Electronic tubes come in packages, each containing N tubes. Let Pk denote the probability that a package contains k defective tubes (0 ~ k ~ m). A sample of n tubes is taken from a package and it is observed that r ~ m tubes are defective.

(a) What is the probability that the selected package actually contains k (:;::: r) defective tubes?

(b) If according to customer demand a package is considered not acceptable whenever it contains d :;::: r defective tubes, what is the probability that the package is not accepted?

159. The probability Pn that n customers visit a supermarket in one day is Pn = pnq, n = 0, 1, .... Two out of three customers, on average, buy a certain type of item. The probability that an item is defective is 1/4.

(a) What is the probability that a customer buys a nondefective item? (b) Given that k nondefective items were sold, show that the conditional

probability an that n customers visited the shop is given by

an = G)pn- k (2 - p)k+l/2n+1.

160. A manufacturer sells an item for $1. If the weight of the item is less than Wo, it cannot be sold and it represents a complete loss. The weight Wof an item follows the normal distribution N(Il, 1); the cost c per item is given by c = a + f3W (a, f3 positive constants). Determine the mean Il so that the expected profit is maximized.

161. A clerk lives at A and works at C and he starts work at 9 a.m. The clerk always takes the train from A to B which is supposed to reach B at 8: 40 a.m. Buses from B leave for C every 15 minutes and the bus which leaves at 8: 45 a.m. is supposed to arrive at 8: 56 a.m.

The train on average experiences delays of 2 minutes and has a standard deviation of 4 minutes. The bus always leaves on time, but arrives on average after a 2-minute delay and a standard deviation of 3 minutes. What is the probability that the cleark arrives late? The clerk's employer drives to his office; he leaves at 8: 45 a.m. and the driving time to the office has a mean value of 12 minutes and a standard deviation of2 minutes. Find the probability that:

(a) both, clerk and employer, arrive late; (b) the employer arrives earlier than the clerk. Assume that the distributions involved are normal.

162.* Quality control. In a continuous manufacturing process the percentage of defective articles is p. To maintain the quality of the product at a certain


standard, the articles are examined, one by one, until a sequence of fixed length r appears with no defective article. Then total inspection terminates and only a certain fraction f of the product is chosen at random for inspection until a defective article appears when the above process of 100% inspection continues. Under the described sampling scheme find:

(a) the probability of a defective sequence, i.e., that a defective article is not followed by r successive good articles;

(b) the expected number of articles in a defective sequence. Hence deduce the expected number, (i) of defective sequences, (ii) of inspected articles, after a defective article;

(c) the expected proportion of the product subject to inspection; (d) the expected percentage p of defective articles going out for sale provided

that any discovered defective article is replaced by a good one; (e) for given values of f and r, what value p*, say, of p maximizes p.

163.* The dilemma of the convict. Three convicts A, B, and C appeal for parole and the board decides to free two of them. The convicts are informed of this decision but are not told who the two to be set free are. The guard, who is A's friend, knows which convicts are going to be set free. A sees that it would not be right to ask the guard about his own fate, but thinks that he might ask for the name of one of the other two who will be set free. He supposes that before he asks the probability of his being set free is 2/3, and when the guard answers that B is pardoned, for example, the probability that he will be set free is lessened to 1/2, because either A and B will be set free or Band C. Having thought about this, A is afraid to ask the guard. Is his fear justified?

164.* Collection of coupons. In each box of a given product there is a coupon with a number from 1 to 6. If a housewife succeeds in getting the series 1-6 she receives a free box of the product. How many boxes must she buy, on average, before she gets a free one?

165.* In a row of 19 seats in an amphitheatre 10 male students and 9 female students sit at random. Find the expected number of successive pairs of seats in which a male student and a female student sit.

166.* A, B, and C are about to play to following fatal game. Each one has a gun and will shoot at his target-A, B, and C taking turns at each other in that order-and it is assumed that anyone who is hit will not be shot at again. The shots, one after another, continue until one of the players remains unhit. What strategy should A follow?

167. Samuel Pepys (who was about to place a bet) asked Newton which of the following events A, B, or C is more probable: A -at least one six when 6 dice are thrown; B-at least two sixes when 12 dice are thrown; and C-at least three sixes when 18 dice are thrown. Which answer (the correct one) did Newton give?

168.* Birthday holidays. The Worker's Legal Code in Erehwon specifies as a holiday any day during which at least one worker in a certain factory has a

36 4. General Problems

birthday. All other days are working days. How many workers must the factory employ so that the number of working man-days is maximized during the year?

169. Bertrand's paradox. A chord AB is randomly chosen in a circle of radius r. What is the probability that the length of AB is less than r?

170.* The neophyte at the horseraces. At a horserace a neophyte better, who is about to bet a certain amount, wants to choose the best horse. It is assumed that no two horses are the same. The better looks at the horses as they pass by, one after another, and he can choose as the winner anyone of the horses, but he cannot bet on a horse he has already let pass. Moreover, he can tell whether any horse passing by is better or worse compared with the preceding ones. Suppose n horses take part in the race. How should he proceed to choose the horse to bet on in order to maximize his probability of winning? What proportion of horses should he wait to pass by before he makes his bet when n is large? (It is assumed that horses parade in front of the neophyte in a random order.)

PART II

ADVANCED TOPICS

CHAPTER 5

Multivariate Distributions

Elements of Theory

1. Multidimensional or vector random variable: A real-valued vector function defined on the sample space Q (more strictly, measurable with respect to the (J' algebra on Q).

The following definitions concerning bivariate distributions extend easily to more than two variables.

2. Thejoint distribution function of the random variables X and Y is defined by

F(x, y) == P[X ::; x, Y::; y] for every point (x, y) E R2.

3. The distribution of the random pair (X, Y) is called discrete if there exists a denumerable set of points (Xi' y) such that

P[X = Xi' Y = Yj] = Pij > 0, and

The probabilities Pij (i = 1, 2, ... , j = 1, 2, ... ) define the so-called joint probability distribution or frequency function of X and Y. Clearly we have

F(x, y) = L Pij' XjS;:X

Yj";Y

4. The distribution of the pair (X, Y) is called continuous if there exists a function f(x, y) such that for every (x, y) the joint distribution function F(x, y)

40 5. Multivariate Distributions

of X and Y can be written as

F(x, y) = f:oo foo f(u, v) du dv.

The function f(x, y) is called the joint density function of X and Y. In this case, for every continuity point (x, y) of f (almost all points are such), we have

o2F(x, y) . P[x < X ::s; x + Ax, y < Y::s; y + Ay] f(x, y) = = hm .

ox oy ~X~O Ax Ay ~y~O

5. Marginal distributions. The distribution of X, as obtained from the joint distribution of X and Y, is referred to as the marginal distribution of X. Similarly for Y. Thus for jointly discrete variables the marginal probability distribution of X is given by

Pi = P[X = xJ = L Pij' j

while for continuous variables the marginal density fx of X is given by

fx(x) = t: f(x, y) dy.

(5.1 )

(5.2)

6. Conditional distributions. The conditional distribution function of X, given Y E B with P[Y E B] > 0, is defined by

P[X::S; x, Y E BJ Fx(xl Y E B) = P[X ::s; xl Y E B] = P[Y E B]· (5.3)

The conditional probability distribution of X, given Y = yj' is defined by

P[X = x· Y = y.J p .. fX(Xil Y = y) = "J =---'1,

P[Y=yJ qj (5.4)

where we set qj = P[Y = Yj] (j = 1,2, ... ). The conditional density of X, given Y = y, for the continuous pair (X, Y) is

defined by

f f(x, y) 1· f ( A x(xl Y = y) = -r-- = 1m x xly < Y ::s; y + LlY) JY(Y) ~y~O

= lim P[x < X ::s; x + Ax, y < Y::s; y + Ay].

~x~o Ax·P[y < Y::s; y + AyJ

(5.5)

~y~O

Hence, and by virtue of (5.1) and (5.2), we deduce the following.

7. Formula of total probability for random variables:

Pi = P[X = xJ = L P[X = x;l Y = yj]P[Y = Yj] = L fx(x;l Y= y)qj' j j

fx(x) = f: fx(xl Y = y)fy(y) dy.

Elements of Theory

Similarly, we obtain the

8. Generalized Bayes formula:

fx(xl Y = y) = 00 fx(x)fy(Ylx = x)

Loo fy(Ylx = x)fx(x) dx

41

(5.6)

9. The expected or mean value of a function 9 of two random variables is defined by

{L: L g(Xi' Yj)Pij i j

E{g(X, Y)} = f: f: g(x, y)f(x, y) dx dy

(for discrete),

(for continuous).

We say it exists if the (double) series or integral is absolutely convergent.

10. The mixed central moment of orders r + s of X and Y is defined by

Jlx, = E[(X - JlxY(Y - Jly )'],

where Jlx = E(X), Jly = E(Y), and e.g., for continuous (X, Y), E(X), is given by

E(X) = L: f: xf(x, y) dx dy = L: x (L+oooo f(x, y) dY) dx

f+oo

= -00 xfAx) dx.

11. The mixed central moment of the second order

JlII = E(X - Jlx)(Y - Jly ) == Cov(X, Y)

is called the covariance of X and Y. The number p (-1 ~ p ~ 1), defined by

( ) Cov(X, Y) p X, Y = ,

(lx(ly

where (I; = Var(X), (I; = Var(Y), is called the correlation coefficient between X and Y.

The following properties can easily be established. For all constants a, b, c, and d we have:

(i) Cov(X + a, Y + b) = Cov(X, Y) (in variance under translations); (ii) Cov(cX, dY) = cd Cov(X, Y);

(iii) p(aX + B, c Y + d) = sg(ac)p(X, Y) where

{ 1 for x> 0,

sg(x) = _ 1 for x < ° (iv) Var(X ± Y) = Var(X) + Var(Y) ± 2 Cov(X, Y).

12. The regression (function) of Y on X, m2(x) say, is defined as the conditional expected value of Y given X = x, that is,

00 t: yf(x, y) dy

m2(x) = E(YIX = x) = f yfy(ylX = x) dy = f) -00 x(x

The curve y = m2 (x) is called the mean regression curve of Yon X. Similarly, we can define the regression of X on Y.

13. The dispersion or variance-covariance matrix of a random vector X =

(X I' ... , X.), usually denoted by L, is defined by

where (Jij = COV(Xi' X), i,j = 1, ... , n. (5.7)

We shall write D(X) = L. If a = (ai, ... , a.), is a vector of constants then

• • Var(alXI + ... + a.X.) = Var(a'X) = a'D(X)a = a'La = L L aiaj(Jij'

i=1 j=1

If the rank r of L is less than n, then the distribution of X is called singular. This is equivalent to the distribution being concentrated in an r-dimensional subspace of the n-dimensional Euclidean space E·. For example, if n = 2 and r = 1 the distribution is concentrated on a straight line in the (X I' X 2) plane and Ip(XI' X2 )1 = 1.

14. The random variables XI'"'' X. are called completely stochastically independent or simply independent if one of the following holds:

(I) For all Borel sets of the real line A I' ... , A.

P[XI E AI' ... , X. EA.] = P[XI E AI] ... P[X. EA.].

(II) The joint distribution function F of X I, ... , X. can be written as

for every (XI' ... , x n ),

where Fi denotes the marginal distribution function of Xi' (III) The joint frequency (density) function f can be written as

f(x l , ... , x n ) = fl(X1) .. ·!.(xn ),

where /; denotes the marginal frequency function of Xi'

15. The main multivariate distributions (a) Discrete distributions

Exercises

(i) Double hypergeometric: R, W, B integers N = R + B + W Then

Pij = (~)(~)C -~ -J/(~)' Osi+jsn,

(a generalization of the simple hypergeometric, see Exercise 175). (ii) Multinomial (k dimensional)

n! P[X - X - ] - "I ". ".+1 1 - n 1 , •.• , k - nk -, , ,PI ",Pk Pk+1'

n1····nk • nk +1· where it was set

(iii) Negative multinomial (k dimensional)

( k+1 )-5-V res + v) k+1 ()?

P[X 1 = n1 , ••• , X k = nk ] = 1 + ,L ()j r() n I' J=l S j=1 nj .

where s > 0, ()j > ° (nj = 0, 1,2, ... ), and nk+1 = n - (n1 + ... + nd.

(b) Continuous distributions (i) Uniform in a bounded set S of E" with density

f(x) = c, XE S,

where c-1 = measure (length, area, volume, etc.) of S. (ii) n-dimensional (nonsingular) normal N(I1, L) with density

43

(5.8)

(5.9)

f(x) = [(2n)PI!:lr1/2 exp[ -!(x - 11),!:-l(X - 11)], (5.10)

where 11 = (111' ... , I1n)' = E(X) denotes the mean vector and L denotes the (positive definite) covariance matrix of the normal random vector X with density (5.10).

(iii) Dirichlet distribution with density

k+1 L Xi = 1, Xi ~ 0. (5.11) i=l

Exercises

171. The joint distribution of (X, Y) is defined by P[X = 0, Y = 0] =

P[X = 0, Y = 1] = P[X = 1, Y = 1] = 1/3. (a) Find the marginal distribution functions of X, Y. (b) Examine whether the points PI: (-1/2,0), P2 : (0, 1) are continuity points

of F(x, y).

172. Show that the function

F(x, y) = {01 for X + Y < 1, for x + y ~ 1,

is not a joint distribution function.

173. We consider a family with two children. Let Xn = 1 if the nth child is a boy for n = 1, 2 and X 3 = 1 if there is only one boy, otherwise Xi = ° (i = 1, 2, 3). Show that the Xi are pairwise independent but not completely independent.

174. X and Y have the joint density

for ° < x < y < 00.

Find (a) the constant c, (b) the marginal distributions of X and Y.

175. The k-dimensional hypergeometric distribution has probability func-tion

where

Pk+l = 1 - (Pl + P2 + ... + Pk), ni > 0, Pi> 0.

Show that the conditional distribution of X k given Xl' ... , X k - l is hypergeometric.

176. If Xl' X 2' ... , X k are distributed according to the multinomial distribution, the conditional distribution of Xl' given X 2 = n2 , ••• , X k = nk , is binomial with parameters n - (n2 + ... + nd and pd(Pl + Pk+l).

177. If Xl' X 2 , ••• , X k have a Dirichlet distribution, show that the conditional distribution of XdSk' for Xi = Xi' where Sk = 1 - (Xl + .... + Xk-d is {3(nk, nk+d.

178. Let X = (Xl' X 2, X 3) be uniformly distributed in the subset of the positive orthant defined by Xl ~ 0, X2 ~ 0, X3 ~ 0, Xl + X2 + X3 ::;; c. Find, (a) the density of X, (b) the marginal distribution of (Xl , X 2 ).

179. A bivariate normal distribution has density

f(x, y) = c exp[ _x2 + xy _ y2].

(a) Find the constant c and the moments of order 2. (b) Find the curves of constant density f(x, y) = c*.

180. If (X, Y) is uniform in the triangle X ~ 0, y ~ 0, X + y ::;; 2, find, (a) the density of (X, Y), (b) the density of X, (c) the conditional density of Y for X = x, (d) E(YIX = x).

181. A die is thrown 12 times. (a) Find the probability that every face appears twice. (b) Let X be the number of appearances of 6 and Y the number of appear

ances of 1. Find the joint distribution of X, Y. (c) Find Cov(X, Y).

Exercises 45

182.* Show that the most probable value (n?, ng, ... , nr+l) of the multinomial distribution satisfies the relations

npi - 1 < n? s (n + k)Pi'

Hint: Show that

i = 1, 2, ... , k + 1.

for i -# j, i,j = 1,2, ... , k + 1.

183. * Multiple Poisson distribution. If the number oftrials n is large and the Pi small so that the npi = Ai are moderate, show that the multinomial distribution can be approximated by the so-called multiple Poisson distribution

184. Given that the density of X and Y is

2 f(x, y) = (1 + x + y)3' x> 0, y > 0,

find, (a) F(x, y), (b) fx(x), (c) fy(YIX = x).

185. Find the density function f(x, y) of the uniform distribution in the circle x2 + y2 S 1. Find the marginal distributions of X and 1'. Are the variables X and Y independent?

186. The joint density of the variables X, Y, Z is

f(x, y, z) = 8xyz,

Find P[X < Y < Z].

0< x, y, z < 1.

187. For each of the following densities f(x, y), find F(x, y), Fx(x), Fy(y), fx(x), fy(y), fx(xl Y = y), fY(ylX = x).

(a) f(x,y)=4xy, O<x, y<l,

(b) f(x, y) = i(x2 - y2)e- X , 0 < x < 00, Iyl < x.

188. The joint probability function P(Xi' y) of X, Y is given in the following table:

~ -1 0 2 p(x)

-1 0.1 0.15 0 0.1 0.35 0 0.15 0 0.1 0.2 0.45

0.05 0.05 0 0.1 0.20 q(y) 0.3 0.2 0.1 0.4 1.00

(a) Calculate the regression lines of X on Y and of Yon X.


(b) Hit is possible to observe only X, what is the best estimate of Yin terms of X in the sense of the mean square error?

189.* Show that the discrete variables X and Y with joint probability function Pij = P[X = Xi' Y = yJ are independent if and only if the matrix of probabilities P = (Pij) has rank 1.

190. Show that two orthogonal (uncorrelated) variables each taking two values are independent.

CHAPTER 6

Generating Functions. Characteristic Functions

Elements of Theory

1. The probability or factorial moment generating function P(t) of a nonnegative integer-valued random variable X is defined by

00

P(t) = E(t X ) = I P[X = k]tk = I Pk tk. k=O

It exists at least for -1 ~ t ~ 1. If the factorial moment n" say, of order r exists,

nr = E[X(X - 1) ... (X - r + 1)] = E[(X)r],

then this is given by nr = p(r) (1 ), r = 1,2, ... , (6.1)

where p(r)(t) denotes the derivative of order r of P(t). Thus we obtain

n 1 = E(X) = P'(I),

and hence, Var(X) = P"(l) + P'(1) - [P'(I)Y (6.2)

2. Distribution of the sum of a random number of random variables. Let

(6.3)

where the random variable N = 0, 1, ... and the Xi are completely independent. If the Xi have the same distribution with probability generating function Px(t), then the probability generating function of SN

(6.4)

48 6. Generating Functions. Characteristic Functions

Moreover, if E(X) and E(N) exist, then

E(SN) = E(X)E(N). (6.5)

3. Compound Poisson distribution. This is the distribution of SN when N has the Poisson distribution. Therefore its probability generating function is

(6.6)

where A. = E(N). If the Xi are Bernoulli random variables, then S~ is Poisson with parameter A.p, where p = E(X;).

4. The moment generating function M(t) of the random variable X is defined by

M(t) = E(etX )

provided it exists for - 0 < t < 0 and some 0 > O.

Proposition. If Mx(t) exists, then it has derivatives of every order for It I < 0 (0 > 0) and the moments of every order exist. In fact, we have for every r = 1,2, ...

E(xr) = M(r)(o),

where M(r)(t) denotes the rth derivative of M. Furthermore, it determines uniquely (characterizes) the distribution of X.

5. Complex random variable: Z = X + iY where the X and Y have a joint probability function. Its expected value is defined by

E(Z) = E(X) + iE( Y), i=J=1.

6. The characteristic function of a random variable X is defined by

cp(t) = E(e i/X ) = E(cos tX) + iE(sin tX).

It exists for every value of the real parameter t since lei/xI = 1.

Properties of a characteristic function cp(t) (i) cp(O) = 1, Icp(t)1 ::; 1.

(ii) cp( - t) = cp(t) where a denotes the conjugate of a complex number a. (iii) cp(t) is uniformly continuous everywhere. (iv) If the rth moment J1.~ = E(xr) exists, then cp(t) has a derivative of order

r, cp(r)(t), and

thus J1.'1 = E(X) = i-1cp'(0), J1.2 = -cp"(O). (v) If cp(t) has a derivative of order r at t = 0, then all moments up to order

r exist if r is even and up to order r - 1 if r is odd.

Elements of Theory 49

(vi) If E(xr) exists then qJ(t) has the Mac Laurin expansion

r , (it)k k qJ(t) = kf:O J1.k k! + O(t ), (6.7)

where O(x) denotes a quantity such that O(x)/x -+ 0 as x -+ O. Moreover, the so-called cumulant generating function or second charac

teristic function

I/!(t) = log qJ(t)

has the expansion

r Ct)i I/!(t) = L Kj -;- + O(tr),

j=O ). (6.8)

where Kj is called the cumulant of order j of X. Note: For a proof of (iv)-(vi) we refer, e.g., to Cramer (1946).

7. The characteristic function of a random variable X determines uniquely (characterizes) the distribution of X. In fact, we have the following inversion formula of characteristic functions (Fourier transforms).

If x + h and x - h (h > 0) are continuity points of the distribution function F of X then

1 fC sinh t . F(x + h) - F(x - h) = lim - __ e-·xtqJ(t) dt.

c-oo n -c t

For a discrete random variable we have

In the special case when X is integer valued we have

1 flt -ikt Pk = P[X = k] = -2 e qJ(t) dt. n -It

(6.9)

If the characteristic function qJ(t) is absolutely integrable in ( - 00,(0), then the corresponding distribution is continuous with density f(x), given by the inversion formula

1 foo . f(x) = -2 e-·xtqJ(t) dt. n -00

(6.10)

8. Common property of the probability generating function, the moment generating function, and the characteristic function. This is very useful for the study of the distribution of sums of independent random variables. Let G(t) denote any of these functions. Then, if Sn = Xl + X2 + ... + Xn where the Xi


are independent we have n

GsJt) = n Gx.(t). (6.11) i=l

9. If P(t), M(t), and ((J(t) exist for a random variable X, then the following relations hold:

Mx(t) = Px(e' ) = ({Jx( - it),

((Jx(t) = Mx(it) = Px(e it ).

10. In addition to characterizing a distribution, a characteristic function facilitates the study of the asymptotic behavior of a sequence of random variables as indicated by the following.

Continuity theorem for characteristic functions (Levy-Cramer). Let {Xn} bea sequence of random variables and let {Fn(x)}, {({In(t)} be the corresponding sequences of distribution functions and characteristic functions. Then a necessary and sufficient condition that the sequence Fn(x) of distribution functions converges to a distribution function F(x) for every continuity point x of F (weak convergence) is that ({In(t) --+ ({J(t), as n --+ 00 for every t and ({J(t) is continuous at t = O. Whenever this holds the limiting characteristic function ({J(t) is the characteristic function of the limiting distribution function F.

11. Infinitely divisible distributions. The random variable X or its distribution is called infinitely divisible if for every integer n, X can be represented as the sum of n independent and identically distributed random variables, i.e., X = Xl + ... + Xn.

Hence the characteristic function ({J(t) of X must satisfy

for every n,

where ((In(t) is a characteristic function.

12. Generating functions for multivariate distributions. The definitions of probability generating functions, moment generating functions, and characteristic function(s) of univariate distributions extend easily to multivariate distributions of (Xl' ... , Xn) = X' by replacing t by the row vector t' = (tl' ... , tn) and X by X, so that tX is replaced by t'X = t 1 X 1 + ... + tnXn.

Thus, for example, the probability generating function of X and Y with joint probability function

is defined by

Pjk = P[X = j, Y = k], j, k = 0,1, ... ,

P(tl' t 2 ) = E(tftD = L Pjkt{t~, j,k

and the characteristic function of X = (Xl' ... , Xn)' by

({Jx(t) = ({Jx(t 1, ... , tn) = E(e it •X ) = Eei(t,X,+"·+lnXn).

Exercises

It can easily be shown that: (i) Px(td = P(tl' 1), Py(t2) = P(1, t2).

(ii) Px+y(t) = P(t, t}. (iii) X and Yare independent if and only if

P(tl' t2} = P(tl' 1)P(1, t2) = PX(tl)Py(t 2)

<P(tl' t2) = <P(tl, 0)<p(0,t2} = <PX(tI)<Py(t2}

(iv) The mixed moment P-}k = E(Xiyk} is given by

I _ ai+k<p(O, O} .-U+k) P-ik - at{ at~ I .

for every (tl' t2),

for every (t I' t 2 )·

51

By analogy to the inversion formula for univariate distributions, we have the following:

If the vertices of the generalized rectangle (or n-cell) Xi - hi ~ Xi ~ Xi + hi (i = 1, ... , n) (hi> 0) are continuity points of the distribution function F(Xl' ... , xn }, then

P[Xi - hi < Xi ~ Xi + hi' i = 1, ... , n]

1 fC fC n sin h·t· = lim n ... <P(tl,···, tn } Il--'-' exp( -iXiti} dti· c-oo 1t -c -c i=l ti

In the special case of an absolutely integrable <p(t I' ... , tn }, the distribution is continuous with density

The continuity theorem also holds, i.e.,

when <p is continuous at (0, ... , O).

Exercises

191. Find the probability generating function and hence the mean and variance ofthe following distributions: (a) binomial, (b) Poisson, (c) geometric, (d) Pascal. Moreover, deduce the corresponding moment generating functions and characteristic functions.

192.* Let Xl = X 2 = X when X has the Cauchy density

1 1 f(x} =;. 1 + X2' -00 < X < 00.

Prove that <PX 1 +X2 (t) = <Px 1 (t)<PX 2 (t),

while X I' X 2 are not independent.

193. From the probability generating function Px(t) of X = 0, 1,2, ... , find the probability generating function of Y = 3X + 2.

194. Using the probability generating function prove that the random variable SN of(6.3) has variance given by Var(SN) = E(N) Var(X) + Var(N)E2(X).

195. A woman continues to have children until she has a boy! Suppose that the probability of having a blond child is P; find the probability that the woman will have k blond children. (Assume that the color is independent of sex.)

196. The number N of visitors to a shop has the Poisson distribution. On average, 150 women visit the shop every day. Of these, 100PI% buy mini, 100p2% buy midi, 100p3% buy maxi, and the remaining buy nothing. (No woman buys more than one dress!!) Let Xl' X 2, X3 be the corresponding numbers of mini, midi, and maxi that are bought in a day. Find the joint probability function of Xl' X 2, X 3. Are they independent?

197. * Equal characteristic functions in a finite interval do not necessarily define the distribution uniquely. Let X, Y be independent with densities

f ( ) = a + b - a cos(ula) - b cos(ulb) xU 2 2 ' nu

-oo<u<+oo. r ) _ c - c cos(ulc) lY(U - 2'

nu

Prove that <Px(t) = <py(t) for It I :<;:;; min{lla, lib, lie}, 2c = a + b.

198.* Let Xl' X 2, ... , Xn be independent random variables having the same normal distribution with mean Jl and variance 1. We set

Moreover, let the random variable T = aYv where a, v are suitable constants, and Yv follows the X2 distribution with v degrees of freedom. Using characteristic functions, find the constants a and v so that the random variables Sn and T have the same means and the same variances.

199. Prove that SN of (6.3) has the characteristic function

<PsN(t) = PN(<Px.(t)).

200. Prove that <p(t) = exp[2(e- I'1 - 1)] is the characteristic function of a Poisson distribution compounded with (generalized by) a Cauchy distribution. Then, using the inversion formula, find its density.

Exercises 53

201. * Let Z. be a random variable following the binomial distribution with parameters nand p = ),/n p, > 0). Moreover, let the random variable Z have a Poisson distribution with parameter A. Prove that limn _", cpzJt) = cpz(t) (t E R), where cpzJt) and cpz(t) are the characteristic functions of Z. and Z, respectively. Then deduce the approximation of the binomial by the Poisson.

202. Calculate the characteristic function of a Gamma distribution with density

f(x) = /(:) e-Axx s - 1, x> 0 (s > 0).

Then deduce the characteristic function of x;. 203. At the nth toss of a coin a player receives or pays 2-' of a drachma

depending on whether heads or tails appears. Let y" be the gain of the player after n tosses. Show that the distribution of y" as n -4 00 tends to the uniform distribution in the interval (-1, 1). First find the characteristic function of Yn'

204. X has distribution function F(x) which is a mixture of two distribution functions Fl , F2 as follows:

F(x) = ),Fl (x) + (l - ),)F2(x), A> O.

If Fl is N (111' aD and F2 is N (112' aD find the characteristic function of F and then the E(X) and Var(X).

205. Find the second characteristic function (cumulant generating function) t/!(t) of the exponential with density

x> 0,

and prove that the cumulant of order r Kr is given by

(r - 1 )! Kr =

(]' r = 1,2, ....

206. (Continuation). Let Y be a discrete random variable defined as follows. For h > 0 and k = 0, 1, 2, ... ,

P[Y = (2k + 1)h/2] = P[kh :$; X:$; (k + 1)h].

Find the probability generating function of Y and then show that E(X) > E( Y), Var(X) < Var(Y).

207.* Suppose that the density f(x, y) = g(X2 + y2), i.e., it has circular symmetry about the origin. Show that the characteristic function of f, cp(t, u), is a function of t 2 + u2 (only). Hence show that the only distribution of independent random variables with the above property is the normal. (Consider an orthogonal transformation of t, u.)

208. Consider two independent random samples Xl' X 2 , ... , X. and

Yl , Y2 , ••• , Yn from the Laplace distribution

-00 < x < 00.

Show that the means of the differences Xi - 1; and Xi + 1; have the same distribution but are not independent. Are they orthogonal?

209. If cp(t) is a characteristic function show that, (a) I cp(tW, (b) e).(q>(t)-l) are also characteristic functions.

210.* Calculate the probability generating function of the multinomial distribution and then show that

211. (Continuation). If, as n -> 00, npj -> Aj (j = 1,2, ... , k + 1), then the multinomial distribution tends to the k-dimensional Poisson Y = (Yl , Y2 , •.. ,

~) where the 1; are independent Poisson, 1j with parameter Aj (cf. Exercise 183).

212.* Calculate the characteristic function of the negative multinomial distribution (5.9) and hence

(a) Verify that E(Xj) = s(}j, Var(X) = sej(1 + e), Cov(Xj, X k ) = seA. (b) Show that if the conditional distributions of the Xj given V = v are

independent Poisson with parameters vAj , respectively (j = 1,2, ... , r) and V has the Gamma distribution with density

AS /(v) = r(s) e-}.V-l, S > 0,

then X = (Xl' X 2 , ••• , X k ) has the above negative multinomial distribution with parameters ej = Ajl A.

213. X, Y have a distribution of the continuous type with characteristic function cp(t l' t2). Show that the conditional characteristic function of X given Y= y is

f_OOoo e-it2Ycp(tl' t 2) dt2

CPx(tll Y = y) = foo -00 e- it2Ycp(0, t2) dt2

214. A random sample Xl' X 2, ... , X n,+n2-n on a continuous random variable X with - 00 < X < 00, is divided at random into three subsets of (nl - n), (n2 - n), and n observations (nl > n, n2 > n). Let Sl' S2' and S3 denote the corresponding sums. Find the characteristic function of

and E(YIZ = z).

215. Show that the following distributions are infinitely divisible: (a) Pascal, (b) Cauchy, (c) Laplace, (d) Gamma.

CHAPTER 7

Distribution of Functions of Random Variables

Elements of Theory

In finding the distribution of a function of a scalar or vector random variable the following formulas turn out to be very useful.

1. Distribution of monotone functions. Let X be a continuous random variable with distribution function Fx and y = g(x) a differentiable and monotone (increasing or decreasing) function with inverse function x = g*(y) (i.e., g'(x) is everywhere either> ° or < 0). Then the random variable Y = g(X) is also continuous with distribution function Fy given by

F ( ) = {Fx(g*(y», y y 1 - Fx(g*(y»,

g'(x) > 0,

g'(x) < 0,

and density function

I dg* I 1 fy(y) = fx(g*(y» ""dY = fx(g*(y» Ig'(x)I'

2. Under the preceding assumptions, except that the equation

y = g(x)

has roots Xl' ... , xn , •.. , i.e., for given y we have

and the derivatives

g'(x;) #- 0, i = 1,2, ... ,

(7.1 )

56 7. Distribution of Functions of Random Variables

Y = g(X) is also continuous with density

fx(x 1) fx(xn) fy(y) = Ig'(xdl + ... + Ig'(xn}l + .... (7.2)

3. Functions of two random variables. An interesting case is when we have a function of two continuous random variables X and Y with joint density f(x, y).

(a) Sum of two random variables. The distribution function Fz(z) of the sum

Z=X+Y

is given by

Fz(z) = f L+Y~z f(x, y) dx dy = t: dy r:Y f(x, y) dx

= f:oo dx r:x f(x, y) dy = toooo dx roo f(x, y - x) dy

= t: dy roo f(x - y, y) dx.

Differentiating with respect to Z we have the density of Z

fz(z) = t: f(z - y, y) dy = f: f(x, z - x) dx.

In the special case of independence of X and Y, the density fz(z) of the sum is called the convolution of the density functions fx and fy, and we have

fz(z) = t"'oo fx(z - y)fy(y) dy = f: fx(x)fy(z - x) dx. (7.3)

The distribution function Fz is given by

Fz(z) = f: Fx(z - y)fy(y) dy = re", Fy(z - x)fx(x) dx. (7.4)

Note. If X and Yare discrete, the integrals are replaced by sums. (b) Distribution of the quotient of two continuous random variables. The

distribution function Fz of the ratio

is given by

X Z=

Y

100 fYZ fO r'" Fz(z) = ° dy -00 f(x, y) dx + -ro dy JZY f(x, y) dx, (7.5)

Elements of Theory

and the density by

fz(z) = L'Xl yf(yz, y) dy - f:oo yf(yz, y) dy.

In the special case of independence of X and Y, (7.5) and (7.6) become

Fz(z) = too Fx(zy)fy(y) dy + f:oo [1 - Fx(zy)Jfy(y) dy,

fz(z) = f:oo IYlfx(zy)fy(y) dy.

57

(7.6)

(7.7)

4. Functions of several random variables. This is the general case in which, given the joint density function f(x 1, ... , xn) of the random variables Xl' ... , X n• we want the distribution of

j = 1, ... ,m. (7.8)

Then the joint distribution function G(Yl' ... ' Ym) of Yl , ... , Ym can be obtained from

G(Yl' ... , Ym) = fi>· f f(x l , ... , xn) dx l , ... , dxn,

where the region of integration D is defined by the relation

D = {(Xl' ... , Xn): gj(X l , ... , Xn)::;; yj,j = 1, ... , m}.

For discrete random variables the distribution is obtained by replacing the multiple integral by a multiple sum over D. Of special interest is the case m = n for which the problem, under certain conditions, has a specific solution.

If the functions gj of (7.8) (j = 1, ... , n) have continuous partial derivatives of the first order so that the Jacobian J of the transformation (7.8) is #0, i.e.,

Ogl Ogl

oX l oXn

#0, (7.9)

ogn ogn

oXl oXn

then there exists an inverse transformation

i = 1, ... , n, (7.10)

and the random variables Yl , ... , y" are continuous with joint density (cf. (7.1 ))

f*(Yl' ... , Yn) = f(gf(Yl' ... , Yn), ... , g:(Yl, ... , Yn))IJ(Yl' ... , Yn)l, (7.11)

where J(Yl' ... , Yn) denotes the Jacobian of the inverse transformation (7.10),

I.e.,

For example, in the case of linear transformations

n

Yi = I aijXj' j=l

i = 1, ... , n,

the Jacobian equals the determinant of A = (a i),

J(XI' ... , xn) = IAI· When A is nonsingular, i.e., I A I i= 0, the density of the random vector

Y=AX,

where Y = (Yl> ... , Y"),, X = (Xl' ... , Xn)' is given by

f*(YI"'" Yn) = f(A-Iy)IAI- I.

Exercises

(7.12)

216. The radius of a circle is approximately measured so that it has the uniform distribution in the interval (a, b). Find the distribution:

(a) of the length of the circumference of the circle; (b) of the area of the circle.

217. Suppose that X and Yare independent random variables with the same exponential density

X> 0.

Show that the sum X + Y and the ratio X/Yare independent.

218. Suppose that X and Yare independent uniform variables in (0, 1). Find the probability that the roots of ).2 + 2X A + Y = ° are real.

219. The angle cp at which a projectile is fired with initial velocity v follows the uniform distribution in the interval (0, n/2). Find the distribution of the horizontal distance d between the firing point and the point at which the projectile falls.

220. * Suppose that X and Yare normal variables with mean 0, variance (f2, and correlation coefficient p.

(a) Find the distribution of X/Y. (b) Using the above result prove that

1 1 arc sin p P[X<0,Y>O]=2 P [XY<O]=4- 2n .

Verify this by integrating the density of (X, Y).

Exercises 59

221. Let A = (X, Y) be the point of impact of a shot on a vertical target with center at origin 0 and suppose the distribution of A is that of Exercise 220 with p = ° (called circular normal). Show that the distance R = (OA) between the point of impact and the centre of the target and the angle of OA with the horizontal axis (i.e., the polar coordinates of A) are independent variables. Find the distribution of R2.

222. If X has the uniform distribution on the interval (0, 1), then: (a) Prove that Y = aX + b is also uniformly distributed (a and b constants). (b) Find the distribution of Y = AX2 + BX + C (A, B, C constants).

223. If X I, X 2 are independent and uniform on the interval (0, 1), find the densities of: (a) Xl + X 2 , (b) Xl - X 2 , (c) IX I - X21, (d) X I /X2.

224. Two friends A and B agree to meet between 12 (noon) and 1 p.m. at a restaurant. Supposing that they arrive at random between 12 and 1 p.m. independently of each other and the lunch lasts 30 minutes, what is the probability that they meet in the restaurant? Let T be the instant of their meeting. What is the conditional distribution of T, (a) given that they meet, (b) given that they meet and A arrives first?

225. Let X and Y be independent with densities

1 1 fx(x) = - f1---:z'

n v 1 - x

Show that X Y is N(O, (J2).

Ixl < 1, I' ( ) _ Y -y/2a 2 JYY -ze .

(J

226.* Given n independent random numbers Xl' X 2 , ... , Xn from (0, 1), called pseudorandom numbers, show that:

(a) Y = - :L?=l 2 log X I has the X2 distribution with 2n degrees offreedom; (b) the variables

~ = -J=2~ cos(2nX2 ),

1/ = J - 2 log X I sin(2nX 2)'

are independent N(O, 1). Thus (a) generates samples of X2 distributions with an even number of

degrees of freedom, while (b) generates a pair of independent normal variables.

227.* Let X be a continuous random variable X with distribution function F( . ); consider the random variable

Y = F(X) = f:oo f(u) duo

This transformation is called the probability integral transformation. Show that the distribution of Y is uniform in the interval (0, 1), that is,

Fy(y) = y, ° < y < 1.

Given a random sample X I' X 2, ... , Xn from the uniform distribution, show


thatthesolutionsYl>Y2, ... ,YnofYl = F- l (xd'Y2 = F- l (X2), ... ,Yn = F-l(xn) form a random sample from the F distribution.

228. (Continuation). Suppose the height X of men has the normal distribution N(167 cm, 9 cm 2 ) and a random sample of 10 men is taken with heights Xl' X 2 , .•. , X IO • What is probability that the (random) interval (j = [min(Xl , ... , X 10), max(Xl , ... , X 10)J covers at least 95% of the population of heights? That is, the P{P[X E (jJ ~ 0.95} is to be calculated; note that the P[X E (jJ, depending on the random interval (j, is a random variable.

229. X is called a lognormal variable, if the log X = Y has a normal distribution N(fJ., (]"2).

(i) Find, (a) the density of X, (b) E(X) and Var(X). (ii) If the Xi are independent lognormal random variables, their product

X 1 X 2'" Xn is also lognormal.

230. Show that if F(x, y) is the distribution function of X and Yand

Z = max(X, Y), W = min (X, Y),

then

Fz(z) = F(z, z), Fw(w) = Fx(w) + Fy(w) - F(w, w).

If F(x, y) is continuous find the densities of Z and W

231. (Continuation). If X and Yare independent N(O, I), show that

1 E{max(X, Y)} = In'

232.* Let X = (Xl' X 2 , •.• , Xd have the Dirichlet distribution. Find the distribution, (a) of 1;. = Xl + X 2 + ... + Xb and (b) of 1'; = Xl + X 2 + ... + Xi (i = 1, 2, ... , k).

233.* (Continuation). If Xi (i = 1,2, ... , k + 1) are independent Gamma variables with parameters nl , ... , nk+l and )., respectively, show that the

x. y=. I

, Xl + ... + Xk+l i = 1,2, ... , k,

have the k-dimensional Dirichlet distribution (5.11).

234. Let X and Y be independent Poisson variables with parameters A and fJ.. Show that:

(a) the sum X + Y is also Poisson (use the convolution formula, cf. (7.3); (b) the conditional distribution of X given that X + Y is binomial.

235. Let X and Y be independent binomial with parameters N, p and M, p, respectively. Show that:

(a) the sum of X + Y is binomial (use the convolution formula); (b) the conditional distribution of X given X + Y is hypergeometric, inde

pendent of p.

