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Unformatted text preview: Instructor's Solutions Manual THIRD EDITION Fundamentals of P ROBABILITY
WITH STOCHASTIC PROCESSES Saeed Ghahramani Instructor's Solutions Manual Third Edition Fundamentals of
With ProbabilitY
Stochastic Processes SAEED GHAHRAMANI
Western New England College Upper Saddle River, New Jersey 07458 C ontents
1
1.2 1.4 1.7 Axioms of Probability 1 Sample Space and Events 1 Basic Theorems 2 Random Selection of Points from Intervals Review Problems 9 7 2
2.2 2.3 2.4 2.5 Combinatorial Methods 13 Counting Principle 13 Permutations 16 Combinations 18 Stirling' Formula 31 Review Problems 31 3 Conditional Probability and Independence 35 3.1 Conditional Probability 35 3.2 Law of Multiplication 39 3.3 Law of Total Probability 41 3.4 Bayes' Formula 46 3.5 Independence 48 3.6 Applications of Probability to Genetics Review Problems 59 56 4 Distribution Functions and Discrete Random Variables 63 4.2 Distribution Functions 63 4.3 Discrete Random Variables 66 4.4 Expectations of Discrete Random Variables 71 4.5 Variances and Moments of Discrete Random Variables 4.6 Standardized Random Variables 83 Review Problems 83 77 iv Contents 5
5.1 5.2 5.3 Special Discrete Distributions
87 87 Bernoulli and Binomial Random Variables Poisson Random Variable 94 Other Discrete Random Variables 99 Review Problems 106 6
6.1 6.2 6.3 Continuous Random Variables 111 Probability Density Functions 111 Density Function of a Function of a Random Variable Expectations and Variances 116 Review Problems 123 113 7
7.1 7.2 7.3 7.4 7.5 7.6 Special Continuous Distributions 126 Uniform Random Variable 126 Normal Random Variable 131 Exponential Random Variables 139 Gamma Distribution 144 Beta Distribution 147 Survival Analysis and Hazard Function Review Problems 153 152 8
8.1 8.2 8.3 8.4 Bivariate Distributions
157 157 Joint Distribution of Two Random Variables Independent Random Variables 166 Conditional Distributions 174 Transformations of Two Random Variables Review Problems 191 183 9
9.1 9.2 9.3 Multivariate Distributions
200 200 Joint Distribution of n > 2 Random Variables Order Statistics 210 Multinomial Distributions 215 Review Problems 218 Contents v 10
10.1 10.2 10.3 10.4 10.5 More Expectations and Variances
222 222 Expected Values of Sums of Random Variables Covariance 227 Correlation 237 Conditioning on Random Variables 239 Bivariate Normal Distribution 251 Review Problems 254 11
11.1 11.2 11.3 11.4 11.5 Sums of Independent Random Variables and Limit Theorems 261 MomentGenerating Functions 261 Sums of Independent Random Variables 269 Markov and Chebyshev Inequalities 274 Laws of Large Numbers 278 Central Limit Theorem 282 Review Problems 287 12
12.2 12.3 12.4 12.5 Stochastic Processes 291 More on Poisson Processes 291 Markov Chains 296 ContinuousTime Markov Chains Brownian Motion 326 Review Problems 331 315 Chapter 1 A xioms
1.2 of Probability SAMPLE SPACE AND EVENTS 1. For 1 i, j 3, by (i, j ) we mean that Vann's card number is i, and Paul's card number is
j . Clearly, A = (1, 2), (1, 3), (2, 3) and B = (2, 1), (3, 1), (3, 2) . (a) Since A B = , the events A and B are mutually exclusive. (b) None of (1, 1), (2, 2), (3, 3) belongs to A B. Hence A B not being the sample space shows that A and B are not complements of one another. 2. S = {RRR, RRB, RBR, RBB, BRR, BRB, BBR, BBB}. 3. {x : 0 < x < 20}; {1, 2, 3, . . . , 19}. 4. Denote the dictionaries by d1 , d2 ; the third book by a. The answers are
{d1 d2 a, d1 ad2 , d2 d1 a, d2 ad1 , ad1 d2 , ad2 d1 } and {d1 d2 a, ad1 d2 }. 5. EF : One 1 and one even.
E c F : One 1 and one odd. E c F c : Both even or both belong to {3, 5}. 6. S = {QQ, QN, QP , QD, DN, DP , N P , N N, P P }. (a) {QP }; (b) {DN, DP , N N}; (c) .
1 3 1 3 7. S = x : 7 x 9 1 ; x : 7 x 7 4 x : 7 4 x 8 4 x : 8 4 x 9 1 . 6 6 8. E F G = G: If E or F occurs, then G occurs.
EF G = G: If G occurs, then E and F occur. 9. For 1 i 3, 1 j 3, by ai bj we mean passenger a gets off at hotel i and passenger b gets off at hotel j . The answers are {ai bj : 1 i 3, 1 j 3} and {a1 b1 , a2 b2 , a3 b3 }, respectively. 10. (a) (E F )(F G) = (F E)(F G) = F EG. 2 Chapter 1 Axioms of Probability (b) Using part (a), we have (E F )(E c F )(E F c ) = (F EE c )(E F c ) = F (E F c ) = F E F F c = F E. 11. (a) AB c C c ; (b) A B C; (c) Ac B c C c ; (d) ABC c AB c C Ac BC; (e) AB c C c Ac B c C Ac BC c ; the sample space, as well. Thus (f) (A  B) (B  A) = (A B)  AB. 12. If B = , the relation is obvious. If the relation is true for every event A, then it is true for S,
S = (B S c ) (B c S) = B c = B c , showing that B = . 13. Parts (a) and (d) are obviously true; part (c) is true by DeMorgan's law; part (b) is false: throw
a foursided die; let F = {1, 2, 3}, G = {2, 3, 4}, E = {1, 4}. 14. (a) n=1 An ; (b) 37 n=1 An . 15. Straightforward. 16. Straightforward. 17. Straightforward. 18. Let a1 , a2 , and a3 be the first, the second, and the third volumes of the dictionary. Let a4 , a5 ,
a6 , and a7 be the remaining books. Let A = {a1 , a2 , . . . , a7 }; the answers are S = x1 x2 x3 x4 x5 x6 x7 : xi A, 1 i 7, and xi = xj if i = j and x1 x2 x3 x4 x5 x6 x7 S : xi xi+1 xi+2 = a1 a2 a3 for some i, 1 i 5 , respectively. 19. m=1 n=m An .
n1 i=1 20. Let B1 = A1 , B2 = A2  A1 , B3 = A3  (A1 A2 ), . . . , Bn = An  Ai , . . . . 1.4 BASIC THEOREMS 1. No; P (sum 11) = 2/36 while P (sum 12) = 1/36. 2. 0.33 + 0.07 = 0.40. Section 1.4 Basic Theorems 3 3. Let E be the event that an earthquake will damage the structure next year. Let H be the event that a hurricane will damage the structure next year. We are given that P (E) = 0.015, P (H ) = 0.025, and P (EH ) = 0.0073. Since P (E H ) = P (E) + P (H )  P (EH ) = 0.015 + 0.025  0.0073 = 0.0327, the probability that next year the structure will be damaged by an earthquake and/or a hurricane is 0.0327. The probability that it is not damaged by any of the two natural disasters is 0.9673. 4. Let A be the event of a randomly selected driver having an accident during the next 12 months.
Let B be the event that the person is male. By Theorem 1.7, the desired probability is P (A) = P (AB) + P (AB c ) = 0.12 + 0.06 = 0.18. 5. Let A be the event that a randomly selected investor invests in traditional annuities. Let B be
the event that he or she invests in the stock market. Then P (A) = 0.75, P (B) = 0.45, and P (A B) = 0.85. Since, P (AB) = P (A) + P (B)  P (A B) = 0.75 + 0.45  0.85 = 0.35, 35% invest in both stock market and traditional annuities. 6. The probability that the first horse wins is 2/7. The probability that the second horse wins
is 3/10. Since the events that the first horse wins and the second horse wins are mutually exclusive, the probability that either the first horse or the second horse will win is 2 3 41 + = . 7 10 70 7. In point of fact Rockford was right the first time. The reporter is assuming that both autopsies
are performed by a given doctor. The probability that both autopsies are performed by the same doctorwhichever doctor it may beis 1/2. Let AB represent the case in which Dr. A performs the first autopsy and Dr. B performs the second autopsy, with similar representations for other cases. Then the sample space is S = {AA, AB, BA, BB}. The event that both autopsies are performed by the same doctor is {AA, BB}. Clearly, the probability of this event is 2/4=1/2. 8. Let m be the probability that Marty will be hired. Then m + (m + 0.2) + m = 1 which gives
m = 8/30; so the answer is 8/30 + 2/10 = 7/15. 9. Let s be the probability that the patient selected at random suffers from schizophrenia. Then
s + s/3 + s/2 + s/10 = 1 which gives s = 15/29. 10. P (A B) 1 implies that P (A) + P (B)  P (AB) 1. 11. (a) 2/52 + 2/52 = 1/13;
(b) 12/52 + 26/52  6/53 = 8/13; (c) 1  (16/52) = 9/13. 4 Chapter 1 Axioms of Probability 12. (a) False; toss a die and let A = {1, 2}, B = {2, 3}, and C = {1, 3}.
(b) False; toss a die and let A = {1, 2, 3, 4}, B = {1, 2, 3, 4, 5}, C = {1, 2, 3, 4, 5, 6}. 13. A simple Venn diagram shows that the answers are 65% and 10%, respectively. 14. Applying Theorem 1.6 twice, we have
P (A B C) = P (A B) + P (C)  P (A B)C = P (A) + P (B)  P (AB) + P (C)  P (AC BC) = P (A) + P (B)  P (AB) + P (C)  P (AC)  P (BC) + P (ABC) = P (A) + P (B) + P (C)  P (AB)  P (AC)  P (BC) + P (ABC). 15. Using Theorem 1.5, we have that the desired probability is
P (AB  ABC) + P (AC  ABC) + P (BC  ABC) = P (AB)  P (ABC) + P (AC)  P (ABC) + P (BC)  P (ABC) = P (AB) + P (AC) + P (BC)  3P (ABC). 16. 7/11. 17.
n i=1 pij . 18. Let M and F denote the events that the randomly selected student earned an A on the midterm
exam and an A on the final exam, respectively. Then P (MF ) = P (M) + P (F )  P (M F ), where P (M) = 17/33, P (F ) = 14/33, and by DeMorgan's law, P (M F ) = 1  P (M c F c ) = 1  Therefore, P (MF ) = 22 11 = . 33 33 17 14 22 3 +  = . 33 33 33 11 19. A Venn diagram shows that the answers are 1/8, 5/24, and 5/24, respectively. 20. The equation has real roots if and only if b2 4c. From the 36 possible outcomes for (b, c),
in the following 19 cases we have that b2 4c: (2, 1), (3, 1), (3, 2), (4, 1), . . . , (4, 4), (5, 1), . . . , (5, 6), (6, 1), . . . , (6, 6). Therefore, the answer is 19/36. if and only if it is neither divisible by 3 nor by 7. Let A and B be the events that the outcome 21. The only prime divisors of 63 are 3 and 7. Thus the number selected is relatively prime to 63 Section 1.4 Basic Theorems 5 is divisible by 3 and 7, respectively. The desired quantity is P (Ac B c ) = 1  P (A B) = 1  P (A)  P (B) + P (AB) =1 21 9 3 4  + = . 63 63 63 7 22. Let T and F be the events that the number selected is divisible by 3 and 5, respectively.
(a) The desired quantity is the probability of the event T F c : P (T F c ) = P (T )  P (T F ) = 333 66 267  = . 1000 1000 1000 (b) The desired quantity is the probability of the event T c F c : P (T c F c ) = 1  P (T F ) = 1  P (T )  P (F ) + P (T F ) =1 200 66 533 333  + = . 1000 1000 1000 1000 23. (Draw a Venn diagram.) From the data we have that 55% passed all three, 5% passed calculus
and physics but not chemistry, and 20% passed calculus and chemistry but not physics. So at least (55 + 5 + 20)% = 80% must have passed calculus. This number is greater than the given 78% for all of the students who passed calculus. Therefore, the data is incorrect. 24. By symmetry the answer is 1/4. 25. Let A, B, and C be the events that the number selected is divisible by 4, 5, and 7, respectively.
We are interested in P (AB c C c ). Now AB c C c = A  A(B C) and A(B C) A. So by Theorem 1.5, P (AB c C c ) = P (A)  P A(B C) = P (A)  P (AB AC) = P (A)  P (AB)  P (AC) + P (ABC) = 50 35 7 172 250   + = . 1000 1000 1000 1000 1000 26. A Venn diagram shows that the answer is 0.36. 27. Let A be the event that the first number selected is greater than the second; let B be the
event that the second number selected is greater than the first; and let C be the event that the two numbers selected are equal. Then P (A) + P (B) + P (C) = 1, P (A) = P (B), and P (C) = 1/100. These give P (A) = 99/200. i=1 28. Let B1 = A1 , and for n 2, Bn = An 
mutually exclusive events and Ai = i=1 n1 i=1 Ai . Then {B1 , B2 , . . . } is a sequence of Bi . Hence 6 Chapter 1 Axioms of Probability P
n=1 An = P
n=1 Bn =
n=1 P (Bn ) n=1 P (An ), since Bn An , n 1. 29. By Boole's inequality (Exercise 28), P
n=1 An = 1  P
n=1 Ac 1  n n=1 P (Ac ). n 30. She is wrong! Consider the next 50 flights. For 1 i 50, let Ai be the event that the ith
mission will be completed without mishap. Then
50 i=1 Ai is the event that all of the next 50 50 missions will be completed successfully. We will show that P i=1 Ai > 0. This proves that Mia is wrong. Note that the probability of the simultaneous occurrence of any number of Ac 's is nonzero. Furthermore, consider any set E consisting of n (n 50) of the Ac 's. It is i i reasonable to assume that the probability of the simultaneous occurrence of the events of E is strictly less than the probability of the simultaneous occurrence of the events of any subset of E. Using these facts, it is straightforward to conclude from the inclusionexclusion principle that, 50 50 50 1 P Ac < P (Ac ) = = 1. i i 50 i=1 i=1 i=1 Thus, by DeMorgan's law,
50 50 P
i=1 Ai = 1  P
i=1 Ac > 1  1 = 0. i 31. Q satisfies Axioms 1 and 2, but not necessarily Axiom 3. So it is not, in general, a probability
on S. Let S = {1, 2, 3, }. Let P {1} = P {2} = P {3} = 1/3. Then Q {1} = Q {2} = 2 1/9, whereas Q {1, 2} = P {1, 2} = 4/9. Therefore, Q {1, 2, } = Q {1} + Q {2} . R is not a probability on S because it does not satisfy Axiom 2; that is, R(S) = 1. 32. Let BRB mean that a blue hat is placed on the first player's head, a red hat on the second
player's head, and a blue hat on the third player's head, with similar representations for other cases. The sample space is S = {BBB, BRB, BBR, BRR, RRR, RRB, RBR, RBB}. This shows that the probability that two of the players will have hats of the same color and the third player's hat will be of the opposite color is 6/8 = 3/4. The following improvement, Section 1.7 Random Selection of Points from Intervals 7 based on this observation, explained by Sara Robinson in Tuesday, April 10, 2001 issue of the New York Times, is due to Professor Elwyn Berlekamp of the University of California at Berkeley.
Threefourths of the time, two of the players will have hats of the same color and the third player's hat will be the opposite color. The group can win every time this happens by using the following strategy: Once the game starts, each player looks at the other two players' hats. If the two hats are different colors, he [or she] passes. If they are the same color, the player guesses his [or her] own hat is the opposite color. This way, every time the hat colors are distributed two and one, one player will guess correctly and the others will pass, and the group will win the game. When all the hats are the same color, however, all three players will guess incorrectly and the group will lose. 1.7 1. 2. RANDOM SELECTION OF POINTS FROM INTERVALS
2 30  10 = . 30  0 3 0.0635  0.04 = 0.294. 0.12  0.04 A = (0, 1)  {1/2}. A is not the sample space but P (A) = 1. 1 (b) False; in the same experiment P {1/2} = 0 while { 2 } = . 3. (a) False; in the experiment of choosing a point at random from the interval (0, 1), let 4. P (A B) P (A) = 1, so P (A B) = 1. This gives
P (AB) = P (A) + P (B)  P (A B) = 1 + 1  1 = 1. 5. The answer is
P {1, 2, . . . , 1999} = 1999 1999 P {i} =
i=1 i=1 0 = 0. 6. For i = 0, 1, 2, . . . , 9, the probability that i appears as the first digit of the decimal represeni i+1 tation of the selected point is the probability that the point falls into the interval , . 10 10 Therefore, it equals i+1 i  1 10 10 . = 10 10 This shows that all numerals are equally likely to appear as the first digit of the decimal representation of the selected point. 8 Chapter 1 Axioms of Probability 7. No, it is not. Let S = {w1 , w2 , . . . }. Suppose that for some p > 0, P {wi } = p, i = 1, 2,
. . . . Then, by Axioms 2 and 3, i=1 p = 1. This is impossible. 8. Use induction. For n = 1, the theorem is trivial. Exercise 4 proves the theorem for n = 2.
Suppose that the theorem is true for n. We show it for n + 1, P (A1 A2 An An+1 ) = P (A1 A2 An ) + P (An+1 )  P (A1 A2 An An+1 ) = 1 + 1  1 = 1, where P (A1 A2 An ) = 1 is true by the induction hypothesis, and P (A1 A2 An An+1 ) P (An+1 ) = 1, implies that P (A1 A2 An An+1 ) = 1. 9. (a) 1 1 1 1 1 1 1 1 1 Clearly,  , + . If x  , + , then, for all n 1, 2 n=1 2 2n 2 2n 2 2n 2 2n n=1 1 1 1 1  <x< + . 2 2n 2 2n Letting n , we obtain 1/2 x 1/2; thus x = 1/2. (b) Let An be the event that the point selected at random is in 1 1 1 1  , + ; then 2 2n 2 2n A1 A2 A3 An An+1 . Since P (An ) = 1 , by the continuity property of the probability function, n P {1/2} = lim P (An ) = 0.
n 10. The set of rational numbers is countable. Let Q = {r1 , r2 , r3 , . . . } be the set of rational
numbers in (0, 1). Then P (Q) = P {r1 , r2 , r3 , . . . } =
i=1 P {ri } = 0. Let I be the set of irrational numbers in (0, 1); then P (I) = P (Qc ) = 1  P (Q) = 1. 11. For i = 0, 1, 2, . . . , 9, the probability that i appears as the nth digit of the decimal representation of the selected point is the probability that the point falls into the following subset of (0, 1):
10n1 1 m=0 10m + i 10m + i + 1 , . 10n 10n Chapter 1 Review Problems 9 Since the intervals in this union are mutually exclusive, the probability that the point falls into this subset is 10m + i + 1 10m + i  10n1 1 1 1 10n 10n . = 10n1 n = 10 10 10 m=0 This shows that all numerals are equally likely to appear as the nth digit of the decimal representation of the selected point. 12. P (Bm ) n=m P (An ). Since n=1 P (An ) converges, m lim P (Bm ) lim m P (An ) = 0.
n=m This gives limm P (Bm ) = 0. Therefore, B1 B2 B3 Bm Bm+1 implies that P
m=1 n=m An = P
m=1 Bm = lim P (Bm ) = 0.
m 13. In the experiment of choosing a random point from (0, 1), let Et = (0, 1)  {t}, for 0 < t < 1.
Then P (Et ) = 1 for all t, while P
t(0,1) Et = P () = 0. 14. Clearly rn (n , n ). By the geometric series theorem, (n  n ) =
n=1 n=1 2n+1 = 1 4 1 1 2 = < . 2 REVIEW PROBLEMS FOR CHAPTER 1 1.
3.25  2 = 0.54. 4.3  2 2. We have that
S= , {1} , , {2} , , {1, 2} , {1}, {2} , {1}, {1, 2} , {2}, {1, 2} . 10 Chapter 1 Axioms of Probability The desired events are (a) (c) , {1} , , {2} , , {1, 2} , {1}, {2} ; (b) , {1, 2} , {1}, {2} ; , {1} , , {2} , , {1, 2} , {1}, {1, 2} , {2}, {1, 2} . 3. Since A B, we have that B c Ac . This implies that (a) is false but (b) is true. 4. In the experiment of tossing a die let A = {1, 3, 5} and B = {5}; then both (a) and (b) are
false. 5. We may define a sample space S as follows.