Exercises 61

236.* Show that the mean

_ 1 n

X=- LXi n i~l

and the variance

2 1 ~ - 2 S = ~~ L... (Xi - X)

n - 1 i~l

of the random sample X I' X 2' ... , Xn from N (tt, (T2) are independent. Hence deduce the distribution of (n - l)s2.

Hint: (a) Use any orthogonal transformation Y = HX such that

Y,. = Jnx, Y = (Yl , ... , Y,.)" x = (Xl' ... , Xn)'· - -

(b) Cov(X, Xi - X) = 0. Show that Cov(X, S2) = tt3/n for samples from nonnormal populations, where tt3 is the population central moment of order 3.

237.* Test for the Behrens-Fisher problem. To test whether two heteroscedastic normal populations have the same means one uses the statistic

t = [m(m - 1)]1/2(X - .Y)/L~ (uj - U)2T2

,

where X, yare the sample means of the independent samples from N (ttl' (Tn and from N(tt2' (Tn, respectively, m ::;; nand uj = Xj - (~)Yj (j = 1,2, ... , m). Show that t has Student's distribution with m - 1 degrees offreedom when ttl = tt2 = tt·

238.* Find the joint density of the ordered statistics X(l) < X(2) < ... < X(n) from a distribution of the continuous type.

239. If Xl' X 2 , ... , Xn is a random sample from the continuous distribution F. Find:

(a) the joint density function of the random variables

Y = F( min Xi)' Z = F( max Xi); l~t~n l~l~n

(b) E(R) = E(Z - Y) where R is the range of the sample.

240. Let X(l) < X(2) < ... < X(n) be the ordered sample from a continuous distribution F(x). Show that

E[X(k+l) - X(ka = G) r+",OCl F"-k(X) [1 - F(X)]k dx.

Hint: Use the joint density of X(kl' X(k+l)'

241.* A random sample of size n is taken from a uniform distribution in the interval (0, 1). Find, (a) the mean E[X(k)]' (b) the distribution of R = X(n) - X(l)'

242. If Yl < Y2 < Y3 are the ordered observations of a random sample of size 3 from a distribution with density

{ I, f(x) = 0,

e - 1 < x < e + 1, elsewhere.

For rJ. = f3 = 0.4 show that p[e - rJ. < Y2 < e + f3] = 0.944.

243. Let Yl < Y2 < Y3 be ordered observations of a random sample of size 3 from a distribution with density

{2X.

f(x) = 0,' 0< x < 1,

elsehwere.

Show that Zl = Yl /Y2 , Z2 = Y2 /Y3 , and Z3 = Y3 are completely independent.

244. If Xl' X z are independent f3 variables with densities f3(n l , nz) and f3(n l + 1/2, nz), respectively, where for n l > 0, nz > 0,

1 f3(m, n) = --xm - l (1 - x)n-l,

B(m, n) O<x<l.

Show that Y = J XIX z has also the f3 distribution f3(2n l , 2nz)·

245. The random pair (X, Y) has density

1 f(x, y) = r(m}r(n) xm-l(y - x)n-l e- y , 0< x < y < 00.

Find the distribution of Z = Y - X. Are Z and X independent? (cf. Exercise 174).

246. Show that Z = 2ft¥, where X and Yare two independent r variables with parameters ()ol' n) and (2 z, n + 1/2), respectively, has the r distribution with parameters (2, 2n) where 2 = J2 l ).z.

247. For the pair (X, Y) of Exercise 245 show that W = X/Y has the density f3(m, n) of Exercise 244.

248. If X and Yare independent N(O, 0"2), show that

XY Z = -r===;===;c

JXz+y z '

are also independent normal.

249.* Let X have an n-variate normal distribution N(fl, I:). Find the density of Y = AX where A is a nonsingular matrix.

250. The points of impact of two players AI' A z on a vertical target follow the circular normal distribution with mean the center of the target and variances O"f and 0"1, respectively. Find the probability that the impact point of Al is closer to the center than that of A z.

CHAPTER 8

Limit Theorems. Laws of Large Numbers. Central Limit Theorems

Elements of Theory

1. Convergence of a sequence of random variables. A stochastic sequence, i.e., a sequence {Xn} of random variables defined on a probability space {Q, B, P} may converge in several ways.

(a) In probability or stochastically or weakly to a random variable X denoted by

if for every e > 0

lim P[IX. - XI> e] = o. (8.1)

(b) With probability one or almost surely (a.s.) or strongly to a random variable X denoted by

if

p[lim Xn = X] = 1, .-00 or, equivalently, if for every e > 0

lim p[sup IXn - XI> e] = o. N-oo n~N

(c) In quadratic mean to a random variable X, denoted by

Xn~X

(8.2)

(8.3)

64 8. Limit Theorems. Laws of Large Numbers. Central Limit Theorems

if

n~oo

(d) In law or distribution to X, denoted by

Xn~X

if at every continuity point x of the distribution function F of X

lim Fn(x) = F(x),

where Fn denotes the distribution function of Xn.

2. Relations between the modes of convergence. For every constant e the following hold:

(i) Xn ~ X => Xn ~ X;

(iii) Xn ~ X => Xn ~ X; 00

( 1·1·) X P X L n --> e ¢> n --> e;

(iv) Xn~X => Xn~X;

and L E(Xn - e)2 < 00 => Xn ~ e. n=1

(8.4)

3. Laws of large numbers (LLN). They refer to the weak or strong convergence of a sample mean Xn = (lin) L?=1 Xi to a corresponding population (distribution) mean /1. Thus we have the

Weak LLN (WLLN): - p

If Xn --> /1;

Strong LLN (SLLN): If Xn ~ /1.

The basic tool in the proof of the WLLN is the so-called

Chebyshev-Bienayme inequality. If E(X) = /1, Var(X)= (1"2, then for every constant e or A. (A. > I)

(1"2

P[IX - /11;:::: eJ ::; 2" e

This follows from

or (8.5)

Markov's inequality. If Y;:::: 0, i.e., P[Y;:::: OJ = I and E(Y) < 00, then for every constant e > 0

P[Y;:::: eJ ::; E(Y) . e

(8.6)

The Chebyshev-Markov WLLN. If the random variables Xb ... , X n , •••

are independent and E(XJ = /1i, Var(XJ = (1"l with

Var(Xn ) --> 0 as n --> 00,

Elements of Theory

then

1 n

where [in = - L l1i; n i=l

65

(8.7)

as a consequence we have the WLLN for a random sample (i.e., independent and identically distributed random variables) from a population with mean 11 and variance (J2, i.e.,

(8.8)

When the Xi are Bernoulli variables then (8.8) gives, the Bernoulli WLLN, i.e., the proportion Pn = x/n = Xn of successes!.. p, the probability of success in a trial (see (2.5)).

Khinchine's WLLN. For a random sample Xl, ... , X n, .•. from a population with mean 11 = E(XJ (i = 1, ... ,) the WLLN holds. This follows by the continuity theorem for characteristic function(s) (see Chapter 6) and relation (ii) of (8.4) since

q>- (t) ----+ q> (t) = eil't. Xn n-co Il

The classical Poisson LLN: If the Xi are Bernoulli with

then

is a special case of (8.7).

i = 1, ... ,

where 1 n

Pn = - L Pi' n i=l

Strong laws of large numbers (SLLN). E. Borel (1909) first showed the SLLN for Bernoulli cases (as above)

X a.s. n ---> p.

This means that for every e > 0 and b > 0 there exists an integer N = N(e, b) such that

P[IXn - pi < e for every n :2: N] :2: 1 - b.

Later Kolmogorov showed that under the conditions of (8.7)

Moreover for a random sample

Xn ~ 11 ~ E(XJ = 11, (8.8)*

i.e., the SLLN holds if and only if the population mean exists. The proofs of these (see, e.g., Gnedenko (1962)) rest on the following

ineq uali ties.

Kolmogorov's inequality. Let X I, ... , Xn be independent and let Sk =

XI + ... + X k with E(Sk) = mk, Var(Sd = sf (k = 1,2, ... ). Then for every e>O

1 P[ISk - mkl < esn, k = 1,2, ... , n] z 1 - 2'

e (8.9)

The Hajek-Renyi inequality. Under the conditions of (8.9), for every nondecreasing sequence of positive constants Cn and any integers m, n, 0 < m < n,

p[ max cklSk - mkl Z eJ ~ -i(c;s; + f cf var(xk »). (8.10) m<;k<;n e k=m+1

Central limit theorems (CLT). They refer to the convergence of sums of random variables (or equivalently of sample means) in law. The limiting law is usually the normal. The proofs of CL T's, as a rule, rely on the continuity theorem for characteristic functions (see Chapter 6):

L Xn ~ X ¢:> <f>n(t) ~ <f>(t) ¢:> Fn(x) ~ F(x), (8.11)

where <f>n denotes the characteristic function of Xn and Fn the distribution function of Xw

The Levy-Lindeberg CLT. If XI' ... , X n, ... is a random sample with E(X;) = 11 and Var(X;) = (J2, then

n

r,; X _ .L (Xi - 11) S* = v' f/( n 11) = 1=1 L N(O 1) n- r:. -.,'

(J v n(J

i.e., for every x,

lim PES: ~ x] = (x) = _1_ IX e-u2/2 duo (8.12) n~oo fo -00

Lyapunov's CLT. Let X I, ... , Xn, ... be a sequence of independent random variables with

E(X;) = Ili,

and set

( n )1/3

Bn = .L Pi ; 1=1

if

as n ~ 00,

then n

L (Xi - Ili) S: = i=1 ~ N(O, 1).

s n-+co n

(8.13)

Exercises 67

Necessary and sufficient conditions for the validity of the CL T are given by the

Lindeberg-Feller CLT. Let Xn, )-tn' (Tn' Sn, S: be defined as above, and let Fn be the distribution function of Xn. Then for every e > 0,

i max ---0 1 ~i ~n Sn n-oo

and S:.!+ N(O, 1)

The classical De Moivre- Laplace CL T for Bernoulli variables (the normal approximation to the binomial distribution) is a special case of (8.12).

Exercises

251. If the independent random variables Xl, ... , Xn, ... satisfy the condition

V(X;) ~ C < 00, i = 1,2, .. _,

then the SLLN holds.

252. In a sequence of random variables Xl' _ .. , Xn, ... suppose that X k

depends only on X k - 1, Xk+1' but that it is independent of all the other random variables (k = 2, 3, ... ). Show that if V(XJ ~ N < 00 (i = 1, 2, .. _), then the WLLN holds_

253.* If for every n, V(XJ ~ C < 00 and Cov(X;, X) < ° (i,j = 1,2, ... , n), then the WLLN holds.

254.* Let {Xn} be a sequence of random variables so that V(XJ ~ C < 00

(i = 1, 2, __ .) and Cov(X;, Xj) -+ ° when Ii - jl -+ 00 then the WLLN holds (Theorem of Barnstein).

255.* If Xn .!+ X, Yn ~ c, then L

(a) X. + ¥" -+ X + c; L

(b) Xn Yn -+ cX; L

(c) X./¥" -+ X/c.

256.* Let ~n = (X1n , X 2n , ... , X k.)' (n = 1,2, ... ) be a sequence of random vectors and ~ = (X l' X 2, ... , X k )' a random vector with distribution function F(x l' .. - , x k ). If

for all constants c 1, ... , Ck , then the joint distribution function F~n of X In' _ •• ,

X k•

has a limit and

F~JXI' ... , Xk) --+ F(XI' ... , xk)·

257. Let {Xn} be a sequence of independent random variables uniformly bounded, that is, there is a constant M such that P[Xk ~ M] = 1 for all k = 1, 2, ... , and suppose that Var(Xd oF 0 for every k.1f s; --+ 00, then the CLT holds (cf. (8.13)).

258. Show that

. -n n nk 1 hm e L ~=-n~CX) k=O k! 2·

259. If X and Yare independent Poisson variables with parameters Al and )'2' respectively, show that

X - Y - (AI - A2 ) L

(X + y)I/2 ~ N(O, 1). A2 -'00

260. Show that for the sequences {Xn} of independent random variables with

1 - rn (a) P[Xn = ± 1] = 2 '

n 1 P[Xn = ± 2 ] = 2n+1 ' n = 1,2, ... ,

(b) P[Xn = ±nA] = 1/2, the CL T holds.

261. Compare the results given by Chebyshev's inequality and the CLT for the probabilities

P[ -k < S: ~ k] for k = 1,2, 3.

262. A variable X has the Pareto distribution with density

x ~ 2.

Give the graph of the function g«(j) = P[IX -Ill> (j] where 11 = E(X), and compare this with the upper bound given by Chebyshev's inequality.

263. Given that Var(X) = 9, find the number n of observations (the sample size) required in order that with probability less than 5% the mean of the sample differs from the unknown mean 11 of X more, (a) than 5% of the standard deviation of X, (b) than 5% of 11 given that 11 > 5. Compare the answers obtained by using Chebyshev's inequality and the CL T.

264. In a poll designed to estimate the percentage p of men who support a certain bill, how many men should be questioned in order that, with probability at least 95%, the percentage of the sample differs from p less, (i) than 1 %, (ii) than 5%, given that, (a) p < 30%, and (b) p is completely unknown.

265. Each of the 300 workers of a factory takes his lunch in one of three

Exercises 69

competing restaurants. How many seats should each restaurant have so that, on average, at most one in 20 customers will remain unseated?

266. The round-off error to the second decimal place has the uniform distribution on the interval (- 0.05,0.05). What is the probability that the absolute error in the sum of 1,000 numbers is less than 2?

267. The daily income of a card player has uniform distribution in the interval ( - 40, 50).

(a) What is the probability that he wins more than 500 drachmas in 60 days?

(b) What amount c (gain or loss) is such that at most once in 10 times on average the player will have income less than c during 60 days?

268. In the game of roulette the probability of winning is 18/37. Suppose that in each game a player earns 1 drachma or loses 1 drachma. How many games must be played daily in a casino in order that the casino earns, with probability 1/2, at least 1,000 drachmas daily? Then find the percentage of days on which the casino has a loss.

269. Show that the r distribution of Exercise 202 tends to the normal distribution as s ~ 00.

CHAPTER 9

Special Topics: Inequalities, Geometrical Pro ba bili ties, Difference Equations

Elements of Theory

A. Inequalities

In Chapter 8 several inequalities such as the Chebyshev, Markov, and Kolmogorov inequalities were given. Here we give some inequalities concerning expectations.

1. A function g, defined over the interval ((l(, [3), is called convex in ((l(, [3) if for each A (0 :;:; )" :;:; 1), and every pair Xl' X 2 E ((l(, [3), we have

(9.1)

Remark. (a) If 9 is continuous in ((l(, [3) then the validity of (9.1) for A = 1/2 implies the convexity of g.

(b) If the second derivative g"(x) exists, then (9.1) is equivalent to

g"(x) 2 o. (9.2)

The function 9 is called concave if - 9 is convex.

2. Jensen's inequality. Let g(x) be a convex function in ((l(, [3) and P[X E

((l(, [3)] = 1. If E[g(X)] < 00, then

E[g(X)] 2 g(EX). (9.3)

The proof is based on the observation that there exists a line of support of the curve y = g(x) through the point (11, g(l1» with equation y = g(l1) + A(X - 11)

Elements of Theory

such that

3. Cauchy-Schwarz inequality. If E(X2) and E(y2) exist then

[E(XY)J2 :s;; E(X2)E(y2);

equality holds only if Y = AX (A constant). See Exercise 103.

71

(9.4)

4. Holder's inequality. Let p > 0, q > 0, such that p-l + q-l = 1 and X, Y positive random variables. Then

E(XY) :s;; [E(XP)JI/P[E(yq)Jl/q. (9.5)

This gives (9.4) when p = q = 1/2.

Proof The function

g(X) = log x is concave for x > 0, i.e., for each A (0 :s;; A :s;; 1)

), log Xl + (1 - A) log X2 :s;; IOg(AXI + (1 - A)X2)·

Hence

xtx1-). :s;; AX I + (1 - A)X2,

and setting A = p-l, 1 - ), = q-l, Xl = XP/E(XP), X2 = yq/E(P), we obtain

X Y XP yq [E(XP)] lip + [E(yq)Jllq :s;; A E(XP) + (1 - A)E(yq)'

Taking expectations gives (9.5).

5. Minkowski's inequality. For every integer k ;::::: 1 and any random variables X and Y with E(Xk) < 00, E(yk) < 00,

[EIX + Ylkr lk :s;; [EIXn l/k + [EI YlkJl/k. (9.6)

If we define the norm of X by IIXII = [EIXlkJl/k, then (9.6) becomes the triangle inequality

IIX + YII :s;; IIXII + II YII. Proof For k = 1, obvious. For k > 1, replacing Y by IX + Yl k - l in (9.5),

we obtain

EIX + Ylk:s;; E(lXIIX + Yl k - l ) + E(IYIIX + Yl k - l )

:s;; [(E IXlk)l/k + (E I Ylk)l/kJ [E IX + YI(k-l)'r /', (9.7)

where k-l + S-l = 1 and therefore (k - l)s = k. Dividing both sides of (9.7) by (E IX + YI)I/' we obtain (9.6).

72 9. Special Topics: Inequalities, Geometrical Probabilities, Difference Equations

B. Geometrical Probabilities

Geometrical probabilities refer to events corresponding to subsets of a bounded set S of the Euclidean n-space En under the assumption that the distribution over S is uniform. Thus the probability of an event

is given by

A = {x: x E SA C S}

P(A) = /1(SA) /1(S) ,

where /1(R) denotes the Lebesgue measure (geometric size) of the set R, i.e., its length, area, volume, etc., depending on the dimensionality of R. The solution of problems in geometrical probability contributed a lot to the understanding of the role played by the specification of the random experiment related to a certain event. This is demonstrated, e.g., in the classical Bertrand paradox (see Exercise 169).

C. Difference Equations

In many cases a problem in probability can be solved more easily by constructing and solving a related difference equation, i.e., an equation of the form

g(x, f(x), 8f(x), ... , 8 rf(x» = 0,

where f is the unknown function and 8 denotes the forward difference operator defined by

and

N(x) = f(x + h) - f(x)

8 rf(x) = 8(8r- 1f(x»

If x = 0, 1,2, ... , then we set

Yn = f(n),

and we have, e.g.,

8Yn = Yn+1 - Yn'

for some fixed h > 0,

for r ~ 1 (8 Of == f).

n = 0,1,2, ... ,

Of special interest in applications are the linear difference equations with constant coefficients lXi' i.e., of the form

Yn+r + 1X1Yn+r-1 + ... + IXr- 1 Yn+1 + IXrYn = Zn' (9.8)

where n usually ranges over a set of positive integers. If IXr '# ° the difference equation is said to be of order r. The difference equation is called nonhomogeneous (complete) or homogeneous according to whether Zn '# ° or Zn = 0.

Elements of Theory

The characteristic polynomial tp(A) of (9.8) is defined by

tp(A) = Ar + 1X1 Ar - 1 + ... + IXr - 1 A + IXr •

The solution of the homogeneous equation

Yn+r + 1X 1Yn+r-1 + ... + IXr- 1Yn+1 + IXrYn = 0

is based on the

73

(9.9)

Proposition. (i) With each simple root A of the characteristic equation tp(A.) = 0 let us associate the special solution cA. n (c any constant) of the homogeneous difference equation (9.9).

(ii) With each real root A of multiplicity k of tp(A) = 0 associate the special solution

(co + c1 n + C2n2 + ... + Ck_l nk- 1)An,

where co, c 1, •.. , Ck - 1 are arbitrary constants. (iii) With each pair of (conjugate) complex roots oflength (absolute value)

p and argument () associate the special solution

pn[Ao cos(n() + Bo) + ... + Ak_1 nk-1(cos(n() + Bk-dJ,

where k is the mUltiplicity of the complex root and Ao, ... , A k - 1 , Bo, ... , Bk - 1

arbitrary constants. The sum of the above special solutions of (9.9) determine the general

solution of (9.9) containing r arbitrary constants whic~ are determined by the initial or boundary conditions satisfied by (9.9).

General solution ofthe complete equation (9.8). This can be found by adding a special solution of (9.8) to the general solution of the homogeneous (9.9).

Example. Solve the nonhomogeneous equation of the second order

Yn+2 - Yn+l - 6Yn = 2n, n = 0,1, .... (9.10)

The roots of the characteristic equation

tp(A.) = A2 - A. - 6 = 0

are Al = 3, A2 = - 2, and by the preceding proposition the general solution of the homogeneous equation

Yn+2 - Yn+1 - 6Yn = 0,

IS

Y: = c 1 3n + c2 (-2)",

with c1 , C2 arbitrary constants. For the general solution of (9.10) we must first find a special solution y~

which is to be added to y:. We consider the special solution

(9.11 )

where e has to be determined to satisfy (9.10); substituting (9.11) into (9.10) gives

i.e., y" = -(t)2".

Thus the general solution of (9.10) is

y" = y: + y~ = e1 3" + e2(-2)" - (t)2".

Exercises

A. Inequalities

270. If g(x) ~ 0 for every x and g(x) ~ e for x E (a, P), then

P[X E (a, P)] s e- l E[g(x)].

271. (Continuation). Show that for every constant t > 0

1 P[X > t] S )2 E(X + e)2.

(t + e

272. If Xl and X 2 . are independent and identically distributed random variables then for every t > 0

P[IX l - X 2 1 > t] s 2P[IXl l > !t].

273. Cantelli's inequality. One tail Chebyshev's inequality. If E(X) = 11, Var(X) = (J2, then

(J2

P[X - 11 s e] S 2 2' (J + e

e < 0,

e ~ 0,

(see Exercise 271).

274. * Gauss inequality. If Xo is the mode of a continuous unimodal distribution and ,2 = E(X - xof then

4 P[IX - xol ~ .h] s 9A. 2 '

275. If g(x) ~ 0 and even, i.e., g(x) = g( -x) and in addition g(x) is nondecreasing for x > 0, show that for every e > 0

P[IXI > ] < Eg(x) - e - gee) .

Exercises 75

276. (Continuation). If g(x) of Exercise 275 satisfies I g(x)1 ::;; M < 00, then

P[IXI :;::.: e] :;::.: Eg(X~ g(e).

277. (Continuation). Let g be as in Exercise 275 and P[IXI ::;; M] = 1 then

P[IXI > c] > Eg(X) - g(c). - - geM)

278.* Berge's inequality. Let E(Xi) = Ili' Var(X;) = (1? (i = 1,2) and p(x l , x 2 ) = p, then

p[ (IXI - Ilil IX2 - 1l21) ] 1 + Ji=P2 max , > c::;; 2 • (11 (12 C

279.* Jensen's inequality for symmetric matrix functions (Cacoullos and Olkin, Biometrika, 1965). The symmetric matrix function G(X) of the matrix X of order m x n is called convex if, for every pair X I' X 2 of m x n matrices and 0 ::;; ). ::;; 1, by analogy to (9.1), it satisfies

G(..1.XI + (1 - ).)X2 ) ::;; ..1.G(Xd + (1 - ..1.)G(X2 ),

where for symmetric matrices X, Y we write X :;::.: Y if X - Y is a nonnegative definite matrix. Let X be a random matrix, i.e., a matrix whose elements are random variables and G a convex matrix function. Then G(EX) ::;; EG(X).

280. For a positive random matrix show that

E(X- I ) :;::.: (E(X)fI.

281. Show that if E(Y) < 00 and Y> 0 then

E 10g(Y) ::;; log E(Y).

282. For independent and identically distributed random variables Xl' X 2'

... , Xn with P[Xi > 0] = 1 and Var(log X;) = (12 show that for every 8> 0

Hence deduce

B. Geometrical Probabilities

283. A point is chosen at random on each of two consecutive sides of a rectangle. What is the probability that the area of the triangle bounded by the two sides of the rectangle and the line segment connecting the two points is less than 3/8 of the area of the rectangle?

284. Two points are selected at random on segment AB which is thus divided into three parts. What is the probability that the three line segments can form a triangle?

285. What is the probability that three points selected at random on the circumference of a circle lie on a semicircle?

286. A line segment of length 1 is divided into five parts by four points selected at random on it. What is the probability that each part is less than 1/2?

287.* A line segment AB is divided by a point C into two parts AC = a and CB = b. Two points X and Yare taken on A C and CB, respectively. What is the probability that AX, XY, BY can form a triangle?

288.* BufJon's needle problem. A smooth table is given with equidistant parallel lines at distance 2a. A needle of length 2J1. (J1. < a) is dropped onto the table. What is the probability that the needle crosses a line?

289. * A closed convex curve with diameter less than 2a is dropped onto the table of Exercise 288. Find the probability that a line is crossed by the curve.

C. Difference Equations

290. A has a 1 O-drachma coins and B has b 5-drachma coins. Each of them takes a coin out of his pocket and gives it to the other. This is repeated n times. What is the probability that A gives a to-drachma coin to B on the nth exchange?

291. A tosses a die until an ace or a six appears twice. (a) What is the probability that the game terminates with the appearance

of an ace after no more than n tosses? (b) What is the probability that the game ends with the appearance of an

ace?

292. A professor has three lecture rooms A, B, and C available each year for teaching a course. Since he wishes to use a different classroom every year, at the end of the year he chooses one of the other two rooms for the next academic year as follows. The first year he teaches in A and selects between Band C by tossing a coin. If heads appears he chooses B, otherwise he selects C. At the end of each year he makes his selection in the same way. Find the probability Pn that he teaches in A during the nth year. Similarly for Band C.

293. Find the probability P. that in a sequence of n Bernoulli trials, with probability of success (S) p the pattern SF does not appear at all (F == failure).

294. Let Yk be the expected number of occupied cells when placing (at random) k balls into n cells. Find Yk:

(a) by using difference equations; (b) by referring to the expected value of a binomial.

Exercises 77

295. A pair of dice is thrown n times; at each throw we observe whether a double six appears or not. Show that the probability Po of an even number of double six's satisfies the difference equation

17 1 Po -18 Po - 1 = 36'

Hence find Po'

296. * Each of n urns contains w white balls and b black balls. One ball is transferred from the first urn to the second urn, then another ball from the second urn is transferred to the third one, and so on. Finally, we draw a ball from the nth urn. What is the probability that it will be white?

297. A and B play the following game. A throws a coin I with probability of heads P until he gets tails; then B begins throwing another coin II with probability of heads pi until he gets tails. Then A begins again and so on. Given that A begins the game, what is the probability of heads at the nth throw?

298. A throws two dice and he gets r points if r aces appear (r = 0, 1,2); the game stops when no ace appears for the first time. Find the probability Po that he has n points at stopping. If the game continues indefinitely, what is the probability that A has n points at some point of the game?

299. A box contains a good lamps and b bad lamps. At each draw of a lamp, if it is good it is kept in the box, otherwise it is replaced by a good one (from another box). What is the expected number of good lamps in the box after n trials? Hence calculate the probability of selecting a good lamp at the nth trial.

300. Let Pk denote the probability that no cell takes more than two balls when placing k balls in n cells at random. Find, (a) Pk (b) limk _ oo Pk and n = constant, (c) lim Pk for k = constant and n ~ 00.

CHAPTER 10

General Exercises

301.* Using the Cauchy-Schwarz inequality show that

get) = log E(IXlt)

is a convex function. Then show that [E(lxlt)]l/t is an increasing function of t.

302. * Every real characteristic function qJo(t) satisfies the inequality

1 - qJo(2t) :s; 4(1 - qJo(t)).

303. * Every real characteristic function CJJo(t) satisfies

1 + qJo(2t):2: 2 {qJo (t)} 2.

304. If Xl' X 2 , X 3 , X4 are jointly normal with COV(Xi' X) = E(XiX) = (1ij' show by the use of moment generating functions, that

E(X 1 X 2 X 3 X 4 ) = (112(134 + (114(123 + (113(124·

305. Let Xl' X 2 , ••• , Xn be a random sample from the uniform distribution in (0, 1) and let y" = n min(X1, X 2 , •.• , Xn). Using characteristic functions, show that the asymptotic distribution (n --> 00) of Yn is the exponential e-Y,

y > 0.

306. Let (Xl' Yd, ... , (Xn' y") be a random sample from the bivariate normal with means 0, variances 1, and correlation p. Let the random variable

Find E(2).

Z. = {I if (Xj - ~)(lj - ~) > 0, J ° if (Xj - X)( lj - Y) < 0.

307.* n tickets are numbered 1 through n. Of these, r tickets drawn at random are the lucky ones. A lucky ticket receives an amount equal to the

10. General Exercises 79

number it bears. What is the expected amount given on every draw? What is the variance of this amount?

308. In the generalized binomial with P[Xi = 1] = Pi (i = 1,2, ... , n), show that as n --+ 00 the distribution of the number of successes X = L7=1 Xi tends to the Poisson with parameter A where A = limn_oo L7=1 Pi assuming maxi Ai --+ 0.

309. Let X be the number offailures preceding the nth success in a sequence of independent Bernoulli trials with probability of success p. Show that if as n --+ 00, q = 1 - P --+ 0, so that nq --+ A, then

310. If the continuous random variable X is such that for some integer k > 0, kX has the Gamma distribution and the conditional distribution of Y given X is Poisson, then Y has a negative binomial distribution.

311. Let Xl' X 2 , .•• , Xn be a random sample from the Laplace distribution

-00 < x < 00.

By using characteristic functions show that the random variable

is distributed as (0"/2n)X~n' Hence deduce that E(Y) = 0".

312. Let Xl' X 2, ... , Xn be a random sample from a distribution with E(Xi ) = ° and Var(X;) = 1. Show that

j~sn Y" = -n--~ N(O, 1),

L Xl i=l

Sn Zn = JFi ---+ N (0, 1), n n-a)

L xl i=l

where Sn = Xl + X2 + ... + Xn.

313. Let X(1)' ... , X(n) be the ordered sample from a uniform distribution in (0, 1) and Y,. = X(1) + X(2) + '" + X(r)' Show that the conditional distribution of W = y"/l(r+1) is the distribution of the sum of r observations of a random sample from the same uniform distribution.

314.* N points Xl' X 2 , ... , Xn are selected at random in (0,1). Show that

80 10. General Exercises

the distribution of the sum Sn = Xl + X 2 + ... + Xn has density

f,,(X) = 1 {xn-I _ (n)(X _ l)n-1 + ... } (n - I)! 1

= n {(n - x)n-I _ (n - 1) (n - 1 - x)n-I + ... } (n - 1)1 n 1 n - 1 '

where the sum extends as long as x-I, x - 2, ... and (n - 1 - x), (n - 2 - x), ... are positive.

315. Let X I, X 2, ... , Xn be independent discrete variables with the same uniform distribution on the set of integers 0, 1, ... , m - 1, i.e.,

1 P[XI =j] =

m for j = 0, ... , m - 1.

Find the probability generating function of the sum Sn = Xl + ... + Xn and hence the probability function of Sn (Feller, 1957, Exercise 18, page 266).

316. (Continuation). A die is thrown three times. What is the probability that the sum of the outcomes is 1O?

317.* Generalized Banach match box problem. A chain smoker has m + 1 match boxes; each time he lights a cigarette he chooses one of the boxes at random. Suppose that (to begin with) every box contains N matches. Let Xl' X 2 , ••• , Xrn be the numbers of matches left in the remaining m boxes, when:

(a) a box is found empty for the first time; (b) any box is emptied first. Note: The case m = 1 is known as the Banach match box problem (Feller,

ibid, p. 157).

318. (Continuation). Suppose that box Ck (k = 1, ... , m) is chosen with probability Pk. What is the probability that Ck empties first?

319. Show that Xl' X 2, X3 with ajoint discrete distribution are independent if and only if

P[X3 = x31X2 = X2' XI = Xl] = P[X3 = X3]

for all (Xl' X2) with P[XI = XI' X 2 = x 2] > 0 and

P[X2 = x2 1XI = Xl] = P[X2 = x 2] for all Xl with P[XI = Xl] > o. 320. Let X I' X 2, ... , Xn be independent and identically distributed positive

random variables with E(X;) = /l, E(Xi-l) = r. Let Sn = XI + X 2 + ... + Xn. Show that E(S;;I) exists and

( Xi) _ 1 E - --, Sn n

i = 1, ... , n.

(Feller, ibid, Exercise 36, page 226.)

10. General Exercises 81

321. (Continuation). Show that

(Sm) m E Sn = -;;- for m ::;; n,

322. A caterpillar is moving on the edges of a tetrahedron ABCD on whose top there is glue. In a unit of time the caterpillar goes from any vertex (except D) to any other vertex with the same probability 1/3. Suppose that the caterpillar at time t = 0 is on the vertex A. What is the probability that

(a) the caterpillar finally gets stuck; (b) the caterpillar finally gets stuck coming from vertex B?

323. A flea moves at random on a plane with leaps of constant length a and in a random direction at each step. If it starts from the origin, let (Xn' y"f be the position of the flea after n jumps. Where

n

Xn = L a cos ()i' i=l

n

Y" = L a sin ()i' i=l

The angles ()i are independently uniform in (0, 2n). (a) Find, (i) E(Xn ), E(Y,,), Var(Xn ), Var(y"), (ii) E(R;) where R; = X; + y"2. (b) Show that the X n , Y" are uncorrelated but not independent. (c) Find the distribution of (Xn' y") for large n (n -+ 00) and the density of

Rn. What is the expected distance of the flea from the origin for large n?

324. * A non normal distribution with all marginals normal. Let the random variables XI' ... , Xn have the joint density

1 (1 ~ 2) [On 1 2 ] f(x I' ... , Xn) = (2 nl2 exp --2.L... Xi 1 +. (Xi exp( -2"Xi » . n) ,=1 ,=1

Show that any n - 1 of the XI' ... , Xn are independent normal with mean 0 and variance t. Are the X I, ... , Xn jointly normal?

325.* In Exercise 288 suppose the plane is divided into squares by equidistant lines parallel to the axes. Let 2a be the side of each square and 21l the length of the needle. Find the average number of lines crossed by the needle, (a) Il > 0(, (b) when Il is arbitrary.

326. * Let X, Y be independent X2 variables each with n degrees of freedom. Show that

z=JnX-Y 2 JXY

has the Student distribution with n degrees of freedom. Hence deduce that:

(a) tn=f(ft:-k). where Fm,n denotes an F variable with m and n degrees of freedom.

82 10. General Exercises

(b) If tn(a) and Fn,n(a) are the upper percentile points of tn and Fn,n, respectively, express one in terms of the other.

327. If for a given value A. of A the random variable X(A) is Poisson with parameter A. and A is a random variable with a gamma density

aP r(A.) = r(p) A.P-le-a..,

show that X(A) has the negative binomial distribution

Pk = P[X(A) = kJ = ( ~p) ( _ p)kqP, k = 0,1, ... ,

where p = (1 + a)-l, q = a(a + 1)-1.

328. * Show that for any two random variable(s) X and Y with Var(X) < 00,

Var(X) = E[Var(XI Y)] + Var[E(XI Y)].

329.* Using the distribution of X(/) where Xli)' ••• , X(n) is the ordered sample from the uniform in (0, 1), show the following relation between the incomplete B function lx(nl' n2) and the binomial distribution function:

t (n)pkqn-k = Ip(r, n - r + 1) k=r k

= 1 fp xr-1(1 _ x)n-r dx. B(r, n - r + 1) Jo

SUPPLEMENTS

SUPPLEMENT I

Miscellaneous Exercises

I-I. For any two events A, B show that the following relationships are equivalent: A c B, A' :=J B', A u B = B, and A n B' = 0.

1-2. Show that (A.6.. BY = (AB) u (A' B') and P(A.6.. B) = P(A) + P(B) -2P(AB), where .6.. is the symmetric difference, defined in Exercise 34.

1-3. Let n be the number of elements in the sample space Q and N(E) the number of elements in the event E. Show that the data: N = 1,000, N(A) =

525, N(B) = 312, N(C) = 470, N(AB) = 42, N(AC) = 147, N(BC) = 86, and N(ABC) = 25 are inconsistent.

Hint: Check N(A u B u C).

1-4. (Lewis Caroll, A Tangled Tale, 1881). In a fierce battle, at least 70% of the soldiers lost an eye, at least 74% lost an ear, at least 80% lost an arm, and at least 85% lost a leg. What is the smallest possible percentage of soldiers that each lost an eye, an ear, an arm, and a leg?

Answer: 10%.

1-5. In a market research survey, from a sample of 1,000 smokers, 811 said that they prefer brand "A", 752 prefer "B", and 418 prefer "C". Furthermore, 570 said that they like both "A" and "B", 356 that they like both "A" and "C", 348 that they like both "B" and "C", and 297 that they like all three brands. How did the researcher discover that the results were inconsistent?

Hint: How many prefer at least one of the three brands?

1-6. We throw two dice once. Let A: doubles, B: at least one die shows 5 or 6, C: the sum of the outcomes of the two dice is less than 6.

(a) Find P(A), P(B), P(C), P(AB), and P(AC). (b) Using only the results of (a) calculate: P(A u B), P(B u C), and P(A u C). Answer: (a) 1/6,5/9,5/18, 1/18, 1/18. (b) 2/3,5/6, 7/18.

86 Supplement I. Miscellaneous Exercises

1-7. Three athletes are participating in the Marathon. Their chances of finishing under 2 hours are 1/3, 1/5, and 1/12, respectively. Their times for the duration of the course are independent. What is the probability that at least one finishes in less than 2 hours?

Answer: 23/45.

1-8. A horse has probability p of jumping over a specific obstacle. Given that in 5 trials it succeeded 3 times, what is the conditional probability that it had succeeded in the first trial?

Answer: 3/5.

1-9. In a train car 3 seats are facing the front and 3 are facing the rear. Two women and three men enter the car and sit at random. What is the probability:

(a) that the two women are facing each other? (b) that two men are facing each other? Answer: (a) 1/10. (b) 3/10.

1-10. On the basis of the following data, examine the relationship between the father having brown eyes (event A) and the son having brown eyes (event B). The percentages observed were: AB: 5%, AB': 7.9%, A'B: 8.9%, and A'B': 78.2%.

Answer: P(B/A) = 0.39 and P(B/A') = 0.10.

1-11. We throw three dice. What is the probability that at least one 6 appears given that the three outcomes were completely different. Generalize the result for the case of a regular polyhedral die.

Answer: G) I G)' 1-12. Ninety percent ofa population have a television set (event A), while

80% have a car (event B). What is the minimum percentage of the population with a car that also have a television?

Answer: P(A/B) ~ 87.5%.

1-13. Let P(A) = (I. and P(B) = 1 - I: where I: is a small number. Give the upper and lower bound of P(A/B).

Answer: ((I. - 1:)/(1 - 1:) ~ P(A/B) ~ (1 - 1:).

1-14. Let Xl' X 2 be independent random Bernoulli variables with the same parameter p and let Y = 0 if Xl + X 2 = even and Y = 1 if Xl + X 2 = odd.

(a) For what values of p are Xl and Y independent? (b) Using (a), define three events A, B, and C such that they are pairwise

independent but not completely independent. Answer: (a) p = 1/2. (b) A = {Xl = O}, B = {X2 = O}, C = {Y = O}.

1-15. We throw three dice twice. What is the probability of having the same result the second time when the dice are, (a) not distinguishable? (b) distinguishable?

Answer: (a) 1/56. (b) (6 + 90 x 3 + 120 X 6)/66 •

Supplement I. Miscellaneous Exercises 87

1-16. At a party, A is introduced to 6 women and 4 men. What is the probability that he identifies all the couples correctly if he knows that there are, (a) 4 couples, (b) 3 couples?

Answer: (a) 1/360. (b) 1/480.

1-17. A young person driving a motorcycle on a highway with 4 lanes changes lane every 15 seconds. What is the probability p of finding himself in the lane in which he started 1 minute later?

Answer: The answer depends on whether the initial lane is an interior or an exterior one. In the first case p = 5/8, and in the second case p = 1/5.

1-18. In city A it rains half of the days of the year and the weather forcast is correct with probability 2/3. Mr. Sugar worrying about getting wet always takes his umbrella when the forcast calls for rain and with probability 1/3 otherwise. Find the probability that:

(a) he is caught in the rain without his umbrella; (b) he has his umbrella with him when there is no rain. Answer: (a) 2/9. (b) 5/9.

1-19. The telegram signals (.) and (-) are sent with proportion 3:4. Due to faults in transmission a dot (.) becomes (-) with probability 1/3. If a signal has arrived as ( . ) what is the probability that it was transmitted correctly?

Answer: 43/84 (Bayes formula).

1-20. A die is thrown 10 times and let the maximum outcome be M and the minimum m. Find the P(m = 2, M = 5).

Hint: Consider the probability P(m ~ 2, M ~ 5). Answer: (4/6)10 - 2(3/6)10 + (2/6)10.

1-21.* A theorem of Poisson. An urn contains b black balls and w white balls. n balls are drawn from the urn at random, their color unnoticed, then m additional balls are drawn. Show that the probability that there are k black balls among the m balls is the same as if the m balls were chosen from the beginning (i.e., before choosing the n balls).

In other words, in sampling from a finite population the final (hypergeometric) probabilities are not affected by the content of an initial sample when we do not know that content. Obviously, if we know the composition of the first sample (i.e., the colors of the n balls), then, in general, the probabilities are affected.

Hint: Verify first the case m = 1, calculating the probability P(An) of choosing a black ball on the n + 1 trial; show that P(An) = b/(b + w) with O~n~b+w-l.

1-22. After an earthquake the books of a large library fall on the floor and the servant, who does not know how to read, res helves them in any order.

(a) What is the probability that exactly k books are placed back in their initial position?

(b) How many books are expected to be reshelved in their initial position? Answer: (a) e-1/kL (b) 1.

1-23. John Smith's problem. John Smith in 1693 posed the following problem: Is the probability for each of the three players to win the same if: the first must obtain at least one ace throwing 6 dice, the second at least two aces throwing 12 dice, and the third at least three aces throwing 18 dice? Newton and Toilet found that the first player has a better chance than the second one, and the second player has a better chance than the third one.

1-24. A homogeneous and regular die with v faces is thrown n times. Let p(n, k) be the probability that a given face appears less than k times. Show that:

(a) p(vn, n) is a decreasing function of v for given n; (b) p(vn, n) < 1/2; (c) p(2n, n) --+ 1/2 as n --+ 00 (De Moivre central limit theorem).

1-25. The Massachusetts State Lottery issues 1 million tickets. The first winning ticket wins $50,000, the next 9 winning tickets win $2,500 each, the next 90 tickets win $250 each, and 900 tickets win $25 each. What is the expected gain of a person buying, (a) one ticket? (b) ten tickets?

Answer: E(I?=l Xi) = I?=l E(XJ (a) $0.1175. (b) $1.175.

1-26. (Continuation). What is the answer to (a) and (b) if: (i) the state issues 2 million tickets?

(ii) the winning tickets win twice as much?

1-27. In a survey on the housing problem, 2 apartments are chosen from each of 4 buildings (having both owner-occupied as well as rented apartments). If we know that:

Building No. 1 Owner-occupied 5 Rented 8

234 482

10 9 10

What is the expected number of owner-occupied apartments in the above sample?

Answer: 2(5/13 + 4/14 + 8/17 + 2/10). Use additivity of expectations.

1-28. Out of 365 people: (a) What is the probability that at least two have their birthday on 2 April? (b) What is the expected number of people having their birthday on 2

April? (c) What is the expected number of days in a year that are birthdays: (i) for

at least one person? (ii) for at least two persons? How can you approximate the answers to the above questions?

Supplement I. Miscellaneous Exercises

Answer: (a) 1 - (364/365)365 - 365(364)364/365365 ~ 1 - 2e-1 = 0.26. (b) 1.

(c) (i) 365 {1 - G~:r65} ~ 365(1 - e-1 ) = 231. .

(ii) 365p where p was given in (a).

89

1-29. The duration T of a telephone call satisfies the following equation:

P[T> t] = p·exp[ -At] + (1 - p)·exp[ -,ut] for t > 0,

where 0 :s; p :s; 1, A > 0, ,u > O. Find E(T) and Var(T). Answer: E(n = piA + (1 - p)/,u. Var(n = 2p/A2 + 2(1 - p)/,u2 - [E(n]2.

1-30.* The generalized problem of points by Pascal. Two gamblers A and B playa series of independent games in each of which A has probability p of winning and B has probability q = 1 - p of winning. Suppose that at a given moment the series of games is interrupted and A needs m additional points and B needs n additional points to be the winner. The bet should be split between A and B in the ratio P(A):P(B), where P(A) = 1 - P(B) is the probability that A eventually wins. Show that

m+n-l (m + n - 1) n-l (m + k - 1) PA = L pk - qm+n-l-k = L pmqk.

k=m k m=O k

Note: The above solution was first obtained by Montmort (1675-1719). Explain the idea behind each term in the expressions for PA • Try to show the equality between those two expressions for PA using the properties of binomial coefficients (see Chapter 1).

1-31. An employee leaves between 7:30 a.m. and 8:00 a.m. to go to his office, and he needs 20-30 minutes to get to the railway station. The departure time and the duration of travel are distributed independently and uniformly in the corresponding intervals. He can take either of the following two trains: the first leaves at 8: 05 a.m. and takes 35 minutes; and the second leaves at 8: 25 a.m. and takes 30 minutes.

(a) What time, on average, does he arrive at his office? (b) What is the probability that he misses both trains? (c) How many times is he expected to miss both trains in 240 working days? Answer: (a) 8: 50. (b) 1/24. (c) 10.

1-32. Show that if X and Yare independent, then (cf. Exercise 328)

Var(XY) = Var(X)·Var(Y) + Var(X)·[E(y)]2 + Var(Y)·[E(X)Y

Conclude that if Xl, ... , Xn are completely independent with means zero, then

Var (11 X) = 11 Var(X;).

1-33. We toss a fair coin until two consecutive heads or tails appear. What is the probability that an even number of tosses will be required?

Answer: 2/3.

1-34.* Maxwell-Boltzmann statistics. s distinguishable balls are placed randomly into n different cells. Find the probability that exactly r cells will be occupied.

Answer:

Hint: The probability that exactly k events occur among n events A 1 ,

... , An is given by (see, e.g., Feller, 1957),

Sr - (r: 1)Sr+1 + ... + (-l yG)Sn (cf. (Ll)).

1-35. A large board is divided into equal squares of side a. A coin of diameter 2r < a is thrown randomly onto the board. Find the probability that the coin overlaps, (i) with exactly one square, (ii) with exactly two squares.

Answer: (i) (1 - 2r/af. (ii) 1 - 4(r/a)2.

1-36. A square of side a is inscribed into a circle. Find the probability that: (i) any randomly selected point of the circle will be an interior point of the

square. (ii) of eight randomly selected points in the circle, three fall inside the

square, two in one of the four remaining parts of the circle, and one in each of the other three parts of the circle.

Answer: (i) 2/n. (ii) (8! 4/3! 2! 1! 1! 1!)(2/n)3(1 - 2/n)SCl/4)5.

1-37. The distance between two successive cars on a highway follows the exponential distribution with mean 20m. Find the probability that there are 90 to 110 cars on a 2-km stretch of the highway.

Answer: e-100 It280 100k/k! ~ 0.68.

1-38. The number of cars crossing a toll bridge follows the Poisson distribution with an average of 30 cars per minute. Find the probability that in 200 seconds no more than 100 cars will pass through.

Answer: The sum of independent exponentials is Erlang (i.e., gamma with s = 100, see (2.11));

1 fOO 99 -t/2 _ ~ -100 lOok "'--'-"""'-;1'""0"""0 e dt - L.. e . 99! 2 200 t k=O k!

1-39. (Continuation). Establish the relation

An f.00 -AX n-1 d n~l -At (At)k e x x = L.. e --,

(n - I)! t k=O k!

between the Erlang distribution function and the Poisson distribution function.

Hint: The probability that less than n events occur in time t, according to a Poisson distribution (process) X(t) with

is equal to the probability that the waiting time w,. until the nth event exceeds t. Hence, also conclude that the density of w,. is the Erlang density on the left side of (*) above.

1-40. (Continuation). Show that

1 fOO "-1 Ak e- x x"-1 dx = L e- A_.

(n - I)! A k=O k!

1-41. Define (a; b) = (b) - (a) for ° < a the distribution function of N(O, 1). Show that <1>(0; 1) > <1>(1; 2). Generalize this result for any equal intervals.

Answer: (a + x) - (a) > (b + x) - (b); x > 0.

1-42. Using the fact that if X follows the normal distribution N(O, 1), then

I P[X2 ::s; x] = (2/n)I/2 I x e- u2/2 du,

conclude that

r(1j2) = I() x(i/2)-le- x dx = In.

1-43. The probability of hitting a target with one shot is PI and the probability of destroying the target with k(k ~ 1) shots is 1 - p~. Find the probability of destroying the target after n shots.

Answer: 1 - [1 - Pl(1 - P2)]".

1-44. Let X follow a Poisson distribution with parameter A. We select N random points from the interval [0, 1]. Let Xi be the number of points in the interval [(i - 1)/n, i/n] (i = 1, ... , n). Show that the random variables Xi are independent and identically distributed.

1-45. If X and Yare independent with P[X = 0] = P[X = 1] = 1/2 and P[y::s; y] = y (0 ::s; y ::s; 1), find the distribution ofthefollowing random variables W, Z where:

(i) W = X + Y; (ii) Z = XY. Answer: (i) few) = 0.5, 0< w < 2;

(ii) fez) = {0.5, 0.5,

z = 0,

o<z<1.

1-46. Let X be a uniform random variable on the interval (-a, a) and Ya random variable independent of X, with distribution function F(y). Show that the distribution function of G = X + Y is

1 fa G(z) = - F(z - x) dx. 2a -a

1-47. Let X be uniformly distributed on the interval (0, 1). Find the prob-ability density function of:

(i) Y = aX + b; a, b constants (a > 0); (ii) Z = I/X;

(iii) W = g(X), where g is a continuous and monotone function of x in (0, 1). Answer:

(i) Uniform; (ii) fez) = l/z2, 1 < z < 00;

(iii) few) = 1/lg'(w)l·

In addition to Problems 1-23 (John Smith, 1963) and 1-30 (points of Pascal), it is worth giving here some other problems which are of some historical interest, and show the difficulty in solving probability problems encountered by even well-known figures in the history of mathematics.

1-48. Luca dal Bargo or Pacciali, 1494. Each of three players A, B, and C has probability 1/3 of hitting a target better than the other two. At each round of the game each shoots at the target once. The winner is the first player to win six rounds. They bet 10 ducats each. At a certain point, however, when A has 4 victories, B has 3 victories, and C has 2 victories, they have to terminate the game. How should they split their bet? (cf. Problem 30).

1-49. Huyghens,* 1657. A and B playa game in which each of them throws two ordinary dice. A wins if he scores 6 and B wins if he scores 7. First, A throws the dice once, then B throws the dice twice, and they continue alternating after each throw until one of them wins. What is the probability of winning for each of them?

Note: Huyghens and Spinoza gave the same answer:

P(A)/P(B) = 10355/1227 ~ 0.84.

(cf. Exercise 67).

I-50. Huyghens, 1657. A deck of forty cards consists of 10 red, 10 black, 10 blue, and 10 green cards. Four cards are drawn at random and without replacement. What is the probability that all the colors will show up?

(Huyghens gave the answer 1,000/9,129 = 0.109; is it correct?)

1-51. Huyghens, 1657. An urn contains 8 black balls and 4 white balls. Each of three players A, B, and C (and in that order) draws a ball without replace-

* He is the author of the first book on probability, De Ratiociniis in Ludo Aleae, 1654.


ment and the first one to draw a white ball wins. What is the probability of winning for each of them?

Hint: A can win on the first, fourth or seventh draw, etc.

1-52.* Montmort, 1708. Three players A, B, and C play the following game. They play three rounds for each of which, each of them has probability 1/3 of winning. A is the winner of the game if he wins a round before B or C wins two rounds; B is the winner if he wins two rounds before player A wins one round, or C wins two rounds; C wins like B. What is the probability of winning for each of them?

(Montmort gave P(A) = 15/27, P(B) = P(C) = 5/27).

1-53.* Daniel Bernoulli, 1760. A container has 2N cards, numbered in such a way that each of the numbers 1, 2, ... , N appears twice. 2N - n cards are drawn at random. What is the expected number of pairs of cards remaining in the container?

(Bernoulli found n(n - 1)/(4N - 2)).

I-54. Euler, 1763. N lottery tickets are numbered 1 to N. n tickets are selected at random. What is the probability that r consecutive numbers are

selected (r = 1,2, ... , n)? /(10) Example: For N = 10, n = 4, r = 4, the probability is 7 4 = 1/30.

I-55. * Condorcet, 1785. The probability of event A is p. What is the probability that:

(a) in n trials event A occurs k times consecutively? (b) in n trials event A occurs k times consecutively and then AC also occurs

k times consecutively? (c) in n + m trials event A occurs n times given that in r + k trials it

occurred r times? Condorcet gave the following answer for (c):

(r + k + I)! (r + n)! (k + m)!jr! k(r + k + m + n + I)!. Laplace and G. Polya disagreed with this solution. What do you say?

1-56.* Laplace, 1812. In a lottery with N tickets, n numbers are chosen as winners. What is the probability p that in k lottery games all the N numbers will appear?

(Laplace for N = 90, n = 5, and k = 85 gave p = 1/2).

SUPPLEMENT II

Complements and Problems

1. Multivariate Distributions

1.1. The density of the impact point (X, Y) of a circular target is given by

f(x, y) = c(R - Jx2 + y2)

Find, (a) the constant c, (b) the probability p that the impact point falls in a circle with radius a < R centered at the origin.

Answer: (a) c = 3/R2. (b) p = 3a2/R2(1 - 2a/3R).

1.2. The random variables X, Y satisfy the linear relation

aX - bY = c,

where a, b, c are constants. Find, (a) the correlation p(X, Y), and (b) the ratio ux/uy of the standard deviations.

Answer: (a) p = sg(ab). (b) ux/uy = Ib/al.

1.3. The dispersion (variance~covariance) matrix of a three-dimensional normal distribution is

2

6

3

-2] 3 . 8

If /lx = /ly = /lz = 0, find the den~ f(x, y, z) and its maximum. Answer: f(x, y, z) = 1/(2nj230n) exp{ - 2jo(39x2 + 36y2 + 26z2

- 44xy + 36xz - 38yz)}

fmox = f(O, 0, 0) = (2nJ230n)-1 = 0.00595.

1. Multivariate Distributions 95

1.4. The joint density of X and Y is given by

f(x, y) = ye- y(x+l), x> 0, y > 0.

Find, (a) the marginal densities of X and Y, and (b) the conditional distribution function Fx(xIY) of X given Y = y.

Answer: (a) fx(x) = l/(x + 1)z, fy(y) = e-Y• (b) 1 - e-YX •

1.5. Suppose that the random variable X coincides with the random variable Xi with probability Pi' that is,

P[X = XJ = Pi

Show that

and E(X;) = mi ,

V(X) = I Pi V(Xi) + V(M), i

where P[M = mJ = Pi (i = 1,2, ... ).

i = 1,2, ....

1.6. The random vector (X, Y, Z) has density f(x, y, z). X, Y, Z can only be observed simultaneously. An observation gave the values u, v, w without knowing the correspondence between them and X, Y, Z. Find the probability that X = u, Y = v, and Z = w.

Answer:

f(u, v, w)

{j(u, v, w) + f(u, w, v) + f(v, u, w) + f(v, w, u) + f(w, u, v) + f(w, v, u)} .

1.7. The joint distribution function F(x, y) of the random variables X and Y is given by

F(x, y) = sin x sin y, ° s x S n/2, ° s y s n/2.

Find, (a) the density function f(x, y), and (b) the dispersion matrix. Answer: (a) f(x, y) = cos x cos y. (b) (J'; = (J'; = n - 3, (J'xy = 0.

1.8. The normal density ofa point A = (Xl' Xz, X 3) is given by

What is the probability that A will fall into an ellipsoid, with principal semiaxes E 1 , Ez, E3 along the axes Ox, Oy, Oz, where E 1 , Ez, E3 are the probable errors of X, Y, Z, respectively, i.e.,

P[IXI < E 1 ] = P[I YI < Ez] = P[IZI < E3] = 1/2.

Answer: (1/2) - (J2Eo/Jn)e-l/ZE6, where Eo is the probable error of N(O, 1).

1.9. The vector (X, Y) has the uniform distribution in the ellipse

XZ yZ aZ + bZ S 1.

96 Supplement II. Complements and Problems

Find: (a) the marginal distributions of X and Y; (b) the conditional distributions of X given that Y = y, and the conditional

distributions of Y given that X = x; (c) the Cov(X, Y) = axy . Are X, Y independent?~--:--::;-,-::;-:-Answer: (a)fx(x) = 2J1 - (x2/a2)/na,fy(y) = 2J1 - (y2/b2)/nb (Ixl < a,

Iyl < b). (b) fr(ylx) = (2bJ1 - (x2/a2»-1 for Ixl < a, Iyl < bJ1 - (x2/a2);fx(xIY)

similar. (c) axy = 0 but they are dependent since f(x, y) "# fx(x)fy(y).

1.10. The point (X, Y, Z) is uniformly distributed in the sphere x 2 + y2 + Z2 ~ R2. Find the density of Z and the conditional density f(x, Ylz). Generalize for an n-dimensional sphere.

Answer: fz(z) = 3(R2 - z2)/4R 3, Izl < R, f(x, ylz) = 1/n(R2 - Z2).

1.11.* Given fy(y), E[Xly], V(Xly), find E(X) and V(X). Answers:

E(X) = f: E(Xly)fy(y) dy = Ey[E(XI Y = y)],

V(X) = E[V(XI Y)] + V[E(XI Y)] (Exercise 328)

= f: V(Xly)fy(y) dy + t: [E(Xly) - E(X)]2fy(y) dy.

1.12. The vector (X, Y) has normal density

f(x, y) = ce-4X2_6xy-9y2.

Find the densities fx(x), fy(y), fx(xly), fy(Ylx) and the constant c. Answer:

3)3 c=--,

n

fx(xly) = }.e-<2X+1.5Y)"

Note that they are all normal.

3)3 {27 2} fy(y) = -- exp -- y , 2Jn 4

1.13.* Suppose that X has a spherically symmetric n-dimensional distribution, i.e., with density f(xI + ... + x;). Prove that the density g of the "generalized X2 variable" U = X'X is given by

1. Multivariate Distributions

where Cn is the "surface" area of the n-dimensional sphere, i.e.,

2nn/2

Cn = r(n/2)·

97

Hint: Make use of spherical (polar) coordinates r, 81 , ... , 8n - l : XI = r sin 81 ,

k-I n-l

Xk = r sin 8k [] cos 8i , i=l

k = 1, ... , n - 1, Xn = r [] cos 8i , i=l

where -tn < 8i :::;; tn (i = 1, ... , n - 2 and - n < 8n - 1 :::;; n). Prove that the Jacobian is r [];;=f (r cOSn-k-1 8k ) and make use of J':i~/2 COSk- 1 8 d8 = B(t, k/2).

1.14.* (Continuation). For n = 2, show that the polar coordinates (R, 8), where

x = R cos 8, Y = R sin 8,

are independent, 8 is uniformly distributed on [0, 2n), and R has density (cf. Exercise 221)

h(r) = 2nrf(r2 ).

1.15. * (Continuation). Show that the correlation coefficient based on the single observation (X, y), i.e.,

has density

1 1 f(ro) = - -I 2 '

n vi 1 - ro -1 < ro < 1.

Hint: ro = sin 28 has the same distribution as under a spherical normal (centered at the origin). Hence conclude that tan 28 = ro/J1 - r6 has the Cauchy distribution (t with one degree of freedom).

1.16. * If X = (X I' ... , Xn)' is N (0, In) and X is partitioned into k subvec-tors X(l)' ... , X(k) with VI' V2 , ... , Vk components, respectively, show that (X(l)X(1), ... ,X(k)X(k))/IXI 2 has the Dirichlet distribution (see (5.11)), with density

r -(n) 2 k-l

f(t t) - (1 t ... t )(vk /2)-1 [] t(v;/2)-1 I,···, k-I --k-~() - 1- - k-I . i , n Vk ,=1 r -

i=1 2

Hence, for VI = V2 = ... = Vs = 1, vs + I = n - s (k = s + 1),

(lXII, IX2 1,·· ., IXkl)

IXI

has the density

Vi ~ 0,

k

if L u'f < 1. i=l

This is the distribution of any k-dimensional subvector of an n-dimensional vector

U=(U1 ,···,U.)

uniformly distributed on the (surface of the) unit sphere

xi + ... + x; = 1.

1.17. * (Continuation). In general, if X has a spherical distribution (see Problem 1.13), since this is invariant under rotations in n-space, the distribution of the direction X/IXI is the same as that of V and is independent of its length IXI (for any spherical distribution with P[IXI = 0] = 0).

Hence conclude that the results of Problem 1.16 hold if X is any spherically symmetric random vector.

1.18. Let X, Y be random variables with

E(X) =}1, E(Y) = '1, VeX) = (Jf, V(Y) = (Ji, Cov(X, Y) = (J12·

Find an approximate formula for

E[g(X, Y)], V[g(X, Y)],

using the linear expansion of g(x, y) around (}1, '1) (up to second-order moments). Generalize for n random variables.

Answer: 1 [02g 2 02g 02g 2J

E[g(X, Y)] ~ g(}1, '1) + 2 ox2 (J1 + 2 oxoy (J12 + oy2 (J2 ,

( Og)2 2 og og (Og)2 2 V[g(X, Y)] ~ ox (J1 + 2 ox oy (J12 + oy (J2'

where the partial derivatives are evaluated at the point (}1, '1). Let X = (Xl' ... , X.)" E(X) = (P1' ... , }1.)' = p, Cov(X) = 1: = «(Ji)' the

covariance matrix of X (see (5.7)), and

Vg = (::1' ... ' ::.)'.

1. Multivariate Distributions 99

where the derivatives of g(x 1, ... , xn ) are evaluated at p. Then

1 n n

E[g(X)] ~ g(p) + "2 i~ j~ gij(lij'

V[g(X)] ~ (Vg)'~(Vg).

1.19. X and Yare independent and identically distributed random variables with density given by

2 f(x) = r.--::i'

lty 1 - x O~x~1.

Using Problem 1.18 find the E(Z) and V(Z) where

X Z=arctan Y'

Answer:

E(X) It E(Z) = arc tan -- ~ -

E(Y) 4'

1.20. The fundamental frequency of a chord is given by the formula

Q=~ fF 2 '-l;;L'

where F is the tension, m is the mass of the chord, and L is its length. If the mass is assumed to be constant and F, L are random variables with E(F) = f, E(L) = I, V(F) = (lJ, V(L) = (I,z, and Cov(F, L) = (III' find an approximation for the variance of Q.

Answer:

f [(lJ (I,z 20j lJ V(Q) ~ 16ml p + r - fl (see Problem 1.13).

1.21. Let the vector (X, Y) have density f(x, y). We define the complex variables

Z = X + iY,

and Zt = Zeit. Prove that, in order for all Zt to have the same distribution, it is necessary that f(x, y) = g(X2 + y2) for some density g (see Problem 1.13).

Hint: The distribution of Zt has to be invariant under orthogonal transformations (rotations).

1.22. If X has density f as in Problem 1.13, show that

y = p + AX

has density of the form

(a nonsingular n x n constant matrix)

cIAr I/2f((y - p)'A -l(y - p», A=AA',


and characteristic function of the form

e;t'''I/I(t' At) for some function 1/1.

1.23. Let Pij = p(X;, XJ be the correlation coefficient between X;, Xj for i, j = 1, ... , n. For n = 3 and Pl2 = P13 = P23 = p, what are the possible values that P can take?

Answer: - 1/2 ::s; P ::s; 1. Prove that the correlation matrix P = (p;J is nonnegative definite, i.e., t' Pt 2 ° for all t' = (t I' ... , tn).

1.24. If the p vector X is N(J1, A), show that the quadratic form

has the X~ distribution, with p degrees of freedom, and density given by (cf. Problem 1.13)

1.25. * The entropy H of a multivariate distribution with density f(x) is defined by the formula

H(f) = - E[log f(X)] = - f. .. f f(x) log f(x) dx.

Prove that for the normal distribution of Problem 1.24

1.26. If X is N(J1, A) with density f, find the density g of the random variable

Y = f(X).

Hint: See Problem 1.24.

[ 1 ](PI21-1

g(y) = 2Jc log cy2 ' c = (2nVIAI.

1.27. If X = (X I' ... , Xn) is uniformly distributed in the simplex T:

Xl + X2 + ... + Xn ::s; 1, X; 2 0, i = 1, ... , n,

show that the density of X is

f(x) = n! for x E T

Show that the (n - I)-dimensional marginals are also special cases of the Dirichlet distribution (5.11).

2. Generating Functions 101

2. Generating Functions

2.1. Two persons shoot at a target, each has n shots. Using generating functions, find the probability P that they have the same number of successes, given that each of them has a probability 0.5 of success (hitting the target) at each shot.

Answer: The coefficient of the constant term in the expansion of the generating function

1 1 ( l)n (1 + U )2n G(t) = 2n (t + 1)" 2n 1 + t = 4nu n '

that is,

= ~(2n). P 4n n

2.2.* To obtain the title of champion in chess, an opponent has to win at least 12.5 points out of the maximum of 24. In the case of a tie (12: 12), the defending champion maintains the title. The probability that each of the two players wins a game is the same, and equals half the probability of a tie. Find:

(a) the probability Pch that the defending champion will maintain his title, as well as the probability Pop that the opponent will become the champion;

(b) the probability P that 20 games will be played in the contest. Answer: (a) The probability Pch equals the sum of the coefficients of the

nonnegative powers of t in the expansion of

(1 1 1)24 (1 + t)48

G(t) = 4 t + 4t + "2 = 448t24 '

Pch = 4;4 k~4 (:8) = ~ 4;4 [248 + G!) ] = 0.5577, Pop = 1 - Pch·

(b) The probability of the complementary event equals the sum of the coefficients of t, with exponents between -4 up to 3 in the expansion of

1 (1 + t)20 G(t) = 420 t20 ;

1 23 (40) P = 1 - 20 L k = 0.22. 4 k=16

2.3. In a lottery, tickets are numbered from 000000 to 999999 and they are all equally probable. What is the probability P that the sum of the digits of a chosen number be 21?

Answer: The probability P is the coefficient of t 21 in the expansion of

1 1 (1 - t 1 0)6 G(t) = -(1 + t + t 2 + ... + t 9 )6 = - --~

106 106 (1 - t)6

where we made use of the equality

-~~~ = 1 + t + t 2 + .... 1 ( n) (n + 1) (1 ~ tt n ~ 1 n ~ 1

So

2.4. Prove that the Pascal distribution, for r -+ 00 and rq -+ A, converges to the Poisson with parameter ) ..

2.5. Let X, Y, Z be independent random variables, each taking the values 1,2, ... , n with the same probability (uniformly). Find P[X + Y = 2Z] from their joint generating function G(t1' t 2 , t3)'

Answer: Equal to the constant term 1/2n of the generating function Gw of W=X + Y~2Z.

-2 (1 ~ S·)2(1 ~ S2.) Gw(s) = G(s, s, s ) = n3(1 ~ sf(1 ~ S2)S2' 2'

2.6. In Exercise 315, let

q.,k = peS. ::;; k].

Find the generating function Q(t) of q •. k for a given n and hence q •. k'

Answer: 1 ~ P(t) = (1 ~ t)Q(t) (where P(·) is the generating function of S.). So

1" . (n) (k + n ~ mj ) q.,k = 1 ~ n L.... (~1)J . . m j<k/m ] n

2.7. (Continuation). Suppose k and m tend to infinity so that kim -+ x. Find the limit of q •. k'

Answer:

1 ~~ ~ (~l)j(~)(X~j)'= 1 ~F(x), n. J<X ]

where F is the distribution function of the sum of n independent random variables from the uniform distribution (0,1) (see Exercise 314).

2.8. Find the densities of the distributions with characteristic functions

1 + it <P1(t)=1+t2'

1 ~ it <P2(t) = ~1 -~2'

+t

Answer: 11 (x) = e-x , x > 0, 12 (x) = eX, x < 0.

2.9. Find the moments of the Laplace distribution with characteristic function

2. Generating Functions

Answer: Using the Taylor series expansion

conclude that

1 OC!

" Ct)2k 1 + t 2 = k-f-O I ,

I {k! J1k = o

for even k,

for odd k.

2.10. Determine the distribution with characteristic function

1 cp(t) = 2e-il _ 1.

Answer: Discrete random variable with probability function

k = 1,2, ....

103

Make use of the expansion of cp(t) in powers of !e il and the expression of the Dirac function J

J(x) = - e i1x dt. 1 fOC! 2n -00

2.11. Find the characteristic function of Y = aF(X) + b, where X is a continuous random variable and F is its distribution function.

Hint: Y is uniform on (a, b); see Exercise 227.

2.12. A vector I; = (X l' ... , X.) follows the n-dimensional normal distribution with E(XJ = a, V(XJ = (J2 (i = 1, ... , n) and covariances

i = 1,2, ... , n - 1,

COV(Xi' Xj) = 0

Find the characteristic function of 1;.

for li-jl>l.

Answer:

{• (J2. .-1}

cp(t l' ... , t.) = exp ia L tk - - L tf - p(J2 L tktk+1 . k=1 2 k=1 k=1

2.13. Let cp(t) be the characteristic function of a random variable X with E(X2) = J1~.

(a) Prove that

is a characteristic function. (b) Conclude that

is a characteristic function.

1 t/!(t) = ---,- cp"(t)

J12

Answer: (a) !/I(t) is the characteristic function of the random variable with distribution function

1 IX G(x) = ---;- u2 dF(u). J12 -co

(b) Application of (a) with X normal N(O, 1).

2.14. Show that

) { I - (Itlln) for It I < n, aCt = ° forltl>n,

is the characteristic function of a random variable X, and determine the corresponding distribution.

Answer: Since la(t)1 is integrable, X is continuous with density

f - ~ In -iIX( -l!l)d _ 1 - cos(nx) (x) - 2 e 1 t - 2 2 . n -n n n x

2.15. Let Xl' X 2 , X3 be independent N(O, 1). Find the characteristic function of the pair Zl' Z2' where

Hence conclude the characteristic function of Zl' Are Zl' Z2 independent? Answer:

(f1(tl' t2) = (1 + d + tn- 1/2 ,

(f1z,(t) = (f1(t, 0) = (1 + t2)-1/2 = (f1z,(t).

No, because (f1(tl' t 2) =F (f1z,(t l )(f1Z,(t2)·

2.16. Find the distributions corresponding to the characteristic functions

Answer:

I (f1l(t) = -h t' cos

fl(X) = , nx

2 cosh 2

I (f12(t) = -'-h-' sm t

2.17. Bivariate negative binomial. Show that

I (f13(t) = -h2 •

cos t

. 1 f3(X) = ----.

2 . h nx sm 2

P(Sl' S2) = p~(1 - P1Sl - P2 S2)-"

with positive parameters and Po + Pl + P2 = 1 is the probability generating function of a pair (X, Y), so that the marginal distributions of X and Yand X + Yare negative binomials.

2.18. Problems related to a game of billiards. Let a "round" consist of a

2. Generating Functions 105

sequence of Bernoulli trials before the first failure. Find the generating function and the probability function of the total number Sk of successes in k rounds.

Answer: Pascal distribution.

2.19. (Continuation). Let G be the number of successive rounds up to the nth success (i.e., the nth success occurs during the Gth round). Find the distribution of G and the

Answer:

E(G), V(G).

"-I

P[G = k] = I P[Sk-1 = np[Xk ~ n - k] j=O

"~I k-I .(k+ j -2) _.(k+n-2) = L... q pl . p" 1 , j=O ] n - 1

E(G) = 1 + nq , p

nq V(G) = 2".

P

2.20.* (Continuation). Consider two independent sequences of Bernoulli variables with parameters PI and P2' respectively, or two players with "skills" PI and P2' respectively. Prove that each player will need the same number of rounds (trials) until the Nth success (a tie) with probability

(PIP2)N l (N : ~ ~ 2}qlq2r l

N I_ZN N- I (N-1) k = (PIP2) (1 - qlqz) k~O k (qlqz) .

2.21. Show that the characteristic function of the f3 density (2.1 0) is

r(p + q) 00 r(p + j)(it)j

r(p) j~O r(p + q + j)ru + 1)

2.22. * Show that the characteristic function of the density in Problem 1.13 is of the form cp(ti + t~ + ... + t~) (see Problem 1.21). If

" cp(d + ... + t~) = Il cp(t;), i=l

that is, X I' X z, ... , X" are independent, prove that X is N(O, 1). In other words, a spherically symmetric random vector has independent components if and only if it is multivariate normal.

Hint: Equation (*) is known as Hamel's equation and its only continuous solution is

cp(t) = eel

for some constant c (which for a characteristic function must be < 0).


3. Transformation of Random Variables

3.1. Find the density of

y= +j[XT, where X - N(O, 1).

Answer:

4y {y4} fy(y) = .j2n exp -2 ' y ~ 0.

3.2. X has a uniform distribution on (0,1). Find the density of the random variable Y defined by

X = !{1 + _2 IY exp {-~} dt}. 2 fo 0 2

Answer: Y is N(O, 1) (see Exercise 227).

3.3. The random variables X and Yare related by

X = f:oo f(t) dt,

where f(t) ~ ° satisfies the

t: f(t) dt = 1.

Prove that if X is uniform on (0, 1), then f(t) is the density of Y (see Exercise 227).

3.4. Suppose X and Yare independent random variables. Prove that the product XY has the same distribution if:

(a) X and Yare N(O, 1); or (b) X is N(O, 1) and Y has density

(X2 density; see (2.11)).

3.5. The roots of x 2 + Yx + Z = ° follow the normal distribution (-1, 1). Find the density of the coefficients Y and Z (cf. Exercise 222).

Answer: fy(y) = ±(2 - Iyl)

fz(z) = -t log Izl

for Iyl s 2,

for Izl s 1.

3.6. Let f(x, y, z) be the density of a point (X, Y, Z) in the three-dimensional space and

8 . Y = arc sm R'

What is the density of the pair (R, 8)?

3. Transformation of Random Variables 107

Answer: f(r, e) = r2 cos e J5" f(r cos e cos cp, r cos e sin cp, r sin e) dcp (cf. Problem 1.11).

3.7. X and Yare independent normal N(O, 0'2). Find the distribution of (a) Z = X/I YI; (b) W = IXIII YI.

Answer: (a) Cauchy Distribution (cf. Exercise 220). (b) f(w) = (2/n) x [1/(1 + w2 )] (folded Cauchy).

3.8. X and Yare independent random variables with uniform distribution on (0, a). Find the distribution of Z = X/Y and examine the existence of moments of Z.

Answer:

{

.l 2'

f(z) = _1 2Z2'

z ~ 1,

z ~ 1.

There are no ordinary moments.

3.9. The joint density of X, Y is

f(x, y) = e-Y for ° ~ x ~ y < 00.

(a) Find the marginal distributions of X, Y and the conditional distribution of Y for given X.

(b) Ofthe variables X, Y, Y - X, and X/Ywhich are pairwise independent? (c) In the conditional distribution of Y, given X = x, determine the interval

(y, y + a) so that

Q(a) = P[y ~ Y ~ Y + a]

can be maximized. These intervals with variable x define a zone B. What is the probability of the zone, i.e.,

Answer: (a)

(b)

(c)

P[(X, Y) E B] = P(B)?

fx(x) = Loo e-Y dy = e-x ,

fy(Ylx) = eX-Yo

(X, Y - X) and (Y, X/y)o

f u +a

Q(a) = sup eX - Y dy = sup ex-u(l - e-a )

u~x u u~x

(for u = x).

P(B) = Q(a) as also concluded from the independence of X and Y - X.

3.10. * X, Yare independent exponentials each with density

t ~ O. (1)

(a) Find the joint conditional density of X, Y, given that X ::;; Y (ordered sample of two observations).

(b) Let Xl ::;; Xl::;; X3 ::;; X 4 be an ordered sample from (1). What is the density fl of(Xb Xl) and what is the conditional density fl of (X3, X 4 ) for a given (Xl' Xl)? What do you conclude?

Answer: (a)

f(x, y) = 201e-8(x+ y ),

fy(Ylx) = Oe-8(y-X\ O<x<y (see Exercise 238),

i.e., Y - X, given that X = x, has the density (1).

foo 1 m(x) = E[YIX = x] = x yfy(ylx) dy = x + o·

This also follows from the fact that Y - X, given X = x, follows (1), whose mean is I/O.

(b)

fl (Xl' X2) = 1201e-8(x 1 +3x2 ),

f2(X 3 , x4) = 201e-8(x3+x4-lx2),

We observe that, for Xl = Xl, the pair (X3 - Xl' X 4 - Xl) is distributed as (X, Y) in (a), i.e., the distribution is not affected by the knowledge that X exceeds a given value, because of the "lack of memory" of the exponential.

3.11.* If the random variables Xi are independent normal N(a + bti, (Jl) (i = 1, ... , n), where ti are constants with 2:7=1 ti = 0, find:

(a) the joint distribution of X I' ... , Xn; (b) the joint distribution of

_ 1 n

X =- 2: Xi n i=I

and n I n b = i~ tiXi i~ tl.

Answer: X is N(a, (Jl/n) and b is also N(b, (Jl/2:7=I t?).

3.12. * (Continuation). By a proper orthogonal transformation of X = (X I' ... , Xn)', say Y = OX (0 n x n orthogonal matrix), prove that the random variables X, band 2:7=1 (Xi - X - btY are completely independent. Conclude that 2:7=1 (Xi - X - btY/(Jl has the Xl distribution with n - 2 degrees of freedom.

3.13. X and Yarejointly normal with E(X) = E(Y) = 0, V(X) = V(Y) = 1, and p(X, Y) = p, Ipi < 1. Consider the polar coordinates R = JXl + yl

4. Convergence of Random Variables 109

and () = tan- 1(y/X). Show that () has density (cf. Exercises 220, 221)

f(()) =.JI=P2 0 < () < 2n. 2n( 1 - P sin 2())'

3.14. An elliptically symmetric t distribution. The random vector X =

(X 1, ... , X k ) has the density

f(x) = r((k + n)/2) [1 + (x _ )'A -1( _ )]-(n+k)/2 r(k/2)(kn)kI2 JI x JI .

If A = I and k = 1 this is the Student distribution with n degrees of freedom. If A = h, then X has a spherically symmetric t distribution (see Problem 1.13). Show that:

(a) E(X) = JI. (b) X has the same distribution as

JI + A 1/2 Z/X~,

where Z is N(O, Id independent of the X~. Hence conclude that (cf. Problem 1.24)

(c) As n -+ 00,

1 F = -2 (X - JI)' A-I (X - JI) is Fk,n'

kXn

X~Z.

3.15. Let f(x) denote the density of a random variable X. If f( .) is monotone and bounded, then f(X) has a uniform distribution if and only if X has the (negative) exponential distribution.

Hint: See (7.1) or Exercise 227.

4. Convergence of Random Variables

4.1. A Geiger-M iiller counter registers a particle emitted by a radioactive source with probability 10-4 .

(a) Assuming that during a certain period, the source emitted 3 x 104 , what is the probability that more than two particles will be registered.

(b) How many particles have to be emitted so that at least four particles will be registered with probability 0.99?

Answer: (a) 1 - e-J.[l - ), - (A,2/2)] with A, = 3 X 104 X 10-4 = 3. (b)

X ::= 10.7, n = X' 104 ::= 107,000.


4.2. In some of the countries of western Europe, from the seventeenth century up to World War I, the following state lottery was played: 5 numbers out of 90 were drawn and a player had the right to bet, say, an amount a, on one or more numbers; if all the numbers he had chosen were among the 5 numbers drawn, then he would win an amount determined as follows: 15a if he had bet on one number, 270a if he had bet on 2 numbers, 500a if he had bet on 3 numbers, 75,OOOa if he had bet on 4 numbers, and 1,OOO,000a if he had bet on 5 numbers.

(a) What is the expected gain of a player who is betting on 1,2, 3,4, and 5 numbers?

(b) Suppose that 100,000 players bet on 3 numbers; what is the probability that at least 10 players win?

Answer: (a) Let Pk be the probability that a player who bets on k numbers will win, and that Ek will be his expected gain. Then

( 90 - k) 5-k 1 2 1

Pk = C~) ; P1 = 18' P2 = 801' P3 = 11 748' ,

1 1 P4 = 511,038' Ps = 43,949,268'

1 1 E1 = 15a 18 - a = -(ja,

(b) 0.24.

4.3. * Making use of the Stirling formula prove that, as A -+ 00 and for a constant k, we have the local limit theorem for the approximation of the Poisson probability function by the normal

I1IAk;. 1 {I 2}1 v)- -e- - -- exp --(k - A) -+ O. k! foi 2A.

4.4.* If Xn.!. c and f is continuous at c, then f(Xn) .!. f(c).

4.5.* Suppose {Xn, n;::: I} is a sequence of random variables with E(Xn) = o and V(Xn) = u2 (O) where the parameter 0 takes values in some interval. Prove that if u;(O) -+ 0, then

for any bounded and continuous function g. Answer: Examine the

f: Ig(x) - g(O)1 dFn(x)

4. Convergence of Random Variables 111

for Ix - 01 < b where Ig(x) - g(O)1 < e, and for Ix - 01 ~ b where Ig(x) - g(O)1 :$;

M < 00, and make use ofthe Chebyshev inequality (Fn(x) denotes the distribution function of Xn).