S = x1 x2 xn : n 1, xi {H,T}; xi = xi+1 , 1 i n  2; xn1 = xn . 6. A venn diagram shows that 18 are neither male nor for surgery. 7. We have that ABC BC, so P (ABC) P (BC) and hence P (BC)  P (ABC) 0. This
and the following give the result. P (A B C) = P (A) + P (B) + P (C)  P (AB) + P (AC) + P (BC)  P (ABC) P (A) + P (B) + P (C). 8. If P (AB) = P (AC) = P (BC) = 0, then P (ABC) = 0 since ABC AB. These imply that
P (A B C) = P (A) + P (B) + P (C)  P (AB)  P (AC)  P (BC) + P (ABC) = P (A) + P (B) + P (C). Now suppose that P (A B C) = P (A) + P (B) + P (C). This relation implies that P (AB) + P (BC) + P (AC)  P (ABC) = 0. (1) Since P (AC)  P (ABC) 0 we have that the sum of three nonnegative quantities is 0; so each of them is 0. That is, P (AB) = 0, Now rewriting (1) as P (AB) + P (AC) + P (BC)  P (ABC) = 0, the same argument implies that P (AB) = 0, Comparing (2) and (3) we have P (AB) = P (AC) = P (BC) = 0. P (AC) = 0, P (BC) = P (ABC). (3) P (BC) = 0, P (AC) = P (ABC). (2) Chapter 1 Review Problems 11 9. Let W be the event that a randomly selected person from this community drinks or serves white wine. Let R be the event that she or he drinks or serves red wine. We are given that P (W ) = 0.40, P (R) = 0.50, and P (W R) = 0.70. Since P (W R) = P (W ) + P (R)  P (W R) = 0.40 + 0.50  0.70 = 0.20, 20% percent drink or serve both red and white wine. 10. No, it is not right. The probability that the second student chooses the tire the first student
chose is 1/4. 11. By De Morgan's second law,
P (Ac B c ) = 1  P (Ac B c )c = 1  P (A B) = 1  P (A)  P (B) + P (AB). 12. By Theorem 1.5 and the fact that A  B and B  A are mutually exclusive,
P (A  B) (B  A) = P (A  B) + P (B  A) = P (A  AB) + P (B  AB) = P (A)  P (AB) + P (B)  P (AB) = P (A) + P (B)  2P (AB). 13. Denote a box of books by ai , if it is received from publisher i, i = 1, 2, 3. The sample space
is
S = x1 x2 x3 x4 x5 x6 : two of the xi 's are a1 , two of them are a2 , and the remaining two are a3 . The desired event is E = x1 x2 x3 x4 x5 x6 S : x5 = x6 . 14. Let E, F , G, and H be the events that the next baby born in this town has blood type O, A, B,
and AB, respectively. Then P (E) = P (F ), P (G) = These imply P (E) = P (F ) = 20P (H ). Therefore, from P (E) + P (F ) + P (G) + P (H ) = 1, we get 20P (H ) + 20P (H ) + 2P (H ) + P (H ) = 1, which gives P (H ) = 1/43. 1 P (F ), P (G) = 2P (H ). 10 15. Let F , S, and N be the events that the number selected is divisible by 4, 7, and 9, respectively. We are interested in P (F c S c N c ) which is equal to 1  P (F S N ) by DeMorgan's law. 12 Chapter 1 Axioms of Probability Now P (F S N ) = P (F ) + P (S) + P (N )  P (F S)  P (F N )  P (SN ) + P (F SN ) = 142 111 35 27 15 250 3 + +    + = 0.429. 1000 1000 1000 1000 1000 1000 1000 So the desired probability is 0.571. 16. The number is relatively prime to 150 if is not divisible by 2, 3, or 5. Let A, B, and C be the
events that the number selected is divisible by 2, 3, and 5, respectively. We are interested in P (Ac B c C c ) = 1  P (A B C). Now P (A B C) = P (A) + P (B) + P (C)  P (AB)  P (AC)  P (BC) + P (ABC) 50 30 25 15 10 5 11 75 + +    + = . = 150 150 150 150 150 150 150 15 Therefore, the answer is 1  4 11 = . 15 15 c c c c c c 17. (a) Uic Dic ; (b) U1 U2 Un ; (c) (U1 D1 ) (U2 D2 ) (Un Dn ); (d) (e) c c c c c c c c (U1 D2 U3 D3 ) (U1 U2 D2 D3 ) (D1 U2 U3 D3 ) (D1 U2 D2 U3 ) c c c c c c c c c c (D1 U1 D2 U3 ) (D1 U1 U2 D2 ) (D1 U1 D2 U2 D3 U3 ); c c c D1 D2 Dn . 18. 103 199  96 = . 199  0 199 cases in which a < c, a > c, and a = c, it can be checked that there are 73, 73, and 27 cases in which b2 < 4ac, respectively. Therefore, the desired probability is 173 73 + 73 + 27 = . 216 216 19. We must have b2 < 4ac. There are 6 6 6 = 216 possible outcomes for a, b, and c. For Chapter 2 C ombinatorial Methods
2.2 COUNTING PRINCIPLES 1. The total number of sixdigit numbers is 91010101010 = 9105 since the first digit
cannot be 0. The number of sixdigit numbers without the digit five is 8 9 9 9 9 9 = 8 95 . Hence there are 9 105  8 95 = 427, 608 sixdigit numbers that contain the digit five. 55 = 3125. (b) 53 = 125. 2. (a) 3. There are 26 26 26 = 17, 576 distinct sets of initials. Hence in any town with more than
17,576 inhabitants, there are at least two persons with the same initials. The answer to the question is therefore yes. 4. 415 = 1, 073, 741, 824. 5.
2 1 = 22 0.00000024. 23 2 2 6. (a) 525 = 380, 204, 032. (b) 52 51 50 49 48 = 311, 875, 200. 7. 6/36 = 1/6. 8. (a) 9.
4322 1 = . 12 8 8 4 64 (b) 1  8562 27 = . 12 8 8 4 32 1 0.00000000093. 415 10. 26 25 24 10 9 8 = 11, 232, 000. 11. There are 263 102 = 1, 757, 600 such codes; so the answer is positive. 12. 2nm . 13. (2 + 1)(3 + 1)(2 + 1) = 36. (See the solution to Exercise 24.) 14 Chapter 2 Combinatorial Methods 14. There are (26  1)23 = 504 possible sandwiches. So the claim is true. 15. (a) 54 = 625. (b) 54  5 4 3 2 = 505. 16. 212 = 4096. 17. 1 
48 48 48 48 = 0.274. 52 52 52 52 (a) 9 9 8 7 = 4536; (b) 5040  1 1 8 7 = 4984. 18. 10 9 8 7 = 5040. 19. 1 
(N  1)n . Nn 20. By Example 2.6, the probability is 0.507 that among Jenny and the next 22 people she meets
randomly there are two with the same birthday. However, it is quite possible that one of these two persons is not Jenny. Let n be the minimum number of people Jenny must meet so that the chances are better than even that someone shares her birthday. To find n, let A denote the event that among the next n people Jenny meets randomly someone's birthday is the same as Jenny's. We have 364n P (A) = 1  P (Ac ) = 1  . 365n To have P (A) > 1/2, we must find the smallest n for which 1 or 1 364n > , n 365 2 364n 1 < . 365n 2 log This gives 1 2 = 252.652. n> 364 log 365 Therefore, for the desired probability to be greater than 0.5, n must be 253. To some this might seem counterintuitive. 21. Draw a tree diagram for the situation in which the salesperson goes from I to B first. In this situation, you will find that in 7 out of 23 cases, she will end up staying at island I . By symmetry, if she goes from I to H , D, or F first, in each of these situations in 7 out of 23 cases she will end up staying at island I . So there are 4 23 = 92 cases altogether and in 4 7 = 28 of them the salesperson will end up staying at island I . Since 28/92 = 0.3043, the answer is 30.43%. Note that the probability that the salesperson will end up staying at island I is not 0.3043 because not all of the cases are equiprobable. Section 2.2 Counting Principle 15 22. He is at 0 first, next he goes to 1 or 1. If at 1, then he goes to 0 or 2. If at 1, then he goes to 0 or 2, and so on. Draw a tree diagram. You will find that after walking 4 blocks, he is at one of the points 4, 2, 0, 2, or 4. There are 16 possible cases altogether. Of these 6 end up at 0, none at 1, and none at 1. Therefore, the answer to (a) is 6/16 and the answer to (b) is 0. 23. We can think of a number less than 1,000,000 as a sixdigit number by allowing it to start with
0 or 0's. With this convention, it should be clear that there are 96 such numbers without the digit five. Hence the desired probability is 1  (96 /106 ) = 0.469.
e e e 24. Divisors of N are of the form p11 p22 pkk , where ei = 0, 1, 2, . . . , ni , 1 i k. Therefore, the answer is (n1 + 1)(n2 + 1) (nk + 1). 25. There are 64 possibilities altogether. In 54 of these possibilities there is no 3. In 53 of these
possibilities only the first die lands 3. In 53 of these possibilities only the second die lands 3, and so on. Therefore, the answer is 54 + 4 5 3 = 0.868. 64 26. Any subset of the set {salami, turkey, bologna, corned beef, ham, Swiss cheese, American
cheese} except the empty set can form a reasonable sandwich. There are 27  1 possibilities. To every sandwich a subset of the set {lettuce, tomato, mayonnaise} can also be added. Since there are 3 possibilities for bread, the final answer is (27  1) 23 3 = 3048 and the advertisement is true. 27. 11 10 9 8 7 6 5 4 = 0.031. 118 P (Ac Ac Ac ) = 1  P (A1 A2 A3 ). Now 1 2 3 28. For i = 1, 2, 3, let Ai be the event that no one departs at stop i. The desired quantity is
P (A1 A2 A3 ) = P (A1 ) + P (A2 ) + P (A3 )  P (A1 A2 )  P (A1 A3 )  P (A2 A3 ) + P (A1 A2 A3 ) = 1 1 1 7 26 26 26 + 6 + 6  6  6  6 +0= . 6 3 3 3 3 3 3 27 Therefore, the desired probability is 1  (7/27) = 20/27. 29. For 0 i 9, the sum of the first two digits is i in (i + 1) ways. Therefore, there are (i + 1)2
numbers in the given set with the sum of the first two digits equal to the sum of the last two digits and equal to i. For i = 10, there are 92 numbers in the given set with the sum of the first two digits equal to the sum of the last two digits and equal to 10. For i = 11, the corresponding numbers are 82 and so on. Therefore, there are altogether 12 + 22 + + 102 + 92 + 82 + + 12 = 670 16 Chapter 2 Combinatorial Methods numbers with the desired probability and hence the answer is 670/104 = 0.067. 30. Let A be the event that the number selected contains at least one 0. Let B be the event that it contains at least one 1 and C be the event that it contains at least one 2. The desired quantity is P (ABC) = 1  P (Ac B c C c ), where P (Ac B c C c ) = P (Ac ) + P (B c ) + P (C c )  P (Ac B c )  P (Ac C c )  P (B c C c ) + P (Ac B c C c ) = 8 9r1 8 9r1 8r 8r 9r + +   9 10r1 9 10r1 9 10r1 9 10r1 9 10r1  7 8r1 7r + . 9 10r1 9 10r1 2.3 PERMUTATIONS
1 1 = 0.0417. 4! 24 1. The answer is 2. 3! = 6. 3.
8! = 56. 3! 5! 4. The probability that John will arrive right after Jim is 7!/8! (consider Jim and John as one
arrival). Therefore, the answer is 1  (7!/8!) = 0.875. Another Solution: If Jim is the last person, John will not arrive after Jim. Therefore, the remaining seven can arrive in 7! ways. If Jim is not the last person, the total number of possibilities in which John will not arrive right after Jim is 7 6 6!. So the answer is 7! + 7 6 6! = 0.875. 8! 5. (a) 312 = 531, 441. (b) 6. 6 P2 = 30. 7. 8. 12! = 924. 6! 6! (c) 12! = 27, 720. 3! 4! 5! 20! = 3, 491, 888, 400. 4! 3! 5! 8! (5 4 7) (4 3 6) (3 2 5) = 50, 400. 3! Section 2.3 Permutations 17 9. There are 8! schedule possibilities. By symmetry, in 8!/2 of them Dr. Richman's lecture
precedes Dr. Chollet's and in 8!/2 ways Dr. Richman's lecture precedes Dr. Chollet's. So the answer is 8!/2 = 20, 160. 10. 11! = 92, 400. 3! 2! 3! 3! 11. 1  (6!/66 ) = 0.985. 12. (a)
11! = 34, 650. 4! 4! 2! 10! = 6300. 4! 4! 8! = 840. 4! 2! 7! = 210. 4! (b) Treating all P 's as one entity, the answer is (c) Treating all I 's as one entity, the answer is (d) Treating all P 's as one entity, and all I 's as another entity, the answer is (e) By (a) and (c), The answer is 840/34650 = 0.024. 13. 14. 15. 8! 2! 3! 3! 9! 3! 3! 3! m! . (n + m)! 68 = 0.000333. 529 = 6.043 1013 . 16. Each girl and each boy has the same chance of occupying the 13th chair. So the answer is
12/20 = 0.6. This can also be seen from 12 19! 12 = = 0.6. 20! 20 17. 12! = 0.000054. 1212 5! 18! = 0.00068. 22! 18. Look at the five math books as one entity. The answer is 19. 1  20.
9 P7 97 = 0.962. 2 5! 5! = 0.0079. 10! 21. n!/nn . 18 Chapter 2 Combinatorial Methods 22. 1  (6!/66 ) = 0.985. 23. Suppose that A and B are not on speaking terms.
committees can be formed in which neither A serves nor B; 4 134 P3 committees can be formed in which A serves and B does not. The same numbers of committees can be formed in which B serves and A does not. Therefore, the answer is 134 P4 + 2(4 134 P3 ) = 326, 998, 056. (b)
m Pn . 134 P4 24. (a) mn . 25. 3 26. (a)
(b) (c) n!. 8! 2! 3! 2! 1! 68 = 0.003. 20! = 7.61 106 . 39 37 35 5 3 1 1 = 3.13 1024 . 39 37 35 5 3 1 27. Thirty people can sit in 30! ways at a round table. But for each way, if they rotate 30 times
(everybody move one chair to the left at a time) no new situations will be created. Thus in 30!/30 = 29! ways 15 married couples can sit at a round table. Think of each married couple as one entity and note that in 15!/15 = 14! ways 15 such entities can sit at a round table. We have that the 15 couples can sit at a round table in (2!)15 14! different ways because if the couples of each entity change positions between themselves, a new situation will be created. So the desired probability is 14!(2!)15 = 3.23 1016 . 29! The answer to the second part is 24!(2!)5 = 2.25 106 . 29! 28. In 13! ways the balls can be drawn one after another. The number of those in which the first
white appears in the second or in the fourth or in the sixth or in the eighth draw is calculated as follows. (These are Jack's turns.) 8 5 11! + 8 7 6 5 9! + 8 7 6 5 4 5 7! + 8 7 6 5 4 3 2 5 5! = 2, 399, 846, 400. Therefore, the answer is 2, 399, 846, 400/13! = 0.385. Section 2.4 Combinations 19 2.4 COMBINATIONS
20 = 38, 760. 6
100 1. 2.
i=51 100 = 583, 379, 627, 841, 332, 604, 080, 945, 354, 060 5.8 1029 . i 25 = 6, 864, 396, 000. 6 40 2 52 5 3. 20 6 12 3 4. = 0.066. 5. 6. 7. 8. 9. N 1 n1 5 3 8 3 N n = n . N 2 = 10. 2 5 2 3 = 560. 3 18 18 + = 21, 624. 6 4 10 5 12 = 0.318. 7 12 . Therefore, the coefficient of x 9 9 10. The coefficient of 23 x 9 in the expansion of (2 + x)12 is
is 23 12 = 1760. 9 11. The coefficient of (2x)3 (4y)4 in the expansion of (2x  4y)7 is
of x 3 y 2 in this expansion is 23 (4)4 7 = 71, 680. 4 7 . Thus the coefficient 4 12. 9 3 6 6 +2 4 3 = 4620. 20 Chapter 2 Combinatorial Methods
10 13. (a) 10 5 210 = 0.246; (b)
i=5 10 i 210 = 0.623. 14. If their minimum is larger than 5, they are all from the set {6, 7, 8, . . . , 20}. Hence the answer
is 15 5 6 2 20 = 0.194. 5 28 4 34 6 6 6 10 12 + + + 6 6 6 6 34 6 15. (a) = 0.228; (b) = 0.00084. 16. 50 5 150 45 200 50 n = i n = i = 0.00206. n 17.
i=0 n i=0 2i n i=0 n i=0 n i ni 2 1 = (2 + 1)n = 3n . i n i ni x 1 = (x + 1)n . i xi 18. 6 4 5 2 66 = 0.201. 19. 212 24 = 0.00151. 12 4 52 5 36 52 5 = 0.0000015. 20. Royal Flush: Straight flush: = 0.000014. 13 12 Four of a kind: 52 5 4 1 = 0.00024. Section 2.4 Combinations 21 13 Full house: 4 4 12 3 2 52 5 = 0.0014. 4 Flush: 13  40 5 = 0.002. 52 5 10(4)5  40 = 0.0039. 52 5 13 4 12 2 4 3 2 = 0.021. 52 5 4 4 11 2 1 52 5 Straight: Three of a kind: Two pairs: 13 2 4 2 = 0.048. 13 One pair: 4 12 3 4 2 3 = 0.42. 52 5 1 the sum of all of the above cases = 0.5034445. None of the above: 21. The desired probability is
12 6 24 12 12 6 = 0.3157. 22. The answer is the solution of the equation x 3 x(x  1)(x  2) = 120 and its solution is x = 6. = 20. This equation is equivalent to 22 Chapter 2 Combinatorial Methods 23. There are 9103 = 9000 fourdigit numbers. From every 4combination of the set {0, 1, . . . , 9},
exactly one fourdigit number can be constructed in which its ones place is less than its tens place, its tens place is less than its hundreds place, and its hundreds place is less than its 10 thousands place. Therefore, the number of such fourdigit numbers is = 210. Hence 4 the desired probability is 0.023333. 24.
(x + y + z)2 =
n1 +n2 +n3 n! x n1 y n2 zn3 n1 ! n2 ! n3 ! =2 = 2! 2! 2! x 2 y 0 z0 + x 0 y 2 z0 + x 0 y 0 z2 2! 0! 0! 0! 2! 0! 0! 0! 2! + 2! 2! 2! x 1 y 1 z0 + x 1 y 0 z1 + x 0 y 1 z1 1! 1! 0! 1! 0! 1! 0! 1! 1! = x 2 + y 2 + z2 + 2xy + 2xz + 2yz. 25. The coefficient of (2x)2 (y)3 (3z)2 in the expansion of (2x  y + 3z)7 is
coefficient of x 2 y 3 z2 in this expansion is 22 (1)3 (3)2 7! = 7560. 2! 3! 2! 7! . Thus the 2! 3! 2! 26. The coefficient of (2x)3 (y)7 (3)3 in the expansion of (2x  y + 3)13 is
the coefficient of x 3 y 7 in this expansion is 23 (1)7 (3)3 13! 3! 7! 3! 13! . Therefore, 3! 7! 3! = 7, 413, 120. 27. In 52! 52! ways 52 cards can be dealt among four people. Hence the sample = 13! 13! 13! 13! (13!)4 space contains 52!/(13!)4 points. Now in 4! ways the four different suits can be distributed among the players; thus the desired probability is 4!/[52!/(13!)4 ] 4.47 1028 . 28. The theorem is valid for k = 2; it is the binomial expansion. Suppose that it is true for all
integers k  1. We show it for k. By the binomial expansion, (x1 + x2 + + xk ) =
n1 =0 n n n n n1 x (x2 + + xk )nn1 n1 1 (n  n1 )! n x n2 x n3 xk k n2 ! n3 ! nk ! 2 3 =
n1 =0 n n1 x n1 1 n 2 +n3 ++nk =nn1 =
n1 +n2 ++nk =n n (n  n1 )! n x n1 x n2 xk k n1 n2 ! n3 ! nk ! 1 2 Section 2.4 Combinations 23 =
n1 +n2 ++nk n! n x n1 x n2 xk k . n1 ! n2 ! nk ! 1 2 =n 29. We must have 8 steps. Since the distance from M to L is ten 5centimeter intervals and the
first step is made at M, there are 9 spots left at which the remaining 7 steps can be made. So 9 the answer is = 36. 7 2 1 98 98 + 49 48 100 50 30. (a) = 0.753; (b) 250 100 = 1.16 1014 . 50 31. (a) It must be clear that
n 2 n1 + nn1 n2 = 2 n2 + n2 (n + n1 ) n3 = 2 n3 + n3 (n + n1 + n2 ) n4 = 2 . . . nk1 nk = + nk1 (n + n1 + + nk1 ). 2 n1 = (b) For n = 25, 000, successive calculations of nk 's yield, n1 = 312, 487, 500, n2 = 48, 832, 030, 859, 381, 250, n3 = 1, 192, 283, 634, 186, 401, 370, 231, 933, 886, 715, 625, n4 = 710, 770, 132, 174, 366, 339, 321, 713, 883, 042, 336, 781, 236, 550, 151, 462, 446, 793, 456, 831, 056, 250. For n = 25, 000, the total number of all possible hybrids in the first four generations, n1 + n2 + n3 + n4 , is 710,770,132,174,366,339,321,713,883,042,337,973,520,184,337, 863,865,857,421,889,665,625. This number is approximately 710 1063 . 32. For n = 1, we have the trivial identity
x+y = 1 0 10 1 1 11 x y + x y . 0 1 24 Chapter 2 Combinatorial Methods Assume that (x + y)n1 = This gives (x + y)n = (x + y)
n1 n1 i=0 n1 i=0 n  1 i n1i . xy i n  1 i n1i xy i
n1 i=0 =
i=0 n n  1 i+1 n1i x y + i n  1 i ni xy + i1
n1 i=1 n1 i=0 n  1 i ni xy i =
i=1 n  1 i ni xy i x i y ni + y n
n i=0 =x +
n1 i=1 n n1 n1 + i1 i n i ni xy + yn = i = xn + n i ni xy . i 33. The desired probability is computed as follows.