4.6. * (Continuation). If Fn is the "binomial" with

6;(0) = 0(1 - 0), n

and g is defined in [0, 1], then the Bernstein polynomial

f g (~) (n) Ok(1 - Ork -----:::-> g(O) k=O n k n 00

(1)

uniformly in ° :$; 0 :$; 1. So we have a simple proof of the Weierstrass Approximation Theorem, according to which such a function g can be uniformly approximated by a polynomial. Moreover, (1) results in approximating polynomials.

4.7.* (Continuation). If nXn is Poisson with parameter nO, then

e-n8 I g (~) (n~)k ---. g(O) k=O n k.

(2)

and the convergence is uniform in every finite interval of values of O. Equation (2) also holds true for nonintegral values of n.

4.8. Suppose XI' X 2 , ... is a sequence of independent normal random variables with E(X;) = ° and V(Xi) = 1 and let

In XI + ... + Xn ¢n = n ,

X 2 + ... + X 2 I n

X +"'+X (= I n

n JXf+"'+X;'

Prove that ¢n and (n ---. N(O, 1). Hint: Xf + ... + X; !. n; use (3.4)(i).

4.9. (Continuation). Find the asymptotic (n ---. 00) distribution of

and !' = nXn+1 'on 2'

Xn

Answer: (X; - n)/fo and ¢n are asymptotically N(O, 1).

4.10. n numbers are chosen at random from the integers 1,2, ... , N, say kl' k2' ... , kn' We set Xi = ° if ki = ° modulo 3, Xi = 1 if ki = 1 modulo 3, and Xi = - 1 if ki = 2 modulo 3. Set Sn = I?=l Xi' Prove that when nand N ---. 00 so that n/N ---. c > 0, Sn/Jn ~ N(O, 2/3).

4.11.* Suppose SN = Xl + X 2 + ... X N is the sum ofa random number N of random variables Xi' where Xi and N are independent, IXil < c, E(X;) = 11, V(Xi) = 6 2, E(N) = n, and V(N) :$; n l-<, e > 0. Make use of Exercise 328 and

the CL T, and prove that as n -+ 00

SN - nil r:. -+ N(O, 1). (Jy n

4.12. Let {Xn' n ;:::.: 1} be a sequence of independent random variables and

n = 1,2, ....

Prove that the sum

Sn = Xl + ... + Xn

(a) converges in law, and (b) find its asymptotic distribution. Answer: (a) The characteristic function of Sn

<fJn(t) = Jl cos Uk). because cOS(t/2k) - 1 behaves in the same way as - t 2/22k+1.

(b)

sin t t/2n sin t <fJn(t) = -t - sin(t/2n ) ~ -t-'

i.e., <fJn(t) converges to the characteristic function of the uniform in [ -1,1].

4.13. If a sequence of normal random variable(s) Xn converges in law to the random variable X, then X is also normal or degenerate.

Answer: The characteristic function of Xm <fJn(t) = exp(il1nt - J:(J;t2) -+ <fJx(t) uniformly for It I < <5, <5 > 0. So

l<fJn(t)1 2 = e-O'~t -+ I <fJx(tW,

where, if {(J;} is bounded, then (J; -+ (J2 ;:::.: 0; hence

eil'nt = <fJn(t)e(I/2)0'~t2 -+ <fJ(t)e(1/2)0'2t2.

Therefore I1n converges, say, to 11 and finally

<fJn(t) -+ eil't-(1/2)0'2t2.

If lim (J; = (J2 = 0, then X degenerates (shrinks) to the point 11.

4.14. Prove that the negative binomial

( r + k - 1) P[X = kJ = k prq\ k = 0,1,2, ... , (1)

with r -+ 00 and q -+ ° so that rq -+ A. > 0, converges to a Poisson with parameter A..

4. Convergence of Random Variables

Answer: The generating function of (1),

(1 - q)' ).(5-1) -----e . (1 - qs)' rq~).

113

4.15. If the stochastic sequence {Xn} converges in probability to the random variable X, where P[X i= 0] = 1, then the sequence {1/Xn } ~ 1/X.

4.16. Consider the sequence {Xn} of independent random variables with

n = 1,2, ....

If the series La; converges, prove that the sequence of partial sums

Sn = Xl + ... + X n,

also converges in each of the four modes of convergence (see Chapter 8). Answer: The Chebyshev inequality for ISn - Sml and the Cauchy criterion

imply the convergence of Sn' It should also be noted that the Xn are uniformly bounded, since an --+ 0, and then the modes of convergence are equivalent.

4.17. Let {Xd be a sequence ofrandom variable(s) with

P[Xk = ± k a ] = 1/2.

(a) Find the values of a for which

in probability, almost surely, and in quadratic mean. (b) Examine the convergence of Xn in law for a = 1/2. Answer: (a) E(Xk ) = 0, V(Xd = at = k 2a,

_ 1 n

V(Xn) = 2 2: k2a = O(n 2a - l ) --+ 0. n k=l

Therefore 2a - 1 < 0, i.e., a < 1/2. This also suffices for

X- Lm. ° n ---+ .

For the Strong Law of Large Numbers (SLLN) it is enough that

n a 2

k~l k~ < 00 => a < 1/2.

(b) Let <Pn(t) be the characteristic function of Xn. We have

because a = 1/2. For It I ~ b, log <Pn(t) -+ -t4 /4 uniformly. Therefore

- (.)2) Xn-+ N 0'2 .

4.18.* By the so-called Monte Carlo method, how many trials are needed to estimate the integral

("/2 I = J 0 sin x dx,

so that the absolute error of the esimtate will not exceed 0.1 % of I, with probability p ~ 0.99?

Answer: The integral

2 21"/2 - I = - sin x dx, 71: 71: 0

can be considered as the mean value of the sin x, where the random variable X is uniform on (0, 71:/2). So an approximate value of I is

where Xn are (pseudo) random numbers in (0,71:/2). Now, using

I - I ~( ) '" N(O, 1),

(1 In

where

we find:

n ~ 1.55 x 106 = 1,550,000.

4.19. Suppose Vis a region ofa plane with area one andf defined in V with If(x, y)1 ~ c. For the calculation of 1= Hvf(x, y) dx dy, by the Monte Carlo method, we randomly choose n points (Xl' yd, (X2' Y2)' ... , (Xn' Yn) in V and we calculate I according to the formula

1 n

In = - L f(x;, yJ n ;=1

Prove that

5. Miscellaneous Complements and Problems 115

5. Miscellaneous Complements and Problems

5.1. Bore/-Cantelli Lemma. In the study of sequences of events A l , A z, ... with Pk = peAk]' a significant role is played by the Borel-Cantelli lemma:

(a) If the series LPk converges, then a finite number of the events Ak occurs with probability 1.

(b) If the events are (completely) independent and the series L Pk diverges, then an infinite number of the events Ak occur with probability 1.

In the study of the divergence of series of random variables the following theorem is very helpful.

5.2. Kolmogorov's Three-Series Theorem. Suppose {Xm n ~ 1} is a sequence of independent random variables, and for any c > 0, consider the truncated random variables

Then the series

X' = {Xn if IXnl ~ c, n ° otherwise.

00

S = L Xn n=l

converges with probability 1 if and only if the following three series converge:

L P(IXnl > c) < 00, and L E(X~). Otherwise, S converges with probability 0.

The following "laws" are also worth mentioning:

5.3. The Law of the Iterated Logarithm. This law concerns the frequency of appearance of large values of the standardized number of successes,

S* = Sn - np n fiN'

in an infinite sequence of Bernoulli trials. According to the De Moivre-Laplace Central Limit Theorem (CL T), we have

peS: > x] "" 1 - <l>(x),

where "" means that the ratio of the two sides tends to one. Therefore, for a particular n, large values (> 4) of S: are improbable, but,

obviously, for large n, it is possible for S: to exceed, sooner or later, any large value. How soon this may occur is given by the Law of the Iterated Logarithm (Khintchine; see, e.g., Feller, 1957). With probability 1

I. S: 1m sup = 1,

n-oo J2 log log n (1)

i.e., for A > 1, only a finite number of the events

Sn> np + ;..FnMJ210g log n (2)

occurs with probability 1, while for )0 < 1, (2) holds for infinitely many n with probability 1.

For reasons of symmetry (of the distribution of Sn, (1) implies

S* lim inf n = - 1. n~OC! J2 log log n

The behavior of Sn is further illustrated by the following:

(a) There exists a constant c > 0 that depends on p, but not on n, so that for all n

P[Sn > np] > c.

Hint: The probability, according to the binomial distribution, is always positive and, according to the CL T of Laplace, it approaches 1/2.

(b) Suppose x (0:::; x < 1), with decimal expansion

(3)

where each ai is one of the digits 0, 1, ... , 9. Let ai = 0 (instead of 0 we could have chosen any other digit) define a success with p = 1/10. Therefore, (3) corresponds to an infinite sequence of Bernoulli trials with p = 1/10, and all the limit theorems concerning Bernoulli trials with p = 1/10 can be translated into theorems on decimal expansions. Sn(x), the number of zeros among the first n decimal digits of x, takes the place of Sn. Show that

(i) Sn(x)lx --+ 0.1 in measure (Lebesgue) (in probability). (ii) Sn(x)!x --+ 0.1 almost everywhere (with probability 1). .... Sn - nllO

(111) hm sup I 1/2· n~OC! (n log og n)

Answer: (i) WLLN, (ii) SLLN, (iii) Application of (1).

5.4. The Zero-One Law of Kolmogorov. Let {Xn , n ;?: I} be a sequence of independent random variables and A an event independent of (any event defined in terms of) Xl' ... , X k for every k. Then, either P(A) = 0 or P(A) = 1.

5.5.* The tail of the normal N(O, 1). For large x(x --+ 00), the approximation

1 1 - <1>(x) '" - o/(x) (4)

x

is valid; more precisely, for every x > 0, the double inequality

(1 1) 1 o/(x) - - - < 1 - <1>(x) < o/(x)-X x 3 X

(5)

is valid. Moreover, for every constant a > 0 show that, as x -+ 00,

Hint: Verify that

1 foo foo - cp(x) = cp(y)(1 + y-2) dy > cp(y) dy = 1 - (x) x x x

and, similarily, make use of the integral of

5.6. Consider a sequence {An' n ~ I} of independent events. Then a finite or infinite number of events Ak occur with probability 1, according to whether the series Ln P(An) converges or diverges, respectively (see the above BorelCantelli Lemma). Show this by using the Three-Series Theorem.

5.7. Let {Xn' n ~ I} be a sequence of random variables with lin = E(Xn) < 00 and X a random variable with V(X) < 00. If, for each k, Xl' ... , X k and X - (Xl + ... + X k ) are independent, prove that V(Xk ) < 00 for every k, and that the series

converges with probability 1 (almost everywhere or almost surely).

5.8. In a sequence of (independent) Bernoulli trials with probability of success p, the event Ak is realized if k consecutive successes appear between the 2kth and 2k +1 th trial. Prove that if p ~ 1/2, then an infinite number of the events Ak occur with probability 1, whereas for p < 1/2 with probability 1 a finite number of the events Ak occur.

5.9.* Bochner-Khintchine Theorem. A continuous function cp(t) with cp(O) =

1 is a characteristic function if and only if it is nonnegative definite, i.e., if for each n ~ 1 and for each n-tuple of real numbers t l , ... , tn and complex numbers Zl' ... , Zn it satisfies the

v v

L L cp(tj - tk)ZjZk ~ O. j=l k=l

Prove that (*) is necessary. Hint: The left side of(*) is equal to

where F is some distribution function. For a proof of the sufficiency of (*) see, e.g., Gnedenko (1962).

5.10. Prove that the functions, (a) e- i1tl, (b) 1/(1 - iltl), (c) cos t 2 , are not characteristic functions.

5.11. If tp(t) is a characteristic function, prove that g(t) = e'P(t)-1 is also a characteristic function.

Hint: See (6.6).

5.12. Show that, for a real characteristic function tp(t), the following inequalities are valid:

(a) 1 - tp(nt) :s;; n2 (1 - tp(t)), n = 0,1,2, ...

(b) 1 + tp(2t) ~ 2[tp(t)Y

5.13. Let X be the number of Bernoulli trials required until we have observed r successive successes. Find the generating function of X and E(X).

Answer: 00 • 00 • p't'(l - pt)

P(t) = I t P[X = n] = L t P." = , ,+1' .;, .;, 1 - t + p qt

1 ' E(X) = P I (l) = ~.

p'q'

5.14.* Let M. be the maximum number of consecutive successes (the maximum length of a run) observed in n Bernoulli trials. If

p." = P[M. :s;; r],

show that

p." = 1 - Pl., - P2" - ... - P.,r>

where P." was defined in the preceding problem, and therefore, the generating function of P." is

Also, show that

00 • 1 - P(t) 1 _ p't' "\' P t = = ~~--=--~cc-.';:, .,' 1 - t 1 - t + p'qt,+1 .

E(M) = log n + 0(1), -logp

V(M) = 0(1).

5.15. Find the distributions with characteristic functions, (a) cos t, (b) cos 2 t, (c) Ir;o Pk cos kt where Pk ~ 0, I Pk = 1.

Answer: (a) cos t = Heit(-l) + eit(l)]. Therefore

if x < -1,

if -1 :s;; x < 1,

if x ~ 1.

5. Miscellaneous Complements and Problems

{o 1/4

F(x) = ~/4

if x < -2, if -2.:s; x < 0, if O.:s; x < 2,

if x ~ 2.

(c) LPk cos kt = tLPke-ikt + tLPkeikt.

119

Therefore this is a discrete distribution with jumps tPk at the points ± k (k = 0, 1, ... ).

5.16. Find the discrete distributions with generating functions, (a) i(1 + S)3, (b) t(1 - tS)-l, (c) e(S-l), (d) (is + i)lOO.

Answer: (a) Discrete with probabilities 1/8, 3/8, 3/8, 1/8 at the points 0, 1, 2, 3, respectively.

(b) The probability of the value k is r k - 1 (k = 0,1,2, ... ). (c) Poisson with A = 1. (d) Binomial with P = 1/4 and n = 100.

5.17. Using the Kolmogorov inequality (8.9), show that if the series

00

L V(Xd/k2 < 00, k=l

then the SLLN holds, i.e.,

5.18. A fair but loss-incurring game (a probability "paradox"). The probability that, in each play of a game, the player receives 2k dollars is

Pk = 2kk(k + 1)' k = 1,2, ... , (1)

and that he receives ° dollars is Po = 1 - (PI + P2 + .. '). So, the expected (average) gain in each game is

If, in each game, the player pays a I-dollar fee, the net profit ofthe player after n games is equal to

n

L X k - n = Sn - n, k=I

with E(Sn - n) = 0,

i.e., the game is fair (Xk is a random variable with distribution (1». For every e > 0, however, the probability that, in n games, the player will lose more than (1 - e)n/log2 n dollars, approaches 1, i.e., it can be proved (truncating the

random variables X k , see Feller (1957) or Gnedenko (1962» that

. [ (1 - e)n] hm P Sn - n < 1 = 1. n~CX) og2 n

5.19. In a sequence {Xn' n ~ l} of Bernoulli random variables, suppose that

y" = 0 if XnXn+l = 1 or Xn = 0 and Xn+1 = 0,

y" = 1 if Xn = 1 or Xn+ 1 = 1.

Find E(Zn) and V(Zn) where Zn = I?=l 1';. Answer:

E(Zn) = nE(Y1 ) = 2npq,

V(Zn) = 2npq(1 - 2pq) + 2(n - l)pq(p ~ q)2.

5.20. We randomly choose two numbers in (0, 1). What is the probability p that their sum be smaller than 1 and their product smaller than 3/16?

Answer: p = area S, where S = {(x, y): x ~ 0, y ~ 0, x + y ~ 1, xy ~ 3/16}.

1 3 p = 4 + 16 log 3 = 0.456.

5.21. A bus on line A arrives at a bus station every 4 minutes and a bus on line B every 6 minutes. The time interval between an arrival of a bus for line A and a bus for line B is uniformly distributed between 0 and 4 minutes. Find the probability:

(a) that the first bus that arrives will be for line A; (b) that a bus will arrive within 2 minutes (for line A or B). Answer: Suppose x is an instant of time, where 0 ~ x ~ 12 minutes. The

times of arrival of the buses for line A are x = 0, 4, 8, and of buses for line B the arrival times are y, y + 6 with 0 ~ y ~ 4.

(a) The favorable cases are: for 0 < y ~ 2 we have y < x ~ 4 or 6 + y ~ x ~ 12; and for y > 2 we have y < x < 8 or y + 6 < x < 12. Therefore p = 2/3.

(b) Favorable cases: 2 ~ x ~ 4, 6 ~ x ~ 8, 10 ~ x ~ 12,4 + Y ~ x ~ 6 + y. For y < 2 we have 0 < x < y, and for y > 2 we have y - 2 ~ x ~ y. Therefore p = 2/3.

5.22. N stars are randomly scattered, independently of each other, in a sphere of radius R.

(a) What is the probability that the star nearest to the center is at a distance at least r?

(b) Find the limit of the probability in (a) if

and N/R 3 -> 4nA/3.

Answer: (a) (l - r3 / R 3)N. (b) exp{ -41!Ar3/3} (close to the sun A ~ 0.0063, when R is measured in

parsecs).

5.23. A satellite moving on an orbit between two parallels 60° north and 60° south (latitude) is equally likely to land at any point between the two parallels. What is the probability P that the satellite will land in the north above 30°?

Answer:

P = {R2 I21r de 1~:3 cos <p d<P} I {2R2 I21r de I 1r13 cos <p d<P} = 0.21.

5.24. In the equation A 3 + 3X A + Y = 0, the coefficients X, Yare uniformly distributed in the rectangle IXI s a, I YI s b. What is the probability P that the equation has real roots?

Answer: It is required that y2 + X 3 sO, i.e., for X sO, when y2 s _X3. If a3 S b2 , we have

1 fa a312 P = 2ab 0 x 312 dx = 5b'

1 1 fb 1 ( 3b213 ) P = 2 - 2ab 0 y213 dy = 2 1 - -a- .

5.25. * The binomial distribution via difference equations. Let Pk,n denote the probability of k successes in n independent Bernoulli trials. By using the generating function Gn(t) of PII • k (k = 0, 1,2, ... ), deduce the binomial distribution.

Hint: Pk,n satisfies the partial-difference equation

Pk,n = pPk-I,n-1 + qPk,n-I'

from which we can find the difference equation for Gn :

Gn(t) = (pt + q)Gn- 1 (t),

which has the solution

i.e., the probability generating function of the binomial.

5.26. A and B play the following game. They flip a coin, if the outcome is a head A gets 1 dollar from B, otherwise A pays 1 dollar to B. Initially, each of them has 3 dollars. The game terminates when either A or B loses all his money. What is the probability Pn that n flips are needed?

Answer: Pn > 0 only for n = 2k + 1 (k = 1, 2, ... ). Suppose qk is the probability that the game will not finish after 2k + 1 flips. Then

. _ ~(~)(n-3)12 I.e., Pn - 4 4

5.27. Let y" = max{XI' ... , Xn} where Xl' ... , Xn are independent and identically distributed random variables with a uniform distribution on (0, 1).


Show that the distribution of

Zn = n(1 - y,,)

converges (as n --+ OC!) to the exponential distribution with distribution function F(z) = 1 - e- z •

Answer: Examine the limit of the sequence {Fn(z)} of the distribution functions of {Zn}'

5.28. A discrete random variable Xn that appears in the theory of extremes has the distribution function

Fn(n) = 1 _ (n)r , (n + nX)r

1 ::;; r ::;; n,

and Xn takes the values l/n, 2/n, .... Show that the sequence {Xn} converges in law to a continuous distribution with distribution function

{O, F(x) = 1 _ (1 + xrr,

x::;; 0,

x> 0. p

5.29. If Xn --+ X and E(Xn - y"f --+ 0, show that the sequence {Y,,} also converges to X in probability.

PART III

SOLUTIONS

Solutions

The set of female students in the freshman and sophomore years.

C' F = F - C: The set of non-Cypriot female students. A 1 F' C = A 1 C - F: The set of Cypriot male freshmen students. A3FC' = A3F - C: The set of non-Cypriot junior female students. (A 1 u A 2 )CF: The set of Cypriot female students in the freshman

and sophomore years.

2. (a) (A u B)(A u C) = A u (BC). (b) (A u B)(A' u B) = B u (AA') = B u 0 = B. (c) (A u B)(A' u B)(A' u B') = B( A' u B') = B(AB)' = B - A (by de Mor

gan's laws and (b».

3.

B

(a) The shaded area represents AuBuC.

B

(b) The shaded area represents (AB'C') u (A' BC') u (A' B'C) u(A'B'C').

126

(c) The shaded area represents (A u B u C)' = A'B'C.

A B

(e) The shaded area represents (AB'C) u (A'BC) u

(A'B'C).

(g) The shaded area represents AuA'B'.

Solutions

B

(d) The shaded area represents ABC.

B

(f) The shaded area represents ABC = AB - C.

4. (a) A = {(HHT), (THH), (HTH)},

B = {(HHT), (THH), (HTH), (HHH)},

C = {(THH), (THT), (TTH)}.

(b) (i) A' B = {(HHH)}: exactly three heads. (ii) A' B' = (A u B)' = B' = {(HTT), (TTT). (TTH), (THT)}: at least two

tails. (iii) AC = {(THH)}. The first is tails, and the second and third are heads.

5. Denoting by M and F a male and a female child, respectively, we have

A = {MFMF, FMFM},

B = {MMMM, MMFM, MFMM, MFFM},

C = {MMFF, MFFM, MFMF, FFMM, FMFM, FMMF},

D = {MMMM, MMMF, FMMM, FFFF, FFFM, MFFF}.

Solutions 127

6. A = {abebea. aebeba}.

7. 44.

A = {aaaa, bbbb, ecce, dddd};

B = {aabe, aaeb, abbe, ... }, that is, G) G) G) G) = 144 points,

where a, b, e, and d denote the four types of cellular disorders, and the position of the letter denotes the patient.

A = {ab*, ba*, a*b, b*a, *ab, *ba}, B = {ab*, a*b, ba*, b*a},

where a, b denote the two persons, * means "none", and the position of a, b, * indicates the floor, e.g., a*b means: a at first floow and b at third floor.

9. (\0) G) G) = 100 (by the multiplication rule).

10. 4!. (3!)2(4!f (i.e., permutations of the four works times permutations of volumes within each of the four works.

11. Identify the n persons with n cells and the r objects with r balls to be placed in the n cells:

(i) if the objects are distinguishable there are n' ways;

(ii) if the objects are indistinguishable there are (n + ~ - 1) ways.

12. The 5 boys define consecutive gaps (positions) between any two of them; the 5 girls can be placed in the positions in as many ways as the number of permutations of 5 things taken all at a time, i.e., 5! = 120.

The 5 boys can be seated at a round table in as many ways as the number of cyclical permutations of 5, i.e., (5 - I)! = 4! = 24. Therefore, the total number of ways is equal to 4! 5! = 2,880.

13. As many chords as the number of combinations (~), i.e., 28. As many

tr!angles as the number of combinations (~), i.e., 56. As many hexagons as

(6) = 28.

14. Cyclical permutations of n, i.e., (n - 1)!. The probability that any fixed person sits right or left of any given person is the same for every person; and hence is equal to l/(n - 1). Therefore, the probability sought is 2/(n - 1). This also follows from the fact that out of the (n - I)! total cases there are 2(n - 2)! favorable cases.

15. By virtue of (1.5) the 20 recruits can be distributed into 4 camps in 20! (5!)4 ways. However, to each division in 4 groups, there correspond 4! distribu-

128

tions to 4 camps. Hence, there are

20! (5!)44!

distributions into 4 groups of 5 each.

16. According to the multiplication formula, we can form

G)·G)·7!

Solutions

words (every permutation of the seven letters is supposed to be a word).

17. (i) Pascal triangle. (ii) Use (i) repeatedly.

18. The n things are divided into two groups of m and n - m. Any r of the n will contain k of the m and r - k of the n - m where k = 0, 1,2, ... , r. Hence

the expression for (~). 19. (i) The expansion of (1 - 1)" = O.

(ii) Differentiate (1 + x)" once and put x = 1. (iii) Differentiate (1 + x)" twice and put x = 1.

20. (i) Expand the right-hand side. (ii) Use 17(i).

21.

25. Examine the inequality ak+tiak ~ 1. It gives k ::; (nx - l)/(x + 1) and the required maximum value of ak is the ak*+1 with

k*=[~J x + 1 '

where [a] denotes the integer part of a.lf(nx - l)/(x + 1) is an integer, r say, then there are two maxima, namely, the ar and ar+1 • On the other hand, if x < lin then ak+1 < ak for every k and max ak = ao = 1. For e = 2.72, n = 100 we have

k*=[lOOe-lJ. e + 1

26. peA') = 1 - peA) = 1 - t = i, peA' u B) = peA') + PCB) - peA' B)

= 1 - peA) + PCB) - {PCB) - P(AB)}

= 1 - peA) + P(AB) = 1 - t + i = i,

Solutions

peA u B') = peA) + PCB') - P(AB')

= peA) + 1 - PCB) - {peA) - P(AB)}

= 1 - PCB) + P(AB) = 1 - t + i = g, P(A'B') = P[(A u B)'] = 1 - peA u B)

= 1 - peA) - PCB) + P(AB) = 1 - t - t + i = 172'

peA' u B') = P[(AB)'] = 1 - P(AB) = 1 - i = i.

129

27. From A ~ A u B, B ~ A u B, we conclude that peA) ~ peA u B), PCB) ~ peA u B). Hence

peA u B) ~ max{P(A),P(B)}, (1)

and as peA) = 3/4 > PCB) = 3/8 it follows that peA u B) ~ 3/4. Because of A ;2 AB, B ;2 AB we have P(AB) ~ peA), P(AB) ~ PCB) and so

P(AB) ~ min(P(A), PCB)). (2)

Consequently, P(AB) ~ 3/8. By the addition theorem we have P(AB) + peA u B) = peA) + PCB) and,

because of peA u B) ~ 1, P(AB) ~ peA) + PCB) - 1, but P(AB) ~ ° and hence

P(AB) ~ max{O, peA) + PCB) - I}.

P(AB) ~ i + i-I = l From (1), (2), and (3) for peA) = 1/3, PCB) = 1/4 we have the inequalities

peA u B) ~ t and ° ~ P(AB) ~ l 28. We have P(A'B) = PCB) - P(AB) and peA) = 1 - peA'). Hence

P(AB) - P(A)P(B) = P(AB) - [1 - P(A')]P(B)

= P(A')P(B) - [PCB) - P(AB)]

= P(A')P(B) - P(A'B).

P(AB) - P(A)P(B) = P(AB) - P(A)[1 - PCB')]

= P(A)P(B') - [peA) - P(AB)]

= P(A)P(B') - P(AB').

The required relation follows from (1) and (2).

29. By the addition theorem, we have

(3)

(1)

(2)

P(A 1 A 2 ) = peAl) + P(A 2 ) - peAl u A 2 ) ~ peAl) + P(A 2 ) - 1. (1)

That is, for n = 2 the inequality holds. Suppose that it holds for n = k, i.e.,

k

P(A 1 A 2 ••• Ad ~ L P(A;), - (k - 1). i~l

130 Solutions

It is sufficient to show that it also holds for n = k + 1. Using (1) and (2) we get

P(A 1A 2 ···Ak+1) = P[(A 1A 2 · •• Ak)Ak+1]:::;; P(A 1A 2 • .. Ak) + P(Ak+d-l

k k+l

~ L: P(Ad - (k - 1) + P(Ak+l) - 1 = L: P(AJ - k. i=l i=l

30. We have

PEA u B u C] = PEA u (B u C)] and by the addition theorem

= P(A) + P(B u C) - P[A(B u C)]

= P(A) + P(B u C) - P[(AB) u (AC)]

= P(A) + P(B) + P(C) - P(BC) - P(AB) - P(AC)

+ P(ABC).

PEat least one] = P(A u B u C) = 0.5 + 0.4 + 0.3 - 0.35 - 0.25 - 0.2 + 0.15 = 0.55, that is, 55%.

31. (a) Let Pn(k) denote the probability that the integer n is divisible by k. Then Pn(k) = (l/n) [n/k] where [x] denotes the integer part of x. Hence the required probability equals

(b) For n = 100 we get from (*)

1 ([100J [100J [100J) 1 1 100 3 + 4 - 12 = 100 (33 + 25 - 8) = 2:'

(c) Taking the limits as n --+ 00 in (*) we obtain

1 1 1 1 3 + 4 -12 = 2:'

32. (i) Let Pk = P[exactly k of the events A, B, C occur] (k = 0, 1,2,3), then

Po = P[A'B'C'] = P[(A u B u C)'] = 1 - P(A u B u C) = 1 - P(A) - P(B)

- P(C) + P(AB) + P(AC) + P(BC) - P(ABC),

P1 = P[AB'C' u A'BC' u A'B'C] = P[A(B u C)'] + P[B(C u A)']

+ P[C(A u B)']

= P(A) - P[AB u AC] + P(B) - P(BC u BA) + P(C) - P[CA u CB]

= P(A) - P(AB) - P(AC) + P(ABC) + P(B) - P(BC) - P(BA)

+ P(ABC) + P(C) - P(CA) - P(CB) + P(ABC)

= [P(A) + P(B) + P(C)] - 2[P(AB) + P(AC) + P(BC)] + 3P(ABC),

Solutions

P2 = P[ABC u AB'C u A' BC] = P(ABC) + P(ACB') + P[BCA']

= P(AB) - P(ABC) + P(AC) - P(ABC) + P(BC) - P(ABC)

= [P(AB) + P(AC) + P(BC)] - 3P(ABC),

P3 = P(ABC).

131

(ii) PEat least k of the events A, B, C occur] = L~=kPn the Pi as in (i); see also Exercise 30.

33. By the addition theorem, for n = 2, (1.1) holds. Suppose that it holds for n = r. We shall show that it also holds for n = r + 1. By the addition theorem we have

By hypothesis for n = r it follows that

p[~ (A iA'+1)] = ktl (_I)k-lS:

where S: = L P(Ai,Ai2···AikA'+I)· l~il <i2< ... <ik5:r

Let

Sr = L P(Ai, Ai2 ··· Ai)' k= 1, ... ,r+ 1. (2) 1 ~il < ... <ik 5:r+l

Then , ,+1

SI + P(A,+d = L P(A i) + P(A,+d = L P(A;) = st*, (3) i=1 i=1

and, in general,

Sk + S:_I = L P(A 1 A2 ••• AiJ 1 ~il < .-. <ik 5:r

+ L P(Ai,Ai2···Aik_,A'+1) 15:il < ... <ik- I $r

= S:*, k = 2, ... , r,

By virtue of (2)-( 5), (1) becomes

(4)

(5)

132 Solutions

r

= SI + P(Ar+t> + L (_l)k-l [Sk + S:-I] + (-I)'S: k=2

r+l = L (_I)k- 1S:*,

k=1

i.e., (1) holds for n = r + 1.

34. For any two events A and B we have: (i) d(A, B) = P(A t::,. B) = P[AB' u BA'] ~ 0 and d(A, B) = 0 if A = B, that

is, d is a nonnegative function. (ii) d(A, B) = P(A t::,. B) = P(B t::,. A) = d(B, A) (symmetric property).

To complete the proof we have to show that for any events A, B, and C the triangle inequality holds, i.e.,

(iii) d(A, C) ::;; d(A, B) + d(B, C),

d(A, B) + d(B, C) - d(A, C) = P(AB' u BA') + P(BC' u CB')

- P(AC' u CA')

= P(AB') + P(BA') + P(BC') + P(CB')

- P(AC') - P(CA')

= P(AB'C u AB'C') + P(A'BC u A'BC')

+ P(ABC' u A'BC') + P(AB'C u A'B'C)

- P(ABC' u AB'C') - P(A'BC u A'B'C)

= P(AB'C) + P(AB'C') + P(A'BC) + P(A'BC')

+ P(ABC') + P(A'BC') + P(AB'C)

+ P(A'B'C) - P(ABC') - P(AB'C')

- P(A'BC) - P(A'B'C)

= 2[P(AB'C) + P(A'BC')]

~O.

35. (a) The probability Po of no lucky ticket is

and the probability of winning is 1 - Po. (b) The probability of winning is

Solutions 133

37. Let Xj denote the number of persons born on the jth day of the week (among the 7 persons). Then (Xl' ... , X 7) has the multinomial distribution

7

L nj = 7. j=l

(a) P[Xj = 1, j = 1,2, ... , 7J = 7! (1/7)7. (b) peat least two on the same dayJ = 1 - P[Xj = 1, j = 1,2, ... , 7J =

1 - 7! (lj7r.

7' (1)4(5)3 (c) P[two on Sunday and two on TuesdayJ = 2! 2; 3! 7 7 .

38.

39. For real roots we must have

b2

ac<-. - 4 (1)

From the values of the product ac we observe that inequality (1) holds as follows:

Value of b: 234 5 6 Favorable cases: 3 8 14 17

Thus we have 43 favorable cases out of a total of 63 = 216. Hence the probability of real roots is 43/216 and of complex roots is 173/216.

4

40. (a) C52)

(b) 13 x 48

C52)

5 X 45

(c) C52)

(a favorable case for each of the 4 suits).

(to each choice of 4 cards of the same face value (ace, 2, 3, etc.) there correspond 48 ways of choosing the fifth card).

(there are 5 quintuplets with successive face values, e.g., (2, 3, 4, 5, 6), (3, 4, 5, 6, 7), etc., and each of the 5 cards can be chosen in 4 ways corresponding to the 4 suits).

134 Solutions

41. (i) Multiple hypergeometric:

C~)C:)C:)(13 - VI 1~ V2 - vJ I GD· (ii) (~)(134~ V) I G~)' V = 0,1,2,3,4.

(iii) (~)(~)(13 ~ v~ ~ vJ I G~)' (a) Let the event Aj be the jth player who gets an ace (j = 1,2,3,4). Then

4! 48!/(12!)4 P[AIA2 A 3A4] = 52!/(13!)4

(b) Let Bj be the event when the jth player gets all the aces (j = 1,2,3,4). Then

Pea player gets all the aces] = 4P[Bj ] = 4 (:8) I G~)' (c) Pea player gets VI aces and his partner V2 aces]

(4)( 48 )(4-VI)(35+VI) = 2 X VI 13 - VI V2 13 - V2

G~)G~) (i.e., of the two hands one has VI aces and the other has V2 aces), if VI -:f. V2 ,

when VI = V2 the factor 2 above should be deleted.

42. (a) The ten boys define ten successive intervals (arcs). The six girls can

choose six of the intervals in C60) ways. The event A that a certain boy

remains between two girls can be realized in (!) ways. Hence

(b) The six girls can enter in (10)6 ways. The event B that certain girl stands by certain boy can be realized in 2 x (9)5 ways. Hence

P[ ] = 2 X (9)5 = ! B (10)6 5'

This is immediate since the girl can choose 2 favorable positions out of the 10. We cannot conclude that her choice is not random.

Solutions

43. (a) (i) Spq25 with p = 1 - q = 1/26. (ii) (3/1W.

(iii) 1/265•

(b) (i) C:) Ie:) = ;6·

(ii) (D I Cs6). (iii) 1/(26k

135

44. Let us represent the r balls by r stars, and the n cells by n spaces between n + 1 bars. Each distribution starts and ends with a bar and between these extreme bars there are r stars and n - 1 bars. To each selection, either of the r places for the stars or of the n - 1 places for the bars, there corresponds a distribution. Consequently, there are

such distributions (Feller, 19S7, p. 36). An alternative proof is by induction on n as follows: Suppose (*) holds for n = k, i.e., the number of solutions Xi ~ 0, Xi integers, of the equation

is given by

( k + r - 1) s(k, r) = k _ 1 .

Then clearly the solutions of

Xl + ... + Xk+l = r,

or equivalently of

are the solutions of

X I + ... + Xk = r - i, i = 0, 1, ... , r,

that is,

s(k + 1, r) = s(k, r) + s(k, r - 1) + ... + s(k, 0)

(k + r - 2) (k - 1) = (k + r - 1) + + ... + , k-l k-l

which, by Exercise 17(ii), gives

( k + r) s(k + 1, r) = k '

i.e., (*) holds for n = k + 1.

136 Solutions

45. The number of distributions with exactly m empty cells is (:).

( r - 1 ). Indeed, if we represent the r balls with r stars and the n cells n - m-1

by spaces between n + 1 bars, then we can select m cells (empty cells) from the

n in (:) ways. Without any loss of generality, let the empty cells be the first

m, that is, the spaces between the first m + 1 bars. The r stars leave r - 1 spaces of which n - m - 1 are to be occupied by bars and hence we have

( r - 1 ) choices. Using the result of Exercise 44, we conclude that the n - m-l

required probability equals

46. (a) There are (52). ordered samples of n cards and of these 4· (48).-1 have an ace as the nth card. Thus the required probability is

( 52 - n) 4·(48).-1 3

P(A.) = (52). = C:) , n = 1, 2, ... , 49.

(b) Let E. the event that the first ace will appear after the nth card. This is equivalent to the event that the first n cards will contain no ace

n = 0, 1, ... ,48.

Another proof Let An the event in (a). Then E. = An+1 u A.+2 U··· A49 and, as the Aj are mutually exclusive,

where for the sum

49 (52 - i) i=~1 3

the result of Problem 17(ii) was used.

Solutions 137

47. (a) Ignoring leap years, there are 365v ways of distributing v persons in 365 days. The event Av that no two persons have the same birthday can be realized in (365)v ways ( = 0 for v > 365). Hence

[A ] = (365)v = nV (1 _ k - 1) ~ 1 _ v(v - 1) P v 365v k=l 365 730 '

where the approximation ~ holds for small v. Thus

v(v - 1) P[A~] = 1 - P[Av] ~ 730 .

(b) Solving the inequality P[A~] ;? 1/2 gives v = 23 with P[A;3] = 0.507. For v = 60, it turns out that P[A~o] = 0.994, that is, in a class of60 it is almost certain that at least two have a common birthday.

48. Let Ak be the event that the kth letter is placed in the kth envelope (k = 1,2, ... , N). Then

p = P[each letter is placed in a wrong envelope]

= 1 - P[at least one letter is placed in the right envelope]

= I-P[U AkJ. k=l

By Poincare's theorem we have

p[ U AkJ = Sl - S2 + ... + (_l)N- 1SN' k=l

where

Sk = L P[Ai1 ... AiJ. 1 Sit < ... <ikSN

But

[ ] (N - I)! P Ak =-N"!'

(N - 2)! P[AiAj] = " N.

(N - 3)! P(AiAjAd = N! '

so that, in general,

_ (N)(N - k)! _~. Sk - k N! - k!'

the last term is

138 Solutions

and so

P=I-P[U Ak]=I-[I-~+~- ... +(-lt-l~]= f (_I)k~. k~l 21 31 N. k~2 k.

This probability, even for moderate values of N (N > 4), approaches 1 - e-1 =

0.632. Note: The problem of at least one match (here in terms of envelops and

letters) is referred to as the problem of rencontres.

49. (a) Let Ak be the event that the number k does not appear (k = 1, 2, ... , n). The probability Pm that exactly m among the n events AI' ... , An occur simultaneously is given by

(m + 1) (m + 2) m(n) Pm = Sm - m Sm+l + m Sm+2 - ... + (_I)n- m Sn'

where Sk is defined as in Problem 48* and is here given by

The required probability is given by

p! = Pn-m·

(b) P[each of the numbers 1,2, ... , n will appear at least once]

= 1 - Pn[at least one of the numbers 1,2, ... , n will not appear]

= 1 - p[ U Ak] = 1 - ± (-I)k- 1Sk = 1 _ ± (_l)k-l(n)([n - k]r)N k~l k~l k~l k (nr)N

= ± (_l)k(n)([n - k]r)N. k~O k (nr)N

50. P[m balls will be needed] = (;)p[n - 1 among the numbers 1,2,

... , n will appear at least once after m - 1 trials, and at the mth trial the nth number will appear]

= G) [1 - :t: (_I)k-l] Sk nr - ~ + 1

= (n)[nf (_I t (n - 1)[([n - k - l]r)m-l]. r 1 k~O k (nr)m-l nr - m + 1

= ± (_I)k-l (n = 1) ([n -_k]r)m-l . k~l k 1 (nr 1 )m-l

Solutions 139

51. (a) Let Ai be the event that the ith man takes the ith coat and the ith hat. Then

P = P[no one takes both his coat and hat]

N

= 1 - L (_l)k- 1Sk' k=l

where

Sk = L P[A i ,··· AiJ (Poincare's theorem), k = 1, ... , 2. lSi 1 <"'<ik s,N

But

[ (N - 1)!J2 PEAk] = N! '

N [(N - 1)!J2 (N - 1)! Sl = L P(Ad = N = ,

k=l N! N! 1!