12 6 30 2 28 2 26 2 24 2 22 2 10 1 20 2 18 3 15 3 12 3 9 3 6 3 3 3 1230 0.000346. 34. (a) 10 6 2 6 = 0.347; 20 6 10 2 (b) 9 4 2 4 = 0.520; 20 6 10 3 20 6 (c) 8 2 2 2 = 0.130; (d) 20 6 = 0.0031. 35. 26 13 52 26 26 13 = 0.218. Section 2.4 Combinations 25 36. Let a 6element combination of a set of integers be denoted by {a1 , a2 , . . . , a6 }, where a1 <
a2 < < a6 . It can be easily verified that the function h : B A defined by h {a1 , a2 , . . . , a6 } = {a1 , a2 + 1, . . . , a6 + 5} is onetoone and onto. Therefore, there is a onetoone correspondence between B and 44 A . This shows that the number of elements in A is . Thus the probability that no 6 44 49 consecutive integers are selected among the winning numbers is 0.505. This 6 6 implies that the probability of at least two consecutive integers among the winning numbers is approximately 1  0.505 = 0.495. Given that there are 47 integers between 1 and 49, this high probability might be counterintuitive. Even without knowledge of expected value, a keen student might observe that, on the average, there should be (49  1)/7 = 6.86 numbers between each ai and ai+1 , 1 i 5. Thus he or she might erroneously think that it is unlikely to obtain consecutive integers frequently. 37. (a) Let Ei be the event that car i remains unoccupied. The desired probability is
c c c P (E1 E2 En ) = 1  P (E1 E2 En ). Clearly, P (Ei ) = P (Ei Ej ) = P (Ei Ej Ek ) = (n  1)m , nm (n  2)m , nm (n  3)m , nm 1 i n; 1 i, j n, i = j ; 1 i, j, k n, i = j = k; and so on. Therefore, by the inclusionexclusion principle,
n P (E1 E2 En ) =
i=1 (1)i1 n (n  i)m . i nm
n i=0 So
c c c P (E1 E2 En ) = 1  n i=1 (1)i1 (1)i n (n  i)m = nm i (1)i n (n  i)m nm i 1 = m n (b) n i=0 n (n  i)m . i Let F be the event that cars 1, 2, . . . , n  r are all occupied and the remaining cars are n unoccupied. The desired probability is P (F ). Now by part (a), the number of ways m r 26 Chapter 2 Combinatorial Methods passengers can be distributed among n  r cars, no car remaining unoccupied is
nr i=0 (1)i nr (n  r  i)m . i So 1 P (F ) = m n and hence the desired probability is
nr i=0 nr i=0 (1)i nr (n  r  i)m i 1 n nm r (1)i nr (n  r  i)m . i 38. Let the n indistinguishable balls be represented by n identical oranges and the n distinguishable cells be represented by n persons. We should count the number of different ways that the n oranges can be divided among the n persons, and the number of different ways in which exactly one person does not get an orange. The answer to the latter part is n(n  1) since in this case one person does not get an orange, one person gets exactly two oranges, and the remaining persons each get exactly one orange. There are n choices for the person who does not get an orange and n  1 choices for the person who gets exactly two oranges; n(n  1) choices altogether. To count the number of different ways that the n oranges can be divided among the n persons, add n  1 identical apples to the oranges and note that by Theorem 2.4, the total (2n  1)! number of permutations of these n  1 apples and n oranges is . (We can arrange n! (n  1)! n  1 identical apples and n identical oranges in a row in (2n  1)!/ n! (n  1)! ways.) Now (2n  1)! 2n  1 each one of these = permutations corresponds to a way of dividing the n! (n  1)! n n oranges among the n persons and vice versa. Give all of the oranges preceding the first apple to the first person, the oranges between the first and the second apples to the second person, the oranges between the second and the third apples to the third person and so on. Therefore, if, for example, an apple appears in the beginning of the permutation, the first person does not get an orange, and if two apples are at the end of the permutations, the (n  1)st and the nth 2n  1 persons get no oranges. Thus the answer is n(n  1) . n 39. The left side of the identity is the binomial expansion of (1  1)n = 0. Section 2.4 Combinations 27 40. Using the hint, we have
n n+1 n+2 n+r + + + + 0 1 2 r = n n+2 n+1 n+3 n+2 +  +  0 1 0 2 1 + = n+4 n+3 n+r +1 n+r  + +  3 2 r r 1 n n+1 n+r +1 n+r +1  + = . 0 0 r r 41. The identity expresses that to choose r balls from n red and m blue balls, we must choose
either r red balls, 0 blue balls or r  1 red balls, one blue ball or r  2 red balls, two blue balls or 0 red balls, r blue balls. 1 n 1 n+1 = . Hence i+1 i n+1 i+1 1 n+1 n+1 n+1 n+1 + + + 1 2 n+1 = 1 (2n+1  1). n+1 42. Note that The given sum = 5 3 3 2 43. 45 = 0.264. t m N t nm . N n PN (N  t)(N  n) = . PN1 N (N  t  n + m) 44. (a) PN = (b) From part (a), we have This implies PN > PN1 if and only if (N  t)(N  n) > N (N  t  n + m) or, equivalently, if and only if N nt/m. So PN is increasing if and only if N nt/m. This shows that the maximum of PN is at [nt/m], where by [nt/m] we mean the greatest integer nt/m. 45. The sample space consists of (n + 1)4 elements. Let the elements of the sample be denoted by x1 , x2 , x3 , and x4 . To count the number of samples (x1 , x2 , x3 , x4 ) for which x1 + x2 = x3 + x4 , let y3 = n  x3 and y4 = n  x4 . Then y3 and y4 are also random elements from the set {0, 1, 2, . . . , n}. The number of cases in which x1 + x2 = x3 + x4 is identical to the number of cases in which x1 + x2 + y3 + y4 = 2n. By Example 2.23, the number of nonnegative integer 28 Chapter 2 Combinatorial Methods 2n + 3 . However, this also counts the solutions in which one 3 of x1 , x2 , y3 , and y4 is greater than n. Because of the restrictions 0 x1 , x2 , y3 , y4 n, we must subtract, from this number, the total number of the solutions in which one of x1 , x2 , y3 , and y4 is greater than n. Such solutions are obtained by finding all nonnegative integer solutions of the equation x1 + x2 + y3 + y4 = n  1, and then adding n + 1 to exactly one of x1 , x2 , y3 , and y4 . Their count is 4 times the number of nonnegative integer solutions of n+2 x1 + x2 + y3 + y4 = n  1; that is, 4 . Therefore, the desired probability is 3 solutions to this equation is 2n + 3 n+2 4 3 3 (n + 1)4 2n2 + 4n + 3 . 3(n + 1)3 = 46. (a) The n  m unqualified applicants are "ringers." The experiment is not affected by their
inclusion, so that the probability of any one of the qualified applicants being selected is the same as it would be if there were only qualified applicants. That is, 1/m. This is because in a random arrangement of m qualified applicants, the probability that a given applicant is the first one is 1/m. (b) Let A be the event that a given qualified applicant is hired. We will show that P (A) = 1/m. Let Ei be the event that the given qualified applicant is the ith applicant interviewed, and he or she is the first qualified applicant to be interviewed. Clearly,
nm+1 P (A) =
i=1 P (Ei ), 1 (n  i)! . n! where P (Ei ) = Therefore,
nm+1 nm Pi1 P (A) =
i=1 nm+1 nm Pi1 (n  i)! n! =
i=1 nm+1 (n  m)! (n  i)! (n  m  i + 1)! n! 1 m! 1 m 1 (n  i)! (m  1)! (n  m  i + 1)! (m  1)! n! m! (n  m)! 1 n m ni m1 =
i=1 nm+1 =
i=1 Section 2.4
nm+1 i=1 Combinations 29 = 1 m 1 n m ni . m1 (4) nm+1 To calculate
i=1 ni ni , note that is the coefficient of x m1 in the expansion m1 m1
nm+1 i=1 nm+1 i=1 nm+1 of (1 + x) ni . Therefore, ni is the coefficient of x m1 in the expansion of m1 (1 + x)n  (1 + x)m1 . x (1 + x)ni = This shows that
i=1 ni m1 is the coefficient of x m in the expansion of n . So (4) implies that m 1 m 1 n m n 1 = . m m 6 equal 1 (1 + x)  (1 + x) n m1 , which is P (A) = 47. Clearly, N = 610 , N (Ai ) = 510 , N (Ai Aj ) = 410 , i = j , and so on. So S1 has
terms, S2 has 610  6 equal terms, and so on. Therefore, the solution is 2 6 10 6 10 6 10 6 10 6 10 6 10 5 + 4  3 + 2  1 + 0 = 16, 435, 440. 1 2 3 4 5 6 n3 , 3 A1  = 1 n 2 3 3 1 n3 , 2 A2  = 1 n 2 3 3 2 n3 . 1 48. A0  = 1 n 2 3 The answer is A0  (n  4)(n  5) = . A0  + A1  + A2  n2 + 2 2n . Its coefficient in (1 + x)n (1 + x)n is n n n + n1 2 n 1
2 49. The coefficient of x n in (1 + x)2n is
n 0 n n + n 1 = n 0
2 n n + + n2 n n n
2 n 0 + + n 2 2 + + , 30 Chapter 2 Combinatorial Methods since n n = , 0 i n. i n1 50. Consider a particular set of k letters. Let M be the number of possibilities in which only
n M n!. All k we got to do is to find M. To do so, note that the remaining n  k letters are all addressed incorrectly. For these n  k letters, there are n  k addresses. But the addresses are written on the envelopes at random. The probability that none is addressed correctly on one hand is M/(n  k)!, and on the other hand, by Example 2.24, is these k letters are addressed correctly. The desired probability is the quantity
nk 1
i=1 (1)i1 = i!
n i=2 n n i=2 (1)i1 . i! So M satisfies M = (n  k)! (1)i1 , i! (1)i1 . i! and hence M = (n  k)! i=2 The final answer is n M k = n! n (n  k)! k
n i=2 (1)i1 i! n! 1 = k! n i=2 (1)i1 . i! 51. The set of all sequences of H's and T's of length i with no successive H's are obtained either
by adding a T to the tails of all such sequences of length i  1, or a TH to the tails of all such sequences of length i  2. Therefore, xi = xi1 + xi2 , i 2. Clearly, x1 = 2 and x3 = 3. For consistency, we define x0 = 1. From the theory of recurrence i Br i relations we know that the solution of xi = xi1 + xi2 is of the form xi = Ar1 + 2 , where 1+ 5 1 5 and r2 = and so r1 and r2 are the solutions of r 2 = r + 1. Therefore, r1 = 2 2 1+ 5 i 1 5 i xi = A +B . 2 2 5+3 5 53 5 Using the initial conditions x0 = 1 and x2 = 2, we obtain A = and B = . 10 10 Section 2.5 Stirling's Formula 31 Hence the answer is xn 1 n = n 2 2 = 5+3 5 10 1 10 22n 1+ 5 n 53 5 1 5 n + 2 10 2 n n 5+3 5 1+ 5 + 53 5 1 5 . 52. For this exercise, a solution is given by Abramson and Moser in the October 1970 issue of the
American Mathematical Monthly. 2.5 STIRLING's FORMULA 4 n (2n)2n e2n 2n 1 (2n)! 1 1 = . 2n 2n 2n e2n 22n n! n! 2 (2 n) n n 2 n 3 3 4 n (2n)2n e2n (2n)! 2 = n. 2 4n e4n (2 n) n2n e2n (4n)! (n!) 4 8 n (4n) 1. (a)
(b) REVIEW PROBLEMS FOR CHAPTER 2 1. The desired quantity is equal to the number of subsets of all seven varieties of fruit minus 1
(the empty set); so it is 27  1 = 127. 2. The number of choices Virginia has is equal to the number of subsets of {1, 2, 5, 10, 20} minus
1 (for empty set). So the answer is 25  1 = 31. 3. (6 5 4 3)/64 = 0.278. 4. 10 5.
10 = 0.222. 2 9! = 7560. 3! 2! 2! 2! 6. 5!/5 = 4! = 24. 7. 3! 4! 4! 4! = 82, 944.
23 6 30 6 8. 1  = 0.83. 32 Chapter 2 Combinatorial Methods 9. Since the refrigerators are identical, the answer is 1. 10. 6! = 720. 11. (Draw a tree diagram.) In 18 out of 52 possible cases the tournament ends because John wins
4 games without winning 3 in a row. So the answer is 34.62%. 12. Yes, it is because the probability of what happened is 1/72 = 0.02. 13. 9 8 = 43, 046, 721. 14. (a) 26 25 24 23 22 21 = 165, 765, 600;
(b) (c) 26 25 24 23 22 5 = 39, 468, 000; 5 3 2 1 26 25 24 23 = 21, 528, 000. 2 1 1 1 2 1 2 1 15. 6 6 6 6 + + + 3 1 1 1 10 3 = 0.467. Another Solution: 6 6 + 3 1 10 3 4 2 = 0.467. 16. 8 4 6 P4 = 0.571. 8 P6 278 = 0.252. 288 17. 1  18. (3!/3)(5!)3 = 0.000396. 15!/15 19. 312 = 531, 441.
4 1 48 12 3 1 36 12 2 1 24 12 1 1 12 12 20. 52! 13! 13! 13! 13! = 0.1055. Chapter 2 Review Problems 33 21. Let A1 , A2 , A3 , and A4 be the events that there is no professor, no associate professor, no
assistant professor, and no instructor in the committee, respectively. The desired probability is P (Ac Ac Ac Ac ) = 1  P (A1 A2 A3 A4 ), 1 2 3 4 where P (A1 A2 A3 A4 ) is calculated using the inclusionexclusion principle: P (A1 A2 A3 A4 ) = P (A1 ) + P (A2 ) + P (A3 ) + P (A4 )  P (A1 A2 )  P (A1 A3 )  P (A1 A4 )  P (A2 A3 )  P (A2 A4 )  P (A3 A4 ) + P (A1 A2 A3 ) + P (A1 A3 A4 ) + P (A1 A2 A4 ) + P (A2 A3 A4 )  P (A1 A2 A3 A4 ) = 1  34 6 28 28 24 22 22 18 16 18 + + +     6 6 6 6 6 6 6 6 16 12 12 6 10 6  + + + +  0 = 0.621. 6 6 6 6 6 6 Therefore, the desired probability equals 1  0.621 = 0.379. 22. (15!)2 = 0.0002112. 30!/(2!)15 N . n 40 1 52 13 23. (N  n + 1)
4 2 48 24 52 26 39 8 52 13 24. (a) = 0.390; (b) = 6.299 1011 ; (c) 13 5 8 8 39 13 31 5 = 0.00000261. 25. 12!/(3!)4 = 369, 600. 26. There is a onetoone correspondence between all cases in which the eighth outcome obtained
is not a repetition and all cases in which the first outcome obtained will not be repeated. The answer is 65555555 5 7 = 0.279. = 6 66666666 27. There are 9 103 = 9, 000 fourdigit numbers. To count the number of desired fourdigit
numbers, note that if 0 is to be one of the digits, then the thousands place of the number must be 34 Chapter 2 Combinatorial Methods 0, but this cannot be the case since the first digit of an ndigit number is nonzero. Keeping this in mind, it must be clear that from every 4combination of the set {1, 2, . . . , 9}, exactly one fourdigit number can be constructed in which its ones place is greater than its tens place, its tens place is greater than it hundreds place, and its hundreds place is greater than its thousands 9 place. Therefore, the number of such fourdigit numbers is = 126. Hence the desired 4 probability is = 0.014. 28. Since the sum of the digits of 100,000 is 1, we ignore 100,000 and assume that all of the numbers
have five digits by placing 0's in front of those with less than five digits. The following process establishes a onetoone correspondence between such numbers, d1 d2 d3 d4 d5 , 5 di = 8, i=1 and placement of 8 identical objects into 5 distinguishable cells: Put d1 of the objects into the first cell, d2 of the objects into the second cell, d3 into the third cell, and so on. Since 8+51 12 this can be done in = = 495 ways, the number of integers from the set 51 8 {1, 2, 3, . . . , 100000} in which the sum of the digits is 8 is 495. Hence the desired probability is 495/100, 000 = 0.00495. Chapter 3 C onditional Probability and I ndependence
3.1 CONDITIONAL PROBABILITY 1. P (W  U ) = P (U W ) 0.15 = = 0.60. P (U ) 0.25 2. Let E be the event that in the blood of the randomly selected soldier A antigen is found. Let
F be the event that the blood type of the soldier is A. We have P (F  E) = P (F E) 0.41 = = 0.911. P (E) 0.41 + 0.04 3. 0.20 = 0.625. 0.32 4. The reduced sample space is (1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4) ; therefore, the
desired probability is 1/7. 2 30  20 = . 30  15 3 5. 6. Both of the inequalities are equivalent to P (AB) > P (A)P (B). 7.
2 1/3 = . (1/3) + (1/2) 5 8. 4/30 = 0.133. 36 Chapter 3 Conditional Probability and Independence = 0.239. 40 65 2 8i i 1 105 i=0 8 1/19 if i = 0 10. P ( = i  = 0) = 2/19 if i = 1, 2, 3, . . . , 9 0 if i = 10, 11, 12, . . . , 18. 9. 40 65 2 6 105 8 11. Let b gb mean that the oldest child of the family is a boy, the second oldest is a girl, the youngest
is a boy, and the boy found in the family is the oldest child, with similar representations for other cases. The reduced sample space is S = ggb , gb g, b gg, b bg, bb g, gb b, gbb , bgb , b gb, b bb, bb b, bbb . Note that the outcomes of the sample space are not equiprobable. We have that P {ggb } = P {gb g} = P {b gg} = 1/7 P {b bg} = P {bb g} = 1/14 P {gb b} = P {gbb } = 1/14 P {bgb } = P {b gb} = 1/14 P {b bb} = P {bb b} = P {bbb } = 1/21. The solutions to (a), (b), (c) are as follows. (a) P {bb g} = 1/14; (b) P {bb g, gbb , bgb , bb b, bbb } = 13/42; (c) P {b bg, bb g, gb b, gbb , bgb , b gb} = 3/7. 12. P (A) = 1 implies that P (A B) = 1. Hence, by
P (A B) = P (A) + P (B)  P (AB), we have that P (B) = P (AB). Therefore, P (B  A) = P (B) P (AB) = = P (B). P (A) 1 Section 3.1 Conditional Probability 37 13. P (A  B) = P (AB) , where b P (AB) = P (A) + P (B)  P (A B) P (A) + P (B)  1 = a + b  1. P (AB) 0. P (B) 14. (a) P (AB) 0, P (B) > 0. Therefore, P (A  B) =
(b) P (S  B) = P (SB) P (B) = = 1. P (B) P (B) P i=1 Ai B = (c) P
i=1 Ai B = P i=1 Ai B P (B) P (Ai B) P (B) = i=1 P (B) =
i=1 P (Ai B) = P (B) P (Ai  B).
i=1 Note that P ( Ai B) = P (Ai B), since mutual exclusiveness of Ai 's imply that of i=1 i=1 Ai B's; i.e., Ai Aj = , i = j , implies that (Ai B)(Aj B) = , i = j . 15. The given inequalities imply that P (EF ) P (GF ) and P (EF c ) P (GF c ). Thus
P (E) = P (EF ) + P (EF c ) P (GF ) + P (GF c ) = P (G). 16. Reduce the sample space: Marlon chooses from six dramas and seven comedies two at random.