[ (N - 2)!J2 P(A.A.) =

I J N! '

and in general

(N)[(N - 2)!J2 S2 = L P(AiAj) = l";i<j";N 2 N!

(N - 2)! N!2! '

[ (N - k)!J2 P[Ai, A i2 ··· AiJ = N! '

S = (N)[(N - k)!J2 = (N - k)!. k k N! N! k! '

the last term is

Therefore

= 1 - f (-1t-1 (N - k)! = f (-1t(N - k)! P k=l N! k! k=O N! k! .

(b) P[each man takes the wrong coat and the wrong hat]

= P[each takes a wrong coat]· P[each takes a wrong hat]

= [t (-1)kk\J [t (-1)kk1,J = [t (-1tk\J2 (see Exercise 48). k-2 . k-2 . k-2 .

52. (a) Let Pk be the probability that the housewife gets k free packets. Then

(1)

with Si (i = 1,2, 3,4), defined as in Problem 48*, and let Ai be the event

140 Solutions

that the ith letter of the word TIDE does not appear, (i = 1,2,3,4). We have

Substituting into (1) we get

On the other hand,

P2 = (~~4(~y (multinomial distribution). (2)

53. Let A be the event that both the selected balls are white. We have:

and because of

N N1 -l 1 > -,---=-__ --,--Nl + N2 Nl + N2 - 1

for N2 > 0,

equation (1) gives the inequalities

( Nl )2 1 ( Nl - 1 )2 Nl + N2 > "2 > Nl + N2 - 1 '

the first of which gives

1 Nl > j2 (Nl + N2),

and the second

Hence, it follows that

that is, Nl > (j2 + I)N2'

(1 + j2)N2 < Nl < (1 + j2)N2 + 1.

(1)

(2)

For N2 = 1 we get 2.41 < Nl < 3.14, that is, Nl = 3. Then we observe that

P(A)=~'~=~ 4 3 2

and the required minimum value of Nl is 3.

Solutions

(b) If N2 is even, then, from (1) and (2), we have

2 4 6

Nl between

4.8 and 5.8 9.7 and 10.7

14.5 and 15.5

Thus the minimum value of Nl is 15.

possible Nl

5 10 15

P(A)

10/21 45/91

1/2

141

(c) The minimum value of N corresponds to the minimum value of Nl since Nl is an increasing function of N2 • Thus in (a) N = 3 + 1 = 4, and in (b) N = 15 + 6 = 21.

54. Since A and B are independent we have

P(AB) = P(A)P(B), and hence

P(AB') = P(A) - P(AB) = P(A) - P(A)P(B) = P(A)[1 - P(B)]

= P(A)P(B'),

which shows that A and B' are independent. Similarly, it is shown that

P(A' B) = P(A')P(B),

P(A' B') = P(A u B)' = 1 - P(A u B) = 1 - P(A) - P(B) + P(AB)

= 1 - P(A) - P(B) + P(A)P(B) = [1 - P(A)] [1 - P(B)]

= P(A')P(B').

(1)

(2)

55. Since the events A, B, and C are completely independent, the following relations hold:

P(AB) = P(A)' P(B),

P(BC) = P(B)P(C),

P(CA) = P(C)P(A),

P(ABC) = P(A)' P(B)P(C).

From (2) and (4) we obtain

P(AA*) = P(ABC) = P(A)'P(BC) = P(A)'P(A*),

(1)

(2)

(3)

(4)

which shows that the events A and A* = BC are independent. The proof of the remainder is similar.

56. x = P(A'B'C) = P(A u B)'C = P(C) - P(A u B)C = P(C)

- P(AC u BC)

= P(C) - P(AC) - P(BC) + P(ABC)

= P(C) - P(A)P(C) - P(B)P(C) + P(A)P(B)P(C).

142 Solutions

Letting P(A) = a we obtain

x = (1 - a)(1 - P(B»P(C), (I)

b = I - P(A u B u C) = P(A'B'C') = P(A')P(B')P(C')

= (I - P(A»(I - P(B»(I - P(C».

b = (1 - a)(1 - P(B»(I - P(C», (2)

c = I - P[ABC] = I - P(A)P(B)P(C) = I - aP(B)P(C). (3)

From (1) and (2) we get

x P(C) -b 1 - P(C)

Substituting (4) into (3) we obtain

::;. P[C] = ~b. x+

P(B) = (I - c)(x + b). ax

Combining (I), (4), and (5) we get

(4)

(5)

x = (1 _ a) (I _ (I - c)(x + b»). _x_ = I - a . ax - (1 - c)(x + b), ax x+b x+b a

ax(x + b) = (1 - a)ax - (I - a)(1 - c)(x + b),

ax2 + abx = (1 - a)ax - (I - a)(1 - c)x - (I - a)(1 - c)b,

ax2 + ab - (I - a)(a + c - I)x + (I - a)(1 - c)b = O. (6)

As x represents probability, both roots of (6) must be positive and therefore their sum as well, i.e.,

(1 - a)(a + c - I) - ab > 0, ab

a-I+c>-I - a'

(1 - a)2 + ab c>--,---

I-a

57. Associate the events Ai with independent Bernoulli trials; hence use the binomial distribution.

58. Let Ai be the event that an ace appears at the ith throw and let Bi be the event that a six appears at the ith throw. Then the required probability equals

59. (a) We have

_ (~Y _ (4)2 _ 16 -GY - 5 - 25·

P[X2 > 0] = I - P[X2 = 0].

Solutions· 143

By the theorem of total probability

P[X2 = 0] = P[Xl = 0]P[X2 = 0IX1 = 0] + P[Xl = I]P[X2 = 0IX1 = 1]

1 1 1 1 1 25 + P[Xl = 2]P[X2 = 0IX1 = 2] =:r 1 + 2'4 + 4· 16 = 64'

and hence

(b) Using Bayes's formula, the required probability equals

P[Xl = 21X2 = 1]

P[Xl = 2]P[X2 = IIX1 = 2]

P[X1 = I]P[X2 = IIX1 = 1] + P[Xl = 2]P[X2 = IIX1 = 2]

1 1

4 4 -

1 1 1 1 5 2'2 + 4·4

60. Let Am denote the event Xmax:::; m (m = 1,2, ... , n). Then

P[An] = P[Xl :::; m, X 2 :::; m, ... , X,:::; m]

m' = P[Xl :::; m]···P[X,:::; m] = ,.

n

Clearly, Am- 1 Am and the event [Xmax = m] = Am - Am- 1 . Consequently,

(a) P[Xmax = m] = P[Am] - P[Am-1Am] = P[Am] - P[Am- i ]

m' - (m - 1)'

n'

m = r, r + 1, ... , n.

61. The number X of successes (wins) ofa team in n games obeys a binomial distribution with probability p = 1/2 for success, that is,

P[X = k] = G)Gr k = 0, 1, 2, ... , n.

(a) P[the series will end in at most 6 games]

= P[in exactly 4 games] + P[in exactly 5 games]

+ P[in exactly 6 games].

144 Solutions

Now we have

P[in exactly 4 games] = PEA wins the 4 games] + P[B wins the 4 games]

P[in exactly 5 games] = 2· PEA wins 3 of the first 4 games and also wins

the 5th (last) game] = 2· G) GY =~. P[in exactly 6 games] = 2·G)GY = 156'

Hence,

[ h . '11 d' 6] 1 1 5 11 Pte senes Wl en In at most games ="8 + 4 + 16 = 16'

(b) The required probability p equals

p = P[B will win the games 3 to 6]

+ P [A will win 2 of the games 3 to 5 and the sixth game]

=(t)4 + 3(t)4 =t.

62. Let Ai be the event when the patient has illness Ai (i = 1,2,3), and let B be the event when the result of the test is positive twice. By hypothesis, the a priori probabilities are

1 P(Ad = 2'

1 P(A 2 ) = 4'

After the three independent repetitions of the test the probabilities that the result will be positive are

(3)(1)23 9 P[BIA 1 ] = 2 4 4 = 64'

P[BIA 2 ] = (D GY =~,

From Bayes's formula we get the required probabilities

2 9 4 64

PEA liB] = -2 -9--1-24--1-2-7

4"64 +4'64 +4"64

24 P[A 2 IB] = 68'

18 18 18 + 24 + 27 69'

Solutions 145

63. Let G be the event that a person is Greek, let T be the event that a person is Turkish, and let E be the event that a person speaks English. Then from Bayes's formula we have

P G E _ P[GJP[EIGJ 0.75 x 0.20 [ I J - P[GJP[EIGJ + P[TJP[EI TJ 0.75 x 0.20 + 0.25 x 0.10

= 0.857,

that is, of the English speaking population of Nicosia, 85.7% are Greeks.

64. Let Ai be the event that A forgets his umbrella in the ith shop, let Bi be the event that B forgets his umbrella in the ith shop, and let Bo be the event that B has left his umbrella at home. Then:

(a) P(they have both umbrellas) = P(A'lA~A~)[P(Bo) + P(BbB~B;B3)]

(b) P(they have only one umbrella)

27 64 + 27 2457 64 2·64 8192

= P(A'lA~A~)[P(BbBd + P(BbB~B2) + P(BbB~B;B3)]

+ [P(Ad + P(A'lA 2) + P(A'lA~)][P(Bo) + P(BbB~B;B3)J = (~)3[~~~~~ + ~(~)2 ~J

4 24244 2 4 4

+[~+~~+(~y ~lG+~GYJ 999 6279 7278

= 8192 + 8192 = 8192

(c) P(B lost his umbrellalthey have only one umbrella)

P(they have one umbrella and B lost his umbrella) P(they have only one umbrella)

999 8192 999 7278 7218 8192

65. The condition for A and B to be independent is

P(AB) = P(A)P(B).

146 Solutions

We have

P(A) = 1 - P(A') = 1 - [P(all boys) + P(all girls)]

( 1 1) 2n- 1 - 1 = 1 - 2n + 2n = -2n---'-1 -,

(l)n (1) (l)n-l n + 1 P(B) = P(all boys) + P(one girl) = 2" + n 2" 2" = ~'

. I n P(AB) = P(one gu) = 2n '

and by condition (*) we must have

n

2n

n + 1 2n- 1 - 1 -- => 2n- 1 = n + 1 => n = 3.

2n 2n - 1

66. Let A be the event that a really able candidate passes the test and let B be the event that any candidate passes the test. Then we have

P[BIA] = 0.8, P[BIA'] = 0.25, P(A) = 0.4, P(A') = 0.6,

and by Bayes's formula

P[AIB] _ P(A)P(BIA) - P(A)P(BIA) + P(A')P(BIA')

0.32 0.32 + 0.15

that is, about 68%.

00

67. PEA] = L PEA wins at the k + 1 throw of the dice] k=O

where

PI = PEA wins in a throw of the dice] = :6'

P2 = P[B wins in a throw of the dice] = 366'

consequently,

5 36

P(A) = -----:3---:-1----:3=-=-0

1 - 36· 36

30 61

32 47'

68. Let Ai be the event that player Pi wins (i = 1,2,3), let Bj be the event

Solutions 147

that a non-ace appears at thejth throw, and let P(AJ = ai (i = 1,2,3). Then

and

because if PI does not win on the first throw then Pz plays "first" and hence P(AzIBI ) = P(Ad = a l • Similarly,

25 a3 = P(A3) = P(A 3BI Bz) = P(BIB2)P(A3IBIBz) = 36al'

and, since al + az + a3 = 1, we get

36 a-

I - 91' 30

a -z - 91'

69. By an argument similar to Exercise 68 we get

k = 1,2, ... , N, q = 1 - p.

and by the condition I,~=I Pk = 1

pqk-I Pk = -1--N ' -q

k = 1,2, ... , N.

70. Let W be the event that finally a white ball is selected from A, and WI be the event that a white ball is selected from A, and let Wz be the event that a white ball is selected from B. Then by the theorem of total probability we get

peW] = P[WI WI Wz]· P[WI W2] + P[WI WI W;JP[WI W;J

+ P[WIW;W2]P[W;W2] + P[WIW;W;JP[W;W;J,

where setting WI + hi = Ni (i = 1,2), we have

, WI - 1 P[WI WI W2] = --,

NI

WI - 1 P[WI WI W2] = --,

NI

P[WI w.'W'] = ~ I 2 N I '

71. (a) Let A be the event that a person has the disease, and let Bi be the

148 Solutions

event that the ith test is positive (i = 1, 2). Then \

PEA] = 0.1, P[B;lA] = 0.9, P[BiIA'] = 1 - P[B;IA'] = 0.1.

P[BIA] = P[B1 IA]P[B2 IA] = 0.81, P[BIA'] = 0.Q1,

P[A]· P[BIA] 81 P[AIB] = P[A]. P[BIA] + P[A']· P[BIA'] = 81 + 9 = 0.9.

(b) If C denotes the event that only one test is positive then

P[ClA] = G)P[BdA]. P[B;IA] = 0.18,

P[ClA'] = G)P[B1IA'].P[BIIA'] = 0.18.

Thus we find

PAC = P[A]P[ClA] = 0.1. [ I] P[A]· P[ClA] + P[A']P[ClA']

72. (a) Let A be the event that coin A is selected, let B be the event that coin B is selected, and let Hk be the event that heads appears k times (k = 0,1,2). Then by the theorem of total probability

We have

PEA] = P[B] =~, P[HkIA] = G)(~YGy-k,

P[HkIB] = G) Gy(~y-k, and therefore

(i)

(ii)

(b) P[heads at least once in strategy (a)] = P[H1 ] + P[H2 ]

11 16'

P[heads at least once in strategy (b)] = G) GY + (D GY 3 4·

Solutions 149

Hence he must follow the second strategy ifhe wants to maximize the probability of at least one head.

73. Let A be the event that a white ball is selected from A, let B1 be the event that two white balls are selected from B, let B2 be the event that one white ball and one black ball are selected from B, and let B3 be the event that two black balls are selected from B. Then by the theorem of total probability we get:

G) P[BI] = (I})' G)(~)

P[B2 ] = -(122) , G)

P[B3] = e22)'

8 P[AIBI ] = j)'

Hence

7 peA IB2 ] = 13'

136 peA] = ---.

429

6 P[AI B3] = 13·

(b) The required probability p say, is given by

Applying Bayes's formula we get

PCB IA] = P[BI]P[AIBIJ = ~ 1 peA] 34'

P[B2 ]P[AIB2 ] 7 PCB I A] =------------- = -.

2 peA] 34

Hence

13 P = ---.

34

74. Let the event A be that the test is positive and the event B that a woman has the disease. Then

P(B)P(AIB) P[BI A] = --------------------------

P(B)P(AIB) + P(B')P(AIB')

I 19 --_._-2000 20

1 19 1999 1 2000· 20 + 2000·20

19

2018'

that is, less than 1 'X. of the women with positive test have the illness. The lady's fear is rather unjustified.

75. Let A, B, C be the events that the clerk follows routes A, B, C, respectively, let L be the event that the clerk arrives late, and let S be the event

150 Solutions

that the day is sunny. Then

P[CSL] P[S]P[CLIS] (a) P[ClSL] = P[SL] = P[S]P[LIS]

P[ClS]P[LICS]

P[AIS]P[LIAS] + P[BIS]P[LIBS] + P[ClS]P[LICs]

Since

P[AIS] = P[BIS] = P[ClS] =~,

P[LIBS] = 0.10,

we get P[ClSL] = 0.5.

P[LICS] = 0.15,

, P[S']P[LIS] (b) P[S IL] = P[S]P[LIS] + P[S']P[LIS'] ,

P[LIAS] = 0.05,

P[LIS] = P[AIS]P[LIAS] + P[BIS]P[LIBS]

1 30 + P[ClS]P[LICS] = 3 100'

P[LIS'] = P[AIS']P[LIAS'] + P[LIBS'] + P[ClS']P[LICS']

1 ( 6 15 20 ) 1 41 = 3 100 + 100 + 100 = 3" 100'

3 , 1 peS) = 4' P(S) = 4·

Consequently,

41 P[S'IL] =131.

76. Let A be the event that the painting is original, and let B be the event that the expert says it is an original. Then according to Bayes's formula

P[A]P[BIA] (a)P[AIB] = P[A]P[BIA] + P[A']P[BIA']

5 9 6 10 45

5 9 1 1 46 «( 10 +"6·10

(b) Let A be the event that the second choice is original, and let H be the event that the expert decides correctly that it is an original. Then by the total probability theorem we have

P[A*] = P[A*IH]P[H] + P[A*IH']P[H'] = ~.1O + ~.~ =~. 10 11 10 11 110

Solutions 151

77. Let A be the event that the 5 cards are clubs, and let B be the event that the 5 cards are black. Then

P[AB] C:) I C52) C:)

P[AIB] = PCB] = C56) I C52) = C56)'

78. Let A 2 denote 2 boys, let B2 denote 2 girls, and let Cn denote n children in the family. Then

(a)

(b)

and since

and

we get

P[C2 IA 2 ] = :[C2 ]P[A 2 IC2] •

L P[Cn ]P[A 2 ICn] n=2

00

peA B ] ~ P[Cn ]P[A 2 B2 ICn ] P[B2 IA 2 ] = 2 2 = _n-_: _____ _

P[A 2 ] L P[Cn ]P[A 2 ICn]

n=2

(l)n-l Pn = P[Cn] = (1 - 2a)"2 ' n = 2, 3, ... ,

P[A 2 ] = n~2 P[Cn ]· P[A 2 ICn ] = (1 - 2a) n~2 G) Gyn-l _ (1 - 2a) ~ _ (~)n-2 _ (1 - 2a) 2 _ 8(1 - 2a) - 42 n4:2 n(n 1) 4 - 42 ( 1)3 - 27 '

1--4

00

P[A 2 B2 ] = L P[Cn ]· P[A 2 B2 ICn ] = P[C4 ]· P[A 2 B2 IC4 ] n::::::4

6(1 - 2a) 3(1 - 2a)

128 64 Finally,

(1 - 2a)

8 27 P[C2 IA 2 ] = 8(1 - 2a) = 64'

27

152 Solutions

3(1 - 2a)

64 81 P[B2IA2J = 8(1 - 2a) S12·

27

79. (a)

P[A3I B3J = P[A3J· P[B3I A3J P[A l JP[B3 IA l J + P[A2JP[B3IA 2J + P[A3JP[B3I A3J

(b)

(c)

6

11

P[A2I B3J = ;[A2JP[B3IA2J

I P[AiJP[B3IAJ i=l

0.2 x 0.1

0.11

P[A 2IBl uB2J = ;[A2J·P[Bl uB21A2J ,

I P[AJ· P[Bl u B21 AiJ i=l

whereby the addition theorem for conditional probabilities

and the conditional probabilities are given in the table.

2

11

80.

(Ia) Pk = G)C70YC30Y-k, k = 0,1, ... , S (binomial).

(Ib)

(II a)

C)(S ~ k) Pk = CSO) , k = 2, 3, 4, S (hypergeometric).

(10 - k)S P[Xmin > kJ = lOs '

P[Xmin = kJ = P[Xmin > k - 1J - P[Xmin > kJ

(10 - k + 1)s - (10 - k)S lOs k = 1, 2, ... , 10.

( Ib) P[X. > kJ = (10 - k)s I mm (lO)s

Solutions 153

(IlIa) P[Xmax ~ k] = C~y, P[Xmax = k]= P[Xmax ~ k] - P[Xmax ~ k - 1]

k = 1,2, ... , 10.

(kh (IIIb) P[Xmax ~ k] = (lOh k = 5, 6, ... , 10.

(IVa) (7)( 3 )k-l Pk = 10 10 ' k = 1,2, ....

(IVb) k = 1,2,3,4.

81. (I) and (IV) as in the preceding problem.

(lIa)

(lIb)

(IlIa)

1[10 - ~~k - I)J, P[Xmin ;::: k] = [7 - (k - I)J5

10 '

k = 1,2,3,

k = 4, 5, 6, 7.

110 - 2(k - 1)5

(10)5 ' P[Xmin ;::: k] =

(7 - (k - 1»5 (10)5 '

{(2k/IO)5,

P[Xmax ~ k] = (k + 3)/(10)5,

k = 1,2,3,

k = 4, 5, 6, 7.

k = 1,2,3,

k = 4, 5, 6,

{ (2k)5/(l0)5' k = 1,2,3, (IIIb) P[Xmax ~ k] = (k + 3)5/(10)5, k = 4, 5, 6, 7.

(IIIc) P[Xmax ~ k] = (2k/IO)5, k = 1,2,3,

P[Xmax ~ k] = (k + 3/10), k = 4, 5, 6, 7.

82. Let X be the number of defective items in a sample of 10. Then X obeys a binomial law

P(X = k) = CkO) (l/IO)k(9/IO)lo-k,

and the required probability is

P(X = 0) = (9/10)10.

k = 0, 1, ... , 10,

83. Let X be the number of persons that incur the accident in a year. Then X obeys a binomial law with n = 5,000 and P = 1/1,000. Since A. = np = 5, the

154 Solutions

Poisson approximation to the binomial (for large n and small p) applies. Therefore, the required probability is given by

( 50 5 52) 31 P(X ::;; 2) = e- 5 O! + IT + 2! = 2e- 5 •

84. Let X be the number of persons making reservations on a flight and not showing up for the flight. Then the random variable X has the binomial distribution with n = 100 and p = 0.05. Hence

P(X ::;; 95) = 1 - P(X > 95) = 1 - if (lOOk ) (0.05)k(0.95)iOO--k. k;96

85. (i) Let X be the number of accidents per week. Then X obeys a Poisson probability law with parameter A = 2 and hence

P(X ::;; 2) = e- 2 (1 + 2/l! + 22/2!) = 5e- 2 •

(ii) If Y denotes the number of accidents in 2 weeks, then Y has A Poisson distribution with ). = 4. Hence

P(Y::;; 2) = e- 4 (1 + 4/l! + 42/2!) = 13e-4 .

(iii) The required probability equals (P[X ::;; 2])2 = 25e- 4 .

86. The number X of suicides per month obeys a binomial probability law with parameters n = 5 x 105 and p = 4 X 10-6 • Since np = 2 « 10) the Poisson probability law with A = 2 provides a satisfactory approximation to the above binomial law. Hence,

4 2k Po == P(X ::;; 4) = L e- 2 , = 7e- 2 .

k;O k.

Let Y be the number of months with more than four suicides. Then

P(Y = k) = Ck2) (1 - PO)k(pO)i2-k,

P(Y 2 2) = 1 - {P(X = 0) + P(X = I)}.

87. Because of the independence of the tosses the required probability is equal to the probability that 20 tosses result in n heads, that is,

( 20) 1 n 220 '

The required conditional probability equals

C03~ n) GYO C03~ n) Jo C03~ k) GYO Jo C03~ k)'

88. Suppose that a family has n children and let X be the number of boys. Then X obeys a binomial law with parameters nand p = 1/2. The probability

Solutions 155

of at least one boy and at least one girl equals

1 1 P(1 :5; X :5; n - 1) = 1 - P(X = 0) - P(X = n) = 1 - 2 2" = 1 - 2"-1'

and since this probability must be (at least) 0.95 we get 2"-1 ~ 20 and hence n = 6.

89. Ak A Ak- 1 A

p(k; A) = e-;' k! = k e-;' (k _ I)! = kP(k - 1; A).

For given A, we conclude from the above relation that p(k; A) is increasing in k for k < A and decreasing for k > A. If A happens to be an integer then p(A; A) = p(). - 1; A). Thus p(k; A) attains its maximum when k is the largest integer not exceeding A.

90. (a) f:oo f(x) dx = I2 (1 - 11 - xl) dx

f1 f2 1 1 = Jo x dx + 1 (2 - x) dx ="2 +"2 = 1.

}'

~-------L------~~-----.x

Triangular distribution

1 foo P dx 1 [ (x - a)Joo (b) -;- -00 p2 + (x _ a)2 = -;- arc tan -p- -00 = 1.

)'

----~------------~------------------~--.x IX

Cauchy distribution

156 Solutions

(c) f(x) dx = - e(X-I'J/u dx + - e-(x-I'l/u dx = 1. fOO 1 fl' 1 100

-00 20- -00 20- I'

y

------------------~--~---------------------x

(d)

y

1/2e

Laplace distribution

f(x) dx = - xe-x/2 dx = 1. fOO If 00

-00 4 0

max

I I I

_____ J. _______ _ I

~----------~----------~----------------__.x o 2 4

X121-distribution

91. (i) We have

foo f(x) dx = a f3 x dx + a f6 (6 - x) dx = 9a, -00 0 3

and the condition

f: f(x) dx = 1

gives

Solutions 157

(ii) P[X > 3] = ~ L6 (6 - x) dx = ~, P[1.5 < X < 4.5] = ! f3 X dx +! f4.5 (6 - x) dx =~.

91.5 9 3 4

(iii) P(AB) = P[3 < X < 4.5] = 3/8 = P(A)' P(B).

Hence A and B are stochastically independent.

92. (a)

(b)

(c)

(d)

93. (a)

(b)

(c)

9

P[X > 5] = - e-x /5 dx = e-1 • 1 fOC! 5 5

P[3 < x < 6] = ! f6 e-x /5 dx = e-3/5 _ e-6/5. 5 3

P[X < 3] =! f3 e-x /5 dx = 1 - e-3/5. 5 0

P[X < 61X > 3] = P[3 < X < 6] = 1 _ e-3/5. P[X> 3]

P[0.1 < X < 0.2] = 0.1.

L P[0.k5 < X < 0.k6] = 10 x 0.01 = 0.1. K=O

P[0.3 s P < 0.4] = P[0.09 s X < 0.16] = 0.07.

94. (I) Let X be the height of a man in centimeters. Then X is N(167.9).

[ X - 167 J 1 (a) P[X> 167] = P 3 > 0 = P[Z > 0] = 2

(Z'" N(O, 1)).

[X - 167 170 - 167J

(b) P[X > 170] = P 3 > 3

= P[Z > 1] = 1 - <l>(l) = 16%,

where <l> denotes the cumulative distribution function of the standard normal N(O, 1).

(II) (i) Let Y be the number of men that have height greater than 170 cm. Then Y obeys a binomial law with n = 4 and p = P[X > 170] = 0.16. Hence

P(Y = 4) = (0.16)4 = 0.0007.

(ii) If Z denotes the number of men that have height greater than the mean Jl = 167, then the random variable Z has the binomial distribution with parameters n = 4 and p = P[X > 167] = 0.5. Thus

P[Z = 2] = G) (0.5)4 = 0.375.

158

95. (a) Let X be the length of a bolt. Then X is N(5, (0.2)2),

p = P[X ¢ (4.8,5.2)] = 1 - P[4.8 < X < 5.2]

= 1 - P[ -1 < Z < 1] = 2(1 - <1>(1)), p = 0.32.

Solutions

(b) If Y denotes the number of defective bolts in the sample, then Y has the binomial distribution with n = 10, p = 0.32. Hence P[Y = 0] = (0.68)10.

96. Let T be the time interval (in minutes) between two successive arrivals. Then T has the exponential distribution with mean 3, i.e.,

f(t) = te-t/3 , t > o. (a) PET < 3] = 1 - e- 1 = 0.37. (b) P[T> 4] = e-4/3 .

(c) Let X be the number of customers per hour and Y be the number of customers who buy the object. Then X has a Poisson distribution with A = 20. On the other hand,

P[Y = k/X = n] = (~}O.l)k(0.9rk, k = 0, 1, ... , n.

Hence

00

P[Y = k] = L P[Y = klX = n]P[X = n] n=k

= f (n)(O.l)k(0.9rke-202~n = e- 22:, n=k k n. k.

k = 0,1, ... ,

that is, Y has a Poisson distribution with parameter A = 2.

97. We have

Fy(y) = P[Y :s; y] = p[X2 :s; y] = P[ -.JY < X < .JY]

= F(.JY) - F( -.JY).

Since F is differentiable so is Fy and Y is a continuous random variable with density

fy(y) = dd Fy(y) = 1 r.: [f(.JY) + f( - .JY)]. Y 2yy

98. Using (*) of the preceding exercise, we obtain:

(a) 1 [ y+Jl2] Y > o. fy(y) = j2;ry exp ---2 - ,

2a 2ny a

(b) fy(y) = _1_ e-Jy, 2.JY

y > O.

1 1 (c) fy(y) = - .JY ' y > O.

n y(l + y)

Solutions 159

99. Fy(y) = P[IXI < y] = P[ -y < X < y] = F(y) - F(-y),

d fr(y) = dy Fy(y) = f(y) + f( - y), y > O.

(a) 1

fr(y) = ;;Cexp[ _(y2 + Jl2], U Y 2n

y > O.

(b) y > O.

(c) 2 1

fy(y) = ~ 1 + y2 ' y > O.

100. Fx(x) = P[X ::; x] = P[log X ::; log x] = P[Y::; log x]

= Fy(log x),

d 1 fx(x) = -d Fx(x) = - fy(log x)

x x

1 [(IOgX- Jl)2] = ;;C exp - 2 2 ' X > O.

uy 2nx U

101. I = f: Ix - mlf(x) dx = - f:oo (x - m)f(x) dx

+ too (x - m)f(x) dx.

Applying Leibnitz's rule of differentiation, d iq(y) iq(y) a -d f(x, y) dx = ~ f(x, y) dx + f(q(y), y)q'(y) - f(p(y), y)p'(y),

y p(y) p(y) uy

we get

dl fm foo dm = -00 f(x) dx - m f(x) dx

and setting this equal to zero gives

f m 1 F(m) = -00 f(x) dx = 2'

that is, m is the median of the distribution.

102. We have

E(X) = t: xf(x) dx = f:oo (x - a)f(x) dx + a f: f(x) dx

= roo (x - a)f(x) dx + Loo (x - a)f(x) dx + a. (1)

160

Since E(X) < 00, the following integrals exist

roo (x - a)f(x) dx = J:oo yf(y + a) dy,

{oo (x _ a)f(x) dx = Loo yf(y + a) dy,

and moreover, since f(a + y) = f(a - y), we have

J:oo yf(y + a) dy = - Loo yf(a + y) dy.

Hence from (1) we obtain E(X) = a.

103. E(tX + y)2 = E(t2 X 2 + 2tXY + y2)

= t2E(X2) + 2tE(XY) + E(y2).

Since E(tX + y)2 20 we get

t2E(X2) + 2tE(XY) + E(y2) 2 0 for every t.

Solutions

Therefore the (constant) coefficients of the quadratic in t must satisfy the relation

E2(XY) - E(X2)E(y2) ~ O.

104. IE(X)I = Itoooo x dF(x) I ~ too", Ixl dF(x) = E(IXI).

105. We have

E[JA(X) = 1· P[X E A] + O· P[X ¢ A] = P[X E A] = P(A) = L dFx(x).

106. Applying Chebyshev's inequality, we get

1 P[IX - E(X)I > ka] ~ k2'

since E(X) = 0 and L\(X) = E(X2) - [E(X)]2 = 0 it follows that

1 P[IXI > 0] ~ k2 for every positive integer k.

Hence for every B > 0 we have

P[IXI > 0] < B.

This implies that X = 0 with probability 1.

107. E(X - C)2 = E(X - jJ. + jJ. - cf = E(X - jJ.)2 + (jJ. - C)2,

so that the minimum is attained when c = jJ. = E(X).

Solutions 161

108. {l = E(X) = L: xf(x) dx = J:oo x dF(x) - Loo x d[I - F(x)]

= [xF(x)]~oo - [x(I - F(x)](joo - J:oo F(x) dx + Loo [1 - F(x)]dx

= Loo [1 - F(x)] dx - J:oo F(x) dx,

where we used

lim xF(x) = lim x[I - F(x)] = 0 (Exercise 113). x--oo x-+oo

109. E(X - e)k = J: (x - e)kf(x) dx

= foo (x - e)k dF(x) - Loo (x - e)k d[I - F(x)]

= [(x - e)kF(x)J_oo - [(x - e)k(I - F(x))]~

where we used

- foo F(x)d(x - e)k

+ 100 [1 - F(x)]d(x - e)k

= k 100 (x - e)k-l[I - F(x)] dx

- k foo (x - C)k-l F(x) dx,

lim (x - c)kF(x) = lim (x - enI - F(x)] = O. x--co

110. The rth central moment is given by

r = 1,2, ... , n.

The last relation shows that the first n moments determine the first n central

162 Solutions

moments. The inverse is also true because

J1~ = E[X'] = E[(X - J1) + J1J' = E [ktO G}X - J1)kJ1r-kJ

= ktO G) J1kJ1,-k

provided the mean J1 is given.

111. Suppose X is continuous with density f(x) (if X is discrete the proof is analogous). Then for every integer r > 0, we have

t: IxI1(x) dx ~ t: K'f(x) dx = K' < 00,

hence E(X') exists.

112. Since the nth order moment exists, the integral f~oo xnf(x) dx converges absolutely, that is,

t: Ixl'i(x) dx < +00.

For every k = 1,2, ... , n - 1, we have

and therefore

f:oo Ixlkf(x) dx < t: Ixlnf(x) dx + 1 < +00.

113. Since the mean of X exists, the integral f~oo Ix I dF(x) also exists. Thus

o = lim x[1 - F(x)] = lim x foo dF(y) ~ lim foo y dF(y) = 0 X-a) x- 00 x x-co x

= lim x[I - F(x)] = o. x-+-oo

Similarly, we can show that

lim xF(x) = O. x-+-oo

114. We have

n

V[SnJ = L V(Xk) + 2 L Cov(Xi, X j ) = nu2 + n(n - I)p0"2, (1)

where

k=l i<j

V(Xd = 0"2, k = 1,2, ... , n,

V(Xk) = E[Xf] - (E[Xk])2.

p = 0"2 COV(Xi' X), i -# j,

Solutions 163

But, since each of the numbers 1,2, ... , N has probability liN of appearing in any selection, we have for k = 1,2, ... , n

1 N N + 1 E(Xk ) = - L k = --

N k=l 2 (2)

and

[ 2 1 ~ k2 (N + 1)(2N + 1) E Xd = N L.. = 6 .

k=l (3)

On the other hand, since

V[Jl xkJ = V(constant) = 0 = Nu2 + N(N - 1)pu2,

we get 1

p=---N-1

(4)

and from (1), (2), (3), and (4) we obtain

n(n2 -1)[ n-lJ V[Sn] = 12 1 - N _ 1 .

Note that if X k (k = 1,2, ... , N) were independent, e.g., when we draw a sample from an infinite population (that is when N ..... (0) then we would have

V[S] = n(n2 - 1) n 12'

The dependence of X k , when we draw a sample from a finite population, reduces the variance by the factor lOO«n - l)/(N - 1))%, referred to as the finite population correction factor.

115. Let X be the number of white balls in the sample. Then: (a) X is binomial b(k, n, p), where p = N 1 /(N1 + N2 ), and hence

(b) X is hypergeometric and

This expectation is easily computed if we represent X as a sum

n

X = L X" r=1

164 Solutions

where

x = {I , 0 if the rth selection is a white ball,

if the rth selection is a black ball, r = 1,2, ... , n,

are identically distributed and therefore

E(X,) = P(white ball at the r selection] = P = NI , NI + N2

r = 1,2, ... , n.

Hence

n

E(X) = L E(X,) = np. ,=1

116. The random variable X takes on the values 0, 1, and 2 with probabilities

2 4 8 Po = P[X = 0] = 3"5 = 15'

1 4 2 1 6 PI = P[X = 1] = 3"5 + 3"5 = 15'

1 1 1 P2 = P[X = 2] = 3" 5 = 15'

respectively. (a) The expected number of correct answers is

6 1 8 J-l = E(X) = O'Po + I'PI + 2'P2 = 15 + 2'15 = 15'

(b) 2 6 4 10

E(X )=PI+4P2=T5+15=T5'

Hence

V(X) = E(X2) _ [E(X)]2 = 10 _ ~ = ~. 15 225 225

Another method. Let X k be the number 0 or 1 of correct answers in problem k (k = 1, 2). Then X = X I + X 2 and so

1 1 8 E(X) = E(Xd + E(X2) = P[XI = 1] + P[X2 = 1] ="3 + 5 = 15'

V(X) = V(XI ) + V(X2) = P[XI = O]P[XI = 1] + P[X2 = 0]P[X2 = 1]

2 4 86 ="9 + 25 = 225'

117. Let X be the gain from the ticket. Then X takes the values 1,000,500,

Solutions 165

and 100 with probabilities 1/3,000, 1/3,000 and 5/3,000, respectively. Hence

1 1 5 2 E(X) = 1,000· 3,000 + 500· 3,000 + 100· 3,000 = 3·

Since a man pays 1 dollar for the ticket the expected (total) gain is 2/3 - 1 = - 1/3, i.e., a loss.

118. (a) Let X be the number of throws. Then X obeys a Pascal (negative binomial) probability law with parameters n = 3 and p = 2/3. Hence the expected number of throws in a performance of the experiment is

n 9 E(X) = P = 2.

(b) In 10 repetitions the expected number of throws is

10· E(X) = 45.

119. Let X be the total gain of the gambler. The random variable X takes the values-I, 0,1, and 7 with probabilities

( 5)3 125 P[X = - 1] = P[no ace] = - = -

6 216'

P[X = 0] = P[l ace] = G)(DGY = ;156'

P[X = 1] = P[2 aces] = (DGYG) = 2\56,

P[X = 2] = P[3 aces] = G)GY = 2~6· Hence

which shows that the game is not fair. The game becomes fair if the gambler gets a dollars when three aces appear where a satisfies the relation

125 15 a-I a-Ill E(X) = - 216 + 216 + 2t6 = 216 = 0 = a = 111 dollars.

120. Let p. be the probability that the gambler obtains heads for the first time at the nth toss and X be the gambler's gain. Then X assumes the values - 15 and 1 with probabilities

P[X = -15] = P[all4 tosses result in tails] = 116 ,

166 Solutions

1 1 1 1 15 P[X = 1] = PI + pz + P3 + P4 ="2 + 4 + 8 + 16 = 16·

Hence E(X) = 0, that is, on the average, the gambler preserves his capital.

121. Let Ai be the event that book i arrives (reaches its destination), let A be the event that both books arrive, and let B be the event that at least one book arrives, and let 11 denote the expected net value; then:

(i) If each book is sent separately, we have: (a) peA) = 0.9 x 0.9 = 0.81. (b) PCB) = 1 - P(none reaches its destination) = 1 - 0.1 x 0.1 = 0.99. (c) 11 = 0 x P(A~A2) + 2 x P(A I A2 u A~Az) + 4 x P(AIA z) - 0.20

= 0 x 0.Q1 + 2 x 0.18 + 4 x 0.81 - 0.20 = 3.4. (ii) If the books are sent in a single parcel, then:

(a) peA) = 0.9. (b) PCB) = 0.9. (c) J1. = 0 x 0.1 + 4 x 0.9 - 0.15 = 3.45.

Thus, according to criteria (a) and (c), a single parcel is preferable, whereas according to (b), method (i) is better.

122. (a) X = 2Xo where Xo is a Bernoulli random variable with parameter p. Hence

E(X) = 2E(Xo)= 2p, Var(X) = 4 Var(Xo) = 4pq.

(b) X = XI + X z where XI and X z are independent Bernoulli with parameters PI and Pz, respectively. Hence

E(X) = E(XI ) + E(Xz) = PI + Pz, Var(X) = Var(Xd + Var(Xz)

123. Y

y = 1

~---------------------------------------.x o

Solutions

F{x) = {01 - 0.8e- x, x ~ 0, x < 0,

P[X = OJ = F{O) - F{O-) = 0.2,

f{x) = {0.8e-x, 0,

x>O, x <0,

E{X) = 0.8 Loo xe- x dx = 0.8.

167

Suppose that the lifetime of a certain kind of bulb obeys an exponential law with parameter ,1= 1, and consider a box containing 80% ofthis kind of bulb and 20% defective bulbs. Then the lifetime of the bulbs in the box has the above distribution function F{x).

E[Y] = foo 1 - t){X) <p{x - J.l) dx -00 {x)] exp[2J.lX; J.l2]d[2J.lX; J.l2]

1 [2J.lX-J.l2]1+00 = p. [1 - {x)] exp 2 -00

1 foo [2J.lX - J.l2] + P. -00 exp 2 <p{x) dx

1 foo 1 = ° + - <p{x - J.l) d{x - J.l) =-. J.l -00 J.l

125. We have

1 00 Ak A 00 Ak- 1

E{X) = -,,- L k- = -,,- L -,----e - 1 k=1 k! e - 1 k=1 (k - I)!