What is the probability that they are both comedies? The answer is 7 2 13 = 0.269. 2 17. Reduce the sample space: There are 21 crayons of which three are red. Seven of these crayons
are selected at random and given to Marty. What is the probability that three of them are red? 18 21 The answer is = 0.0263. 4 7 18. (a) The reduced sample space is S = {1, 3, 5, 7, 9, . . . , 9999}. There are 5000 elements in S. Since the set {5, 7, 9, 11, 13, 15, . . . , 9999} includes exactly 4998/3 = 1666 odd numbers that are divisible by three, the reduced sample space has 1667 odd numbers that are divisible by 3. So the answer is 1667/5000 = 0.3334. (b) Let O be the event that the number selected at random is odd. Let F be the event that it is divisible by 5 and T be the event that it is divisible by 3. The desired probability is calculated as follows. P (F c T c  O) = 1  P (F T  O) = 1  P (F  O)  P (T  O) + P (F T  O) =1 1000 1667 333  + = 0.5332. 5000 5000 5000 38 Chapter 3 Conditional Probability and Independence 19. Let A be the event that during this period he has hiked in Oregon Ridge Park at least once. Let
B be the event that during this period he has hiked in this park at least twice. We have P (B  A) = where P (A) = 1  and P (B) = 1  So the answer is 0.515/0.838 = 0.615. P (B) , P (A) 510 = 0.838 610 510 10 59  = 0.515. 610 610 20. The numbers of 333 red and 583 blue chips are divisible by 3. Thus the reduced sample space
has 333 + 583 = 916 points. Of these numbers, [1000/15] = 66 belong to red balls and are divisible by 5 and [1750/15] = 116 belong to blue balls and are divisible by 5. Thus the desired probability is 182/916 = 0.199. 21. Reduce the sample space: There are two types of animals in a laboratory, 15 type I and 13
type II. Six animals are selected at random; what is the probability that at least two of them are Type II? The answer is 15 13 + 6 1 28 6 15 5 1 = 0.883. 22. Reduce the sample space: 30 students of which 12 are French and nine are Korean are divided
randomly into two classes of 15 each. What is the probability that one of them has exactly four French and exactly three Korean students? The solution to this problem is 12 9 9 4 3 8 30 15 15 15 = 0.00241. 23. This sounds puzzling because apparently the only deduction from the name "Mary" is that one
of the children is a girl. But the crucial difference between this and Example 3.2 is reflected in the implicit assumption that both girls cannot be Mary. That is, the same name cannot be used for two children in the same family. In fact, any other identifying feature that cannot be shared by both girls would do the trick. Section 3.2 Law of Multiplication 39 3.2 LAW OF MULTIPLICATION 1. Let G be the event that Susan is guilty. Let L be the event that Robert will lie. The probability
that Robert will commit perjury is P (GL) = P (G)P (L  G) = (0.65)(0.25) = 0.1625. 2. The answer is 11 10 9 8 7 6 = 0.15. 14 13 12 11 10 9 3. By the law of multiplication, the answer is
52 50 48 46 44 42 = 0.72. 52 51 50 49 48 47 4. (a)
(b) 7 6 8 5 = 0.0144; 20 19 18 17 8 7 12 8 12 7 12 8 7 8 7 6 + + + = 0.344. 20 19 18 20 19 18 20 19 18 20 19 18 5 5 4 4 3 3 2 2 1 1 6 = 0.00216. 11 10 9 8 7 6 5 4 3 2 1 5 4 3 2 1 = 0.00216. 11 10 9 8 7 5. (a)
(b) 6. 5 5 8 5 3 8 5 3 + = 0.0712. 8 10 13 15 8 11 13 16 7. Let Ai be the event that the ith person draws the "you lose" paper. Clearly,
P (A1 ) = 1 , 200 P (A2 ) = P (Ac A2 ) = P (Ac )P (A2  Ac ) = 1 1 1 1 199 1 = , 200 199 200 199 198 1 1 P (A3 ) = P (Ac Ac A3 ) = P (Ac )P (Ac  Ac )P (A3  Ac Ac ) = = , 1 2 1 2 1 1 2 200 199 198 200 and so on. Therefore, P (Ai ) = 1/200 for 1 i 200. This means that it makes no difference if you draw first, last or anywhere in the middle. Here is Marilyn Vos Savant's intuitive solution to this problem: 40 Chapter 3 Conditional Probability and Independence It makes no difference if you draw first, last, or anywhere in the middle. Look at it this way: Say the robbers make everyone draw at once. You'd agree that everyone has the same change of losing (one in 200), right? Taking turns just makes that same event happen in a slow and orderly fashion. Envision a raffle at a church with 200 people in attendance, each person buys a ticket. Some buy a ticket when they arrive, some during the event, and some just before the winner is drawn. It doesn't matter. At the party the end result is this: all 200 guests draw a slip of paper, and, regardless of when they look at the slips, the result will be identical: one will lose. You can't alter your chances by looking at your slip before anyone else does, or waiting until everyone else has looked at theirs. 8. Let B be the event that a randomly selected person from the population at large has poor credit report. Let I be the event that the person selected at random will improve his or her credit rating within the next three years. We have P (B  I ) = P (BI ) P (I  B)P (B) (0.30)(0.18) = = = 0.072. P (I ) P (I ) 0.75 The desired probability is 10.072 = 0.928. Therefore, 92.8% of the people who will improve their credit records within the next three years are the ones with good credit ratings. 9. For 1 n 39, let En be the event that none of the first n  1 cards is a heart or the ace of spades. Let Fn be the event that the nth card drawn is the ace of spades. Then the event of "no heart before the ace of spades" is 39 En Fn . Clearly, {En Fn , 1 n 39} forms a n=1 sequence of mutually exclusive events. Hence
39 39 39 P
n=1 En Fn =
n=1 39 P (En Fn ) =
n=1 P (En )P (Fn  En ) =
n=1 38 n1 52 n1 1 1 = , 53  n 14 a result which is not unexpected. 13 3 52 9 39 6 10. P (F )P (E  F ) = 10 = 0.059. 43 11. By the law of multiplication,
P (An ) = 2 3 4 n+1 2 = . 3 4 5 n+2 n+2 Section 3.3 Law of Total Probability 41 Now since A1 A2 A3 An An+1 , by Theorem 1.8, P
i=1 Ai = lim P (An ) = 0.
n 3.3 LAW OF TOTAL PROBABILITY
1 1 0.05 + 0.0025 = 0.02625. 2 2 1. 2. (0.16)(0.60) + (0.20)(0.40) = 0.176. 3. 4.
1 1 1 (0.75) + (0.68) + (0.47) = 0.633. 3 3 3 1 12 13 13 39 + = . 51 52 51 52 4 11 50 13 2 52 2 12 50 13 1 52 2 39 1 13 50 39 2 52 2 1 = . 4 5. + + 6. (0.20)(0.40) + (0.35)(0.60) = 0.290. 7. (0.37)(0.80) + (0.63)(0.65) = 0.7055. 8.
1 1 1 1 1 1 (0.6) + (0.5) + (0.7) + (0.9) + (0.7) + (0.8) = 0.7. 6 6 6 6 6 6 9. (0.50)(0.04) + (0.30)(0.02) + (0.20)(0.04) = 0.034. 10. Let B be the event that the randomly selected child from the countryside is a boy. Let E be
the event that the randomly selected child is the first child of the family and F be the event that he or she is the second child of the family. Clearly, P (E) = 2/3 and P (F ) = 1/3. By the law of total probability, P (B) = P (B  E)P (E) + P (B  F )P (F ) = 1 1 2 1 1 + = . 2 3 2 3 2 Therefore, assuming that sex distributions are equally probable, in the Chinese countryside, the distribution of sexes will remain equal. Here is Marilyn Vos Savant's intuitive solution to this problem: 42 Chapter 3 Conditional Probability and Independence The distribution of sexes will remain roughly equal. That's becauseno matter how many or how few children are born anywhere, anytime, with or without restriction half will be boys and half will be girls: Only the act of conception (not the government!) determines their sex. One can demonstrate this mathematically. (In this example, we'll assume that women with firstborn girls will always have a second child.) Let's say 100 women give birth, half to boys and half to girls. The half with boys must end their families. There are now 50 boys and 50 girls. The half with girls (50) give birth again, half to boys and half to girls. This adds 25 boys and 25 girls, so there are now 75 boys and 75 girls. Now all must end their families. So the result of the policy is that there will be fewer children in number, but the boy/girl ratio will not be affected. 11. The probability that the first person gets a gold coin is 3/5. The probability that the second
person gets a gold coin is 2 3 3 2 3 + = . 4 5 4 5 5 The probability that the third person gets a gold coin is 3 2 1 3 2 2 2 3 2 2 1 3 3 + + + = , 5 4 3 5 4 3 5 4 5 5 4 3 5 and so on. Therefore, they are all equal. 12. A Probabilistic Solution: Let n be the number of adults in the town. Let x be the number of men in the town. Then n  x is the number of women in the town. Since the number of married men and married women are equal, we have x 7 3 = (n  x) . 9 5 This relation implies that x = (27/62)n. Therefore, the probability that a randomly selected adult is male is (27/62)n n = 27/62. The probability that a randomly selected adult is female is 1  (27/62) = 35/62. Let A be the event that a randomly selected adult is married. Let M be the event that the randomly selected adult is a man, and let W be the event that the randomly selected adult is a woman. By the law of total probability, P (A) = P (A  M)P (M) + P (A  W )P (W ) = 7 27 3 35 42 21 + = = 0.677. 9 62 5 62 62 31 Therefore, 21/31st of the adults are married. An Arithmetical Solution: The common numerator of the two fractions is 21. Hence 21/27th of the men and 21/35th of the women are married. We find the common numerator because the number of married men and the number of married women are equal. This shows that of every 27 + 35 = 62 adults, 21 + 21 = 42 are married. Hence 42/62th = 21/31st of the adults in the town are married. Section 3.3 Law of Total Probability 43 13. The answer is clearly 0.40. This can also be computed from
(0.40)(0.75) + (0.40)(0.25) = 0.40. 14. Let A be the event that a randomly selected child is the kth born of his or her family. Let Bj
be the event that he or she is from a family with j children. Then
c P (A) =
j =k P (A  Bj )P (Bj ), where, clearly, P (A  Bj ) = 1/j . To find P (Bj ), note that there are i N families with j children. Therefore, the total number of children in the world is c i(i N ) of which j (N j ) i=0 are from families with j children. Hence P (Bj ) = j (N j ) = c i=0 i(i N )
c j j . c i=0 ii j j c i=0 ii This shows that the desired fraction is given by
c P (A) =
j =k c P (A  Bj )P (Bj ) =
j =k 1 j =
j =k j c i=0 ii = c j =k j . c i=0 ii 15. Q(E  F ) = Q(EF ) P (EF  B) P (EFB) = = = P (E  FB). = Q(F ) P (F  B) P (FB) P (FB) P (B) P (EFB) P (B) 16. Let M, C, and F denote the events that the random student is married, is married to a student
at the same campus, and is female, respectively. We have that 1 2 P (F  M) = P (F  MC)P (C  M)+P (F  MC c )P (C c  M) = (0.40) +(0.30) = 0.333. 3 3 17. Let p(k, n) be the probability that exactly k of the first n seeds planted in the farm germinated. Using induction on n, we will show that p(k, n) = 1/(n  1) for all k < n. For n = 2, p(1, 2) = 1 = 1/(2  1) is true. If p(k, n  1) = 1/(n  2) for all k < n  1, then, by the law of total probability, p(k, n) = = k1 nk1 p(k  1, n  1) + p(k, n  1) n1 n1 k1 1 nk1 1 1 + = . n1 n2 n1 n2 n1 This proves the induction hypothesis. 44 Chapter 3 Conditional Probability and Independence 18. Reducing the sample space, we have that the answer is 7/10.
10 7 10 8 6 10 8 5 8 8 3 3 2 1 3 1 2 3 3 3 + + + 18 18 18 18 18 18 18 18 3 3 3 3 3 3 3 3 19. = 0.0383. 20. We have that
P (A  G) = P (A  GO)P (O  G) + P (A  GM)P (M  G) + P (A  GY )P (Y  G) 1 1 1 3 1 5 =0 + + = . 3 2 3 4 3 12 21. Let E be the event that the third number falls between the first two. Let A be the event that
the first number is smaller than the second number. We have that P (E  A) = P (EA) 1/6 1 = = . P (A) 1/2 3 Intuitively, the fact that P (A) = 1/2 and P (EA) = 1/6 should be clear (say, by symmetry). However, we can prove these rigorously. We show that P (A) = 1/2; P (EA) = 1/6 can be proved similarly. Let B be the event that the second number selected is smaller than the first number. Clearly A = B c and we only need to show that P (B) = 1/2. To do this, let Bi be the event that the first number drawn is i, 1 i n. Since {B1 , B2 , . . . , Bn } is a partition of the sample space,
n P (B) =
i=1 P (B  Bi )P (Bi ). Now P (B  B1 ) = 0 because if the first number selected is 1, the second number selected i1 cannot be smaller. P (B  Bi ) = , 1 i n since if the first number is i, the second n1 number must be one of 1, 2, 3, . . . , i  1 if it is to be smaller. Thus
n n P (B) =
i=1 P (B  Bi )P (Bi ) =
i=2 i1 1 1 = n1 n (n  1)n n (i  1)
i=2 = 1 1 (n  1)n 1 1 + 2 + 3 + + (n  1) = = . (n  1)n (n  1)n 2 2 22. Let Em be the event that Avril selects the best suitor given her strategy. Let Bi be the event
that the best suitor is the ith of Avril's dates. By the law of total probability,
n P (Em ) =
i=1 P (Em  Bi )P (Bi ) = 1 n n P (Em  Bi ).
i=1 Section 3.3 Law of Total Probability 45 Clearly, P (Em  Bi ) = 0 for 1 i m. For i > m, if the ith suitor is the best, then Avril chooses him if and only if among the first i  1 suitors Avril dates, the best is one of the first m. So m P (Em  Bi ) = . i1 Therefore, n n 1 m m 1 = . P (Em ) = n i=m+1 i  1 n i=m+1 i  1 Now 1 i1 i=m+1 Thus
n n m n 1 dx = ln . x m m n ln . n m To find the maximum of P (Em ), consider the differentiable function P (Em ) h(x) = Since x n ln . n x n 1 1 ln  =0 n x n implies that x = n/e, the maximum of P (Em ) is at m = [n/e], where [n/e] is the greatest integer less than or equal to n/e. Hence Avril should dump the first [n/e] suitors she dates and marry the first suitor she dates afterward who is better than all those preceding him. The probability that with such a strategy she selects the best suitor of all n is approximately h (x) = h n 1 1 = ln e = 0.368. e e e 23. Let N be the set of nonnegative integers. The domain of f is
(g, r) N N : 0 g N, 0 r N, 0 < g + r < 2N . f f = = 0 gives g r g = r = N/2 and f (N/2, N/2) = 1/2. However, this is not the maximum value because on the boundary of the domain of f along r = 0, we find that Extending the domain of f to all points (g, r) R R, we find that f (g, 0) = is maximum at g = 1 and f (1, 0) = 1 N g 1+ 2 2N  g 1 3N  2 1 . 2 2N  1 2 46 Chapter 3 Conditional Probability and Independence We also find that on the boundary along r = N, f (g, N ) = is maximum at g = N  1 and f (N  1, N ) = 1 1 3N  2 . 2 2N  1 2 g 1 +1 2 g+N The maximums of f along other sides of the boundary are all less than 1 3N  2 . Therefore, 2 2N  1 there are exactly two maximums and they occur at (1, 0) and (N 1, N ). That is, the maximum of f occurs if one urn contains one green and 0 red balls and the other one contains N 1 green 1 3N  2 3 and N red balls. For large N , the probability that the prisoner is freed is . 2 2N  1 4 3.4 1. 2. BAYES' FORMULA
(3/4)(0.40) 3 = . (3/4)(0.40) + (1/3)(0.60) 5 8 1(2/3) = . 1(2/3) + (1/4)(1/3) 9 event that the suspect is lefthanded. Since {G, I } is a partition of the sample space, we can use Bayes' formula to calculate P (G  A), the probability that the suspect has committed the crime in view of the new evidence. P (G  A) = P (A  G)P (G) (0.85)(0.65) = 0.87. P (A  G)P (G) + P (A  I )P (I ) (0.85)(0.65) + (0.23)(0.35) 3. Let G and I be the events that the suspect is guilty and innocent, respectively. Let A be the 4. Let G be the event that Susan is guilty. Let C be the event that Robert and Julie give conflicting
testimony. By Bayes' formula, P (G  C) = (0.25)(0.65) P (C  G)P (G) = = 0.607. c )P (Gc ) P (C  G)P (G) + P (C  G (0.25)(0.65) + (0.30)(0.35) 5. (0.02)(0.30) = 0.1463. (0.02)(0.30) + (0.05)(0.70) 6 3 6 3 11 3 11 3 1 2 1 1 +1 2 2 4 . 37 6. = Section 3.4 Bayes' Formula 47 7. (0.92)(1/5000) = 0.084. (0.92)(1/5000) + (1/500)(4999/5000) selected from urn I is a dime. Then P (B  A) = P (A  B)P (B) = P (A  B)P (B) + P (A  B c )P (B c ) 5 3 2 1 + 7 4 7 4 5 3 2 1 4 + + 7 4 7 4 7 4 68 7 = . 83 5 1 3 7 4 7 8. Let A be the event that two of the three coins are dimes. Let B be the event that the coin 9. (0.15)(0.25) = 0.056. (0.15)(0.25) + (0.85)(0.75) card with both sides black is selected. Define RR and RB similarly. By Bayes' Formula, P (RB  R) = = 1 1 6 1000 100 1 6 P (R  RB)P (RB) P (R  RB)P (RB) + P (R  RR)P (RR) + P (R  BB)P (BB) (1/2)(1/3) 1 = . (1/2)(1/3) + 1(1/3) + 0(1/3) 3 = 0.21. 10. Let R be the event that the upper side of the card selected is red. Let BB be the event that the 11. 5 i=0 1000  i 100 12. Let A be the event that the wallet originally contained a $2 bill. Let B be the event that the
bill removed is a $2 bill. The desired probability is given by P (A  B) = P B  A P (A) P B  A P (A) + P B  Ac P (Ac ) 1 1 2 2 = = . 3 1 1 1 1 + 2 2 2 13. By Bayes' formula, the probability that the horse that comes out is from stable I equals
4 (20/33)(1/2) = . (20/33)(1/2) + (25/33)(1/2) 9 The probability that it is from stable II is 5/9; hence the desired probability is 20 4 25 5 205 + = = 0.69. 33 9 33 9 297 48 Chapter 3 Conditional Probability and Independence 2 4 5 2 8 4 2 4 3 2 = 0.571. 14.
0 5 4 8 4 + 1 4 5 3 8 4 3 1 + 5 2 8 4 3 2 + 3 4 5 1 8 4 3 3 15. Let I be the event that the person is ill with the disease, N be the event that the result of the test on the person is negative, and R denote the event that the person has the rash. We are interested in P (I  R): P (I  R) = P (I N  R) + P (I N c  R) = 0 + P (I N c  R). Since {I N, I N c , I c N, I c N c } is a partition of the sample space, by Bayes' Formula,
P (I  R) = P (I N c  R) = = P (R  I N c )P (I N c ) P (R  I N )P (I N ) + P (R  I N c )P (I N c ) + P (R  I c N )P (I c N ) + P (R  I c N c )P (I c N c ) (0.2)(0.30 0.90) = 0.61. 0(0.30 0.10) + (0.2)(0.30 0.90) + 0(0.70 0.75) + (0.2)(0.70 0.25) 3.5 INDEPENDENCE 1. No, because by independence, regardless of the number of heads that have previously occurred,
the probability of tails remains to be 1/2 on each flip. 2. A and B are mutually exclusive; therefore, they are dependent. If A occurs, then the probability
that B occurs is 0 and vice versa. 3. Neither. Since the probability that a fighter plane returns from a mission without mishap is
49/50 independent of other missions, the probability that a pilot who flew 49 consecutive missions without mishap making another successful flight is still 49/50=0.98; neither higher nor lower than the probability of success in any other mission. 4. P (AB) = 1/12 = (1/2)(1/6); so A and B are independent. 5. (3/8)3 (5/8)5 = 0.00503. 6. (3/4)2 = 0.5625. Section 3.5 Independence 49 7. (a) (0.725)2 = 0.526; (b) (1  0.725)2 = 0.076. 8. Suppose that for an event A, P (A) = 3/4. Then the probability that A occurs in two consecutive independent experiments is 9/16. So the correct odds are 9 to 7, not 9 to 1. In later computations, Cardano, himself, had realized that the correct answer is 9 to 7 and not 9 to 1. 9. We have that
4 P (A beats B) = P (A rolls 4) = , 6 P (B beats A) = 1  P (A beats B) = 1  4 P (B beats C) = P (C rolls 2) = , 6 P (C beats B) = 1  P (B beats C) = 1  4 2 = , 6 6 4 2 4 3 + = , 6 6 6 6 2 4 = , 6 6 P (C beats D) = P (C rolls 6) + P (C rolls 2 and D rolls 1) = P (D beats C) = 1  P (C beats D) = 1  2 4 = , 6 6 P (D beats A) = P (D rolls 5) + P (D rolls 1 and A rolls 0) = 3 3 2 4 + = . 6 6 6 6 10. For 1 i 4, let Ai be the event of obtaining 6 on the ith toss. Chevalier de Mr had
implicitly thought that Ai 's are mutually exclusive and so P A1 A2 A3 A4 = 1 1 1 1 1 + + + =4 . 6 6 6 6 6 Clearly Ai 's are not mutually exclusive. The correct answers are 1  (5/6)4 = 0.5177 and 1  (35/36)24 = 0.4914. 11. (1  0.0001)64 = 0.9936. 12. In the experiment of tossing a coin, let A be the event of obtaining heads and B be the event
of obtaining tails. 13. (a) P (A B) P (A) = 1, so P (A B) = 1. Now
1 = P (A B) = P (A) + P (B)  P (AB) = 1 + P (B)  P (AB) gives P (B) = P (AB). (b) If P (A) = 0, then P (AB) = 0; so P (AB) = P (A)P (B) is valid. If P (A) = 1, by part (a), P (AB) = P (B) = P (A)P (B). 14. P (AA) = P (A)P (A) implies that P (A) = P (A) . This gives P (A) = 0 or P (A) = 1. 2 50 Chapter 3 Conditional Probability and Independence 15. P (AB) = P (A)P (B) implies that P (A) = P (A)P (B). This gives P (A) 1  P (B) = 0;
so P (A) = 0 or P (B) = 1. 16. 1  (0.45)6 = 0.9917. 17. 1  (0.3)(0.2)(0.1) = 0.994. 18. There are
(100 10 9 ) (300 10 9 )  1 = 30 10 21  1 other stars in the universe. Provided that Aczel's estimate is correct, the probability of no life in orbit around any one given star in the known universe is 0.99999999999995 independently of other stars. Therefore, the probability of no life in orbit around any other star is (0.99999999999995)30,000,000,000,000,000,000,000 1 . Using Aczel's words, "this number is indistinguishable from 0 at any level of decimal accuracy reported by the computer." Hence the probability that there is life in orbit around at least one other star is 1 for all practical purposes. If there were only a billion galaxies each having 10 billion stars, still the probability of life would have been indistinguishable from 1.0 at any level of accuracy reported by the computer. In fact, if we divide the stars into mutually exclusive groups with each group containing billions of stars, then the argument above and Exercise 8 of Section 1.7 imply that the probability of life in orbit around many other stars is a number practically indistinguishable from 1. 19. 1  (0.94)15  15(0.94)14 (0.06) = 0.226. 20. A and B are independent if and only if P (AB) = P (A)P (B), or, equivalently, if and only if
m M m+w = . M +W M +W M +W This implies that m/M = w/W. Therefore, A and B are independent if and only if the fraction of the men who smoke is equal to the fraction of the women who smoke. 21. (a) By Theorem 1.6,
P A(B C) = P (AB AC) = P (AB) + P (AC)  P (ABC) = P (A)P (B) + P (A)P (C)  P (A)P (B)P (C) = P (A) P (B) + P (C)  P (B)P (C) = P (A)P (B C). (b) P (A  B)C = P (AB c C) = P (A)P (B c )P (C) = P (AB c )P (C) = P (A  B)P (C). 22. 1  (5/6)6 = 0.6651. Section 3.5 Independence 51 23. (a) 1  (n  1)/n . 24. n (b) As n , this approaches 1  (1/e) = 0.6321. 1  (0.85)10  10(0.85)9 (0.15) = 0.567. 1  (0.85)10 events that the point lies in (0, 1/2), (1/4, 3/4), and (1/2, 1), respectively. 25. No. In the experiment of choosing a random number from (0, 1), let A, B, and C denote the 26. Denote a family with two girls and one boy by ggb, with similar representations for other
cases. The sample space is S = {ggg, bbb, ggb, gbb}. we have P {ggg} = P {bbb} = 1/8, P {ggb} = P {gbb} = 3/8. Clearly, P (A) = 6/8 = 3/4, P (B) = 4/8 = 1/2, and P (AB) = 3/8. Since P (AB) = P (A)P (B), the events A and B are independent. Using the same method, we can show that for families with two children and for families with four children, A and B are not independent. 27. If p is the probability of its occurrence in one trial, 1  (1  p)4 = 0.59. This implies that
p = 0.2. 28. (a) 1  (1  p1 )(1  p2 ) (1  pn ). (b) (1  p1 )(1  p2 ) (1  pn ). 29. Let Ei be the event that the switch located at i is closed. The desired probability is
P (E1 E2 E4 E6 E1 E3 E5 E6 ) = P (E1 E2 E4 E6 )+P (E1 E3 E5 E6 )P (E1 E2 E3 E4 E5 E6 ) = 2p4 p 6 . 30. 5 3 2 3 3 1 3 2 = 0.329. 31. For n = 3, the probabilities of the given events, respectively, are
3 2 and 3 1 1 2 1 2
2 1 2 2 1 1 + 2 2 3 2 1 2 3 = 1 , 2 + 2 3 1 = . 2 4 The probability of their joint occurrence is 3 2 1 2
2 3 1 3 1 = = . 2 8 2 4 So the given events are independent. For n = 4, similar calculations show that the given events are not independent. 52 Chapter 3 Conditional Probability and Independence 32. (a) 1  (1/2)n . (b)
(c) n k 1 2 n . Let An be the event of getting n heads in the first n flips. We have A1 A2 A3 An An+1 . The event of getting heads in all of the flips indefinitely is of probability function (Theorem 1.8), its probability is n=1 An . By the continuity property
n P
n=1 An = lim P (An ) = lim
n n 1 2 = 0. 33. Let Ai be the event that the sixth sum obtained is i, i = 2, 3, . . . , 12. Let B be the event that
the sixth sum obtained is not a repetition. By the law of total probability,
12 P (B) =
i=2 P (B  Ai )P (Ai ). Note that in this sum, the terms for i = 2 and i = 12 are equal. This is true also for the terms for i = 3 and 11, for the terms for i = 4 and 10, for the terms for i = 5 and 9, and for the terms for i = 6 and 8. So
6 P (B) = 2
i=2 P (B  Ai )P (Ai ) + P (B  A7 )P (A7 ) 35 36 +
5 =2 34 1 + 36 36
5 5 33 2 + 36 36
5 5 32 3 + 36 36 5 4 36 31 36 5 36 + 30 36 6 = 0.5614. 36 34. (a) Let E be the event that Dr. May's suitcase does not reach his destination with him. We
have P (E) = (0.04) + (0.96)(0.05) + (0.96)(0.95)(0.05) + (0.96)(0.95)(0.95)(0.04) = 0.168, or simply, P (E) = 1  (0.96)(0.95)(0.96) = 0.168. (b) Let D be the event that the suitcase is lost in Da Vinci airport in Rome. Then, by Bayes' formula, P (D) (0.96)(0.05) P (D  E) = = = 0.286. P (E) 0.168 35. Let E be the event of obtaining heads on the coin before an ace from the cards. Let H , T , A, and N denote the events of heads, tails, ace, and not ace in the first experiment, respectively. We use two different techniques to solve this problem. Section 3.5 Independence 53 Technique 1: By the law of total probability, P (E) = P (E  H )P (H ) + P (E  T )P (T ) = 1 where P (E  T ) = P (E  T A)P (A  T ) + P (E  T N )P (N  T ) = 0 Thus P (E) = which gives P (E) = 13/14. Technique 2: We have that P (E) = P (E  H A)P (H A)+P (E  T A)P (T A)+P (E  H N )P (H N )+P (E  T N )P (T N ). 1 1 1 1 1 12 1 12 +0 +1 + P (E) . 2 13 2 13 2 13 2 13 This gives P (E) = 13/14. P (E) = 1 Thus 12 1 1 + P (E) , 2 13 2 12 1 + P (E) . 13 13 1 1 + P (E  T ) , 2 2 36. Let P (A) = p and P (B) = q. Let An be the event that none of A and B occurs in the first
n  1 trials and the outcome of the nth experiment is A. The desired probability is P
n=1 An =
n=1 P (An ) =
n=1 (1  p  q)n1 p = p p = . 1  (1  p  q) p+q 37. The probability of sum 5 is 1/9 and the probability of sum 7 is 1/6. Therefore, by the result of
Exercise 36, the desired probability is 1/9 = 2/5. 1/6 + 1/9 38. Let A be the event that one of them is red and the other one is blue. Let RB represent the
event that the ball drawn from urn I is red and the ball drawn form urn II is blue, with similar representations for RR, BB, and BR. We have that
P (A) = P (A  RB)P (RB) + P (A  RR)P (RR) + P (A  BB)P (BB) + P (A  BR)P (BR) 9 1 14 2 5 1 9 5 + 10 6 8 1 14 2 6 1 9 1 + 10 6 10 4 1 1 14 2 1 5 + 10 6 9 1 14 2 5 1 1 1 10 6 = = 0.495. 54 Chapter 3 Conditional Probability and Independence 39. For convenience, let p0 = 0; the desired probability is
n n 1
i=1 (1  pi ) 
i=1 (1  p1 )(1  p2 ) (1  pi1 )pi (1  pi+1 ) (1  pn ). 40. Let p be the probability that a randomly selected person was born on one of the first 365 days; then 365p + (p/4) = 1 implies that p = 4/1461. Let E be the event that exactly four people of this group have the same birthday and that all the others have different birthdays. E is the union of the following three mutually exclusive events: F : Exactly four people of this group have the same birthday, all the others have different birthdays, and none of the birthdays is on the 366th day. G: Exactly four people of this group have the same birthday, all the others have different birthdays, and exactly one has his/her birthday on the 366th day. H : Exactly four people of this group have their birthday on the 366th day and all the others have different birthdays. We have that P (E) = P (F ) + P (G) + P (H ) = + + 365 1 30 1 30 4 30 4 4 1461
4 364 4 26! 26 1461 29 4 4 1461
4 26 1 365 1461 1 1 1461
4 364 4 25! 25 1461 25 365 4 26! 26 1461 26 = 0.00020997237. If we were allowed to ignore the effect of the leap year, the solution would have been as follows. 365 30 1 4 364 1 26 26! = 0.00021029. 1 1 26 365 365 41. Let Ei be the event that the switch located at i is closed. We want to calculate the probability of E2 E4 E1 E5 E2 E3 E5 E1 E3 E4 . Using the rule to calculate the probability of the union of several events (the inclusionexclusion principle) we get that the answer is 2p2 +2p3 5p4 +p5 . the corresponding events for B and C, respectively. Clearly, + P (ABC  E c F c )P (E c F c ). Now P (ABC  EF G) = P (ABC), (6) 42. Let E be the event that A will answer correctly to his or her first question. Let F and G be
P (ABC) = P (ABC  EF G)P (EF G) + P (ABC  E c F G)P (E c F G) (5) Section 3.5 Independence 55 and P (ABC  E c F c ) = 1. (7) To calculate P (ABC  E c F G), note that since A has already lost, the game continues between B and C. Let BC be the event that B loses and C wins. Then P (ABC  E c F G) = P (BC). Let F2 be the event that B answers the second question correctly; then
C C P (BC) = P (BC  F2 )P (F2 ) + P (BC  F2 )P (F2 ). (8) (9) To find P (BC  F2 ), note that this quantity is the probability that B loses to C given that B did not lose the first play. So, by independence, this is the probability that B loses to C given that C plays first. Now by symmetry, this quantity is the same as C losing to B if B plays first. Thus it is equal to P (CB), and hence (9) gives P (BC) = P (CB) p + 1 (1  p); noting that P (CB) = 1  P (BC), this gives P (BC) = Therefore, by (8), P (ABC  E c F G) = substituting this, (8), and (7) in (5), yields P (ABC) = P (ABC) p 3 + Solving this for P (ABC), we obtain P (ABC) = Now we find P (BCA) and P (CAB). P (BCA) = P (BCA  E)P (E) + P (BCA  E c )P (E c ) p , = P (ABC) p + 0 (1  p) = (1 + p)(1 + p + p 2 ) P (CAB) = P (CAB  E)P (E) + P (CAB  E c )P (E c ) = P (BCA) p + 0 (1  p) = p2 . (1 + p)(1 + p + p 2 ) 1 . (1 + p)(1 + p + p 2 ) 1 (1  p)p 2 + (1  p)2 . 1+p 1 . 1+p 1 . 1+p 56 Chapter 3 Conditional Probability and Independence 43. We have that
P (H1 ) = P (H1  H )P (H ) + P (H1  H c )P (H c ) = 1 1 3 1 +0 = . 2 4 4 8 c c Similarly, P (H2 ) = 1/8. To calculate P (H1 H2 ), the probability that none of her sons is hemophiliac, we condition on H again. c c c c c c P (H1 H2 ) = P (H1 H2  H )P (H ) + P (H1 H2  H c )P (H c ). c c c c Clearly, P (H1 H2  H c ) = 1. To find P (H1 H2  H ), we use the fact that H1 and H2 are conditionally independent given H . c c c c P (H1 H2  H ) = P (H1  H )P (H2  H ) = 1 1 1 = . 2 2 4 Thus
c c P (H1 H2 ) = 3 13 1 1 +1 = . 4 4 4 16 44. The only quantity not calculated in the hint is P (Ui  Rm ). By Bayes' Formula,
P (Ui  Rm ) = P (Rm  Ui )P (Ui ) P (Rm  Uk )P (Uk )
k=0 n = i n
n k=0 m 1 n+1
m k n = i n
n k=0 m 1 n+1 k n m . 3.6 APPLICATIONS OF PROBABILITY TO GENETICS 1. Clearly, Kim and Dan both have genotype OO. With a genotype other than AO for John, it is
impossible for Dan to have blood type O. Therefore, the probability is 1 that John's genotype is AO. k k(k + 1) +k = . 2 2 2. The answer is 3. The genotype of the parent with wrinkled shape is necessarily rr. The genotype of the other parent is either Rr or RR. But, RR will never produce wrinkled offspring. So it must be Rr. Therefore, the parents are rr and Rr. attached earlobes. Let B represent the dominant allele for freckles and b represent the recessive allele for no freckles. Since Dan has attached earlobes and no freckles, Kim and John both must be AaBb. This implies that Kim and John's next child is AA with probability 1/4, Aa 4. Let A represent the dominant allele for free earlobes and a represent the recessive allele for Section 3.6 Applications of Probability to Genetics 57 with probability 1/2, and aa with probability 1/4. Therefore, the next child has free earlobes with probability 3/4. Similarly, the next child is BB with probability 1/4, Bb with probability 1/2, and bb with probability 1/4. Hence he or she will have no freckles with probability 1/4. By independence, the desired probability is (3/4)(1/4) = 3/16. 5. If the genes are not linked, 25% of the offspring are expected to be BbV v, 25% are expected to be bbvv, 25% are expected to be Bbvv, and 25% are expected to be bbV v. The observed data shows that the genes are linked. 6. Clearly, John's genotype is either Dd or dd. Let E be the event that it is dd. Then E c is the event that John's genotype is Dd. Let F be the event that Dan is deaf. That is, his genotype is dd. We use Bayes' theorem to calculate the desired probability. P (E  F ) = = P (F  E)P (E) P (F  E)P (E) + P (F  E c )P (E c ) 1 (0.01) = 0.0198. 1 (0.01) + (1/2)(0.99) Therefore, the probability is 0.0198 that John is also deaf. 7. A person who has cystic fibrosis carries two mutant alleles. Applying the HardyWeinberg
law, we have that q 2 = 0.0529, or q = 0.23. Therefore, p = 0.77. Since q 2 + 2pq = 1  p2 = 0.4071, the percentage of the people who carry at least one mutant allele of the disease is 40.71%. 8. Dan inherits all of his sexlinked genes from his mother. Therefore, John being normal has no
effect on whether or not Dan has hemophilia or not. Let E be the event that Kim is H h. Then E c is the event that Kim is H H . Let F be the event that Dan has hemophilia. By the law of total probability, P (F ) = P (F  E)P (E) + P (F  E c )P (E c ) = (1/2) 2(0.98)(0.02) + 0 (0.98)(0.98) = 0.0196. 9. Dan has inherited all of his sexlinked genes from his mother. Let E1 be the event that Kim is CC, E2 be the event that she is Cc, and E3 be the event that she is cc. Let F be the event that Dan is colorblind. By Bayes' formula, the desired probability is P (E3  F ) = = P (F  E3 )P (E3 ) P (F  E1 )P (E1 ) + P (F  E2 )P (E2 ) + P (F  E3 )P (E3 ) 1 (0.17)(0.17) = 0.17. 0 (0.83)(0.83) + (1/2) 2(0.83)(0.17) + 1 (0.17)(0.17) 10. Since Ann is hh and John is hemophiliac, Kim is either H h or hh. Let E be the event that she
is H h. Then E c is the event that she is hh. Let F be the event that Ann has hemophilia. By 58 Chapter 3 Conditional Probability and Independence Bayes' formula, the desired probability is P (E  F ) = = P (F  E)P (E) P (F  E)P (E) + P (F  E c )P (E c ) (1/2) 2(0.98)(0.02) (1/2) 2(0.98)(0.02) + 1 (0.02)(0.02) = 0.98. 11. Clearly, both parents of Mr. J must be Cc. Since Mr. J has survived to adulthood, he is not cc.
Therefore, he is either CC or Cc. We have P (he is CC  he is CC or Cc) = P (he is Cc  he is CC or Cc) = 1/4 1 P (he is CC) = = . P (he is CC or Cc) 3/4 3 2 . 3 Mr. J's wife is either CC with probability 1  p or Cc with probability p. Let E be the event that Mr. J is Cc, F be the event that his wife is Cc, and H be the event that their next child is cc. The desired probability is P (H ) = P (H EF ) = P (H  EF )P (EF ) = P (H  EF )P (E)P (F ) = 1 2 p p = . 4 3 6 12. Let E1 be the event that both parents are of genotype AA, let E2 be the event that one parent
is of genotype Aa and the other of genotype AA, and let E3 be the event that both parents are of genotype Aa. Let F be the event that the man is of genotype AA. By Bayes' formula, P (E1  F ) = = P (F  E1 )P (E1 ) P (F  E1 )P (E1 ) + P (F  E2 )P (E2 ) + P (F  E3 )P (E3 ) p2 1 p4 = = p2 . 1 p 4 + (1/2) 4p 3 q + (1/4) 4p 2 q 2 (p + q)2 Similarly, P (E2  F ) = 2pq and P (E3  F ) = q 2 . Let B be the event that the brother is AA. We have P (B  F ) = P (B  F E1 )P (E1  F ) + P (B  F E2 )P (E2  F ) + P (B  F E3 )P (E3  F ) = P (B  E1 )P (E1  F ) + P (B  E2 )P (E2  F ) + P (B  E3 )P (E3  F ) = 1 p2 + 1 1 (1 + p)2 (2p + q)2 2pq + q 2 = = . 2 4 4 4 Chapter 3 Review Problems 59 REVIEW PROBLEMS FOR CHAPTER 3 1.
26 12 13 13 12 + = = 0.347. 30 30 30 30 75 2. 1  (0.97)6 = 0.167. 3. (0.48)(0.30) + (0.67)(0.53) + (0.89)(0.17) = 0.65. 4. (0.5)(0.05) + (0.7)(0.02) + (0.8)(0.035) = 0.067. 5. (a) (0.95)(0.97)(0.85) = 0.783; (b) 1  (0.05)(0.03)(0.05) = 0.999775;
(c) 1  (0.95)(0.97)(0.85) = 0.217; (d) (0.05)(0.03)(0.15) = 0.000225. 6. 103/132 = 0.780. 7.
(0.08)(0.20) = 0.0796. (0.2)(0.3) + (0.25)(0.5) + (0.08)(0.20) 26 6 39 6 = 0.929. 8. 1  9. 1/6.