1 00 Ak ,12 00 Ak- 2 A2 E[X{X - I)J = e" _ 1 k~2 k{k - 1) k! = e" - 1 k~2 (k - 2)! = 1 - e ,,'

and so

V(X) = E[X{X - 1)] + E{X) - [E{X)]2 = lA~: A" - (I _A; ")2

A{I - Ae-" - e-") {1 - e-,,)2

126. The elements (simple events) are the following:

AA, ACC, ACBB, ACBACC, ACBAA, ACBACBB, ... ,

BB, BCC, BCAA, BCABCC, BCABB, BCABCAA, ... ,

168 Solutions

and

ACBACBACB. .. (infinite sequence), BCABCABCA ... (infinite sequence).

If we assume that the games are independent, we conclude that the probability of a simple event consisting of k games equals 1/2\ because the probability of each player winning a game is 1/2. Hence

00 1 1 1 1 L 2k"=-+-+-+···= 1. k:2 2 2 4 8

Thus the last two elements of the sample space have zero probability and therefore the game terminates with probability 1.

127. The sample space consists of the elements:

HH, HTT, HTHH, HTHTT, HTHTHH, ... ,

TT, THH, THTT, THTHH, THTHTT, ... ,

and

HTHTHTHT. .. } ... . . up to mfimty, wIth zero probablhty.

THTHTHTH ...

The required probabilities are

P(A) = t 2-; = 1 _ (~)6 = 63, k:2 2 2 64

P(B) = f (~)2k. 2 = ~. k:1 2 3

128. The number of all possible outcomes are 366 = 612. The number of favorable outcomes are 12!/26 and hence the required probability is 12!/(26 . 612 ) = 0.0034.

129. Let Ak be the event that the player k has a complete suit. Then the

required probability is p = P CVI Ak) and by the Poincare formula we get

where

Solutions 169

130. Let the event A be that I win the bet. Then, denoting the result "tails" by T, we have

A = HHHuHHTuTHH.

Tossing the coin C1 first, that is, playing according to the series C1 C2 C1 , we have

peA) = P1P2P1 + P1P2(1 - pd + (1 - P1)P2P1 = P1P2(2 - P1)'

On the other hand, tossing the coin C2 first we have

peA) = P2P1P2 + P2P1(1 - pz) + (1 - P2)P1P2 = P1pz(2 - P2),

and, since P1 > P2'

P1P2(2 - pd < P1P2(2 - P2),

that is, you must select coin C2 for the first toss. Remark. If one has to play against two persons, he should play first against

the stronger player in order to maximize his chances of winning.

131. Method 1. If we denote by Ak the appearance of the kth pair, then the required probability is

Applying Poincare's theorem, since P(Ai! A i2 .. . Ai) = ° for k ~ 3, we get

20· 18 . 16· 14 224 99 P = 1 - P(no one pair) = 1 - 20. 19. 18. 17 = 1 - 323 = 323

or

CO} 24

P ~ 1 - Pf«iection of 'hoo< belonging to 4 pai,,) ~ 1 - (2;) 99

323

For a random selection of 10 shoes the probability of selecting no pair is less than 6%.

170 Solutions

132. (a) (l/N)(l - 1/N)n-1 (geometric with P = liN). (b) (1 - II N)" because the first n balls must fall in the cells 2, 3, ... , N.

133. (a) The n (n :$; N + 1) first balls must fall in different cells. Thus the required probability is (N)nINn.

(b) The probability, say Pn' that n throws will be necessary equals the probability that the first n - 1 balls will fall in different cells and the nth ball in one of the n - 1 occupied cells. Hence

(N)n-1 n - 1 Pn = N n- 1 ---;;;-'

and the expected number of throws is given by the sum

N~l _ ~ n(n - 1)(N)n-1 L... nPn - L... n . n=i n=i N

134. (a) Let E1 be the event that the nth card is the first ace. Then

P(E ) = (48)n-1 (4)1 1 (52)n'

(b) Let E2 be the event that the first ace appears among the first n cards

P(E ) = 1 _ (48)n. 2 (52)n

135. The thirteenth diamond may appear at the nth position (26 :$; n :$; 52). Of the 52! cases (permutations ofthe 52 cards), the number offavorable cases is equal to

52 ( 26 ) L (n - 1)!(52 - n)! .13. n=26 n - 26

136. To each triplet of numbers out of the {I, 2, ... , n} there correspond six permutations (ordered triplets), and of these, three have the first number smaller than the second. Since the (nh ordered triplets are equiprobable, the required probability is 1/2.

This also follows from the fact that the first number is equally likely to be larger or smaller than the second one.

137. Let p(n, m) denote the probability that in placing n balls in m cells all cells are occupied. Then the required probability Pn is given by

Pn = p(n, m) - p(n - 1, m).

To find p(n, m), let Aj be the event that thejth cell is empty. By the Poincare formula (1.1)

p(n, m) = 1 - P(A 1 U'" U Am) = 1 - Sl + S2 - ... + (-lrSm

= .f (-l)j(~)(l _1)n, )=0 ) m

Solutions

since

138. (a) The probability function of X is

-). A k

e k!

where we set

P[X = k] = n Ak L e-).

k=O k! k ' ' c .

n A k

c= L -k'· k=O •

For the random variable Y we have

k = 0,1, ... , n,

P[Y = ka - (n - k)b] = P[Y = k(a + b) - nb] = P[X = k]

1 A k

k=0,1, ... ,n.

Hence we find

(a + b)A n+l E(Y) = A(a + b) - , - nb.

n.c

171

(b) To maximize E(Y), it suffices to minimize, with respect to n, the expression

(a + b)An+1

, + nb. n.c

139. Of the six permutations (cf. Exercise 136) only one is such that its first number is the smallest and the second number the largest. Hence the required probability is 1/6.

140. By the Poincare formula (1.1) we have

P(AB u AC u BC) = P(AB) + P(AC) + P(BC) - 2P(ABC) = 0.104

using also the independence of A, B, C.

141. (a) Let Xn denote the outcome of the nth coin on the first throw and y" the outcome of the second throw (n = 1,2,3,4). Then the required probability is

4 (1)4 P[Xn = Y", n = 1, 2, 3,4] = ]] P[Xn = y"] = 2: .

(b) Let X be the number of heads on the first throw and Y be the number

172 Solutions

of heads on the second throw; then

4 4

P[X = Y] = L P[X = k, Y = kJ = L P[X = kJP[Y = kJ k=O k=O

_ t (4)2 1 _ 35 - k=O k 28 - 128

in virtue also of Exercise 21.

142. (a) P[X = 'J = (b)j_, W ] (n)j'

j = 1,2, ... , b + 1.

b+! p+! J'(b) E[XJ = L jP[X = j] = A L ~.

~I ~I ~1 (b)

(c) The required relation follows from

b+1 p+1 (b) L P[X =j] = A L ~ = 1. j=1 j=1 (n)j

143. Obviously, when this occurs 2j balls will have been drawn (j = 1,2, ... , N), and according to the hypergeometric distribution the required probability is equal to

144. P(AC) p[ 0 ACBj] t P(ACB)

P(AIc) = -- = }-I = <-J-_I __ _

P(C) P(C) P(C)

00

L P(AICBj)P(BjC) . 1 00

J= P(C) = j~ P(AIBjC)P(Bjlc).

145. The probability of success for Rena is 1/3 and the probability offailure is 2/3. The probability of success for Nike is: 2/3' 1/2 = 1/3. The probability of success for Galatea is: 2/3' 1/2' 1 = 1/3. Therefore the process is fair (cf. random sampling from a finite population).

In the second case, the probability of success for Rena is a (say) and the probability for Nike is b = a(2/3) and for Galatea is c = a(2/3)' (2/3). Hence

9 b=~ 4 a = 19' 19' c = 19 since a + b + c = 1.

Therefore Nike and Galatea rightly protest.

146. Let Ai be the event that the ith face of the die appears. Let the given

Solutions 173

sides be iI' i2 , ••• , ik • Then the required probability is

where

SI = JI P[A;J = G)G)" S2 = ~ P[A;nA;m] = G)(~r

Hence

Application. What is the probability that each face will appear at least once? Applying formula (1) for k = 6 we find

P6 =.to (_1)"(~)(6 ~ n)' =.to (-I)"C)(6 ~ n)" (cf. Exercise 137).

147. From

we have c = 2.

f(x) dx = c __ e- x2/2 dx = - = 1 f OO foo 1 c

o 0 fo 2

E(X) = xf(x) dx = 2x __ e- x2/2 dx = --- d(e- x2/ 2 ) f oo foo 1 2 foo o 0 fo fo 0

2 _ 2/2 100 2 J2 = ---e x = __ =_

fo 0 fo .;-;c' 148. (a) Let X denote the lifetime in hours of an electric bulb. Then

[ X - 180 ] q = P[X < 200] = P 20 < 1 = <l>(I) = 0.84.

Therefore the required probability is

(1 - q)4 = p4 = (0.16)4.

174 Solutions

(b) Consider the event Ak where the urn contains k bulbs with lifetime greater than 200 hours, and the event B where the selected bulb has a lifetime greater than 200 hours.

P(B) = kt1 P[(BIAk)]P(Ak) = kt1 ~(:)pkq4-k

= P t ( 3 )pk - 1q4-k = p; k=1 k - 1

that is, the probability of selecting a bulb with lifetime greater than 200 hours is the same whether a bulb is chosen from the population of bulbs or from a random sample of size n (here n = 4) of the population.

149. Let X denote the outcome (sum) in a throw 2 ~ X ~ 12. Then

P[X = 2] = P[X = 12] = 1/36,

P[X = 4] = P[X = 10] = 1/12,

P[X = 6] = P[X = 8] = 5/36,

P[X = 3] = P[X = 11] = 1/18,

P[X = 5] = P[X = 9] = 1/9,

P[X = 7] = 1/6.

Let Xn denote the outcome of the nth throw, let W be the event the gambler (finally) wins, let W1 be the event he wins on the first throw, and let L1 be the event he loses on the first throw, then

P[Wd = P[X = 7] + P[X = 11] = 1/6 + 1/18 = 2/9,

P[L1] = P[X1 = 2] + P[X1 = 3] + P[X1 = 12] = 1/9,

and therefore

P(more than one throw) = 1 - P(W1) - P(Ld = 2/3.

We now have for case (a)

2 10

P[W] = 9 + i~4 P[WIX1 = jJP[X1 = jJ, i#7

and from

ao

Pi = P[WIX1 = jJ = L P[X. = 7]P[Xi ::f: 7, Xi ::f: j, i = 2, ... , v-I] .=2

we find

(1)

Solutions 175

Substituting in (l) we find

PEW] = ~: = 0.618.

Under (b) we similarly obtain

2 10 244 PEW] ="9 + j~4 qjP[Xl = j] = 495 ~ 0.493,

U7

where we set Pj = 1 - qj. We notice that (b) defines an almost fair game. In fact, this is the most usual way of playing the game.

150. The expected profit for B is by Exercise 149

1 2 3 4 5 6 5 4 188 2· 36 + 3 . 36 + 4· 36 + 5· 36 + 6· 36 + 7· 36 + 8· 36 + 9· 36 = 36'

For A this is

3x 2x x 6x 36 + 36 + 36 = 36 .

The game is fair if 188/36 = 6x/36. Hence x = 31 t. 151. Let Ei be the event that one white ball and one red ball are drawn on

the ith draw. Then by the multiplication rule the required probability is

P[E 1 E2··· En] = P[E l ]P[E2IE l ]··· P[EnIE 1 ••• En- 1 ]

Another method. The number of ways in which 2n balls may be divided into n pairs is (2n)!/2n of which (n!)2 are favorable (for pairs of different color) because to every permutation of n white balls there correspond n! permutations of red balls which give pairs of balls of different color.

152. Let Ei be the event that the number of the ball drawn from urn A is i and from B larger than i. Then the required probability is

n-l n-l n - i n - 1 P = PEEl U E2 u··· U En- 1 ] = L P(E;) = L -2- = -2-'

i=l i=l n n

Another method. It is equally likely that the ball drawn from A will bear a number larger or smaller than that of B and the probability of equal numbers is l/n. Therefore

1 p+p+-=1

n n-l

and p=--. 2n

176 Solutions

153. Let Po be the required probability. Then: (i) (a) Every person after the second person chooses one of the other

persons, except the first one. Therefore we find

_ [(N - 1)]n-l _ ( _ ~)n-l Po - - 1 N N

(b) Each ofthe n persons chooses one of those who have not been informed. Hence Po = (N)n/Nn.

Similarly, when it is told to k persons at a time, we have

(ii)

(N - krl [ k]n-l (a) P - - 1 --

0- Nn 1 - N '

(a) n = pN and

(N)nk (b) Po = [(Nh]n

(b) lim Po = lim (1 - ~)(1 - ~) ... (1 -~). N~oo N~oo N N N

154. Each gambler may win the game on the kth throw of the game (event

Ad where k = n, n + 1, ... , 2n - 1 with probability (~= ~)GY because

there are (k - 1) ways of winning n - 1 throws out of k - 1 throws and the n - 1

kth throw must be a success. Thus the probability that each player wins the game is

ZI l (k - 1) (~)k = ~. k=n n - 1 2 2

The probabilities after interruption are

2n-l (v - k - m - 1) (l)v-k-m P(I) = L -,

v=n+k n - k - 1 2

Zn-l (v - m - k - 1) (l)v-k-m P(II) = L -.

v=n+k n - m - 1 2

The amount will be divided in parts proportional to P(I), P(I/).

155. Let A/ B be a rational fraction and a an integer. The possible remainders of the division of A by a are 0, 1, ... , a-I. Hence the probability that A is divisible by a is l/a. Similarly for B. Therefore the probability that both A and B are divisible by a is l/a2• The fraction A/B is an irreducible fraction if and only if both A and B are not divisible by any of the prime numbers 2, 3, 5, .... Hence the required probability is given by

Solutions 177

This is obtained by noting that

P

where the summation extends over all nonnegative integers a, b, c, ... , and taking into account that any positive integer n is of the form n = 2a • 3h • sc· ...

156. Let Ap be the event that the chosen number is divisible by the prime p. If PI' P2' ... are different prime divisors of n the events ApI' Ap2 , .. , are independent. The probability that the chosen number is prime with respect to n is, by definition of cp(n), cp(n)/n.

On the other hand, P(Ap) = lip, because there are q numbers out of 1, 2, 3, ... , n which are divisible by p, where n = pq. Thus

q q 1 P(A ) = - = - = -;

P n p'q p therefore

cp(n) = p[n A~J = fl P(A~) = fl (1 - ~). n ~ ~ ~ p

157. The generating function of the number of insects

where N is the number of colonies and lj is the number of insects in the jth colony, is given by

PSN(t) = exp{A(P(t) - I)},

where P(t) is the generating function of lj. But

and therefore

P(t) = log(l - pt) log(l - p)

PSN(t) = exp {A [11:~~11 ~ ~: - I]},

that is, the probability generating function of the negative binomial distribution.

158. Let Ak be the event that the lot contains k defective tubes, and let E, be the event that in a sample of n tubes r are defective. Then

(a)

P[E,IA k ] = (: = :)G)/(~)' (Bayes's formula).

178 Solutions

(b)

df (N = n)j(N) k=r k r k

1 - . m (N - n)j(N) L Pj. .

j=r ] - r ]

159. Let Ak be the event where k defective articles are bought, let Bn be the event where n customers come into the shop, and let E be the event where a customer buys a defective article. Then we have

and

peE) = P(of buying)P(buying a defective item I buying)

2 3 = 3·4 = 2'

P[AkIBnJ = (~);n'

160. Let W denote the weight of an article. Then the proportion of articles with weight less than Wo is

P = P[W < woJ = -- e-(w-Il)2/2 dw = -- e-w2/2 dw. 1 f~ 1 f~~ fo -00 fo -00

Then the expected profit is

K = (1 - p) - (a + bJl).

From

we have the value of Jl (Jl > wo) which maximizes the expected profit.

161. P[the clerk is lateJ = P[the train arrives after 8:45J + P[the train arrives before 8:45JP[the bus arrives after 9J = 1 - <1>(3/4) + <1>(3/4)[1 -(2/3)J = 1 - <1>(3/4)<1>(2/3) ~ 0.4.

(a) P[that the clerk is lateJP[that the employer is lateJ = 0.4 x 0.07 = 0.028 because we have P[that the employer is lateJ = 1 - <1>(3/2) = 0.07.

(b) P [that the employer arrives before the clerk J = P [that the train arrives before 8:45JP[the bus arrives after the employer's carJ + P[that the train arrives after 8:45J = (3/4)(1/ji3) + 1 - <1>(3/4) ~ 0.5 because the arrival time X of the bus obeys the normal probability law N(8: 58,32 ) and the arrival

Solutions 179

time Y of the private car follows the N(8: 57,22) and therefore

P[X> Y] = P[X _ Y> 0] = p[(X - 8:58) - (Y - 8:57) > __ I_J UX-y UX - y

= $(Jo) ~ 0.61.

162. This is due to Dodge (see Rahman, 1967, p. 159). (a) Let En be the event that the nth article examined is the first defective

article (n = 1, 2, ... , r), and let A be the event that the sequence is defective. Then

r r

a = PEA] = P[E 1 U E2 u··· u Er] = L P(En) = L pqr-1 = 1 - qr. (1) n=l n=l

a is also the probability that among r articles at least one is defective. Hence (1). (b) Let X be the number of examined articles of a defective sequence. Then

1 [ ] f kP[X = k] f kqk-1 A = E XIA = L.. = P L...--

k=l PEA] k=l 1 - qr

1 - qr(1 + rp) p(1 _ qr)

(i) Let Y be the number of defective sequences. Then Y obeys the geometric probability law with parameter 1 - P(A) = 1 - a (probability of success).

00 1 _ qr Jl = E(Y) = L n(1 - a)a n = -r- = q-r - 1.

n=O q

1 _ qr AJl + r = -- = r. pqr

(ii)

(c) Let ({J denote the required percentage. Then

({J = (average of examined articles before a defective article + r)/(r + r*)

= fl[f + qr(1 - f)],

where we put r* as the average of articles which passed before a defective one was found; and f- 1 x the average number of examined articles before a defective one was found.

(d) = 1 _ = pqr(1 - f)

p p( ((J) f + qr(1 - f).

(e) p is maximized when p = p* as obtained from

(1 _ f)(1 _ p*r) = f[(r + l)p* - 1]. 1 - p*

180 Solutions

For a more detailed discussion of Exercises 163-170 we refer to Mosteller (1965).

163. The error in A's syllogism is due to a mistaken sample space. He thinks each of the pairs of convicts AB, AC, BC, who are going to be set free, has probability 1/3. This is correct from the point of view of the Board. But when we keep in mind the answer of the guard, we have the events:

1. A and B free and the guard answers B (Bl event).

2. A and C free and the guard answers C (C1 event).

3. Band C free and the guard answers B (B2 event).

4. Band C free and the guard answers C (C2 event).

Then

PEA f ·f h d J PEA free and the guard says BJ ree 1 t e guar says B = [ h d J Pte guar says B

P(Bd 1/3 1/3 + 1/6 = 2/3.

Similarly, if the guard answers that C is set free. Another method. It is equally likely that the guard is going to say that B (B

event) or C (C event) is set free. Then, if A is the event that A is going to be set free, we have

P[AIBJ = P[ABJ = 1/3 = ~ P[BJ 1/2 3·

164. We obtain a number from the first box. The probability of a new number from the next box is 5/6. Therefore, on average, 1/(5/6) = 6/5 are required for the next new number because the mean of the geometric distribution is l/p.

Similarly, 6/4 boxes will be required for the third number, 6/3 for the fourth, 6/2 for the fifth, and 6 for the sixth. Hence the required number of boxes on average is

For n coupons and large n, the required number is n log n + 0.577n + 1/2 boxes because

1 1 1 1 + - + ... + - ~ log n + - + C,

2 n 2n

where C = 0.57721 denotes Euler's constant.

Solutions 181

165. The probability of a couple at the first two seats is

10 9 91010 19'18+ 19'18= 19'

and the expected number of pairs at the first two seats is also

10 10 9 19 = 19' 1 + 19' O.

The same is true for every pair of successive seats and there are 19 - 1 = 18 such pairs. Therefore the expected mean number of pairs (male-female student or female-male student) is 18 '10/19 = 9199 = 9.47. More generally, for a male students and b female students the mean number of couples is

(a + b - 1)[ ab + ab ] = 2ab . (a + b)(a + b - 1) (a + b - l)(a + b) a + b

Remark. We used the fact that the mean of a sum of random variables is equal to the sum of the mean values of the corresponding random variables, no matter whether the variables are (statistically) independent or not.

166. If A shoots and hits C, then B will certainly hit him; therefore he should not shoot at C. If A shoots at B and misses, then B will shoot at the more dangerous, i.e., C, and A has a chance to shoot at B and hit him with probability 0.4.

Naturally, if A fails then he has no chance. On the other hand, suppose that A hits B. Then C and A shoot alternately at each other until either one of them is hit. The probability that A will win is

(0.5)(0.4) + (0.5)2(0.6)(0.4) + (0.5)3(0.6)2(0.3) + "', (1)

where the nth term expresses the probability that in 2n trials C will fail n times, B will fail n - 1 times, and A will hit in the end. The sum in (1) is

(0.5)(0.4) 2 1 _ (0.5)(0.6) = "7 = 0.28 < 0.4.

Thus if A hits B and continues with C, A has a smaller probability of winning than if the first shot fails. Hence to start, A shoots in the air and tries to hit B in the second round. Then C has no chance (of surviving).

167. The expected appearances of a 6 in the experiments of A, B, Care 1, 2, 3, respectively, but this does not mean that the corresponding probabilities of the events are equal. In fact, we have

P(A) = 1 - P(no 6) = 1 - GY ~ 0.665,

( 5)12 (12)(1)(5)11 P(B) = 1 - P(no 6) - P(one 6) = 1 - 6" - 1 6" 6" = 0.619,

182 Solutions

P(C) = 1 - P(no 6) - P(one 6) - P(two 6's)

Indeed Newton advised Pepys to choose the throw of six dice.

168. Employing one worker secures 364 working days; employing two workers, in all probability having different birthdays, we secure 2 x (365 - 2) = 726 working man-days, whereas if the number of workers is large there is a great probability that every day is a holiday because of somebody's birthday. Hence there is an optimum number of workers to be employed by the factory.

Consider n workers and let N be the number of days in a year. The probability that any day is a working day is equal to

p = P[no birthday on a given day] = [(N - 1)/N]n = (1 - l/N)",

and this day contributes, on average, np man-days. This is true of any day. Hence the total expected number of man-days during the whole period of N days when n workers are employed is equal to nNp. This is maximized for some n = no if

and

from which we obtain

(no + 1) (1 -~) ~ no and

and finally no < N < no + 1.

no = N - 1 and no = N give the same maximum, i.e., N 2 (1 - (I/N))N; hence take no = N - 1. For large N, (1 - N-1 t ~ e-1 and hence the expected number of man-days is N2e-1, while if every day was a working day it would be N 2 . Thus in a usual year (365 days) the no = 364 workers will offer about 49,000 man-days. No doubt the wages must be very low in Erehwon.

169. There are several versions of the problem depending on the way in which a chord is chosen. We consider three versions.

(a) The middle of the chord is chosen at random, i.e., it is uniformly distributed over the circle. In this case, we have

area C 3 1 P[(AB) < r] = P[M ¢ C] = 1 - C = 1 - -4 = -4'

area

where C is a circle with radius (see Fig. 1)

(KM) = r cos(1t/6) = r.j3/2.

Solutions 183

Figure 1

(b) Since A is some point on the circumference, take it fixed and choose B. Clearly now

120 1 P[(AB) < r] = 360 = 3'

since if B is chosen 60° on either side of A on the circumference, (AB) will be smaller than r.

(c) For reasons of symmetry, we suppose that AB has a given direction, say, of the diameter 00'; in this case, the middle M of the chord is uniformly distributed on the diameter KK' which is perpendicular to 00' (Fig. 2). We now have

P[(AB) < r] = P[(TM) > (TL)] = P[(TM) > {} rJ = 1- 2r}3/2 = 1- J3 =0.13.

2r 2

A K B

0' t-----'I'=-----l 0

K'

Figure 2

Remark. The variety of solutions exhibited is due to insufficient description of the corresponding random experiment which determines the choice of the

184 Solutions

chord. Thus in (a) we essentially choose at random a point M in the circle and require the probability that M lies in a smaller homocentric cycle. In (b) the points A and B are randomly chosen on the circumference, whereas in (c) we can imagine a cyliner rolling on a diameter of the circle (say KK') and the equally likely events correspond to equal intervals on the diameter (irrespective of their position on the diameter).

170. This is another version of the beauty contest problem or the dowry problem (how to select the most beautiful candidate or the highest dowry without being able to go back and choose a candidate already gone). The neophyte at a given stage of the race knows the rank (first, second, etc.) of the horse passing by; only the horse having the highest rank (up to that point) may be considered for final selection. Such horses will be called "candidates".

Before we consider the general case, let us see some simple cases. For n = 2 the neophyte has a chance of 1/2 of choosing the better horse. For n = 3, choosing at random, e.g., the first horse he chooses has a probability 1/3 of winning since of the six cases 123, 132*,231 *,213*,312,321 only the last two are the favorable ones. The probability gets higher if he ignores the first horse and chooses the next better one. This procedure gives the three cases marked with an asterisk and therefore the probability becomes 1/2.

In general, it will be shown that the best strategy is to let s - 1 horses pass and choose the first candidate after that. Let ~ denote the event that the ith horse wins. Then the ith horse will be chosen (as the best one) if

k = 1,2, .... (1)

We will show that the right-hand side is a decreasing function of i, whereas the left-hand side is an increasing function of i, so that at some stage the process of observing more horses should stop and a decision taken. We observe that the probability that the ith horse is the best one is equal to the probability that the highest rank is among the first i ones, i.e., i/n, that is, an increasing function of i and at some stage it is larger than the probability of winning later. According to this strategy, the neophyte letting s horses pass has a probability of winning equal to the probability of coming across a single candidate after the nth stage. Thus we have

P[~] = P[kth horse has highest rank]

x P[best of the first k - 1 horses is among the first s - 1 horses]

1 s - 1 n k - l'

for s ~ k ~ n and hence n 1 n s - 1

n(s, n) = P[to win] = L P[~] = - L -k-k=s n k=s - 1

s - 1 n-1 1 =- L-

n k=s-l k' 1 < s ~ n. (2)

Solutions 185

Certainly the first horse is a candidate and n(1, n) = l/n. The optimum value s* of s is the smallest integer for which (1) holds, i.e.,

s s (1 1 1 ) - > n(s + 1, n) = - - + -- + ... + --n ns s+l n-1

or

1 1 1 1 1 1 - + -- + ... + -- < 1 < -- + - + ... + --. (3) s s+1 n-l s-l s n-l

For large n the approximation

1 1 1 + - + ... + - ~ log n + C,

2 n

where C = 0.57721 is Euler's constant, gives

s-1 n-l s n n(s, n) ~ --log -- ~ -log-,

n s-1 n s

and by (3), s* ~ ne- l , i.e., for large n the neophyte must wait until e- l = 37% of the horses pass by and then choose the first candidate. It can be found that for n = 10, s* = 4, and n(s, n) = n(4, 10) = 0.399; similarly, n(8.20) = 0.384, n(38,100) = 0.371 thus approaching e- l as n ~ 00. A random choice ofa horse as the best would give l/n.

171. (a)

Fx(x) = 0, x < 0,

Fy(Y) = 0, Y < 0,

Fx(x) = 2/3, ° ~ x < 1,

Fy(Y) = 1/3, ° ~ Y < 1,

F(x) = 1, x ~ 1,

F(y) = 1, y ~ 1,

{o for x < ° or y < 0,

1/3 for (0 ~ x < 00, ° ~ y < 1), F(x, y) =

2/3 for (0 ~ x < 1, 1 ~ y < 00),

1 for x ~ 1, y ~ 1.

(b) (- 1/2, 0) is a continuity point. (0, 2) is a discontinuity point of F(x, y) because for x < 0, F(x, 2) = 0, while F(O, 2) = 2/3.

172. F(x, y) must satisfy F(X2' Y2) - F(Xl' Y2) - F(X2' Yl) + F(Xl' Yl) ~ ° for all Xl ~ Yl' x 2 ~ Y2 (why?); for (Xl' yd = (0, 0), (X2' Y2) = (2, 2) we have

F(2, 2) - F(2, 0) - F(O, 2) + F(O, 0) = 1 - 1 - 1 + ° = -1;

thus it is not a distribution function.

173. P[Xi = 1] = P[Xi = 0] = 1/2, i = 1,2,3;

P[Xi = ai' Xj = aJ = P[Xi = a;]

for all i i= j (i,j = 1,2,3) and ai' aj = 0, 1; then Xl' X 2 , X3 are pairwise

186 Solutions

independent. But the relation

does not hold, for example, for a l = a2 = a3 = 1; then

3 1 P[XI = 1, X 2 = 1, X3 = 1] = 0 #- n P[X; = 1] =-.

;=1 8

174. Putting x = yu we have,

fy(y) = ce-Y f: X n1 - 1(y - x)n2 -1 dx = ce-Yyn1+n2-1 fOl un,+n2-1(1 - u)"rl du

= cyn, +n2 -le- y r(nl )r(n2) r(nl + n2 )'

and from the equation

we have c- l = r(ndr(n2) and Fy(y) is a Gamma distribution with parameter n l + n2 • Similarly, we find that X has the Gamma distribution with parameter nl :

X> o.

175. For n; ~ 0 and n l + ... + n k - l + nk ~ n, we have

where X k = n - (Xl + X2 + ... + X k - l ). This is hypergeometric.

176. Similarly as Exercise 175.

177. The marginal distribution of Xl' ... , Xr (r < k) is Dirichlet with density

r( + + + ) "k+Nr- l r f( ) - nl n2 ••• nk + l 8r +l n !,,-l (1)

Xl'···'Xr -r( ) r( )r( ) x, , nl ... nr n r + l + ... + nk+l ;=1

where we put

8r +1 = 1 - (Xl + X2 + ... + X r ), Nr = n r +l + ... + nk+l·

Solutions 187

Equation (1) for r = 1 and r = k - 1 gives as density of X k for given Xi = Xi (i = 1,2, ... , k - 1),

Then the conditional distribution of Xk/Rk for Sk = Sk is p(nk, nk+l).

178. The density is

f(x, y, z) = 1/V,

(x, y, z) E S = {(x, y, z), x ~ 0, y ~ 0, z ~ 0, x + y + z ~ c},

where V = volume of (S) = c3 /6. The joint density of (X, Y) is

6 fC- X - Y 6(c - x - y) f(x, y) = 3" dz = .

c 0 c

Note. For c = 1, (1) and (2) are special cases of Exercise 177.

(1)

179. (a) Since the quadraticform (see (5.10)) x2 - xy + y2 is positive definite, f(x, y) is the density of the bivariate distribution N(p., I:) where

_ ~(2 1) I:- 31 2·

Hence

(b) The ellipses x 2 - xy + y2 = constant.

180. (a) As in Exercise 178, we find

f(x, y) = 1/2, x ~ 0, y ~ 0, x + y ~ 2.

(b)

1 f2-X 1 fx(x) = 2 0 dy = 2(2 - x).

(c) f(x, y) 1 fy(YIX = x) = fx(x) = 2 _ x' 0< y < 2 - x.

181. (a) Multinomial Pi = P2 = ... = P6 = 1/6, ni = n2 = ... = n6 = 2, n = 12.

(b)

P[X = i, Y = j] = i!j! (121= i _ j)! G)i+iGY2-i-i, Trinomial with Pi = P2 = 1/6, P3 = 2/3.

° ~ i + j ~ 12.

188 Solutions

(c) Cov(X, Y) = E(XY) - E(X)E(Y),

1 E(X) = npl = 12. 6 = 2,

1 E(Y) = np2 = 12. 6 = 2,

i j n-i-j

E(XY)= L i" P1P2P3 O$;i+j$;n 'J i! j! (n - i - j)!

(n - 2)! p~-l pt1 p~-i- j

= n(n - l)P1P2 L. . .. 2$;i+j$;n (z - I)! (j - I)! (n - I - J)!

m! p~p~p~-S-k = n(n - l)P1P2 L

09+k$;m s! k! (n - s - k)!

= n(n - I)P1P2(Pl + P2 + P3)m = n(n - l)P1P2

(where m = n - 2, S = i-I, k = j - 1). Hence Cov(X, Y) = -nplP2' and in this case Cov(X, Y) = -1/3.

182. We have

P[XI = n1 , •.. , Xi = ni, ... , Xj = nj , ... , X k = nk] ni + I.l!l P[XI =n1,···,Xi=ni + I, ... ,Xj=nj-I, ... ,Xk=nk] nj Pi

for i,j = 1,2, ... , k, i =j.

Thus the vector (n~, n~, ... , n~+l) is the most probable, if and only if, for every pair of indices i, j (i #- j) the inequality

PinJ spin? + 1)

is satisfied. From this, summing over j withj #- i we have

k+l k+l

Pi L nJ s (n? + 1) L Pj' j=l j=l Ni Ni

p;{n - n?) s (n? + 1)(1 - Pi),

npi S n? - Pi + 1 < n?,

npi - 1 s n? - Pi < nf.

Similarly, summing over i withj #- i we get

nJ spin + k), j = 1,2, ... , k + 1.

From (2) and (3) the result follows.

(1)

(2)

(3)

183. Putting npi = ).i (i = 1,2, ... , k), in the multinomial probability function, we have

Solutions 189

Hence from (1) the result follows by taking the limit (n -+ 00).

184. (a)

F(x, y) = 2 LX du J: (1 + u + V)-3 dv = LX [(1 + U)-2 - (1 + u + y)-2] du

1 1 1 =1---+ ---

l+x l+x+y l+y

(b)

J OO Joo 1 fx(x) = 0 f(x, y) dy = 2 0 (1 + x + yf3 dy = (1 + X)2·

}; ( IX = x) = f(x, y) = 2(1 + X)2 . Y Y fx(x) (1 + x + y)3

(c)

185. fx(x) dx '" area of the infinitesimal strip ABB' A' (see figure), so that

1 f(x) dx = -(2~).

11:

_-T-.....:A A'

x x

BB'

190

Hence

2 fx(x)=-~,

1t

1 1 ~ . Fx(x) = - + -(xv 1 - x + arc sm x),

2 1t

-1 < y < 1,

1 f . = 1 - - (area 0 the cIrcular part (AOB)), 1t

Fy(y) = Fx(Y) = 1 - Fx( - y).

X and Yare not independent because, for example,

p[x> f, Y> fJ=o#P[X> fJp[x> fJ that is, the relation

P[X E A, Y E B] = P[X E A]P[Y E B]

does not hold for all A, B.

186. P[X < Y < Z] = II dz IZ dy IY dx =!. 000 6

Solutions

This is so, since every ordering of X, Y, Z has the same probability because of symmetry of the joint density.

187. (a)

Fx(x) = Fy(x) = x 2, F(x, y) = x 2y2, fx(x) = fy(y) = 2x, 0 < x, y < 1. (1)

4xy fx(xl Y = y) = 2y = 2x, fy(YIX = x) = 2y. (2)

From (1), the relations (2) follow because of the independence of X, Y.

Solutions

188. We have

I1x = E(X) = -0.35 + 0.20 = -0.15,

E(X2) = 0.35 + 0.20 = 0.55,

Var(X) = E(X2) - 11~ = 0.55 - 0.0225 = 0.5275,

ax = JVar(X) = 0.726,

l1y = E(Y) = -0.3 + 0.1 + 0.8 = 0.6,

191

E(y2) = 0.3 + 0.1 + 1.6 = 2, 2

Var(Y) = E(y2) - - = 2 - 0.36 = 1.64, Y

ay = JVar( y) = 1.28,

E(X Y) = 0.1 - 0.05 - 0.2 + 0.2 = 0.05,

Cov(X, Y) p=

E(XY) - I1xl1y 0.05 + 0.09 = 0.14 = 0.15. axay 0.726 + 1.286 0.932

(a) The regression lines of X on Y and of Yon X are given by

respectively. Substituting parameters in (1) and (2), we have

x + 0.15 = 0.084(y - 0.6),

y - 0.6 = 0.265(x + 0.15).

(1)

(2)

(3)

(4)

Observe that both lines pass through the point (l1x = -0.15, l1y = 0.6); moreover, line (3) passes through (x = -0.2, y = 0) and line (4) passes through (x = 0, y = 0.64) (see figure).

y

-0.2 o x

192 Solutions

(b)

E(YIX = x) = m(x), m( -1) = 0.29, m(O) = 0.78, m(l) = 0.75.

189. If X and Yare independent, then

Pij = P[X = x;]P[Y = Yj] = Piqj' i,j = 1,2, ... ,

and the nth line is (Pnql' Pnq2' ... ) = Pnq (n = 1,2, ... ). So there is only one linearly independent row. Hence the rank r(p) of the matrix P is 1. Conversely, if r(p) = 1, then there is a vector a = (aI' a2' ... ) so that every row 1n can be written as

n = 1,2, ... (An = constant). But

gives (1)

So a is a multiple of q and there are constants A: so that

n,j = 1,2, ....

Hence X and Yare independent (A: will be equal to Pn necessarily).

190. Consider the random variables X and Y with joint probability function

P[X = Xi' Y = yJ = Pij'

and marginal probability functions

i,j = 1,2, (1)

P[X = x;] = Pi' i = 1,2,

respectively. We have

and P[Y = Yj] = qj' j = 1,2, (2)

2 2

L L Pij = 1, (3a) i=1 j=1

2

L Pij = Pi' i = 1, 2, (3b) j=1

2

L Pij = qj' j = 1,2, (3c) i=1

PI + P2 = 1, ql + q2 = 1. (4)

Since X and Yare supposed to be orthogonal we get

Cov(X, Y) = E(XY) - E(X)E(Y) = 0, (5)

2 2 2 2

E(XY) = L L xiYjPij, E(X) = L XiPi, E(Y) = L Yjqj· i=1 j=1 i=1 j=1

Hence (5) is written

X1Yl(Pll - Plqd + X1Y2(P12 - Plq2) + X2Yl(P21 - P2qd + X2Y2(P22 - P2q2)

= O. (6)

Solutions 193

From (3) we have

PI2 = PI - PII' P21 = ql - PII' P22 = q2 - P12 = Pl1PI + P2· (7)

Replacing them in (6) we get

XIYI(PII - Plqd - XIYI(PII - PI + Plq2) - X2YI(PII - ql + P2ql)

+ X2Y2(PII - PI - q2 - P2q2) = O.

The above equation, because of (4), is written

(XIYI - XI Y2 - X2YI + x 2Yz)(Pl1 - Plql) = 0,

or

(XI - X2)(YI - Yz)(Pll - Plqd = 0,

and since XI "# Xz, and YI "# Yz, it follows that PII = Plql. Equations (7), because of the last relation, are written

P12 = PI - Plql = p(l - qd = PlqZ'

P21 = ql - Plql = ql(1 - PI) = P2ql,

P22 = pz - PZI = P2 - P2ql = P2(1 - qd = pzqz,

that is, Pij = Piqj (i,j = 1,2). So X, Yare independent.

191. (a)

P(t) = (pt + q)n, M(t) = P(e t) = (pet + qt, <p(t) = M(it) = (pe it + qt,

J1 = p'(1) = M'(O) = i-I<p'(O) = np, (Jz = npq.

(b)

(c)

P(t) = 1 ~t qt '

(d)

M(t) = e).(e'-l),

pet M(t) = 1 t' - qe

pe it <p(t) = 1 it' - qe

1 J1 =-,

p

P(t) = C ~ qty, M(t) = C ! qety, <p(t) = C !qeity, J1 = p'

192. The characteristic function of X is

1 foo eitx <p(t) = - --2 dx = e- 1tl•

TC -00 1 + X

Hence

2 rq (J =2.

P

194 Solutions

that is, (6.11) is only a necessary but not sufficient condition for the independence of two random variables.

193.

194.

where

Py(t) = E(t Y ) = E(t3X +2) = t2E(t3X ) = t2PX (t 3 ).

Var(SN) = PsN(1) + PsN(1) - PS)1)2,

(see (6.4»,

psJt) = P5c(t)P;'(Px(t»,

PSN(t) = P;(t)P;'(Px(t» + P5c(tf P~(Px(t».

Hence we have

PSN(t) = E[SNJ = P5c(1)P;'(Px(1)) = P5c(1)P;'(1) = E(X)E(N),

PsN(1) = PZ(1)P;'(1) + [P'(1)J2 P~(l)

= E[X(X - 1)JE(N) + E(N2)E[X(X - 1)].