1 10. 5 6 10  10 5 6
10 9 1 6 5 1 6 = 0.615. 2 4 8 7 7 11. = 0.35. = 23 2 4 5 3 + 7 7 7 7 12. Let A be the event of "head on the coin." Let B be the event of "tail on the coin and 1 or 2 on the die." Then A and B are mutually exclusive, and by the result of Exercise 36 of Section 3.5, 1/2 3 the answer is = . (1/2) + (1/6) 4 13. The probability that the number of 1's minus the number of 2's will be 3 is
P (four 1's and one 2) + P (three 1's and no 2's) = 6 4 1 6
4 2 1 1 6 4 6 + 6 3 1 6 3 4 6 3 = 0.03. 60 Chapter 3 Conditional Probability and Independence 14. The probability that the first urn was selected in the first place is
20 45 20 1 + 45 2 The desired probability is 20 10 10 9 + 0.42. 45 19 25 19 1 10 2 . = 19 10 1 25 2 15. Let B be the event that the ball removed from the third urn is blue. Let BR be the event that
the ball drawn from the first urn is blue and the ball drawn from the second urn is red. Define BB, RB, and RR similarly. We have that P (B) = P (B  BB)P (BB) + P (B  RB)P (RB) + P (B  RR)P (RR) + P (B  BR)P (BR) 4 1 5 5 9 5 6 9 1 5 1 1 38 = + + + = = 0.36. 14 10 6 14 10 6 14 10 6 14 10 6 105 16. Let E be the event that Lorna guesses correctly. Let R be the event that a red hat is placed
on Lorna's head, and B be the event that a blue hat is placed on her head. By the law of total probability, P (E) = P (E  R)P (R) + P (E  B)P (B) 1 1 1 = + (1  ) = 2 2 2 This shows that Lorna's chances are 50% to guess correctly no matter what the value of is. This should be intuitively clear. 17. Let F be the event that the child is found; E be the event that he is lost in the east wing, and
W be the event that he is lost in the west wing. We have P (F ) = P (F  E)P (E) + P (F  W )P (W ) = 1  (0.6)3 (0.75) + 1  (0.6)2 (0.25) = 0.748. 18. The answer is that it is the same either way. Let W be the event that they win one of the nights
to themselves. Let F be the event that they win Friday night to themselves. Then P (W ) = P (W  F )P (F ) + P (W  F c )P (F c ) = 1 2 1 1 2 + = . 3 2 3 3 19. Let A be the event that Kevin is prepared. We have that
P (R  B c S c ) = = P (RB c S c ) P (RB c S c  A)P (A) + P (RB c S c  Ac )P (Ac ) = P (B c S c ) P (B c S c  A)P (A) + P (B c S c  Ac )P (Ac ) (0.85)(0.15)2 (0.85) + (0.20)(0.80)2 (0.15) = 0.308. (0.15)2 (0.85) + (0.80)2 (0.15) Chapter 3 Review Problems 61 Note that P (R) = P (R  A)P (A) + P (R  Ac )P (Ac ) = (0.85)(0.85) + (0.20)(0.15) = 0.7525. Since P (R  B c S c ) = P (R), the events R, B, and S are not independent. However, it must be clear that R, B, and S are conditionally independent given that Kevin is prepared and they are conditionally independent given that Kevin is unprepared. To explain this, suppose that we are given that, for example, Smith and Brown both failed a student. This information will increase the probability that the student was unprepared. Therefore, it increases the probability that Rose will also fails the student. However, if we know that the student was unprepared, the knowledge that Smith and Brown failed the student does not affect the probability that Rose will also fail the student. 20. (a) Let A be the event that Adam has at least one king; B be the event that he has at least
two kings. We have P (B  A) = P (Adam has at least two kings) P (AB) = P (A) P (Adam has at least one king) 48 13 52 13 48 4 12 1 52 13 48 13 52 13 1 =  = 0.3696. 1 (b) Let A be the event that Adam has the king of diamonds. Let B be the event that he has the king of diamonds and at least one other king. Then 48 11 P (B  A) = P (BA) = P (A) 3 48 3 48 + + 1 10 2 9 52 13 51 12 52 13 3 3 = 0.5612. Knowing that Adam has the king of diamonds reduces the sample space to a size considerably smaller than the case in which we are given that he has a king. This is why the answer to 62 Chapter 3 Conditional Probability and Independence part (b) is larger than the answer to part (a). If one is not convinced of this, he or she should solve the problem in a simpler case. For example, a case in which there are four cards, say, king of diamonds, king of hearts, jack of clubs, and eight of spade. If two cards are drawn, the reduced sample space in the case Adam announces that he has a king is {Kd Kh , Kd Jc , Kd 8s , Kh Jc , Kh 8s }, while the reduced sample space in the case Adam announces that he has the king of diamonds is {Kd Kh , Kd Jc , Kd 8s }. In the first case, the probability of more kings is 1/5; in the second case the probability of more kings is 1/3. Chapter 4 D istribution F unctions and Discrete R andom Variables
4.2 DISTRIBUTION FUNCTIONS 1. The set of possible values of X is {0, 1, 2, 3, 4, 5}. The probabilities associated with these
values are x P (X = x) these values are 5 2 15 2 5 1 15 2 0 6/36 1 10/36 2 8/36 3 6/36 4 4/36 5 2/36 2. The set of possible values of X is {6, 2, 1, 2, 3, 4}. The probabilities associated with P (X = 6) = P (X = 2) = P (X = 4) = = 0.095, P (X = 2) = P (X = 1) = P (X = 3) = 5 1 = 0.238. 3. The set of possible values of X is {0, 1, 2 . . . , N}. Assuming that people have the disease
independent of each other, P (X = i) = (1  p)i1 p (1  p)N 1iN i = 0. 4. Let X be the length of the side of a randomly chosen plastic die manufactured by the factory,
then P (X3 > 1.424) = P (X > 1.125) = 1.25  1.125 1 = . 1.25  1 2 64 Chapter 4 Distribution Functions and Discrete Random Variables 5. P (X < 1) = F (1) = 1/2.
P (X = 1) = F (1)  F (1) = 1/6. P (1 X < 2) = F (2)  F (1) = 1/4. P (X > 1/2) = 1  F (1/2) = 1  1/2 = 1/2. P (X = 3/2) = 0. P (1 < X 6) = F (6)  F (1) = 1  2/3 = 1/3. 6. Let F be the distribution function of X. Then 0 1/8 F (t) = 1/2 7/8 1 t <0 0t <1 1t <2 2t <3 t 3. 7. Note that X is neither continuous nor discrete. The answers are (a) F (6) = 1 implies that k(36 + 72  3) = 1; so k = 1/33. (b) F (4)  F (2) = 29/33  4/33 = 25/33. (c) 1  F (3) = 1  (24/33) = 9/33. (d)
9 29  F (4)  F (3) 5 33 33 = P (X 4  X 3) = = . 1  F (3) 6 9 1 33 8. F (Q0.5 ) = 1/2 implies that 1 + ex = 2. The only solution of this question is x = 0. So x = 0 is the median of F . Similarly, F (Q0.25 ) = 1/4 implies that 1 + ex = 4, the solution of which is x =  ln 3. F (Q0.75 ) = 3/4 implies that 1 + ex = 4/3, the solution of which is x = ln 3. So  ln 3 and ln 3 are the first and the third quartiles of F , respectively. Therefore, 50% of the years the rate at which the price of oil per gallon changes is negative or zero, 25% of the years the rate is  ln 3 1.0986 or less, and 75% of the years the rate is ln 3 1.0986 or less. 9. (a)
P (X t) = P (t X t) = P (X t)  P (X < t) = F (t)  1  P (X t) = F (t)  1  P (x t) = 2F (t)  1. (b) Using part (a), we have P (X > t) = 1  P (X t) = 1  2F (t)  1 = 2 1  F (t) . Section 4.2 Distribution Functions 65 (c) P (X = t) = 1 + P (X = t)  1 = P (X t) + P (X > t) + P (X = t)  1 = P (X t) + P (X t)  1 = P (X t) + P (X t)  1 = F (t) + F (t)  1. 10. F is a distribution function because F () = 0, F () = 1, F is right continuous, and
F (t) = 1 t e > 0 implies that F is nondecreasing. 11. F is a distribution function because F () = 0, F () = 1, F is right continuous, and
F (t) = 1 > 0 implies that it is nondecreasing. (1 + t)2 12. Clearly, F is right continuous. On t < 0 and on t 0, it is increasing, limt F (t) = 1, and limt F (t) = 0. It looks like F satisfies all of the conditions necessary to make it a distribution function. However, F (0) = 1/2 > F (0+) = 1/4 shows that F is not nondecreasing. Therefore, F is not a probability distribution function. time of the passenger is a random number from (0, 45). The waiting time is X = 45  Y . We have that for 0 t 45, P (X t) = P (45  Y t) = P (Y 45  t) = t 45  (45  t) = . 45 45 13. Let the departure time of the last flight before the passenger arrives be 0. Then Y , the arrival So F , the distribution function of X is 0 F (t) = t/45 1 t <0 0 t < 45 t 45. 14. Let X be the first twodigit number selected from the set {00, 01, 02, . . . , 99} which is between
4 and 18. Since for i = 4, 5, . . . , 18, P (X = i  4 X 18) = 1/100 1 P (X = i) = = , P (4 X 18) 15/100 15 we have that X is chosen randomly from the set {4, 5, . . . , 18}. 15. Let X be the minimum of the three numbers,
P (X < 5) = 1  P (X 5) = 1  36 3 40 3 = 0.277. 66 Chapter 4 Distribution Functions and Discrete Random Variables 16.
P (X2 5X +6 > 0) = P (X 2)(X 3) > 0 = P (X < 2)+P (X > 3) = 20 2 +0 = . 30 3 17.
F (t) = 0 t <0 0 t < 1/2 t 1/2. t 1  t 1 18. The distribution function of X is F (t) = 0 if t < 1; F (t) = 1  (89/90)n if n t < n + 1,
n 1. Since 89 90 26 is the first quartile. Since F (26) = 1  F (63) = 1 
25 = 0.244 < 0.25 < 1  89 90 26 = 0.252 = F (26), 89 62 89 = 0.4998 < 0.5 < 1  90 90 63 is the median of X. Similarly, 89 124 89 = 0.7498 < 0.75 < 1  90 90 implies that 125 is the third quartile of X. F (125) = 1  63 = 0.505 = F (63), 125 = 0.753 = F (125), 19.
G(t) = F (t) 1 t <5 t 5. 4.3 DISCRETE RANDOM VARIABLES 0 1/15 3/15 1. F , the distribution functions of X is given by if x < 1 if 1 x < 2 if 2 x < 3 F (x) = 6/15 if 3 x < 4 10/15 if 4 x < 5 1 if x 5. Section 4.3 Discrete Random Variables 67 2. p, the probability mass function of X, is given by
x p(x) 1 11/36 2 9/36 3 7/36 4 5/36 5 3/36 6 1/36 F , the probability distribution function of X, is given by 0 if x < 1 11/36 if 1 x < 2 20/36 if 2 x < 3 F (x) = 27/36 if 3 x < 4 32/36 if 4 x < 5 35/36 if 5 x < 6 1 if x 6. 3. The possible values of X are 2, 3, . . . , 12. The sample space of this experiment consists of 36
equally likely outcomes. Hence the probability of any of them is 1/36. Thus p(2) = P (X = 2) = P p(3) = P (X = 3) = P p(4) = P (X = 4) = P Similarly, i p(i) 5 4/36 6 5/36 7 6/36 8 5/36 9 4/36 10 3/36 11 2/36 12 1/36 (1, 1) = 1/36, = 2/36, = 3/36. (1, 2), (2, 1) (1, 3), (2, 2), (3, 1) 4. Let p be the probability mass function of X. We have
x p(x) 2 1/2 2 1/10 4 13/45 6 1/9 5. Let p be the probability mass function of X and q be the probability mass function of Y . We
have p(i) = 9 10
i1 1 , 10 i = 1, 2, . . . .
(j 3)/2 q(j ) = P (Y = j ) = P X = 9 j 1 = 2 10 1 , 10 j = 3, 5, 7, . . . . 6. Mode of p = 1; mode of q = 1. 68 Chapter 4 Distribution Functions and Discrete Random Variables 7. (a)
(b) (c) (d) 5 k=1 kx = 1 k = 1/15. k(1)2 + k + 4k + 9k = 1 k = 1/15. 1/9 = 8. 1  (1/9) x=1 1 2 k(1 + 2 + + n) = 1 k = = . n(n + 1) n(n + 1) /2 k =1k= x x=1 (1/9) 1 9 x 1 =1 (e) k(12 + 22 + + n2 ) = 1 k = 6 . n(n + 1)(2n + 1) 8. Let p be the probability mass function of X; then
18 i 28 12  i 46 12 p(i) = P (X = i) = i = 0, 1, 2, . . . , 12. 9. For x < 0, F (x) = 0. If x 0, for some nonnegative integer n, n x < n + 1, and we have
that
n F (x) =
i=0 3 1 4 4 i = 3 1 1 1+ + 4 4 4
n+1 2 + + 1 4 n = Thus F (x) = 3 1  (1/4)n+1 1 =1 4 1  (1/4) 4 0 1  (1/4)n+1 if x < 0 . if n x < n + 1, n = 0, 1, 2, . . . . 10. Let p be the probability mass function of X and F be its distribution function. We have
p(i) = 5 6
i1 1 , i = 1, 2, 3, . . . . 6 F (x) = 0 for x < 1. If x 1, for some positive integer n, n x < n + 1, and we have that
n F (x) =
i=1 5 6 i1 1 5 5 1 = 1+ + 6 6 6 6
n 2 + + 5 6 n1 = 1 1  (5/6)n 5 =1 6 1  (5/6) 6 . Section 4.3 Discrete Random Variables 69 Hence F (x) = 0 1  5 6
n if x < 1 if n x < n + 1, n = 1, 2, 3, . . . . 11. The set of possible values of X is {2, 3, 4, . . . }. For n 2, X = n if and only if either all of
the first n  1 bits generated are 0 and the nth bit generated is 1, or all of the first n  1 bits generated are 1 and the nth bit generated is 0. Therefore, by independence, P (X = n) = 1 2
n1 1 1 + 2 2 n1 1 1 = 2 2 n1 , n 2. 12. The event Z > i occurs if and only if Liz has not played with Bob since i Sundays ago, and
the earliest she will play with him is next Sunday. Now the probability is i/k that Liz will play with Bob if last time they played was i Sundays ago; hence i P (Z > i) = 1  , k i = 1, 2, . . . , k  1. Let p be the probability mass function of Z. Then, using this fact for 1 i k, we obtain p(i) = P (Z = i) = P (Z > i  1)  P (Z > i) = 1  i 1 i1  1 = . k k k 13. The possible values of X are 0, 1, 2, 3, 4, and 5. For i, 0 i 5,
5 i
6 Pi 9 P5i 10! 15! . P (X = i) = The numerical values of these probabilities are as follows. i P (X = i) 0 42/1001 1 252/1001 2 420/1001 3 240/1001 4 45/1001 5 2/1001 14. For i = 0, 1, 2, and 3, we have
10 i 10  i 62i 2 6  2i . 20 6 P (X = i) = The numerical values of these probabilities are as follows. i p(i) 0 112/323 1 168/323 2 42/323 3 1/323 70 Chapter 4 Distribution Functions and Discrete Random Variables 15. Clearly,
6 P (X > n) = P
i=1 Ei To calculate P E1 E2 E6 , we use the inclusionexclusion principle. To do so, we must calculate the probabilities of all possible intersections of the events from E1 , . . . , E6 , add the probabilities that are obtained by intersecting an odd number of events, and subtract all the probabilities that are obtained by intersecting an even number of events. Clearly, there 6 6 6 are terms of the form P (Ei ), terms of the form P (Ei Ej ), terms of the form 1 2 3 P (Ei Ej Ek ), and so on. Now for all i, P (Ei ) = (5/6)n ; for all i and j , P (Ei Ej ) = (4/6)n ; for all i, j , and k, P (Ei Ej Ek ) = (3/6)n ; and so on. Thus P (X > n) = P (E1 E2 E6 ) = =6 6 1 5 6 5 6
n n  6 2 4 6
n 4 6 n + 3 6 6 3
n 3 6  15 n 
n 6 4 +6 2 6 1 6 n + . 6 5 1 6 n  15 + 20 2 6 n Let p be the probability mass function of X. The set of all possible values of X is {6, 7, 8, . . . }, and p(n) = P (X = n) = P (X > n  1)  P (X > n) = 5 6
n1 5 4 6 n1 + 10 3 6 n1  10 2 6 n1 +5 1 6 n1 , n 6. 16. Put the students in some random order. Suppose that the first two students form the first team,
the third and fourth students form the second team, the fifth and sixth students form the third team, and so on. Let F stand for "female" and M stand for "male." Since our only concern is gender of the students, the total number of ways we can form 13 teams, each consisting of two students, is equal to the number of distinguishable permutations of a sequence of 23 M's 26! 26 and three F 's. By Theorem 2.4, this number is = . The set of possible values of 23! 3! 3 the random variable X is {2, 4, . . . , 26}. To calculate the probabilities associated with these values, note that for k = 1, 2, . . . , 13, X = 2k if and only if one of the following events occurs: A: B: One of the first k 1 teams is a femalefemale team, the kth team is either a malefemale or a femalemale team, and the remaining teams are all malemale teams. The first k  1 teams are all malemale teams, and the kth team is either a malefemale team or a femalemale team. Section 4.4 Expectations of Discrete Random Variables 71 To find P (A), note that for A to occur, there are k 1 possibilities for one of the first k 1 teams to be a femalefemale team, two possibilities for the kth team (malefemale and femalemale), and one possibility for the remaining teams to be all malemale teams. Therefore, P (A) = 2(k  1) . 26 3 To find P (B), note that for B to occur, there is one possibility for the first k  1 teams to be all malemale, and two possibilities for the kth team: malefemale and femalemale. The number of possibilities for the remaining 13k teams is equal to the number of distinguishable 26  2k)! permutations of two F 's and (262k)2 M's, which, by Theorem 2.4, is = 2! (26  2k  2)! 26  2k . Therefore, 2 26  2k 2 2 P (B) = . 26 3 Hence, for 1 k 13, 2(k  1) + 2 P (X = 2k) = P (A) + P (B) = 26 3 26  2k 2 1 2 1 1 k  k+ . 650 26 4 = 4.4 EXPECTATIONS OF DISCRETE RANDOM VARIABLES 1. Yes, of course there is a fallacy in Dickens' argument. If, in England, at that time there were
exactly two train accidents each month, then Dickens would have been right. Usually, for all n > 0 and for any two given days, the probability of n train accidents in day 1 is equal to the probability of n accidents in day 2. Therefore, in all likelihood the risk of train accidents on the final day in March and the risk of such accidents on the first day in April would have been about the same. The fact that train accidents occurred at random days, two per month on the average, imply that in some months more than two and in other months two or less accidents were occurring. 2. Let X be the fine that the citizen pays on a random day. Then
E(X) = 25(0.60) + 0(0.40) = 15. Therefore, it is much better to park legally. 72 Chapter 4 Distribution Functions and Discrete Random Variables 3. The expected value of the winning amount is
30 500 1 4000 + 800 + 1, 200, 000 = 0.86. 2, 000, 000 2, 000, 000 2, 000, 000 Considering the cost of the ticket, the expected value of the player's gain in one game is 1 + 0.86 = 0.14. 4. Let X be the amount that the player gains in one game, then
4 3 10 4 6 1 1 10 4 P (X = 4) = = 0.114, P (X = 9) = = 0.005, and P (X = 1) = 1  0.114  0.005 = 0.881. Thus E(X) = 1(0.881) + 4(0.114) + 9(0.005) = 0.38. Therefore, on the average, the player loses 38 cents per game. 5. Let X be the net gain in one play of the game. The set of possible values of X is {8, 4, 0, 6, 10}.
The probabilities associated with these values are 2 1 5 2 2 1 p(8) = p(0) = 1 5 2 = 1 , 10 p(4) = = 4 , 10 and p(6) = p(10) = 2 1 5 2 = 2 . Hence 10 4 1 2 2 4 1 4 +0 +6 + 10 = . 10 10 10 10 10 5 E(X) = 8 Since E(X) > 0, the game is not fair. 6. The expected number of defective items is
3 i
i=0 5 i 15 5i 20 3 = 0.75. Section 4.4 Expectations of Discrete Random Variables 73 7. For i = 4, 5, 6, 7, let Xi be the profit if i magazines are ordered. Then
E(X4 ) = E(X5 ) = 4a , 3 5a 12 4a 2a 6 + = , 3 18 3 18 3 5 6a 7 19a 6 +a + = , 18 18 3 18 18 E(X6 ) = 0 E(X7 ) =  2a 6 a 5 4a 4 7a 3 10a + + + = . 3 18 3 18 3 18 3 18 18 Since 4a/3 > 19a/18 and 4a/3 > 10a/18, either 4, or 5 magazines should be ordered to maximize the profit in the long run. 8. (a)
x=1 6 6 = 2 2x2 x=1 1 6 2 = 1. = 2 x2 6 x=1 (b) E(X) =
x=1 2 x 6 6 = 2 2x2 1 = . x 9. (a)
i=2 p(x) = E(X) = E(X2 ) = 4 1 4 9 9 + + + + = 1. 27 27 27 27 27 xp(x) = 0, E(X) = x 2 p(x) = 80/27. Hence
2 x=2 (b) 2 x=2 2 x=2 xp(x) = 44/27, E(2X2  5X + 7) = 2(80/27)  5(0) + 7 = 349/27.
10 10. Let R be the radius of the randomly selected disk; then E(2 R) = 2
i=1 i 1 = 11. 10 11. p(x) the probability mass function of X is given by
x p(x) Hence 1 1 1 5 3 +0 +3 +4 = , 8 8 4 4 8 3 1 1 1 77 E(X2 ) = 9 + 0 + 9 + 16 = , 8 8 4 4 8 E(X) = 3 3 3/8 0 1/8 3 1/4 4 1/4 74 Chapter 4 Distribution Functions and Discrete Random Variables 1 1 1 23 3 +0 +3 +4 = , 8 8 4 4 8 23 31 77 E(X2  2X) = 2 = , 8 8 8 3 1 1 1 23 E(XX) = 9 + 0 + 9 + 16 = . 8 8 4 4 8 E(X) = 3 10 10 12. E(X) =
i=1 11 1 = i and E(X2 ) = 10 2 i2 i=1 77 1 = . So 10 2 11 77  = 22. 2 2 E X(11  X) = E(11X  X2 ) = 11 13. Let X be the number of different birthdays; we have
P (X = 4) = 365 364 363 362 = 0.9836, 3654 4 365 364 363 2 = 0.0163, 3654 4 4 365 364 + 365 364 2 3 = 0.00007, 3654 365 = 0.000000021. 3654 P (X = 3) = P (X = 2) = P (X = 1) = Thus E(X) = 4(0.9836) + 3(0.0163) + 2(0.00007) + 1(0.000, 000, 021) = 3.98. 14. Let X be the number of children they should continue to have until they have one of each sex. For i 2, clearly, X = i if and only if either all of their first i  1 children are boys and the ith child is a girl, or all of their first i  1 children are girls and the ith child is a boy. Therefore, by independence, P (X = i) = 1 2
i1 i1 1 1 + 2 2 i1 1 1 = 2 2 i1 , i 2. So E(X) = i=2 1 i 2 i=1 = 1 +
i=1 i 1 2 i1 = 1 + 1 = 3. (1  1/2)2 Note that for r < 1, ir i1 = 1/[(1  r)2 ]. Section 4.4 Expectations of Discrete Random Variables 75 15. Let Aj be the event that the person belongs to a family with j children. Then
c c P (K = k) =
j =0 P (K = kAj )P (Aj ) =
j =k 1 j . j
c Therefore, E(K) = c c c kP (K = k) =
k=1 k=1 k
j =k j = j c k=1 j =k kj . j 16. Let X be the number of cards to be turned face up until an ace appears. Let A be the event
that no ace appears among the first i  1 cards that are turned face up. Let B be the event that the ith card turned face up is an ace. We have 4 P (X = i) = P (AB) = P (BA)P (A) = 52  (i  1) Therefore,
49 48 i1 . 52 i1 i E(X) =
i=1 48 4 i1 52 (53  i) i1 = 10.6. To some, this answer might be counterintuitive. 17. Let X be the largest number selected. Clearly,
P (X = i) = P (X i)  P (X i  1) = Hence
N i N n  i1 N n , i = 1, 2, . . . , N. E(X) =
i=1 i n+1 i(i  1)n 1  = n n n N N N N i=1 i n+1  i(i  1)n
N = For large N , 1 Nn N i=1 N i n+1  (i  1)n+1  (i  1)n = n+1 
i=1 Nn (i  1)n . N i=1 (i  1)n 0 N x n dx = N n+1 . n+1 76 Chapter 4 Distribution Functions and Discrete Random Variables Therefore, E(X) N n+1  N n+1 n + 1 = nN . n N n+1 18. (a) Note that 1 1 1 =  . n(n + 1) n n+1
k n=1 So 1 = n(n + 1) k n=1 1 1 1  =1 . n n+1 k+1 This implies that k p(n) = lim
n=1 k n=1 1 1 = 1  lim = 1. k k + 1 n(n + 1) Therefore, p is a probability mass function. (b) E(X) =
n=1 np(n) =
n=1 1 = , n+1 where the last equality follows since we know from calculus that the harmonic series, 1 + 1/2 + 1/3 + , is divergent. Hence E(X) does not exist. 19. By the solution to Exercise 16, Section 4.3, it should be clear that for 1 k n,
2(k  1) + 2 P (X = 2k) = 2n 3 2n  2k 2 . Hence
n n 4k(k  1) + 4k = 2n 3
n E(X) =
k=1 2kP (X = 2k) =
k=1 n 2n  2k 2 = 4 2n 3 4 2n 3 n 2
k=1 k  (4n  2)
3 k=1 k + (2n  n  1)
2 2 n=1 k = 2 n(n + 1) n(n + 1)(2n + 1) n2 (n + 1)2  (4n  2) + (2n2  n  1) 4 6 2 = (n + 1)2 . 2n  1 Section 4.5 Variances and Moments of Discrete Random Variables 77 4.5 VARIANCES AND MOMENTS OF DISCRETE RANDOM VARIABLES 1. On average, in the long run, the two businesses have the same profit. The one that has a profit
with lower standard deviation should be chosen by Mr. Jones because he's interested in steady income. Therefore, he should choose the first business. 2. The one with lower standard deviation, namely, the second device. 3. E(X) =
3 x=3 xp(x) = 1, E(X2 ) = 3 x=3 x 2 p(x) = 4. Therefore, Var(X) = 41 = 3. 4. p, the probability mass function of X is given by
x p(x) Thus 9 12 3 E(X) =  + = , 8 8 8 99 9 783 Var(X) =  = = 12.234, 8 64 64 E(X2 ) = 27 72 99 + = , 8 8 8 X = 12.234 = 3.498. 3 3/8 0 3/8 6 2/8 5. By straightforward calculations,
N E(X) =
i=1 N i 1 1 N (N + 1) N +1 = = , N N 2 2 1 N (N + 1)(2N + 1) (N + 1)(2N + 1) 1 = = , N N 6 6 E(X 2 ) =
i=1 i2 Var(X) = X = (N + 1)(2N + 1) (N + 1)2 N2  1  = , 6 4 12 N2  1 . 12 6. Clearly,
5 E(X) =
i=0 i 13 i 5 E(X2 ) =
i=0 i2 39 5i = 1.25, 52 5 13 39 i 5i = 2.426. 52 5 78 Chapter 4 Distribution Functions and Discrete Random Variables Therefore, Var(X) = 2.426  (1.25)2 = 0.864, and hence X = 0.864 = 0.9295. 7. By the Corollary of Theorem 4.2, E(X2  2X) = 3 implies that E(X2 )  2E(X) = 3.
Substituting E(X) = 1 in this relation gives E(X 2 ) = 5. Hence, by Theorem 4.3, Var(X) = E(X2 )  E(X) By Theorem 4.5, Var(3X + 5) = 9Var(X) = 9 4 = 36.
2 = 5  1 = 4. 8. Let X be Harry's net gain. Then 2 0.25 X= 0.50 0.75 with probability 1/8 with probability 3/8 with probability 3/8 with probability 1/8. Thus 3 3 1 1 + 0.25 + 0.50 + 0.75 = 0.125 8 8 8 8 1 3 3 1 E(X2 ) = (2)2 + 0.252 + 0.502 + 0.752 = 0.6875. 8 8 8 8 E(X) = 2 These show that the expected value of Harry's net gain is 12.5 cents. Its variance is Var(X) = 0.6875  0.1252 = 0.671875. 9. Note that E(X) = E(Y ) = 0. Clearly,
P X  0 t = 0 1 0 1 if t < 1 if t 1, if t < 10 if t 10. P Y  0 t = These relations, clearly, show that for all t > 0, P Y  0 t P X  0 t . Therefore, X is more concentrated about 0 than Y is. 10. (a) Let X be the number of trials required to open the door. Clearly,
P (X = x) = 1  1 n
x1 1 n , x = 1, 2, 3, . . . . Section 4.5 Variances and Moments of Discrete Random Variables 79 Thus 1 E(X) = x 1 n x=1 We know from calculus that r, r < 1, x=1 x1 1 1 = n n x 1
x=1 1 n x1 . (10) xr x1 = 1 . (1  r)2 (11) Thus x 1
x=1 1 n x1 = 1 1 1 1 n
2 = n2 . (12) Substituting (12) in (10), we obtain E(X) = n. To calculate Var(X), first we find E(X 2 ). We have E(X ) =
2 x=1 x 2 1 1 n x1 1 1 = n n x2 1 
x=1 1 n x1 . (13) Now to calculate this sum, we multiply both sides of (11) by r and then differentiate it with respect to r; we get x=1 x 2 r x1 = 1+r . (1  r)3 Using this relation in (13), we obtain 1 1 n E(X 2 ) = 1 n 1 1 n 1+1 Therefore, Var(X) = (2n2  n)  n2 = n(n  1). (b) Let Ai be the event that on the ith trial the door opens. Let X be the number of trials required to open the door. Then 1 P (X = 1) = , n 3 = 2n2  n. 80 Chapter 4 Distribution Functions and Discrete Random Variables P (X = 2) = P (Ac A2 ) = P (A2 Ac )P (Ac ) 1 1 1 = 1 n1 1 = , n1 n n P (X = 3) = P (Ac Ac A3 ) = P (A3 Ac Ac )P (Ac Ac ) 1 2 2 1 2 1 = P (A3 Ac Ac )P (Ac Ac )P (Ac ) 2 1 2 1 1 = 1 n2 n1 1 = . n2 n1 n n Similarly, P (X = i) = 1/n for 1 i n. Therefore, X is a random number selected from {1, 2, 3, . . . , n}. By Exercise 5, E(X) = (n + 1)/2 and Var(X) = (n2  1)/12. 11. For E(X3 ) to exist, we must have E X3  < . Now 3 xn p(xn ) = n=1 6 2 n=1 (1)n n n 6 = 2 n2 6 = 2 n=1 n=1 (1)n < , n n=1 whereas E X  =
3 3 xn p(xn ) n=1 6 n n = 2 2 n 1 = . n 12. For 0 < s < r, clearly,
xs max 1, xr 1 + xr , x R . Let A be the set of possible values of X and p be its probability mass function. Since the rth absolute moment of X exists, xA xr p(x) < . Now xs p(x) xA xA 1 + xr p(x) p(x) +
xA xA = xr p(x) = 1 +
xA xr p(x) < , implies that the absolute moment of order s of X also exists. 13. Var(X)=Var(Y ) implies that
E(X2 )  E(X)
2 = E(Y 2 )  E(Y ) . 2 Since E(X) = E(Y ), this implies that E(X2 ) = E Y 2 . Let P (X = a) = p1 , P (Y = a) = q1 , P (X = b) = p2 , P (Y = b) = q2 , P (X = c) = p3 ; P (Y = c) = q3 . Section 4.5 Variances and Moments of Discrete Random Variables 81 Clearly, p1 + p2 + p3 = q1 + q2 + q3 = 1. This implies (p1  q1 ) + (p2  q2 ) + (p3  q3 ) = 0. The relations E(X) = E(Y ) and E(X2 ) = E(Y 2 ) imply that ap1 + bp2 + cp3 = aq1 + bq2 + cq3 a p1 + b2 p2 + c2 p3 = a 2 q1 + b2 q2 + c2 q3 .
2 (14) These and equation (14) give us the following system of 3 equations in the 3 unknowns p1 q1 , p2  q2 , and p3  q3 . (p1  q1 ) + (p2  q2 ) + (p3  q3 ) = 0 a(p1  q1 ) + b(p2  q2 ) + c(p3  q3 ) = 0 2 a (p1  q1 ) + b2 (p2  q2 ) + c2 (p3  q3 ) = 0. In matrix form, this is equivalent to 1 1 1 p1  q1 0 a b c p2  q2 = 0 . a 2 b2 c2 0 p3  q3 Now 1 1 1 det a b c = bc2 + ca 2 + ab2  ba 2  cb2  ac2 a 2 b2 c2 = (c  a)(c  b)(b  a) = 0, since a, b, and c are three different real numbers. This implies that the matrix 1 1 1 a b c a 2 b2 c2 is invertible. Hence the solution to (15) is p1  q1 = p2  q2 = p3  q3 = 0. Therefore, p1 = q1 , p2 = q2 , p3 = q3 implying that X and Y are identically distributed. (15) 82 Chapter 4 Distribution Functions and Discrete Random Variables 14. Let
P (X = a1 ) = p1 , P (Y = a1 ) = q1 , Clearly, p1 + p2 + + pn = q1 + q2 + + qn = 1. This implies that (p1  q1 ) + (p2  q2 ) + + (pn  qn ) = 0. The relations E(Xr ) = E(Y r ), for r = 1, 2, . . . , n  1 imply that a1 p1 + a2 p2 + + an pn = a1 q1 + a2 q2 + + an qn ,
2 2 2 2 2 2 a1 p1 + a2 p2 + + an pn = a1 q1 + a2 q2 + + an qn , P (X = a2 ) = p2 , P (Y = a2 ) = q2 , ... , ... , P (X = an ) = pn ; P (Y = an ) = qn . . . .
n1 n1 n1 n1 n1 n1 a1 p1 + a2 p2 + + an pn = a1 q1 + a2 q2 + + an qn . These and the previous relation give us the following n equations in the n unknowns p1  q1 , p2  q2 , . . . , pn  qn . (p1  q1 ) + (p2  q2 ) + + (pn  qn ) = 0 a1 (p1  q1 ) + a2 (p2  q2 ) + + an (pn  qn ) = 0 2 2 2 a1 (p1  q1 ) + a2 (p2  q2 ) + + an (pn  qn ) = 0 ...................................................... n1 n1 n1 a1 (p1  q1 ) + a2 (p2  q2 ) + + an (pn  qn ) = 0 In matrix form, this is equivalent to 1 a1 2 a1 . . . 1 a2 2 a2 . . . 1 an 2 an . . . p1  q1 0 p2  q2 0 p3  q3 0 = . . . . . . . pn  qn 0 (16) n1 n1 a2 a1 n1 an Now det n1 n1 a2 a1 1 a1 2 a1 . . . 1 a2 2 a2 . . . n1 an 1 an 2 an . . . (aj  ai ) = 0, = j =n,n1,... ,2
i<j Section 4.6 Standardized Random Variables 83 since ai 's are all different real numbers. The formula for the determinant of this type of matrices is well known. These are referred to as Vandermonde determinants, after the famous French mathematicianA. T. Vandermonde (17351796). The above determinant being nonzero implies that the matrix 1 1 1 a1 a2 an 2 2 2 a a2 an 1 . . . . . . . . . n1 n1 n1 a2 an a1 is invertible. Hence the solution to (16) is p1  q1 = p2  q2 = = pn  qn = 0. Therefore, p1 = q1 , p2 = q2 , . . . , pn = qn , implying that X and Y are identically distributed. 4.6 STANDARDIZED RANDOM VARIABLES 1. Let X1 be the number of TV sets the salesperson in store 1 sells and X2 be the number of TV sets the salesperson in store 2 sells. We have that X1 = (10  13)/5 = 0.6 and X2 = (6  7)/4 = 0.25. Therefore, the number of TV sets the salesperson in store 2 sells is 0.6 standard deviations below the mean, whereas the number of TV sets the salesperson in store 2 sells is 0.25 standard deviations below the mean. So Mr. Norton should hire the salesperson who worked in store 2. 2. Let X be the final grade comparable to Velma's 82 in the midterm. We must have
82  72 X  68 = . 12 15 This gives X = 80.5. REVIEW PROBLEMS FOR CHAPTER 4 10 = 45. We have 2 i p(i) 1, 2, 16, 17 1/45 3, 4, 14, 15 2/45 5, 6, 12, 13 3/45 7, 8, 10, 11 4/45 9 5/45 1. Note that 84 Chapter 4 Distribution Functions and Discrete Random Variables 2. The answer is
1 5 9 9 4 5 2 +2 +3 +4 +5 +6 = 3.676. 34 34 34 34 34 34 3. Let N be the number of secretaries to be interviewed to find one who knows TEX. We must find the least n for which P (N n) 0.50 or 1  P (N > n) 0.50 or 1  (0.98)n 0.50. This gives (0.98)n 0.50 or n ln 0.50/ ln 0.98 = 34.31. Therefore, n = 35. t et/200 , 200 4. Let F be the distribution function of X, then
F (t) = 1  1 + Using this, we obtain P (200 X 300) = P (X 300)  P (X < 200) = F (300)  F (200) = F (300)  F (200) = 0.442  0.264 = 0.178. t 0. 5. Let X be the number of sections that will get a hard test. We want to calculate E(X). The
random variable X can only assume the values 0, 1, 2, 3, and 4; its probability mass function is given by 8 22 i 4i p(i) = P (X = i) = , i = 0, 1, 2, 3, 4, 30 4 where the numerical values of p(i)'s are as follows. i p(i) Thus E(X) = 0(0.2669) + 1(0.4496) + 2(0.2360) + 3(0.0450) + 4(0.00026) = 1.067. 0 0.2669 1 0.4496 2 0.2360 3 0.0450 4 0.0026 6. (a) 1  F (6) = 5/36. 7. We have that (b) F (9) = 76/81. (c) F (7)  F (2) = 44/49. E(X) = (15.85)(0.15) + (15.9)(0.21) + (16)(0.35) + (16.1)(0.15) + (16.2)(0.14) = 16, Var(X) = (15.85  16)2 (0.15) + (15.9  16)2 (0.21) + (16  16)2 (0.35) + (16.1  16)2 (0.15) + (16.2  16)2 (0.14) = 0.013. E(Y ) = (15.85)(0.14) + (15.9)(0.05) + (16)(0.64) + (16.1)(0.08) + (16.2)(0.09) = 16, Var(Y ) = (15.85  16)2 (0.14) + (15.9  16)2 (0.05) + (16  16)2 (0.64) + (16.1  16)2 (0.08) + (16.2  16)2 (0.09) = 0.008. Chapter 4 Review Problems 85 These show that, on the average, companies A and B fill their bottles with 16 fluid ounces of soft drink. However, the amount of soda in bottles from company A vary more than in bottles from company B. 8. Let F be the distribution function of X, Then F (t) = 0 7/30 13/30 18/30 23/30 1 t < 58 58 t < 62 62 t < 64 64 t < 76 76 t < 80 t 80. k (2t)i = 1. Therefore, k i!
2t i=0 9. (a) To determine the value of k, note that
i=0 (2t)i = 1. This i! implies that ke = 1 or k = e
2t 2t . Thus p(i) = e (2t)i . i! (b)
3 P (X < 4) =
i=0 P (X = i) = e2t 1 + 2t + 2t 2 + (4t 3 /3) , P (X > 1) = 1  P (X = 0)  P (X = 1) = 1  e2t  2te2t . 10. Let p be the probability mass function, and F be the distribution function of X. We have
1 3 p(0) = p(3) = , p(1) = p(2) = , and 8 8 0 t <0 1/8 0 t < 1 F (t) = 4/8 1 t < 2 7/8 2 t < 3 1 t 3. 11. (a) The sample space has 52! elements because when the cards are dealt face down, any
ordering of the cards is a possibility. To find p(j ), the probability that the 4th king 4 will appear on the j th card, we claim that in (j  1) P 48! ways the 4th king 3 1 will appear on the j th card, and the remaining 3 kings earlier. To see this, note that 86 Chapter 4 Distribution Functions and Discrete Random Variables 4 combinations for the king that appears on the j th card, and (j  1) P 3 1 different permutations for the remaining 3 kings that appear earlier. The last term 48!, is for the remaining 48 cards that can appear in any order in the remaining 48 positions. Therefore, we have 4 (j  1) P 48! 3 1 = 52! j 1 3 52! 4! 48! 51 3
52 p(j ) = = j 1 3 . 52 4 (b) (c) The probability that the player wins is p(52) = To find 52 = 1/13. 4 1 j 1 jp(j ) = j , E= 52 3 j =4 j =4 4 the expected length of the game, we use a technique introduced by Jenkyns and Muller in Mathematics Magazine, 54, (1981), page 203. We have the following relation which can be readily checked. j This gives
52 52 j j 1 j 1 4 (j + 1) j = 5 4 4 3 , j 5. j
j =5 j 1 4 = 5 3 52 (j + 1)
j =5 j j 1  j 4 4 j =5 = 11, 478, 736, 52 4 52 4 = 53 5 5 4 4 where the nexttothelast equality follows because terms cancel out in pairs. Thus E= j 1 1 j 1 1 4+ j = j 52 52 3 3 j =4 j =5 4 4 1 (4 + 11, 478, 736) = 42.4. 52 4
52 52 = As Jenkyns and Muller have noted, "This relatively high expectation value is what makes the game interesting. However, the low probability of winning makes it frustrating!" Chapter 5 Special Discrete Distributions
5.1 1. BERNOULLI AND BINOMIAL RANDOM VARIABLES
8 4 1 4
4 3 4 4 = 0.087. 2. (a) 64 1 = 32. 2 1 (b) 6 + 1 = 4 (note that we should count the mother of the family as well). 2 6 3 6 2 5 2 1 6 1 10 10 30
3 3. 4. 5. 5 6
2 3 = 0.054.
4 9 10 20 30 = 0.098. = 0.33. 2 3 6. Let X be the number of defective nails. If the manufacturer's claim is true, we have
P (X 2) = 1  P (X = 0)  P (X = 1) 24 24 =1 (0.03)0 (0.97)24  (0.03)(0.97)23 = 0.162. 0 1 This shows that there is 16.2% chance that two or more defective nails is found. Therefore, it is not fair to reject company's claim. 7. Let p and q be the probability mass functions of X and Y , respectively. Then
p(x) = 4 (0.60)x (0.40)4x , x x = 0, 1, 2, 3, 4; 88 Chapter 5 Special Discrete Distributions q(y) = P (Y = y) = P X = =
8 y1 2 y = 1, 3, 5, 7, 9. 4
y1 2 (0.60)(y1)/2 (0.40)4[(y1)/2] , 8.
i=0 15 (0.8)i (0.2)15i = 0.142. i 11 36 5 0
5 9. 10 5 25 36 1 3
0 5 = 0.108. 2 3
5 10. (a) 1   5 1 1 3 1 2 3 4 = 0.539. (b) 5 2 1 10 2 9 10 3 = 0.073. 11. We know that p(x) is maximum at [(n + 1)p]. If (n + 1)p is an integer, p(x) is maximum at
[(n + 1)p] = np + p. But in such a case, some straightforward algebra shows that n n p np+p1 (1  p)nnpp+1 , p np+p (1  p)nnpp = np + p  1 np + p implying that p(x) is also maximum at np + p  1. 12. The probability of royal or straight flush is 40
the average, n 40 gives n = 52 5 2 3
3 52 5 52 . If Ernie plays n games, he will get, on 5 52 royal or straight flushes. We want to have 40n = 1; this 5 40 = 64, 974. = 0.219. 13. 6 3 1 3 3 14. 1  (999/1000)100 = 0.095. 15. The maximum occurs at k = [11(0.45)] = 4. The maximum probability is
10 (0.45)4 (0.55)6 = 0.238. 4 16. Call the event of obtaining a full house success. X, the number of full houses is n independent
poker hands is a binomial random variable with parameters (n, p), where p is the probability 52 that a random poker hand is a full house. To calculate p, note that there are possible 5 4 4 13! 52 poker hands and = 3744 full houses. Thus p = 3744 0.0014. Hence 3 2 11! 5 Section 5.1 Bernoulli and Binomial Random Variables 89 E(X) = np 0.0014n and Var(X) = np(1p) 0.00144n. Note that if n is approximately 715, then E(X) = 1. Thus we should expect to find, on the average, one full house in every 715 random poker hands. 17. 1  18. 1  6 6 1 4 6 3 4 0  6 5 1 4 5 3 0.995. 4 3000 3000 (0.0005)0 (0.9995)3000  (0.0005)(0.9995)2999 0.442. 0 1 19. The expected value of the expenses if sent in one parcel is
45.20 0.07 + 5.20 0.93 = 8. The expected value of the expenses if sent in two parcels is (23.30 2)(0.07)2 + (23.30 + 3.30) 2 (0.07)(0.93) + (6.60)(0.93)2 = 9.4. 1 Therefore, it is preferable to send in a single parcel. 20. Let n be the minimum number of children they should plan to have. Since the probability of all
girls is (1/2)n and the probability of all boys is (1/2)n , we must have 1(1/2)n (1/2)n 0.95. ln 0.05 This gives (1/2)n1 0.05 or n  1 = 4.32 or n 5.32. Therefore, n = 6. ln(0.5) N p(1  p)N1 = Np(1  p)N1 . 1 21. (a) For this to happen, exactly one of the N stations has to attempt transmitting a message.