Replacing in (1) we have the result.

195. The number of the blond children is

(1)

The number of children N is geometric with parameter Pl = 1/2. The X/s are independent Bernoulli with parameter p. Because of (6.4) and Exercise 191 we have as the probability genrating function of SN

Ps (t) = Pl (pt + q) N 1 - q 1 (pt + q)

at + b 1 - et'

where a = P1P2/(1 - qql), b = qql/(1 - qql), e = qlP/(1 - qql)' The probability Pk of k blond children is given by the coefficient of t k in the expansion of Ps)t). This gives

Po = b, Pk = ek-1(a + be), k = 1,2, ....

This is a geometric distribution with modified first term.

196. Let

where

N

Xjk = L Xjk j=l

and, for every k = 1,2, ... , N, P[Xjk = 1J = Pj = 1 - P[Xjk = OJ (j = 1,2,3). The probability generating function of ~k = (X lk , X 2k, X 3k ) for k = 1, 2, ... ,

Solutions 195

N, is

k = 1,2, ... , N.

The random variable N is, by assumption, Poisson with parameter A = 150. It can be shown that the probability generating function of X = (Xl' X 2 , X 3 ) = ~:J=1 ek· Similarly the probability generating function of SN (cf. (6.4» is given by

PX (t1' t 2 , t 3 ) = PN (Pk (t 1 , t2 , t 3» = e).(Pk-l)

3 = e).(P,t, +P2t 2+P,t'+P4- 1 ) = fl e).p;(t;-l).

i=l

This is a product of Poisson probability generating functions with parameters APi· Then Xl' X 2, X 3 are independent Poisson random variables with parameters AP1' AP2' AP3' respectively. This can be directly shown as follows:

197. We have

where

f ) a - a cos(ula) x,(u = 2 '

nu f b - b cos(ulb) x2 (u) = 2 .

nu

We have

() _ e foo itu 1 - cos(ule) d _ {l- eltl, It I :$; lie, CPyt-- e 2 u-

n -00 u 0 for It I > lie,

and

(t) = {I - altl, It I :$; lla, CPx, 0, It I > lla,

(t)={I-b lt l,l t l:$;l /b, CPx 2 0, I tl > lib;

hence, for 2e = a + b

CPx(t) = cPy(t)

Note. The density

f( ) = I - cos (AU) u 1 2 ' II.nu

-00 <u< 00,

196 Solutions

has characteristic function

cp(t) = {I - Itl/A, It I ~ A, 0, It I > A.

198. n

E(Sn) = I E(Xt) = n(1 + 112 ), i=1

n

Var(Sn) = I Var(xt) = n[E(X:) - E(Xt)2]. k=l

But if Z - N(O, 1), then X k = Z + 11 and

E(X:) = E(Z + 11)4 = E(Z4) + 4IlE(Z3) + 6112 E(Z2) + 6113 E(Z) + 114

since

E(Zk) = i-kcpt)(O) = {O for k = 2m + 1, 1,3, ... ,(2m-l) fork=2m(m= 1,2, ... ),

where

From (1) and (2) we have

Var(Sn) = n[(3 + 6112 + 114) - (1 + 1l2)2J = 2n(1 + 2112).

On the other hand, we have

E(X~) = v,

Setting

we have

E(T) = av, Var(T) = 2a2 v.

Var(T) = Var(Sn)

(1)

(2)

199. By an extension of the theorem of total probability for the expected values, we have

00

CPs/t) = E(e itSN) = I E[eitSNIN = nJP[N = nJ n=O

= I E(eitSn)P[N = nJ = I cpx;(t)P[N = nJ = PN(CPxJt», n

because for given N = n the characteristic function of Sn = Xl + '" + Xn is cpx;(t).

200. The probability generating function of a Poisson distribution is

(1)

Solutions 197

the characteristic function of the Cauchy distribution with density

1 1 f(y) = ~l + y2

IS

(2)

So the characteristic function distribution of SN'

SN = Xl + X 2 + ... + X N,

where Xj(j = 1,2, ... , N), are independent and identically distributed Cauchy random variables and N is Poisson independent from Xj' is

CPs)t) = P(cp(t)) = e).(e- 1tl -1).

The density of SN is

e-). foo ( 00 (Ae-t)n) e-). 00 An foo = - L --,- cos tx dx = - L , cos txe-nt dt

n 0 n=O n. n n=O n. 0

e-).f OO 1 00 An n =- costxdx+-e-). L - 2 2 non n=l n! n + X

1 foo 1 00 An n =-e-). cosxtdt+-e-). L, 2 2· nOn n=l n. n + X

We observe that this is a mixture of Cauchy distributions with weights w", i.e., of the form

where

L w"f,,(x), n

, n _).11,

w" = e -, n!

n = 1,2, ....

The f,,(x) represent Cauchy densities

1 n f,,(x) = - 2 + 2' n n X

1 foo fo(x) = - cos tx dt. n 0

-00 < X < 00, n = 1,2, ... ,

201. We have (see Exercise 191), putting p = Aln,

. [A(e it - 1)Jn it cpzJt) = (pe't + q)n = 1 + -+ e).(e -1),

n n+oo

198 Solutions

that is, the characteristic function of Poisson with parameter A. According to the continuity theorem for characteristic functions, the FzJx) -+ Fz(x). Hence the approximation ofa binomial by Poisson, for p -+ 0, n -+ 00, so that np -+ A, IS

k = 0,1,2, ....

202. We have

(A - it)'

For A = 1/2, S = n/2, the Gamma distribution gives x;. Hence

q>x~(t) = (1 - 2itrn/2.

203. Show that the characteristic function q>yJt) of Y",

n k n sin(t/k) q>yJt) = Il cos(t/2 ) = Il 2 . (/2k+1)'

k=1 k=1 sm t

tends (as n -+ 00) to sin t/t.

204. q>x(t) = roooo eitX[A dFI (x) + (1 - A) dF2(x)]

= Aq>1(t) + (1 - A)q>2(t),

where q>j is the characteristic function of Fj (j = 1,2). For Fj - N(llj' an we have

hence

j = 1,2,

q>x(t) = }, exp(ill1 t - !aft2) + (1 - A) exp(i1l2t - !ait2),

E(X) = i-Iq>~(O) = AIl1 + (1 - A)1l2'

Var(X) = - q>;(0) + [q>~(0)]2 = Aaf + (1 - A)ai.

205. The exponetial is a Gamma distribution (see Exercise 202) with parameters S = 2 and A = 9. Hence

( 't)-l q>(t) = 1 - ~

and the generating function of cumulants 'P(t) is given by

( it) it (it)2 (it)' 'P(t) = log q>(t) = -log 1 - 9 = 9 + 292 + ... + r9' + .. '.

Solutions

Then the cumulant of order r is

(r - 1)! K =---, 8'

206. By the definition of Y we have

P[Y = (2k + 1)h/2] = e-kh.9[1 - e-.9h]

and hence the probability generating function of Y is

Py(t) = E(t Y ) = f t(2k+l)h/2e-kh.9[1 - e-.9h] k=O

= t h/2(1 _ e-.9h)/(1 _ the-.9h).

From the above relation we have

Py(1) = E(Y) = ~ + [~h coth Gh8) - ~ ] > ~ = E(X)

because of 0 < tanh{!8h) < t8h and

Var(Y) = ;2 - ;2 [1 - G82h2/ sinh2 G8h)) ] < ;2 = Var(X)

because of sin(h8/2) > t8h.

207. We observe that by the orthogonal transformation

t* = t cos 8 + u sin 8,

x* = x cos 8 + y sin 8,

with Jacobian -1 we have

u* = t sin 8 - u cos 8,

y* = x sin 8 - y cos 8,

cp(t, u) = t: f: ei(IX+UY)g(x2 + y2) dx dy

= f-: fXoo' e i(/*x"+U*Y*)g(x*2 + y*2) dx* dy*.

199

(1)

That is, cp(t, u) is invariant under the orthogonal transformation (1) and so it must be a function of the invariant (function) of the transformation, i.e., of the length Jt2 + u2 of(t, u). Then we deduce

cp(t, u) = cp*(t2 + u2). (2)

For independent random variable(s) X and Y with density g(X2 + y2) we have

CPx,y(t, u) = CPl(t)CP2(U).

Because of (2) we have

(3)

where CPl' CP2 are the characteristic function(s) of X, Y, respectively. We can

200 Solutions

write relation (3) in the form

cp(r) = CP1 (t) CP2 (u), (4)

then putting t = 0 we deduce that cP = CP2 and putting u = 0 we deduce that cP = CPl' Hence

CP1 (t) = cpt(t2),

So cP satisfies the functional equation

cp(t2 + v2) = cp(t2)cp(U2)

the only solution of which (see, e.g., Feller, 1957) is cp(x) = efJx • Since cP is a characteristic function, I cp(t) I :::;; 1, and hence {3 < O. So for some (12 > 0 we have

cp(t2) = e-a2t2/2,

that is the characteristic function ofthe normal N(O, (12).

208. Let

The characteristic functions of U and V are

CPu(t) = ( t2)2n' 1 +-

n

cPy(t) = ( t2 )2n. 1 +-

n

Hence U, V are identically distributed, having the same characteristic function for every t. The joint characteristic function of U and V is

CP(t1' t2) = E exp[i(t1 U + t2 V)] = E exp i _1 __ 2 L Xj + _1 __ 2 L lj [ ( t + tnt - t n )]

n j=l n j=l

= [1 + C1: t 2yrTI + C1: t2Yrn.

We observe that

CP(tl' t2)"# CP(t1' O)cp(O, t2) = CPU(t1)CPy(t2)'

So U, V are not independent. It is easily shown that Cov(U, V) = 0, that is, U and V are uncorrelated. This can also be shown from the fact that the coefficient of t 1 t 2 in the expansion of log cp(t 1, t 2 ) is O.

209. (a) Let Xl' X 2 be independent and identically distributed random variables with characteristic function cp(t). Then the characteristic function of Xl - X2 is cp(t)qJ(t) = Icp(t)12.

(b) Let Xl' X 2 , ••• , Xn be independent and identically distributed random

Solutions 201

variables with characteristic function q>(t) and let N be Poisson with parameter )., independent ofthe X/so Then, from Exercise 199, the characteristic function of the sum

210. Similarly, as in Exercise 196, the multinomial random variable X is written

k

X = (Xl> X 2, ... , X k) = L ~j' j=l

where the probability generating function of ~j is

Pj(t 1t2, ... , td = P1 t1 + P2 t2 + ... + Pktk + Pk+1·

Hence the probability generating function of X is

P(t1' t 2, ... , tk) = (P1 t 1 + P2 t 2 + ... + Pktk + Pk+1)".

From the above relation we have

j = 1,2, ... , k,

a2p I E(XjXs) = -- , atj ats /,=12=---=1.=1

a2PI E[Xj(Xj - 1)] = at] /,=12=---=1.=1'

j = 1,2, ... , k,

hence the assertion is proved.

211. Putting ).)n instead of Pj into P(t1' ... , tk) of Exercise 210 we have

( ± k(t- _ 1»)n P(t 1, ... ,tk)= 1+ j=1 J J ----:::;-+exp [f Aj(tj -1)]=rl e ).j(lrl),

n n 00 j=l j=l

and hence the assertion is proved.

212. We have

By virtue of (2.13), we find

202

213. The conditional characteristic function of X. given Y = y. is

({Jx(t 11 Y = y) = E[eil,XI Y = y] = f'''oo ell,Xfx(xl Y = y) dx.

while the joint characteristic function of X and Y is

({J(t1• t2) = E[e i(I,X+r2Y)] = E[E(ei(I,X+12Y)1 Y)]

= E[eiI2YE(ell,xIY)] = E[eiI2Y({Jx(t1IY)]

= f~oo eiI2'({Jx(t1IY=y)fy(y)dy.

By the inversion formula we have

and since

the result follows.

214. The joint characteristic function of Y, Z is

Solutions

(1)

({J*(t 1• t2) = E[exp(i(t1 Y + t2Z))] = E[exp(it1(S1 + S3) + it2(S2 + S3))]

= E exp(i[t1S1 + (t1 + t2)S3 + t2S2])

= [({J(tdJ"'-"[({J(t 1 + t2)]"[({J(t2)]"2-",

where ((J(t) is the characteristic function of X. By virtue of (1) of Exercise 213. we have

Thus we find

1 foo . [ a({J*l E(YIZ = z)fz(z) = -2 e-1I2% i-1-a- dt2• 1t -00 t1 ,=0

n E(YIZ = z) = (n1 - n)E(X) + -z.

n2

This can be found more easily. directly (see Exercise 320)

(1)

(2)

E(S1 + S31S2 + S3 = z) = E(Sd + E(S3IS2 + S3 = z) = (n1 - n)E(X) + n/z.

Solutions 203

215. If q>(t) is the characteristic function, it must be shown that for every n E N there is a characteristic function q>n(t) such that

This is satisfied by the Pascal, Cauchy, and Gamma distributions as directly deduced from the corresponding characteristic functions in Exercises 191 and 202. For the Laplace distribution, we observe that the characteristic function is

q>(t) = (1 + t2r1 = [(1 + t2)-1/n]n,

where (1 + t2 )-1/n, as implied by Exercise 202, is the characteristic function of the difference of two independent and identically distributed Gamma variables with parameters A. = 1 and s = l/n.

216. Let X be the length of the circumference and Y the area of the circle. We have, because of (7.1),

1 (x) 1 fx(x) = 2n fR 2n = 2n(p - lX)' 2nlX < x < 2np,

1 (if) 1 1 fy(y) = --fx - = ----, 2J"iY n 2J"iY P - lX

217. Applying (7.10) for the transformation

Z = x + t, w = x/y,

with Jacobian J(z, w) = -z/(w + If, we find the density of z, w:

z (zw z) Z 2 _A f(z, w) = (w + 1)2 fx,y w + 1 'w + 1 = (w + 1)2 A. e z, z >0, w>O.

Since it can be written as a product of a function of z and a function of w, it follows that Z, Ware independent; Z is Gamma and W is F with 2 and 2 degrees of freedom.

218. X 2 - Y must be ~ O. Hence the required probability is

p[X2 - Y ~ 0] = r ff(X, y) dx dy = r fdX dy J x2_y;;O:O J x2;;O:y

I I IX2 1 = dx dy =-. o 0 3

Note. It is recommended that the reader show the above result by finding first the density of

that is,

fz(z) = {Jz+l for - 1 < z ~ 0, 1 - Jz for 0 < z < 1.

204 Solutions

219. We have () = v2 jg sin 29 and the distribution function of () is

F(x) = P [~ sin 29 ~ x J = P [sin 29 ~ ~: J = P l? < - arc sm - + P l? > - - - arc sm -[0 1 . gXJ [0 n 1 . gXJ

- 2 v2 - 2 2 v2

2 . gx =-arcsm-.

n v2

Hence the density is

f(x) = 2g 1 n Jv4 _ g2x2

v2

0< x <-. g

This is also found by using (3.2) directly.

220. (a) The density function of Z = XjYis given by

fz(z) = f: lyl./(x.Y)(Yz, y) dy

_ 1 foo [Z2 - 2pz + 1 2J - r;--;:z exp - 2 2(1 2) y 2y dy

2na2'oj 1 - p2 0 a - p

= _ 'oj 1 - I' e-I dt 1 ~ foo n Z2 - 2pz + 1 °

;- (z - p)2 + (1 _ p2)'

(b) Because of symmetry, we have

-oo<z<oo.

P[X < 0, Y> OJ = P[X > 0, Y < 0].

Since

P[XY < OJ = P[XjY < OJ = P[X < 0, Y> OJ + P[X > 0, Y < OJ,

it follows that

P[X < 0, Y> OJ = iP[XjY < OJ = iP[Z < OJ

1 fO 1 fO J1=pz = 2 -00 fz(z) dz = 2n -00 (z _ p)2 + (1 _ p2) dz

= ;n [arc tan (p)Ioo

Solutions 205

= ~[arc tan( -p ) - (-~)J 2n J1=P2 2

1 arc sin p

4 2n

221. The Cartesian coordinates (X, Y) of a point are connected with the polar coordinates (R, e) by the relations

X = R sin e,

The joint density of (R, e) is given by

Y = R cos e,

f(R,ek, 9) = f(x,n(r sin .9, r cos 9)IJ(r, 3)1,

But

, _ 1 -r2/2u 2 f(x, n(r sm 9, r cos 9) - 2na2 e

and IJ(r, 9)1 = r, Thus we obtain

_ 1 -r2/21<u2 • f(R,e)(r, 9) - --2 e r, 2na

0< r < 00, 0:::;; 9 < 2n.

Hence Rand e are independent. The marginal density fR(r) of R is

f21< r 1 f21< {' ( ) = {' (0) dO = _ -r2/2u2 _ dO = ~ -r2/2u 2

JR r J(R,8)r,~ ~ 2 e 2 ~ 2 e , o a n 0 a

0< r < 00,

that is, the Rayleigh distribution, and e is uniform in the interval (0, 2n).

222. (a) If fy(y) is the density of the random variable Y = aX + b then

1 (y - b) {l/a (a > 0), b < y < a + b, fY(Y)=~fx -a- = -l/a (a <0), a+b<y<b,

that is, Y has uniform density in the interval (b, a + b) if a > 0, or in the interval (a + b, b) if a < O.

(b) Let A > 0, B > 0; then

x = x(y) = _l!.- + JB2 + 4(y - C)A > 0 2A 2A

for C < y < A + B + C,

and thus

1 fy(y) = fx(x) J 2 )'

B + 4A(y - C C<y<A+B+C.

Similarly, for the other cases of the constants.

206

223. (a) The density of Y = XI + Xl is given by

fy(y) = r: fx 1(y - x l )fx2 (x l ) dXl

Solutions

If: dXl = y, O::s;; y ::s;; 1,

= II dx l =I-(y-I)=2-y, 1 ::s;; y::s;; 2, y-I

that is,

fy(y) = I - II - yl, O::s;; y::s;; 2.

(b) The density of Z = X I - X 2 is given by

If I dX l = I + z,

fz(z) = f+oo fx (z + x2)fx (Xl) dX2 = -Z

1 2 11- Z

-00 0 dX 2 = I - z,

-I ::s;; z::s;; 0,

O::s;; z ::s;; I,

that is,

fAz) = I - Izl, -1::s;;z::s;;1.

(c)

We have

FN(W) = P[W::s;; W] = P[IZI::S;; W] = P[ -W::S;; Z::s;; W] = FAw) - FA-w)

and thus

fw(w) = fz(w) + fz( -W).

Because of (*), we have

fw(w) = (l -Iwl) + (I -I-wI) = 2(1 - w), O::s;;w::s;;1.

(d) The density of V = XI/Xl is given by (see (7.7»

00 {II X 2 dX l =~, fv(v) = f IX21fx (xlv)fx (Xl) dX2 =

-00 1 2 1~V 1 Xl dX l = -1'

o 2v

o ::s;; v ::s;; 1,

v ~ 1.

224. Let XI be the time of arrival of A and let Xl be the time of arrival of B, then

fXj(x) = 1,

Fxj(xj ) = Xj

o ::s;; Xj ::s;; I, j = 1,2,

o ::s;; Xj ::s;; 1, j = 1,2.

Solutions 207

Given that the lunch lasts 0.5 hours, the probability p of meeting (event A) is given by

p = P(A) = P[IXI - X2 1 ::; 0.5] = P[ -0.5 < Xl - X 2 < 0.5].

The density of Z = Xl - X 2 , from Exercise 223, is given by

fAz) = 1 - Izl, Izl ::; 1.

Hence

p = (1 + z) dz + (1 - z) dz = -[(I + Z)2]g.5 - -[(1 - Z)2]g.5 fo fO.5 1 1

-0.5 ° 2 2

3 -4

If T is the time of meeting then T = max(X I , X 2) and let fT(t) denote the (unconditional) density of T Then, taking the interval [0, 1] we have

(a)

fT(t) = 1/4

fT(t) = 2t

fT(t) = 1

if t ¢: [0, 1],

if 0::; t ::; 1/2,

if 1/2::; t ::; 1.

Hence, the conditional density of T given A, since P(A) = 3/4, is given by

for 0::; t ::; 1.

(b) Similarly as in (l).

225. The density of the product Z = X Y is given by the formula

fro 1 (z) 1 100 1 1 fz(z) = -I I fy(y)fx - dy = -2 -I I ye-y2/2u2 J 2 2 dy. -00 y Y 7W Izl Y 1 - z /y

(1)

For the limits of integration of y, we have Izl < y < 00, because IXI < 1 and y > O. Thus

-oo<z<+oo,

that is, the density of N(O, (J2).

208 Solutions

226. (a)

n n

Y = L 2 log Xj = L lj j=l j=l

where

lj = - 2 log Xj' j = 1, 2, ... , n, independent variables with density

fy/y) = fx j (e-Yj /2) I d~j e-Yj/2 1 = ~ e-Yj/2, Yj > 0.

Thus lj has the Gamma distribution with parameters A. = 1/2 and s = 1, Exercise 203 (i.e., exponential distribution). But the Gamma distribution has the reproductive property and hence Y, as the sum of n independent and isonomic Gamma variables, has Gamma distribution with parameters A. = 1/2 and s = n, that is,

1 fy(y) = 2nr(n) yn-I e- Y/2, y > 0,

that is, Y has the X2 distribution with 2n degrees of freedom. (b) The joint density of the random variables (see (7.11))

~ = J -2log Xl cos 2nX2' '1 = J -2log Xl sin 2nX2

is given by

But

and hence

f(~, '1) = ~e-~2/2 ~e-~2/2, y2n y2n

-00 < ~ < 00, -00 < '1 < 00.

From the above relation, we deduce that ~, '1 are independent N(O, 1).

227. Let Y = F(X). Then

Fy(y) = P[Y:os:; y] = P[F(X) :os:; y]

{O' y:os:; 0,

= P[X:OS:; F-I(y)] = F[F-I(y)] = y,

1, y;o:: 1. ° =s; y =s; 1, (1)

If Xl' X 2, ... , Xn are pseudorandom numbers, then YI = F-I(Xd, ... , Yn =

F- 1(Xn ) are independent, and by (1) they are isonomic (identically distributed) with distribution function F.

Solutions 209

228. By Exercise 239 we have

P[X E b] = R,

where R has density given by (2) of Exercise 239 with n = 10. Then the required probability is

P[R ~ 0.95] = 10 x 9 f 1 r 8 (1 - r) dr = 1 - 10(0.95)9 + 9(0.95)10. 0.95

229. (a)

fx(X) = fy(lOgX)I~(lOg X)I = fo exp [-~(lOg x - 11)2J, x >0. dx (J 2nx 2(J

(b) We have

E(X) = E[e Y ] = My(l),

Var(X) = E(X2) - [E(X)]2 = My(2) - [My(1)]2,

where My(t) = e/ll+u212/2 is the moment generating functions of the normal random variable Y. Hence

E(X) = e/l+u2/2 and Var(X) = e2/l+ 2u2 _ e2/l+u2.

(c) log Xi is normal; hence the sum log Xl + ... + log Xn = log([17;1 X;) is normal. Hence it follows that Y = [17;1 Xi is lognormal.

230. From the definition of the distribution function we have

Fz(z) = P[Z :s z] = P[X :s z, Y:s z] = F(z, z),

Fw(w) = P[W:s w] = 1 - P[W > w] = 1 - P[X > w, Y> w].

Using the addition theorem

P[A u B] = P(A) + P(B) - P(AB) = 1 - P(A'B')

with A = {X:s w}, B = {Y:s w}, we have

1 - P[X > w, Y> w] = P(X :s w) + P(Y:s w) - P[X :s w, Y:s w] .

= Fx(w) + Fy(w) - F(w, w).

If F is continuous, we obtain the densities of Z, W by taking the derivative of the corresponding distribution function; thus, we have

d d JZ JZ fz(z) = dz F(z, z) = dz -00 -00 f(x, y) dx dy

= foo f(x, z) dx + foo f(z, y) dy, (1)

fw(w) = fx(w) + fy(w) - J:oo f(x, w) dx - J:oo f(w, y) dy.

210 Solutions

231. Because of the independence of X, Y (see Exercise 230(1)) we have

fz(z) = 2<1>(z)q>(z),

where <1> and q> is the distribution function and the density of N(O, 1), respectively. Hence

f OO 1 foo fZ E(Z) = -00 zfz(z) dz =; -00 ze- z2

/2 dz -00 e- x2/2 dx

= - e- x2/2 dx ze- z2/ 2 dz = - e- x2 dx = -. 1 foo foo 1 foo 1 n -00 -x n -00 In

232. The joint density of Y1 = Xl' Y2 = Xl + X 2, ... , Yk = Xl + X 2 + ... + X k is given by

Since Xl = Y1,X2 = Y2 - Y1, ... ,Xk = ¥,,- ¥,,_l,we haveJ(Yl,Y2'''',Yk) = 1 and so

. (1 - Yk)"k+l-t,

Then the marginal density of ¥" is

But

= y"l +"2-1 r(n1 )r(n2). r(n1 + n2)

Thus, by successive integrations, we finally obtain

233. The joint density of Y1 , Y2 , ... , ¥", Y = Xl + X 2 + ... + X U1 ' is given by

f1(YYdf2(YY2) .. ·.f,.(YYd.f,.+1(Y(1 - Y1 - Y2 -'" - Yk))IJ(Y1' ... , Yk, Y)I, (1)

Solutions 211

where

Xi> 0, i = 1,2, ... , k + 1,

and since Xl = YY1' X 2 = YY2' ... , X k = YYk' Xk+l = y(1 - Yl - ... - Yk)' we have

J(Yl'···'YbY)=

Y

° ° .. . Y .. . ° °

Y Yk ° ° -Y -Y -y (1 - Yl - ... - Yk)

Then (1) is written as

which gives the required Dirichlet density.

234. (a) The density of the random variable Z = X + Y is given by

that is, Poisson with parameter ..1. + Jl. (b)

Zk = 0, 1, 2, ... ,

P[X = klX + Y = n] = P[X = k, Y = n - k] P[Z = n]

(n) ( ..1. )k( Jl )"-k = k ..1.+Jl ..1.+Jl '

that is, binomial with parameters nand p = ..1./(..1. + Jl).

k = 0,1, ... , n,

212 Solutions

235. (a) n

P[X + Y = n] = L P[X = k]P[Y = n - k] k=O

= pnqA+M-n f. (A)( M ) k=O k n - k

(A + M) n A+M-n = p q , n

n = 0, 1, ... , M + A.

(b)

P[X = k, Y = n - k] P[X = k]P[Y = n - k] P[X = klX + Y = n] = P[X + Y = n] = P[X + Y = n]

236. We have n n n n-1

(n - 1)s2 = L Xi2 - nX2 = L }? - (JnX)2 = L 112 - y"2 = L 112, (1) i=1 i=1 i=1 i=1

since the transformation y = Hx gives

" " y'y = L yJ = x'H'Hx = x'x = L xf. j=1 i=1

The covariance matrix of Y = (Y1, Y2, ... , y")' is D(Y) = HD(X)H' = (12HIH' = (121, that is, Y1, ... , Y" are independent N(O, (12). Hence Y" = JnX, X is also independent of Y1, ... , y"-I' and therefore of S2, because of (1). Moreover, the distribution of (n - l)s2j(12 is X~-I' The independence of X and S2 is also deduced from the fact that

COV(Xi - X, X) = 0, i = 1,2, ... , n.

This, by the joint normality of X, Xi - X, implies their independence, that is, X is independent of Xl - X, X 2 - X, ... , Xn - X; and hence also ofs2. For the Cov(X, S2), taking E(X;) = 0, without any loss of generality, we have by (1)

- 1 [_ n ] n _ Cov(X, S2) = --E X L Xl - __ E(X3)

n - 1 i=1 n - 1

1 [" " ] 1 [n ] = E LX.' L x 2 - E L x~ n(n - 1) i=1 'i=I' n(n - 1) i=1 '

1 1 1 = --J1.3 - ----J1.3 = -J1.3·

n - 1 n(n - 1) n

Solutions

237. Show that

where

1 m X - Y = - L Vj = V,

m ;=1

1 m Vj = uj + -- L y; - y

fo;=1

are independent N(lll - 1l2' a*2) with

m m

and L (Uj - iif = L (v; - V)2. ;=1 ;=1

Hence for III = 112 we have the required distribution.

213

238. The density of the random sample Xl' X 2, ... , Xn from a continuous distribution with density f is

Given that the n! permutations of Xl' ... , Xn give the same ordered sample XC!), X(2), ... , X(n), it follows that the required density is

n

f*(x l , x 2, ... , xn) = n! f(x l , x 2, ... , xn) = nl n f(xJ ;=1

239. (a) By Exercise 227, Y = min(Y1' ... , Y,,), Z = max(Yl , ... , Y,,), where Yl , Y2 , ••• , y" constitute a random sample from the uniform in (0, 1). Working as in Exercise 238, we find that the joint density of Y, Z is

f(y, z) = n(n - 1Hz - yr2, 0< y < z. (1)

(b) The density of R = Z - Y is easily seen to be

fl-r

fR(r) = n(n - 1) 0 rn- 2 dy = n(n - l)rn- 2 (1 - r), 0< r < 1. (2)

Hence we have

E(R) = n(n - 1) f 1 rn - l (1 _ r) dr = n - 1 . o n + 1

240. The joint density of Xw, X(k) (j < k), is given by

f(xj, Xk) = (j _ 1)1 (k _ ;!_ 1)1 (n _ k)l [F(xj)]i-2[F(Xk) - F(xj)]k-j-1

. [1 - F(xk)]n-kf(xj)f(xk), Xj < Xk'

because j - 1 of the X; must be less than Xj' k - j - 1 between Xj and Xk, n - k must be larger than Xk' one in the interval (Xj' Xj + dxj) and one in the interval (Xk' Xk + dxk) (f(x) and F(x) represent the density and the distri-

214 Solutions

bution function of the original distribution, respectively). Hence we have

E[X(k+l) - X(ka

= f~<Xl f~oo (y - x)f(x, y) dx dy

= (k _ I)! (;!_ k _ I)! {f~<Xl Fk-1(X)f(x) dx {<Xl y[1 - F(y)]n-k-If(y) dy

- f~<Xl xFk-1(X)f(x) dx {<Xl [1 - F(y)]n-k-If(y) dY}.

Integrating by parts gives the result. This can also be found from the relation

E[X(k+l) - X(k)J = E[X(k+l)J - E[X(k)J,

using the density of X(k) in Exercise 241.

241. (a) The density of X(k) is (cf. Exercise 240)

n! k-l n-k h(x) = (k _ I)! (n _ k)! [F(x)] [1 - F(x)] f(x), (1)

where for the uniform f(x) = 1 (0 ~ x ~ 1), F(x) = x (0 ~ x ~ 1). Hence we have

n! I I k n-k k E[X(k)J = (k _ I)! (n _ k)! ° x (1 - x) dx = n + 1 .

(b) The density of R is

fR(r) = n(n - 1) f: f(x)f(x + r) [F(x + r) - F(x)]n-2 dx. (2)

For the uniform density we have

fR(r) = n(n - 1) r-r rn- 2 dx = n(n - l)rn-2(1 - r).

(c) From (3) we find

E(R) = n(n - 1) II rn- l (1 - r) dr = n - 11. ° n +

242. Because of (1) of Exercise 241, the density of Y2 is

[) - ! < x < [) + !. Hence the required probability is

f .9+0.4 1°·9 6 (x - [) + !)([) - x + !) dx = 6 y(1 - y) dy = 0.944 .

.9-0.4 0.1

(3)

Solutions

243. The joint density of Y1, Y2, Y3, by Exercise 238, is

g(Y1' Y2, Y3) = 3! (2yd(2Y2)(2Y3) = 48Y1Y2Y3'

According to (7.11) the density of Zl' Z2, Z3 is

f(Zl, Z2, Z3) = g(Y1, Y2, YJ)IJ(Zl' Z2, z3)1 = 48z1Z~Z~,

because

So we have the independence of Zl, Z2' Z3'

244. Show that the density of Y = J Xl X 2 is given by

f co 1 (y2) fy(Y) = 2y -co ~ fX t ~ fx 2(x) dx

[1 [( y2) Jn2 -1 = cy2nt Jy2 X- 3/2 1 - ~ (1 - x) dx.

215

245. Direct application of(7.11)withz = gl(X, y) = Y - x, w = g2(X, y) = x. Z is Gamma with parameter n and it is independent of X.

246. Apply Exercise 244.

247. Application of (7.6). Y is Gamma with parameter m + n.

248. Setting X = R cos B, Y = R sin B, we have

Z = R sin 2B, W= R cos 2B,

where, by Exercise 221, Rand B are independent with densities

1 f(B) = -, 0 < B < 2n.

2n

Now the joint density of Z, W can easily be found to be

1 (Z2 + W2) f(z, w) = 2n0'2 exp - 20'2 '

which shows the independence of Z and W

249. By applying (7.12) to (5.10) we find that Y is N(AJ1., ALA').

(1)

250. Let (Xi' 1;) be the impact point of Ai (i = 1,2). The required probability is equal to

2 2 lCO O't P[R 1 < R 2] = P[R 1 < R 2] = F1 (r)f2(r) dr = 1 - 2 2' o 0'1 + 0'2

216

since Rf = X? + Y?, by Exercise 221, has density

and distribution function

Fi(r) = 1 - exp ( -~ ;2). i = 1,2.

251. Apply (8.9)

252. Follows from (8.5) and

Var(Xn ) = :2 t~ Var(X;) + 2 :t: COV(Xk' Xk+d}

1 :::; -2 {nM + 2(n - I)M} ---+0. n n-X)

253. Follows from

n

Var(Xn ) < n- 2 L Var(Xi ) :::; n- 1c -+ 0 i=l

as n -+ 00.

254. Show that

Var(Xn ) -+ 0 as n -+ 00.

255. We will show that for every continuity point x of Fx

lim P[Xn + Y" :::; xJ = Fx(x - c).

We have

P[Xn + Y" :::; x] = P[Xn + Y" :::; x, I Y" - ci :::; 8]

+ P[Xn + Y" :::; x, I Y" - ci > 8J, where

Solutions

(1)

P[Xn + Y" :::; x, I Y" - cl > 8J :::; P[I Y" - cl > 8] -+ 0

and the probability tends to F(x - c).

as n -+ 00,

P[Xn + Y" :::; x, I Y" - cl :::; 8]

Similarly, one can show (b) and (c).

256. Let

Then

k

Ln = L CjXjn j=l

and k

L = L CjXj. j=l

CfJLJU) = E(e iuLn ) = CfJ~JUC l' UC 2 , •.. , UCk ),

CfJL(U) = E(e iUL ) = CfJ~(UC1' UC 2 , ..• , UCk )·

(1)

(2)

Solutions 217

Since

we have

that is,

qJ~JCi' C2 ,···, Ck ) -> qJ~(Ci' c2 ,···, cd for every (C l' ... , ck ). Hence by the continuity theorem of characteristic functions for multivariate distributions

L en -> e· 257. From P[IXkl ::; MJ = 1 it follows that X k - E(Xd are uniformly

bounded and since s; -> 00 for every a > 0, there is an N such that for n > N the relation

P[IXk - E(Xdl < aSn' k = 1,2, ... , nJ = 1

holds. Hence the condition of Lindeberg- Feller's theorem is satisfied and the CLT holds.

258. If Xn denotes a Poisson random variable with parameter ), = n =

E(Xn), then we have

e-n f n~ = P[Xn ::; nJ = p[Xn ~ n ::; 0] -> <1>(0) = ~ as n -> 00 k=i k. v n 2

by the normal approximation to the Poisson.

259. We have

X!. Ai' Y!. A2 , X + Y ~ Ai + A2 ,

X + Y ~ N()'i + ..1 2 , Ai + A2 )·

The result follows by application of Exercise 255(c) with C = JA i + A2 •

260. Verify that the Lindeberg-Feller condition is satisfied.

261. According to the CL T

P[ -1 < S: < 1J -> <1>(1) - <1>( -1) = 0.68,

while Chebyshev's inequality gives no information since

P[ISnl ~ c5J ::; c5- 2 = 1

For k = 2, Chebyshev's inequality gives

for c5 = 1.

P[ -2 < S: < 2J ~ 0.75,

while the CL T gives as a limit <1>(2) - <1>( - 2) = 0.9544. Similarly, for k = 3, we have 0.8888 and 0.9974, respectively.

218

262. 11 = E(X) = Loo x24x- 4 dx = 24 Loo x- 3 dx = 3,

q(<5) = P[IX - Ill> <5] = P[IX - 31 > <5]

= 1 - P[IX - 31 < <5]

= 1 - P[3 - <5 < X < <5 + 3] = 1 - f3H 24x-4 dx 3-~

= 1 - 8 [(3 ~ <5)3 - (3 : <5)3]

{8(3 + <5)-3, <5 ;?: 1,

= 1 - 8[(3 - <5)-3 - (3 + <5)-3], <5 < 1.

According to Chebyshev's inequality

Var(X) 3 P[IX - III > <5]:5: <52 = <5 2'

Solutions

The following table and figure show the difference between q(<5) and 3<5- 2 •

Note that Chebyshev's inequality for <5 :5: a = J3 gives the obvious upper bound 1.

y y = 1

1/2 1 2 3 5

q(<5)

0.673 0.125 0.064 0.037 0.016

263. Applying the CL T we find that n will satisfy the relation

(a) P[IX - III > 0.05a] ~ P[IZI > 0.05Jn] < 0.05.

Hence 0.05Jn > 1.96 (Z '" N(O, 1)). Thus n ;?: 1537

By Chebyshev

1 1 0.750 0.333 0.120

Solutions 219

P[IX -Ill> 0.0511]::; p[IX -); > 0.05~5J a/yn a/ n

(b)

[ 0.25 r:.J ~ p IZI > -3- y n .

Hence In > 196· (3/25) and n > 553. According to Chebyshev's inequality n must satisfy the relation

(a) - 1

P[IX - Ill> 0.5a] ::; 2 < 0.05 => n > 203, (0.05) n

(b) _ a2 9

P[IX -Ill > 0.0511] ::; n(0.OWIl2 < (0.25)2n

< 0.05 => n > 2880.

264. (i) According to the De Moivre-Laplace theorem and putting Z '" N(O, 1), the required size n of the sample will satisfy the relation

from which n> 1962pq.

Then we have (a) n > 1962 x 0.3 x 0.7 or n > 8067. (b) The maximum value of pq is attained for p = 0.5. Hence n > 1962 x 0.25 or n > 9604.

(ii) Working as in (i), we find (a) n > 323, (b) n > 384.

265. According to the Moivre-Laplace theorem and putting Z '" N(O, 1), the number of seats k must satisfy the relation

[ k - 300 x 1/3 J 1 P[S300 > k] ~ P Z > < -,

)300 x 1/3 x 2/3 20

k - 100 fiN\ > l.645 or k ~ 112.

y 300/3

266. If the X;'s are independent uniform random variables in ( - 0.05, 0.05), we have, according to the CLT with II = 0, Var(XJ = 0.01/12,

p[lli~O Xi I < 2 J ~ P[IZI < 0.!1 J = P[IZI < 2.19] = 2<1>(2.19) - 1

= 0.97.

267. As in Exercise 266 we have, since E(XJ = 5, Var(XJ = 902/12,

p[I Xi> 500J ~ p[Z > 200J ~ 1 - <1>(0.91) = 0.18. i=l 220

220 Solutions

(b) c must satisfy the relation

[ c - 300 J 1 c - 300 P Z < 201.6 ~ 10 => 201.6 = -1.28 => c = 42.

268. The profit Xi of the casino during the game i, is a two-valued random variable Xi with E(X;) = - 1/37. Thus

[ n J [ 1,000-nI37J 1 P L Xi ~ 1,000 ~ P Z ~ In ~ -,

k=l n(1 2

where Z is N(O, 1), and hence

n 1,000 - 37 ~ ° => n ~ 37,000,

P[ofa loss] = p[37fOO Xi < oJ ~ p[Z < -~J ~ 0. i=l 37,000

269. Use Exercise 215(d) and the CLT, or show directly that the characteristic function of the standardized Gamma tends to exp( - t 2/2).

270. E[g(X)] = J:oo g(x) dF(x) ~ r g(x) dF(x) ~ c r dF(X)

= cP[X E (a, b)].

271. Take as g of Exercise 270, g(x) = (x + C)2 with c > 0. If E(X) = ° and E(X2) = (12, then the minimum of (t + C)-2. E(X + C)2 is obtained when c = (12/t . Then we have the required inequality.

272. From the event {IXI - X 2 1 ~ t} we have that at least one of

{IXll > ~ > t}, {IX21 > ~t} occurs. Thus we have

P[IX1 - X 21 ~ t] ~ P[IX11 > 1-t or IX2 1 ~ 1-t] ~ 2P[IX11 > 1-t].

273. See Exercise 271.

274. If the function g(y) is decreasing for y > 0, then for;' 2 > ° we have

;'2 g(y) dy ~ _ y2g(y) dy. foo 4 foo ). 9 0

(i) In the special case when g is constant

we have

g(y) = {A for ° < y < c, ° for y ~ c,

;'2 J: g(y) dy = A;'2(c - ;').

This is maximized for;' = (2/3)c, as we can easily see by differentiation.

(1)

Solutions

(ii) We define the function

hey) = {go(A-) = const., 0 < y < A- + a, y~ll+a,

where ageA-) = fT g(y) dy. Then we have

221

Let f(x) be the density of the continuous random variable X. Since Xo is the only mode (maximum) of f, it is decreasing for x > xo. Hence taking

g(IX*I) = rf(xo + rIX*I), with X* = X - Xo

r

we have by (1) that

275. Application of Markov's inequality and of the relation

P[IXI ~ e] = P[g(X) ~ gee)].

276. E[g(X)] = E[g(x)IIXI ::; e]P[IXI ::; e]

+ E[g(X)IIXI > e]P[IXI > e]

::; g(e)P[IXI ::; e] + MP[IXI > e]

::; gee) + MP[IXI > e].

277. Since P[g(IXI) ::; geM)] = 1, then from Exercise 276 we have

E[g(X)IIXI > e] ::; geM).

278. p[ max(IX1; 1L11, IX2 ~ 1L21) < e] = p[IXl -1L11 < e, IX2 -1L21 < e]

a l a2

= P[(x l , x 2 ) E R],

where R is the rectangle defined by the lines Xi = lLi ± eai (i = 1,2). The function

g(Xl,X2,t) = e2 0 ~t2)[(Xl;lLly _2t(Xl;1L1)(X2~1L2)

for It I < 1,

222

is nonnegative and larger than 1 for (Xl' X 2 ) ¢ R. We then have,

P[(Xl' X 2 ) E R] = f t I(x l , x 2 ) dX l dX2

= 1 _ 2(1 - tp) c2 (1 - t 2 ) ,

It I < 1.

The expression [2(1 - tp)]/[c2 (1 - t2 )] is minimized when

and thus

1 t = -(1 - j1=?)

p

p[IXl - {Ill < c, IX2 - {l21 < c] ~ 1 _ 1 - ~, ~ ~ c

from which the inequality follows.

279. For every vector t, the function

g/(X) = t'G(X)t

Solutions

is a convex scalar function of the matrix X and by Jensen's inequality we have

g/(EX) :::;; E[g/(X)],

that is, for every t,

t'G(EX)t' :::;; Et'G(X)t.

Hence the relation

G(EX) :::;; E[G(X)]

follows.

280. By Exercise 279, it suffices to show that the matrix function G(X) = X- l is convex. For this it must be shown that for symmetric matrices X > 0, Y> 0, of order r (say) and 0:::;; A :::;; 1, the relation

[AX + (1 - A) y]-l :::;; AX- l + (1 - A) y-l (1 )

holds. There is a nonsingular matrix A such that X = AA', Y = ADA', where D a diagonal matrix with elements d l , ... , dr. Hence (1) is equivalent to

[AI + (1 - A)Dr l :::;; AI + (1 - A)D-l,

which is satisfied by diagonal matrices; the assertion now follows from the convexity of the scalar function y = x- l .

Note. For 0< A < 1, (1) holds with strict inequality.

Solutions 223

281. From (d 2jdy2) log y = _ljy2 < ° it follows that g(y) = log y is a concave function; therefore, by Jensen's inequality,

E[log YJ ~ log E(Y).

282. If we consider the random variable Y,. = I7=11og Xi where Xl' Xl' ... , Xn are independent and isonomic, it follows that E(Y,.) = nE[log Xa and Var(Y,.) = n Var(log XJ. By Chebyshev's inequality, for every 6> 0,

Var( Y,.) P[I Yn - E(y")1 < n6J ~ 1 - 2 2 '

n 6

or

n a l

P[n(E(log X;) - 6)J < L log Xi < n[E(log X;) + 6JJ ~ 1 - -2' i=l n6

or

P[exp{ n(E(log X;) - 6)}J < Xl X 2 .•• X. < exp{ n(E(log X;) + 6)}

Hence, and since by Exercise 281 E[log XJ ~ log E(X), we obtain the required one-sided inequality.

283. The area of the triangle is E = -!X Y, where X and Yare independent and uniform in the intervals (0, a) and (0, b), respectively. (a and b denote the length and width of the rectangle, respectively). The density of E is

2 2 f(x) = ab log(ab) + ab log 2x, 0< x < -!ab.

Hence calculate the required probability.

284. Let A be the origin of the axes and IX the abscissa of the point B. Then the randomly selected points Xl and X 2 have the uniform distribution in (0, IX)

and the density of the ordered sample X < Y (say) is

2 f(x, y) = 2' 0< x < y < IX.

IX

In order that X, Y be such that the segments (AX), (Xy), (YB) form a triangle, they must satisfy

2X < IX, IX

Y - X < - and 2

these define the triangle (CDE) (see figure) Hence

IX Y> _. 2'

2 1 P[(X, Y) E (CDE)] = IXl X area of (CDE) = 4.

224 Solutions

y

~--------~------~L---.X al2 a

285. The first point PI can be selected as any point of the circumference and may be regarded as the origin of measuring the angles which define the positions of the other points on the circumference (see figure). For every 8, the probability (elementary) that 82 , which defines the position of P2 , is in the interval (0,8), and that 83 (for the third point P3 ) is in the interval (8,8 + d8), is equal to

8 d8 I 2n 2n = 4n2 8 d8.

Hence the required probability, say P3' will be

P3 = 3! -8 d8 = _82 =-. I" I 3 I" 3 o 4n2 4n2 0 4

The factor 3! is needed because there are 3! different permutations ofthe three points PI' P2 , P3 •

286. Let P be the required probability and let (AB) be the given line segment. Then

(I)

Solutions 225

where Ai is the event that the ith part is larger than 1/2. Since evidently AiAj = 0 (i #- j), we obtain from (1)

5 (1)4 P ~ 1 - i~ P(AJ = 1 - 5 2" '

because the probability that any part is larger than 1/2 is equal to the probability that each of the four points is contained in a subset of AB with length 1/2, that is,

287. As in Exercise 284, we find that the points X, Y must satisfy the relations

a+b X<--

2 ' a+b

X+ Y>-2- and

which define the triangle 0ZH (shaded area). Hence

[(X ) (0 )] area of (ZH0) P ,Y E -HZ =-----

area of (OrE,1)

y

a+b

a+b -2-k-----~

a+b Y<-2-'

b 2a'

~------~~~----~~--x o a+b a a+b 2

288. Let X be the distance of the nearest line to the middle M of the needle AB and qJ the formed angle. A necessary and sufficient condition for the needle to cross the line is

X:::;; Jl sin qJ

and the required probability p, in view of the uniform distribution of X in

226 Solutions

(0, a) and cp in (0, n), is

1 I" 2Jl P = - Jl sin cp dcp = -. an 0 an

x

2a A

289. At first we consider a convex polygon A 1 A2 ... An (with diameter less than 2a). Then a given line (of the parallel ones) can be crossed either by none of the sides of the polygon or by two sides only. The required probability P is given by

P = L Pij' i, j = 1,2, ... , n, (1) i<j

where Pij = P (of crossing sides i and j). On the other hand, the probability Pi (say) of being crossed by the side i, by Exercise 288, is given by

Jli Pi = L Pij=-'

i#j na (2)

where Jli the length of side i (i = 1,2, ... , n). The probability p, by (1) and (2), is

n

lin i~ Jli P P = L Pij = - L Pij = - L Pi = -- = -,

i<j 2 i#j 2 i=l 2na 2na (3)

where P is the perimeter of the polygon. Observing that P is independent of the number n of the sides of the convex polygon and depends only on its perimeter, going to the limit when n -+ 00 and max Jli -+ 0, we see that the required probability for a convex and closed curve is given by (3), where P is the perimeter (length) of the curve.

290. Let an and bn be the expected numbers of lO-drachma coins of A and B, respectively, after n exchanges. Then

an + bn = a + b

Moreover, the difference equation

and ao = a. (1)

(2)

holds, where En is the event that A gets a lO-drachma coin the nth exchange.

Solutions 227

Since bn - 1

PEEn] = -b- = 1 - P[E~],

relation (2) becomes

n = 1,2, ....

Taking into account (1), we obtain its general solution

a =~+~(l_~_~)n n= 1,2, .... n a+b a+b a b '

Hence the required probability Pn (say) that A gives a 10-drachma coin to B after n exchanges is

Pn = ~ = a : b + a ! b ( 1 - ~ - ~ r 291. (a) The required probability Pn (say) that the game finishes with an

ace after no more than n tosses is

where

Pa = P(of ace) = 1/6, Pb = P(of six) = 1/6,

and Po = P(neither ace nor six) = 2/3.

Relation (1) is found from the fact that the event Ak that the game finishes after k tosses with an ace occurs if: (i) One ace appears on the first k - 1 tosses, neither ace nor six on k - 2 tosses, and ace on the kth toss. (ii) One ace and one six appear on the first k - 1 tosses (hence neither ace nor six during k - 3 tosses), and a six appears on the kth toss.

(b) The probability, P (say), that the game finishes with ace is the limit of Pn as n -+ 00. Thus from (1) we find

p;(1 - Po + 2Pb) 1 P = (1 - Po)3 2

Similarly, the probability that the game terminates with six is 1/2 and therefore the game terminates with probability 1. For fixed n there is a positive probability 1 - 2Pn that the game continues after n tosses.

292. We have

Pn = (1 - Pn-l )P(of selection of A in the nth year) = t(1 - Pn-l)

228

with PI = 1. The general solution of the difference equation

Pn + !Pn-I =! is

n = 1,2, ....

293. Pn satisfies the difference equation of order 2

Pn - Pn-I + PQPn-2 = 0, n 2 2,

(i) For P =I Q it has the solution

(ii) For P = Q = 1/2 we have

Po = PI = 1.

n 2 O.

Pn = (n + l)rn, n 2 O.

294. (a) We have

Yk+1 = Yk + 1 . P(the k + 1 ball is placed in an empty cell)

= Yk + n ~ Yk = (1 - ~) Yk + 1. The resulting difference equation

Yk+1 - (1-~)Yk = 1

has the general solution

where c = - n since YI = 1.

(b)

Then

Yk=n+C(I-~y

where X. = {I , 0 if cell i is occupied,

if cell i is empty.

E(XJ = 1 - (1 - ~y, i = 1,2, ... , n,

and thus we find

Solutions

295. Let A be the event of a double six in a toss of two dice. Then we have

35 1 Pn = Pn-I' P[A C

] + (l - Pn-d' peA] = Pn-I' 36 + (1 - Pn-I) 36'

Solutions 229

Hence the required difference equation, with particular solution Pn = a =

1/2, and general solution of the corresponding homogeneous Pn = c· (17/18t. Thus the general solution is Pn = 1/2 + c(17/18)n, where from the initial condition Po = 1 we find that c = 1/2. Hence

Pn = ~[1 + G~)"l n = 0,1,2, ....

296. Let Pk be the probability that a white ball is drawn bearing the number k (k = 1,2, ... , n). Then Pn satisfies the relation

A+l A Pn = A + M + 1 Pn-l + A + M + 1 (1 - Pn-l)

1 A = A + M + 1 Pn-l + A + M + 1

with initial condition

(1)

We find A 1

Pn = A + M + c (A + M + It'

where, in view of (1), c = 0. Hence

A Pn = A + M' n = 1,2, ....

Remark. If we suppose that the ball transferred first is white, then

Pl = 1 A M A

and P = + ---+. n A + M (A + M)(A + M + It 1 n+co A + M

297. Let Pn be the required probability. Then it will satisfy the relation

Pn = PPn-l + p'(l - Pn-d, Pl = P (n=2,3, ... ).

Its general solution is

pi q ( ')n Pn=--+-~ P-P .

q + pi q + pi

For q = 1 - P = pi it gives

1 1 1 Pn = 2 + 2 (p - q)n -+ 2 as n -+ 00.

298. We have

Pn = Pn-l P(of one ace) + Pn-2P(of double ace)

10 1 = Pn-l 36 + Pn-2 36'

230 Solutions

with initial conditions Po = 25/36, PI = 10/36 x 25/360. Solving this homogeneous difference equation we find Pn' For the probability, say qn' that one obtains n points, when the game is not interrupted, we have the difference equation

Compare Pn and qn'

299. Let An be the expected number of good tubes obtained after n trials and Pn the required probability. Then

An+1 = An + 1· P(of drawing a bad tube during the (n + 1) draw)

= An+1 (1 - a ~n b) and Ao = a. Hence we find

and

300. Define success (S) on the rth draw of a ball to be that the (r - 1) ball was drawn from the same cell r = 2, ... , k. The probability of such a success Sis l/n. If in this sequence there is no sample SS (event A), then in the sequence of trials there is no cell receiving more than two balls successively. If qk is the probability of A in a sequence (S and SC) of n trials, then Pk = qk-I (k > 2) and qk satisfies the relation

n-1 In-I qk = --qk-I + ~--qk-2'

n n n

from which we have

n-l n-I Pk = Pk(n) = --Pk-I + --qk-2' k > 2, (1)

n n

with PI = qo = 1, P2 = ql = 1. If AI' A2 are the roots of the characteristic equation of (1), then Pn is given by

Pk = [(1 - A2))"~ - (I - AI))"~-I]/(AI - A2)'

(a) lim Pk = 0 as k ~ 00, because A~ ~ 0, )"~ ~ O. (b) lim Pk(n) = 1 as n ~ 00, because Al ~ 1, A2 ~ O.

301. Applying Holder's inequality (9.5) with P = q = 1/2 for the random variables IXI(t+h)/2 and IXI(t-h)/2 we have

o ~ h ~ t.

Setting tl = t + h, t2 = t - h, equation (1) becomes

E2 [IXI(t,+t2)/2] ~ E(IXl t')E(IXl t2),

(1)

Solutions 231

or, taking logarithms,

2 log E[IXI(t,+t2)/2] :::;; log E(IXlt,) + log E(lxlt2).

Setting g(t) = log E(lxlt), we have

that is, g(t) satisfies (9.1) with A = 1/2 and hence it is convex since it is also continuous. Since g(O) = 0, the slope g(t)/t of the line that connects the origin with the point (t, g(t)) increases as t increases and hence

eg(t)/t = [E(IXlt)]l/t

is an increasing function of t.

302. CPo(t), as the real characteristic function of a random variable X, is given by

CPo(t) = f: cos txfx(x) dx

and the required inequality for CPo follows by taking the expected value in the inequality

1 - cos 2tX = 2(1 - cos 2 tX) :::;; 4(1 - cos tX).

For the other inequality we observe that if cp(t) is the characteristic function of the random variable X then Icp(tW = cp(t)cp(t) is the characteristic function of Xl - X 2 , where Xl' X 2 are independent having the same distribution as X, and it is a real function. Taking CfJo(t) = Icp(tW in the above inequality, we have the required inequality.

303. Applying the Schwartz inequality

to

cp(t) = f: (cos tx)f(x) dx

with

f(x) = cos txJlW, g(x) = JJ(X),

we have

f oo foo 1 ICPo(tW:::;; -00 cos2 txf(x) dx = -00 2(1 + cos 2tx)f(x) dx.

Hence we have the required inequality.

232 Solutions

304. E(XiXj) = Cov(Xi, Xj) (i -# j), implies that at most one of the Xi' say Xl' has mean -# O. Since the central moments of order 3 are

we have E(X2 X 3 X 4 ) = O. It must be shown now that for jointly normal random variables Xl' X 2 , X 3 , X 4 with mean 0, we have

From the joint characteristic function of Xl' X 2' X 3' X 4

CP(tl, t 2 , t 3 , t4 ) = exp[-! ± t (Jijtitj ] 2 i=l j=l

and from the relation (cf. 6.15)

04cp(t l , t 2 , t 3 , t4)1 ~1234 = , ot l ot2 ot3 ot4 (0,0.0.0)

(1) is readily obtained.

305. By Exercise 241, the density of nX(I) is

( )n-l

fn(x) = 1 - ~ , 0< x < n,

and its characteristic function is

that is, the characteristic function of an exponential with density f(x) = e- X •

306. We have

E(2) = E(Z) = P[(Xj - X)(lj - Y) > 0]

and let p(Xj - X, lj - Y) = p*. Then (see Exercise 220),

( 1 arc sin p* E Zj) = 2" + --n--'

307. Let Xl' X 2 , •.• , Xr be the selected lucky numbers and

The Xi are isonomic random variables with distribution

1 P[Xi = k] =-,

n i= 1,2, ... ,r, k= 1,2, ... ,n.

Solutions

Hence we have

1· n + 1 E(Xi )=- L k=-,

n k=1 2

• 1 1 (n + 1)(2n + 1) E(Xt) = L _k2=_(12+22+"'+n2)= ,

k=1 n n 6

n2 - 1 u2 = Var(XJ = E(Xl) - [E(XJJ2 = -12-'

Therefore we have

E(S) = r(n + 1) r 2'

r

Var(Sr) = L Var(Xi) + r(r - 1) COV(Xi' X) = ru2 + r(r - 1)pu2, i=1

233

(1)

(2)

where p = COV(Xi' Xj) = P(Xi' Xj), the correlation of Xi' lj. Since the joint distribution of Xi' lj is independent of r, taking r = n, relation (2) gives

Var(S.) = varC~ X) = nu2 + n(n - l)pu2 = 0, (3)

because L7=1 Xi = L7=1 i = constant. Hence p = -lj(n - 1) and from (1), (2), and (3) we find that

( ) _ r(n2 - 1) (1 r - 1) VarSr - --- . 12 n - 1

308. The probability generating function of X, because of the independence of the Xi' is

• • p.(t) = n Px,(t) = n (Pit + qi)' (1)

i=1 i=1

Taking logarithms in (1), we obtain

• • log p.(t) = L log(pit + qi) = L log(1 + Pi(t - 1»

i=1 i=1

• (t - 1)2 • = (t - 1) L Pi - L pt + .... i=1 2 i=1

(2)

By hypothesis

as n -+ 00

and the right-hand member of (2) tends to (t - 1).A., that is, p.(t) tends to the probability generating function of the required Poisson.

234 Solutions

309. Putting q = A/n, we have

( n+k-1) k-l( i)Ak( A). P[X = k] = qkp• = n 1 + - , 1 - - , k 1=0 n k. n

and since

(1 +~) ~ 1, as n ~ 00,

it follows that

k = 0,1,2, ....

310. X will have a Gamma distribution and the unconditional distribution of Y is a continuous mixture of the Poisson with parameter A, which follows the Gamma distribution of X (see Exercise 327).

311. The distribution of IX - f.11/(J is exponential with density

f(x) = e- X

and characteristic function (Exercise 202)

cp(t) = (1 - itrl.

Hence

~ Y = ± _I X-"..j_-_f.1_1 (J j=l (J

has characteristic function

CP.(t) = cp·(t) = (1 - it)-·

and (2n/(J) Y has characteristic function (1 - 2iW·, that is, ofax~ •. Finally, from E(X~.) = 2n, it follows that E(Y) = (J. Thus the so-called mean deviation of the sample, that is, Y in the case of a Laplace distribution, gives an unbiased estimator of the standard deviation (J of the distribution.

X 2

312. L _i .!!.. E(Xl) = 1 (WLLN) and S.Jn ~ N(O, 1) (CLT). n

According to Exercise 255(c), L

y" -+ N(O, 1).

Similarly for Z •.

313. The distribution of X(l)"'" X(r), given that X(r+l) = X, is the distribution of the ordered sample of r independent uniform Xl' ... , Xr in (0, x). Moreover, Xl + .,. + Xr = X(l) + ... + X(r) = Y,. and hence the conditional distribution of Y,., given that X(r+l) = x, is the distribution of the sum of observations of a random sample from the uniform in (0, x). Thus, the condi-

Solutions 235

tional distribution of Y,.I X('+l)' given X(,+l) = X, is the distribution of the sum of r independent uniform variables X I lx, ... , X,lx in (0, 1).

314. By induction. For n = 2 we have, by the convolution formula,

if: dy=x,

f2(X) = S:-X dy = 2 - x,

x:s;1

= (2 ~ I)! {(~)x -G)(X - l)}. Suppose the relation holds for n = k, that is,

J;(x) = 1 {(k)Xk-I_(k)(X_l)k-I+ ... }. k (k - I)! 0 1

We shall show that it holds for n = k + 1. Indeed, we have

J;.+l (x) = LX_I

J;.(y)fl (x - y) dy.

For 0 < x < 1 the lower limit of integration is 0, while for k :s; x :s; k + 1 this is equal to x-I; the upper limit is x for J;.(y) > 0 and 0 < y :s; k. Hence the assertion follows.

The second expression of fn(x) follows from the first one if x is replaced by n - x, since if Xl' ... , Xn are uniform in (0,1), then 1 - Xl' ... , 1 - Xn are also uniform in (0, 1) and therefore the distribution of Sn is the same as that ofn - Sn.

315. The probability generating function of Sn is P(t) = P;,(t) where

[ 1 t m In 00 P(t) = (; ) = L tkp[Sn = k].

m - t k=O

From this we find

P[Sn = k] = ~ f (_I)k+i+im(~)( -n .), m i=O I k - ml

(1)

where, in reality, the sum extends only over all i for which the relation k - mi ~ 0 holds.

316. Because of (1) of Exercise 315, the required probability is

P[Sn = k - n] = P[S3 = 10 - 3] = P[S3 = 7]

= ;3 [(_1)7G)( ~3) + (-1)14G)( ~3)J 1 1

= 63 (36 - 9) = s.

236 Solutions

317. Let C l , ... , Cm+l denote the r = m + 1 boxes. Then

r

(a) P[X I = Xl' ..• , Xm = Xm] = L P[Ai' Xl = Xl' ... , Xm = Xm], (1) i=l

where, e.g.,

r(rN - Xo) P[A"X l =Xl> ... 'Xm=xm]= m rxo-rN-l, (2)

r(N + 1) TI (N - x;)! i=l

where we put Xo = Xl + ... + X m, since Cr is found empty (event Ar) and Xl' ... , Xm matches in the remaining m boxes requires that N - Xi matches are taken from Ci (i = 1,2, ... , m) before the (N + l)st selection of Ci (i = 1,2, ... , m) (negative multinomial, see (s.7» and

f .. 1 P (0 selectton of Ci III every draw) = -,

r i = 1,2, ... , r.

But, by symmetry, the r probabilities in (1) are equal and therefore

P[XI = Xl' ... , Xm = Xm]

= rP[Ak , Xl = Xl' ... , Xm = Xm]

r(rN - XO) ____ m ____ rXo-rN, Xi = 0, 1, ... , N. r(N + 1) TI (N - Xi)!

i=l

(b) Now N - Xi matches must be drawn from Ci before the Nth selection of Cr. Thus

(Here Xi = 1,2, ... , N (i = 1,2, ... , m).)

318. The required probability, sayan is equal to

N N ar = L ... L P[XI = Xl'· .. , Xm = Xm],

xI:;:1 xm=l

where, putting Xo = Xl + ... + Xm , we have (cf. Exercise 317)

r(rN - xo) m pf-x ,

P[XI = Xl' ..• , Xm = Xm] = r(N) I] (N - X;)!

In the special case r = 2, az can be written as an incomplete B function (see Exercise 329):

r(2N) f P2 a = -- X N- l (1 - X)N-l dx == I (N N)

2 r(N) 0 pz ' •

Solutions 237

319. By definition of independence the conditions are necessary. That they are sufficient follows from the fact that the second one gives

P[XI = Xl' X 2 = X2] = P[XI = Xl]' P[X2 = x 2]

The first one gives

for all (Xl' x 2 ). (1)

P[XI = Xl' X 2 = X2, X3 = X3] P [X 3 = Xl' X 2 = X 2, Xl = Xl] = ---=---==-::----=--=-----=-=-=----=--:=----=-=

P[XI = Xl' X 2 = X2]

= P[X3 = X3].

This with (1) gives the condition of complete independence

P[XI = Xl' X 2 = X2, X3 = X3] = P[XI = XI]P[X2 = X2]P[X3 = X3]

for all (Xl' X2, X3)' For n variables Xl' ... , Xn, the conditions become

P[Xi = x;lXj = Xj' j = 1,2, ... , i-I] = P[Xi = xJ, i = 1,2, ... , n.

320. E(S;;I) exists because P[Sn- 1 < Xli] = 1. Since E(XJSn) = a (i = 1, 2, ... , n), we have

( n X.) n (X) (X) 1 = E L ~ = L E ~ = nE ~ .

i=l Sn i=1 Sn Sn Hence,

a = E(::) = n- l .

321. By Exercise 320, we have

( Sm) (Xi) m E Sn = mE Sn = ~ for m:S;; n,

while, for m z n, since S;;l and Xn+1' •.• , Xm are independent, we have

E(Sm) = E(I + ~. f Xi) = 1 + (m - n)E(Xi) Sn Sn l=n+1 Sn

= 1 + (m - n)E(X;)E GJ. 322. (a) The required probability P is given by

00 00 1 (2)n-1 P = L P(An) = L - - = 1,

n=l n=l 3 3

where An is the event that it gets stuck exactly after n steps, i.e., at time t = n (geometric distribution with P = 1/3).

(b) Let Pa, Pb, Pc be the probabilities that the caterpillar eventually gets stuck coming from A, B, C, respectively. Then we have

Pa + Pb + Pc = 1,

238 Solutions

since the caterpillar goes in one step to B with probability 1/3 and having gone there it has the same probability of getting stuck (coming from B) as if it had started at A. Hence

Pb = Pc = 1/5, Pa = 3/5.

323. (a) (i) E(Xn) = naE(cos 0i) = na(2n)-1 I2l< cos ° dO = 0.

Similarly, E(Y,,) = 0,

Var(Xn) = E(X;) = na2E(cos 2 0i) + n(n - l)a2E(cos 0i cos OJ)

= na2 E(cos2 0;),

since the independence of 0i gives

E(cos 0i cos 0) = E(cos Oi)E(cos OJ) = 0, i ¥= j.

From (1), we find

Var(Xn) = tna2 •

(ii) From (1) and (2) we find

Similarly,

E(R;) = na2•

(b) Cov(Xn' Yn) = E(Xn Y,,) = na2 [E(sin 0i cos 0i)

+ (n - I)E(sin 0i cos OJ)] = 0.

(1)

(2)

Hence the coordinates X n , Yn of the position of the flea after n jumps are uncorrelated. However, we shall show that they are not independent. Let n = 1. Then we have, e.g.,

that is, Xl' Y1 are not independent. Similarly, by induction, the dependence of X n , Y" is shown.

(c) By the CLT we have

L ( na2) Xn -+N 0'2 '

Hence X n , Y" are asymptotically normal and independent as well, because they are uncorrelated. Then, the asymptotic (as n -+ (0) distribution of Rn, by Exercise 221, has the density

I" () _ 2r -"Ina' JR r - -2e .

n na

From the above relation we conclude that the expected distance of the flea

Solutions 239

from the origin (for large n) is

E(Rn) = fo~.

324. Let ~j = (Xl' "., X j - l , X j +1, "., Xn) (j = 1,2, .'" n). The density of ~j is

where

Hence, any subvector of ~j and therefore any subset of Xl"'" Xn has the multinormal distribution with independent components. Yet, X = (X I' ... , Xn) with density f(x) does not have an n-dimensional normal distribution and X I,

X 2 , ••• , Xn are not completely independent. Remark. If X I' ... , Xn are uncorrelated, that is, P(Xi' X) = 0 (i # j), and all

the marginal distributions are normal, then X I, X 2, ... , Xn are jointly normal if and only if XI' X 2 , "., Xn are completely independent.

325. (a) Because of the uniform distribution of the angle (J, Exercise 288, the mean number of crossings of the needle by horizontal or vertical lines is the same, and equal to the probability of crossing one of the parallels, i.e., 2f1/na. But the mean number of crossings of the lines of the table is equal to the sum of the expected crossings ofthe horizontal and vertical lines, i.e., 4f1/na. Thus for a needle having length equal to the side of the squares, the mean number of crossings is 4/n ~ 1.27.

(b) If the needle has arbitrary length I, let us imagine that this is divided into n equal pieces of length less than 2a. Each of the pieces separately gives mean number of crossings equal to 21/nna. But the mean of the sum is equal to the sum of the means and hence the required mean number of crossings by the needle is 21/na. Throwing the n pieces of the needle separately instead of throwing the needle as a solid body does not influence the mean.

Note. Use the result for the estimation of the value of n, e.g., throwing a toothpick on a flat surface divided into little squares.

326. (Cacoullos, JASA, 1965) (a) We have

G(u) = P [X - Y ] ~ uJ = 2nr 21 2 f (xy)(n/2 l-l e-(x+Yl/2 dx dy, 2JXY (n/) (x-Yl=2uJ~Y

240 Solutions

where putting first x = p COS2 0, Y = P sin2 0, and then w = cot 20, we have by the Legendre duplication formula,

22n-1 r(n)r(n + t) = fir(2n),

r(~)fU dw

G(u) = fir(~) -00 (1 + w2)(n+1)/2'

Hence we obtain the required distribution of Z. Since Z can be written as

we have the relation between tn and Fn,n' (b) For 0 < IX < 0.5 we have (for Fn,n = 1, tn = Z = 0)

P[O < tn < tn(lX)] = P[l < Fn,n < Fa],

where Fa is the solution F > 1 of the equation

tn(lX) = f( JF - ft). that is,

1 + t;(IX). n

From (1) and from the relation P[Fn,n ::; 1] = 1/2, we have

P[Fn,n ::; Fa] = 1 - IX

and hence Fn,n(lX) = Fa. For 0.5 < IX < 1, as is well known, we have

tn(lX) = - tn(1 - IX) and Fn,n(lX) = Fn~~(1 - IX).

(1)

(2)

327. According to the generalized theorem of total probability, we have

Pk = e- A_ f(.A.) d.A. = _a_ .A.k+p-1e-(a+1)A d.A. JOO.A.k P Joo o k! r(p)k! 0

aP r(k + p) (-p) k

= r(p)k! (a + l)k+P = k (- p) qP,

where we made use of r(x + 1) = xr(x) (x > 0). Hence the negative binomial distribution was obtained as a mixture of the Poisson and Gamma distributions.

Solutions

328. By definition we have

and since

E[Var(XI Y)] = E[E(X21 Y) - {E(XI y)}2]

= E[E(X2IY)] - E[E(Xly)]2,

Var[E(XI Y)] = E[E(XI y)]2 - {E[E(XI y)}2,

E[E(XI Y)] = E(X),

Adding (1) and (2), we have

E(X2) - {E(X)}2 = Var(X).

329. The distribution function of X(k), by Exercise 241, is

Fk(p) = P[X(k) :::; p] = f: !xk(x) dx

= 1 fP Xk- 1 (1 - x)n-k. B(k, n - k + 1) 0

241

(1)

(2)

(1)

On the other hand, we have that X(k) :::; P if and only if at least k of the Xi are less than p. Since the Xi are independent and P[Xi :::; p] = p, we have

(2)

From (1) and (2) the assertion follows.

Bibliography

1. Cacoullos, T. (1969). A Course in Probability Theory (in Greek). LeousisMastroyiannis, Athens.

2. Cramer, H. (1946). Mathematical Methods of Statistics. Princeton University Press, Princeton, NJ.

3. Dacunha-Castelle, D., Revuz, D., and Schreiber, M. (1970). Recueil de Problemes de Calcul des Probabilites. Masson Cie, Paris.

4. Feller, W. (1957). Introduction to Probability Theory and its Applications, Vol. I. Wiley, New York.

5. Fisz, M. (1963). Probability Theory and Mathematical Statistics. Wiley, New York. 6. Gnedenko, B.V. (1962). The Theory of Probability. Chelsea, New York. 7. Mosteller, F. (1965). Fifty Challenging Problems in Probability. Addison-Wesley,

Boston, MA. 8. Parzen, E. (1960). Modern Probability Theory and its Applications. Wiley, New

York. 9. Rahman, N.A. (1967). Exercises in Probability and Statistics. Griffin, London.

10. Sveshnikov, A.A. (1968). Problems in Probability Theory. Saunders, Philadelphia.

Index

Addition theorem for probabilities 4 Arc sine distribution 58, 59 Axioms of probability theory 4

Banach match box problem 80 Bayes formula 5,41 Behrens-Fisher problem 61 Bernoulli, Daniel 93 Bernoulli trials 18, 65, 67, 79, 105, 115,

118, 120 Bernstein polynomial III Bertrand's paradox 36 Beta distribution 19

characteristic function of 105 in evaluating the binomial

distribution function 82 Beta function 19 Billiards 104 Binomial coefficients 5, 8, 9 Binomial distribution 18

factorial (probability) generating function 51, 121

normal approximation of 20,21,45, 53

reproductive property of 60 Birthdays 11, 12, 35, 88 Bochner-Khintchine theorem 117 Borel, E. 3, 65 Borel-Cantelli lemma 115,117 Bose-Einstein statistics 12 Bridge 11,30 Buffon's needle problem 76

Cartesian product 6 Cauchy distribution 19, 58

characteristic function of 197 infinite divisibility of 54 mixture of 197

Central limit theorems 66-67 Central moments 25 Characteristic function 48, 50, 117

continuity theorem for 50, 51, 217 inversion formula for 49,51

Chebyshev's inequality. See Inequality Chi-squared (X 2 -) distribution 19

approximation by normal 111 approximation of noncentral X2 52 generalized X2-distribution 96

Circular normal distribution 59, 62 Combinatorics 5, 8 Compound distribution 47,52, 111

Poisson 48 Conditional

characteristic function 54 density 40 distribution 40 probability 4

Condorcet, Marquis De 93 Convergence 63-69, 109-114

in distribution (law) 64, 111-114 in probability 63, 113 in quadratic mean 63, 113 with probability one (strongly, almost

surely) 63, 64, 113, 117 Convex function 70, 75

246

Convolution 56 binomial 60 exponential 58 Poisson 60 uniform 59, 79, 120

Correlation coefficient 41 Coupon collector's problem 35, 52 Covariance 41

matrix 42 Cramer, H. 49, 50, 243 Craps, game of 33 Cumulant generating function 49 Cumulants 49, 53 Cumulative distribution function. See

Distribution function

DeMoivre-Laplace theorem 67, 115 Difference equations 72-73,76-77, 121 Difference of sets 3

probability of 4 symmetric 10

Dirac b-function 103 Dirichlet distribution 43, 44, 60, 97 Dispersion matrix. See Covariance

matrix Distribution

continuous 17, 39, 43 discrete 17, 39, 42 singular 42

Distribution function 17, 39, 40 Distribution of functions of random

variables 22, 23, 55-62, 106-109 Dodge, H.F. 179

Elliptic symmetry 109 Entropy 100 Erehwon 35 Erlang distribution 90,91 Euler's constant 185

function 33 problem 93

Event impossible sure 1

Expected value 24,28,41 approximate 98 geometric interpretation 27 limiting 110

Exponential distribution 19 as a limit of extremes 122 cumulants of 53

F -distribution 20, 109 relation to t 81

Factorial moment 25, 26 generating function 25

Index

Feller, W. 33,67,80,90, 115, 120,200, 243

Finite population correction 163

Gamma distribution 19,53 approximation by normal 69 characteristic function of 53 in evaluating the Poisson distribution

function 91 infinite divisibility of 54 mixing for a Poisson 79, 82

Gamma function 19 Gauss-Laplace distribution. See

Normal distribution Geiger-Muller counter 109 Generating function 47, 50, 10 1-105

factorial moment 25,47 moment 26, 48 probability 25,47

Geometric distribution 18 Geometric probabilities 72, 75, 76,

90 Gnedenko, B.Y. 65, 117, 120, 243

Hamel's functional equation 105 Huyghens, C. 14,92,200 Hypergeometric distribution 6, 18,28,

60 double 43 multiple 44

Independence, of normal sample mean and variance 61

Independent events 5 completely 5

Independent experiments 5 Independent random variables 42,51

pairwise 44 Inequality 70-71, 74-75

Berge 75 Bonferoni 10 Cantelli 74 Cauchy-Schwarz 26, 71 Chebyshev (Binayme) 64, 111 Gauss 74 Hajek-Renyi 66 Holder 71 Jensen 70

for matrix functions 75 Kolmogrov 66, 119 Markov 64 Minkowski (triangle) 71

Index

Infinitely divisible distribution 50 Iterated logarithm, law of 115

Khinchine's, law of the iterated logarithm 115

weak law of large numbers 65 Kolmogorov's, probability axioms 4

three-series theorem 115, 117 zero-one law 116

Laplace, P.S. 4, 93 Laplace distribution 19, 79

characteristic function of 102 moments of 102

Laws oflarge numbers strong 65, 67 weak 64,67

Legendre duplication formula for the r -function 240

Levy, P. 50, 66 Lindeberg-Feller theorem 67,217 Lognormal distribution 23, 60 Lyapunov's theorem 66

Marginal distribution 40 Mathematical induction 7 Maxwell-Boltzmann statistics 90 Mean. See Expected value Median 25, 26 Mixed distribution 29 Mixture of distributions 53 Mode 45,74 Moment-generating function 26 Moments 25

about the origin 25 central 25 condition for existence of 28 mixed 41,51

Monte Carlo method 114 Montmort, P.R. 89, 93 Mosteller, F. 243 Multinomial coefficients 6 Multinomial distribution 43

conditional of 44 moments of 54 multiple Poisson limit of 45

Multiplication formula for probabilities 4

Multiplication principle 6 Multivariate distribution 39

characteristic function of 50 probability density of 40 probability function of 39

247

Multivariate normal distribution 43, 44,94,95,97

characteristic function of 103 conditional distributions in 96 distribution of linear functions 62 distribution of quadratic forms 100 marginal distributions of 81,96 moments of 78 probability in ellipsoid 95

Negative binomial distribution 18 as a compound Poisson 34 as a mixture of Poisson and gamma

79,82 infinite divisibility of 54

Negative multinomial 43 as a mixture of a multiple Poisson

and a gamma distribution 54 moments of 54

Newton 35, 88 Normal distribution 19

approximations by 66 asymptotic evaluation of the

distribution function of 116 characterization of 105 folded 32 moments of 196

Occupancy number 6, 11 Order statistics 61

from exponential 108 from uniform 79, 82

Orthogonal transformation 53, 108

Paccioli 92 Pap test 15 Pareto distribution 68 Pascal, distribution. See Negative

binomial distribution, points 33, 89 triangle 5, 8

Pepys, Samuel 35 Percentiles 25 Permutations 5

cyclical 127 Poincan':'s formula 4 Poisson distribution 18

approximation by normal 110 approximation of binomial 20, 21,

45,54 approximation of Pascal 102 compound 48 in evaluating the gamma distribution

function 90, 91

248

Poisson distribution (cont.) multiple 45, 54 normal approximation of 217 probability generating function 51 truncated 29

Poisson, S., a theorem of 87 Poker II Polya, G. 93 Probability

conditional 4 density function 17, 40 differential 18 function 17, 39 generating function 25,47, 50 integral transformation 59 space 63

Probable error 95 Pseudorandom numbers 59. See also

Random numbers

Quadratic functions of normal random variables, distribution of 100

Quality control 34, 35 Quantiles 25

Rahman, N.A. 179, 243 Random digits 22, 116 Random numbers 22,59, 114 Random variables 17

complex 48 continuous 17,39,43 discrete 17, 39,42 isonomic 208 orthogonal 46 vector 39

Range of a random sample 61 Rayleigh distribution 205 Rectangular distribution. See Uniform

distribution Regression, function 42

linear 108 Relations between modes of convergence

64 Round-off error 69

Index

Runs, maximum length 118

Sampling with replacement 13, 28, 31 without replacement 6, 13,28,31,32

Sigma (Borel) field 3 Singular distribution 42 Smith, John 88 Spherical coordinates 97 Spherically symmetric distribution 53,

96-100 Spinoza, B. 92 Standard deviation 25 Standard normal distribution 29

asymptotic evaluation of 116 Strong law oflarge numbers 65, 67 Student's t-distribution 19

elliptic 109 relation to F 81

Tolerance limits 60 Total probability formula 5, 32, 40 Triangular distribution 21 Trinomial distribution 188 Truncated

Poisson distribution 29 random variable 1I5, 119

Uniform distribution 19 in an ellipse 96 in a simplex 44,100 multivariate 43 n-fold convolution of 79 on the unit sphere 98

Uniqueness theorem for characteristic functions 49

Variance 25 approximation of 98 of a product of random variables 89

Venn diagram 7 .

Weierstrass approximation theorem III

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