The probability of this is (b) Let f (p) = Np(1  p)N 1 . The value of p which maximizes the probability of a message going through with no collision is the root of the equation f (p) = 0. Now f (p) = N (1  p)N1  Np(N  1)(1  p)N2 = 0. Noting that p = 1, this equation gives p = 1/N. This answer makes a lot of sense because at every "suitable instance," on average, Np = 1 station will transmit a message. (c) By part (b), the maximum probability is f 1 N =N 1 N 1 1 N
N1 = 1 1 N N1 . As N , this probability approaches 1/e, showing that for large numbers of stations (in reality 20 or more), the probability of a successful transmission is approximately 1/e independently of the number of stations if p = 1/N . 90 Chapter 5 Special Discrete Distributions 22. The k students whose names have been called are not standing. Let A1 , A2 , . . . , Ank be the students whose names have not been called. For i, 1 i n  k, call Ai a "success," if he or she is standing; failure, otherwise. Therefore, whether Ai is standing or sitting is a Bernoulli trial, and hence the random variable X is the number of successes in n  k Bernoulli trials. For X to be binomial, for i = j , the event that Ai is a success must be independent of the event that Aj is a success. Furthermore, the probability that Ai is a success must be the same for all i, 1 i n  k. The latter condition is satisfied since Ai is standing if and only if his original seat was among the first k. This happens with probability p = k/n regardless of i . However, the former condition is not valid. The relation P Aj is standing  Ai is standing = k1 , n shows that given Ai is a success changes the probability that Aj is success. That is, Ai being a success is not independent of Aj being a success. This shows that X is not a binomial random variable. 23. Let X be the number of undecided voters who will vote for abortion. The desired probability
is n + (b  a) P b + (n  X) > a + X = P X < = 2 = 1 2 [ n+(ba) ] 2 n
i=0 [ n+(ba) ] 2
i=0 n i 1 2 i 1 2 ni n . i 24. Let X be the net gain of the player per unit of stake. X is a discrete random variable with
possible values 1, 1, 2, and 3. We have 3 0 3 P (X = 1) = 1 3 P (X = 2) = 2 3 P (X = 3) = 3 1 6 1 6 1 6 1 6
0 P (X = 1) = 2 3 5 3 125 = , 6 216 75 5 2 = , 6 216 15 5 = , 6 216 5 0 1 = . 6 216 75 15 1 125 +1 +2 +3 0.08. 216 216 216 216 Therefore, the player loses 0.08 per unit stake. E(X) = 1 Hence Section 5.1 Bernoulli and Binomial Random Variables 91 25.
n E(X ) =
2 x=1 n x 2 n x p (1  p)nx = x n (x 2  x + x)
x=1 n n x p (1  p)nx x =
x=1 n x(x  1) n x n x p (1  p)nx + x p (1  p)nx x x x=1 =
x=2 n! px (1  p)nx + E(X) (x  2)! (n  x)!
n = n(n  1)p2
x=2 n  2 x2 p (1  p)nx + np x2
n2 = n(n  1)p2 p + (1  p) + np = n2 p2  np2 + np. 26. (a) A fourengine plane is preferable to a twoengine plane if and only if
1 4 0 4 2 0 p (1  p)4  p(1  p)3 > 1  p (1  p)2 . 0 1 0 This inequality gives p > 2/3. Hence a fourengine plane is preferable if and only if p > 2/3. If p = 2/3, it makes no difference. (b) A fiveengine plane is preferable to a threeengine plane if and only if 5 5 5 4 5 3 3 2 p (1  p)0 + p (1  p) + p (1  p)2 > p (1  p) + p 3 . 5 4 3 2 Simplifying this inequality, we get 3(p  1)2 (2p  1) 0 which implies that a fiveengine plane is preferable if and only if 2p  1 0. That is, for p > 1/2, a fiveengine plane is preferable; for p < 1/2, a threeengine plane is preferable; for p = 1/2 it makes no difference. 27. Clearly, 8 bits are transmitted. A parity check will not detect an error in the 7bit character
received erroneously if and only if the number of bits received incorrectly is even. Therefore, the desired probability is
4 n=1 8 (1  0.999)2n (0.999)82n = 0.000028. 2n 28. The message is erroneously received but the errors are not detected by the paritycheck if for
1 j 6, j of the characters are erroneously received but not detected by the paritycheck, and the remaining 6j characters are all transmitted correctly. By the solution of the previous exercise, the probability of this event is
6 (0.000028)j (0.999)8(6j ) = 0.000161.
j =1 92 Chapter 5 Special Discrete Distributions 29. The probability of a straight flush is 40
1 This gives 52 0.000015391. Hence we must have 5 n 3 (0.000015391)0 (1  0.000015391)n . 0 4 1 (1  0.000015391)n . 4 So n Therefore, n 90, 072. log(1/4) 90071.06. log(1  0.000015391) 30. Let p, q, and r be the probabilities that a randomly selected offspring is AA, Aa, and aa,
respectively. Note that both parents of the offspring are AA with probability (/n)2 , they are 2 both Aa with probability 1  (/n) , and the probability is 2(/n) 1  (/n) that one parent is AA and the other is Aa. Therefore, by the law of total probability, p =1 n q =0 n r =0 n
2 2 1 1 1 2 1 1 1 + 2 + 1 = + , 4 n 2 n n 4 n 2 n 4 2 1 2 1 1 1 2 + 1 1 =  + 2 , 2 n 2 n n 2 2 n 2 1 2 1 2 + 1 +02 . 1 = 1 4 n n n 4 n + The probability that at most two of the offspring are aa is
2 i=0 m i r (1  r)mi . i The probability that exactly i of the offspring are AA and the remaining are all Aa is m i mi pq . i 31. The desired probability is the sum of three probabilities: probability of no customer served and
two new arrivals, probability of one customer served and three new arrivals, and probability of two customers served and four new arrivals. These quantities, respectively, are (0.4)4 4 4 4 4 (0.45)2 (0.55)2 , (0.6)(0.4)3 (0.45)3 (0.55), and (0.6)2 (0.4)2 (0.45)4 . The 2 1 3 2 sum of these quantities, which is the answer, is 0.054. Section 5.1 Bernoulli and Binomial Random Variables 93 32. (a) Let S be the event that the first trial is a success and E be the event that in n trials, the
number of successes is even. Then P (E) = P (ES)P (S) + P (ES c )P (S c ). Thus rn = (1  rn1 )p + rn1 (1  p). Using this relation, induction, and r0 = 1, we find that rn = 1 1 + (1  2p)n . 2 (b) The left sum is the probability of 0, 2, 4, . . . , or [n/2] successes. Thus it is the probability of an even number of successes in n Bernoulli trials and hence it is equal to rn . 33. For 0 i n, let Bi be the event that i of the balls are red. Let A be the event that in drawing
k balls from the urn, successively, and with replacement, no red balls appear. Then P (B0 A) = P (AB0 )P (B0 )
n = 1
n 1 2
k n P (ABi )P (Bi )
i=0 i=0 ni n n i 1 2 n = 1
n i=0 n i ni n k . 34. Let E be the event that Albert's statement is the truth and F be the event that Donna tells the
truth. Since Rose agrees with Donna and Rose always tells the truth, Donna is telling the truth as well. Therefore, the desired probability is P (E  F ) = P (EF )/P (F ). To calculate P (F ), observe that for Rose to agree with Donna, none, two, or all four of Albert, Brenda, Charles, and Donna should have lied. Since these four people lie independently, this will happen with probability 1 4 4 2 2 1 2 2 4 41 + + = . 2 3 3 3 3 81 To calculate P (EF ), note that EF is the event that Albert tells the truth and Rose agrees with Donna. This happens if all of them tell the truth, or Albert tells the truth but exactly two of Brenda, Charles and Donna lie. Hence P (EF ) = Therefore, P (E  F ) = P (EF ) 13/81 13 = = = 0.317. P (F ) 41/81 41 1 3
4 + 1 3 3 2 2 3 2 1 13 = . 3 81 94 Chapter 5 Special Discrete Distributions 5.2 POISSON RANDOM VARIABLES
e3 30 = 1  e3 = 0.9502. 0! 1. = (0.05)(60) = 3; the answer is 1  2. = 1.8; the answer is
3 i=0 e1.8 (1.8)i 0.89. i! e2 20 e2 21  = 1  3e2 = 0.594. 0! 1! e0.7 (0.7)0 e0.7 (0.7)1  0.156. 0! 1! 3. = 0.025 80 = 2; the answer is 1  4. = (500)(0.0014) = 0.7. The answer is 1  5. We call a room "success" if it is vacant next Saturday; we call it "failure" if it is occupied.
Assuming that next Saturday is a random day, X, the number of vacant rooms on that day is approximately Poisson with rate = 35. Thus the desired probability is
29 1
i=0 e35 (35)i = 0.823. i! e10.5 (10.5)10 = 10! 6. = (3/10)35 = 10.5. The probability of 10 misprints in a given chapter is
0.124. Therefore, the desired probability is (0.124)2 = 0.0154. 7. P (X = 1) = P (X = 3) implies that e =
is e 6 6 5 e 3 from which we get = 6. The answer 3! 5! = 0.063. en/k (n/k)0 = en/k . So the answer is 0! 8. The probability that a bun contains no raisins is
4 2n/k e (1  en/k )2 . 2 9. Let X be the number of times the randomly selected kid has hit the target. We are given that
P (X = 0) = 0.04; this implies that Now e 20 = 0.04 or e = 0.04. So =  ln 0.04 = 3.22. 0! e 1! P (X 2) = 1  P (X = 0)  P (X = 1) = 1  0.04  = 1  0.04  (0.04)(3.22) = 0.83. Therefore, 83% of the kids have hit the target at least twice. Section 5.2 Poisson Random Variables 95 10. First we calculate pi 's from binomial probability mass function with n = 26 and p = 1/365.
Then we calculate them from Poisson probability mass function with parameter = np = 26/365. For different values of i, the results are as follows. i 0 1 2 3 Binomial 0.93115 0.06651 0.00228 0.00005 Poisson 0.93125 0.06634 0.00236 0.00006. Remark: In this example, since success is very rare, even for small n's Poisson gives good approximation for binomial. The following table demonstrates this fact for n = 5. i 0 1 2 Binomial 0.9874 0.0136 0.00007 Poisson 0.9864 0.0136 0.00009. 11. Let N (t) be the number of shooting stars observed up to time t. Let one minute be the unit of
time. Then N (t) : t 0 is a Poisson process with = 1/12. We have that P N (30) = 3 = e30/12 (30/12)3 = 0.21. 3! 12. P N (2) = 0 = e3(2) = e6 = 0.00248. 13. Let N (t) be the number of wrong calls up to t. If one day is taken as the time unit, it is reasonable
to assume that N (t) : t 0 is a Poisson process with = 1/7. By the independent increment property and stationarity, the desired probability is P N (1) = 0 = e(1/7)1 = 0.87. 14. Choose one month as the unit of time. Then = 5 and the probability of no crimes during
any given month of a year is P N (1) = 0 = e5 = 0.0067. Hence the desired probability is 12 (0.0067)2 (1  0.0067)10 = 0.0028. 2 15. Choose one day as the unit of time. Then = 3 and the probability of no accidents in one day
is P N (1) = 0 = e3 = 0.0498. The number of days without any accidents in January is approximately another Poisson random variable with approximate rate 31(0.05) = 1.55. Hence the desired probability is e1.55 (1.55)3 0.13. 3! 96 Chapter 5 Special Discrete Distributions 16. Choosing one hours as time unit, we have that = 6. Therefore, the desired probability is
P N (0.5) = 1 and N (2.5) = 10 = P N (0.5) = 1 and N (2.5)  N (0.5) = 9 = P N (0.5) = 1 P N (2.5)  N (0.5) = 9 = P N (0.5) = 1 P N (2) = 9 = 31 e3 129 e12 0.013. 1! 9! 17. The expected number of fractures per meter is = 1/60. Let N (t) be the number of fractures
in t meters of wire. Then et/60 (t/60)n , n = 0, 1, 2, . . . . n! In a ten minute period, the machine turns out 70 meters of wire. The desired probability, P N (70) > 1 is calculated as follows: P N (t) = n = P N (70) > 1 = 1  P N (70) = 0  P N (70) = 1 70 = 1  e70/60  e70/60 0.325. 60 18. Let the epoch at which the traffic light for the leftturn lane turns red be labeled t = 0. Let
N (t) be the number of cars that arrive at the junction at or prior to t trying to turn left. Since cars arrive at the junction according to a Poisson process, clearly, N (t) : t 0 is a stationary and orderly process which possesses independent increments. Therefore, N (t) : t 0 is also a Poisson process. Its parameter is given by = E N (1) = 4(0.22) = 0.88. (For a rigorous proof, see the solution to Exercise 9, Section 12.2.) Thus P N (t) = n = and the desired probability is
3 e(0.88)t (0.88)t n! e(0.88)3 (0.88)3 n! n , n P N (3) 4 = 1 
n=0 0.273. 19. Let X be the number of earthquakes of magnitude 5.5 or higher on the Richter scale during the next 60 years. Clearly, X is a Poisson random variable with parameter = 6(1.5) = 9. Let A be the event that the earthquakes will not damage the bridge during the next 60 years. Since the events {X = i}, i = 0, 1, 2, . . . , are mutually exclusive and {X = i} is the sample i=1 space, by the Law of Total Probability (Theorem 3.4), P (A) =
i=0 P (A  X = i)P (X = i) =
i=0 (1  0.015)i (0.985)(9) i!
i e9 9i i! =
i=0 (0.985)i e 9 9 i i! = e9 = e9 e(0.985)(9) = 0.873716. i=0 Section 5.2 Poisson Random Variables 97 20. Let N be the total number of letter carriers in America. Let n be the total number of dog bites
letter carriers sustain. Let X be the number of bites a randomly selected letter carrier, say Karl, sustains on a given year. Call a bite "success," if it is Karl that is bitten and failure if anyone but Karl is bitten. Since the letter carriers are bitten randomly, it is reasonable to assume that X is approximately a binomial random variable with parameters n and p = 1/N . Given that n is large (it was more than 7000 in 1983 and at least 2,795 in 1997), 1/N is small, and n/N is moderate, X can be approximated by a Poisson random variable with parameter = n/N. We know that P (X = 0) = 0.94. This implies that (e 0 )/0! = 0.94. Thus e = 0.94, and hence =  ln 0.94 = 0.061875. Therefore, X is a Poisson random variable with parameter 0.061875. Now P X>1X1 = P (X > 1) 1  P (X = 0)  P (X = 1) = P (X 1) 1  P (X = 0) 1  0.94  0.0581625 = 0.030625, 1  0.94 = where e 1 = e = (0.061875)(0.94) = 0.0581625. 1! Therefore, approximately 3.06% of the letter carriers who sustained one bite, will be bitten again. P (X = 1) = 21. We should find n so that 1  enM/N (nM/N )0 . This gives n N ln(1  )/M. The 0! answer is the least integer greater than or equal to N ln(1  )/M. 22. (a) For each kcombination n1 , n2 , . . . , nk of 1, 2, . . . , n, there are (n  1)nk distributions
with exactly k matches, where the matches occur at n1 , n2 , . . . , nk . This is because each of the remaining n  k balls can be placed into any of the cells except the cell that has the same n number as the ball. Since there are kcombinations n1 , n2 , . . . , nk of 1, 2, . . . , n, the total k number of ways we can place the n balls into the n cells so that there are exactly k matches is n (n  1)nk n k (n  1)nk . Hence the desired probability is . k nn (b) Let X be the number of matches. We will show that limn P (X = k) = e1 /k!; that is, X is Poisson with parameter 1. We have n (n  1)nk n k = lim n n k n
n n lim P (X = k) = lim = lim n n1 n n (n  1)k 1 n! 1 1 n k! (n  k)! n 1 1 = e1 (n  1)k k! 98 Chapter 5 Special Discrete Distributions Note that limn 1  formula, 1 n n = e1 , and lim n n! = 1, since by Stirling's (n  k)! (n  1)k n! 2 n nn en lim = lim n (n  k)! (n  1)k n 2(n  k) (n  k)nk e(nk) (n  1)k = lim
n n (n  k)k 1 nn n (n  1)k ek n  k (n  k) 1 = 1, ek
n = 1 ek 1 where (n  k)n k nn ek because = 1 n n (n  k) n n ek . 23. (a) The probability of an even number of events in (t, t + ) is n=0 e ()2n = e (2n)! = e n=0 1 ()2n = e (2n)! 2 n=0 ()n 1 + n! 2 n=0 ()n n! 1 1  1 e + e = (1 + e2 ). 2 2 2 (b) The probability of an odd number of events in (t, t + ) is n=1 e ()2n1 = e (2n  1)! = e n=1 ()2n1 1 = e (2n  1)! 2 n=0 ()n 1  n! 2 n=0 ()n n! 1 1  1 = 1  e2 . e  e 2 2 2 24. We have that
P N1 (t) = n, N2 (t) = m =
i=0 P N1 (t) = n, N2 (t) = m  N (t) = i P N (t) = i = P N1 (t) = n, N2 (t) = m  N (t) = n + m P N (t) = n + m = Therefore, et (t)n+m n+m n . p (1  p)m (n + m)! n P N1 (t) = n =
m=0 P N1 (t) = n, N2 (t) = m Section 5.3 Other Discrete Random Variables 99 =
m=0 n+m n et (t)n+m p (1  p)m n (n + m)! = (n + m)! n etp et (1p) (t)n (t)m p (1  p)m n! m! (n + m)! m=0 m etp et (1p) (tp)n t (1  p) = n! m! m=0 = = etp (tp)n n! et (1p) t (1  p) m! m=0 m etp (tp)n . n! It can easily be argued that the other properties of Poisson process are also satisfied for the process N1 (t) : t 0 . So N1 (t) : t 0 is a Poisson process with rate p. By symmetry, N2 (t) : t 0 is a Poisson process with rate (1  p). 25. Let N (t) be the number of females entering the store between 0 and t. By Exercise 24,
e15(2/3) 15(2/3) P N (15) = 15 = 15!
15 N (t) : t 0 is a Poisson process with rate 1 (2/3) = 2/3. Hence the desired probability is = 0.035. 26. (a) Let A be the region whose points have a (positive) distance d or less from the given tree.
The desired probability is the probability of no trees in this region and is equal to
2 ed ( d 2 )0 = e d . 0!
2 (b) We want to find the probability that the region A has at most n  1 trees. The desired quantity is n1 d 2 e ( d 2 )i . i! i=0 27. p(i) = (/ i)p(i  1) implies that for i < , the function p is increasing and for i > it is
decreasing. Hence i = is the maximum. 5.3 OTHER DISCRETE RANDOM VARIABLES 1. Let D denote a defective item drawn, and N denote a nondefective item drawn. The answer
is S = N N N, DN N, N DN, N N D, N DD, DN D, DDN . 100 Chapter 5 Special Discrete Distributions 2. S = ss, f ss, sf s, sff s, ff ss, f sf s, sfff s, f sff s, fff ss, ff sf s, . . . . 3. (a) 1/(1/12) = 12. (b)
11 12
2 1 0.07. 12 4. (a) (1  pq)r1 pq. (b) 1/pq. 5.
7 (0.2)3 (0.8)5 0.055. 2 6. (a) (0.55)5 (0.45) 0.023. (b) (0.55)3 (0.45)(0.55)3 (0.45) 0.0056. 7.
5 1 45 7 50 = 0.42. 8 8. The probability that at least n light bulbs are required is equal to the probability that the first
n  1 light bulbs are all defective. So the answer is p n1 . 9. We have
P (N = n) = P (X = x) n1 x p (1  p)nx x x1 = . n x n p (1  p)nx x 10. Let X be the number of the words the student had to spell until spelling a word correctly. The
random variable X is geometric with parameter 0.70. The desired probability is given by
4 P (X 4) =
i=1 (0.30)i1 (0.70) = 0.9919. 11. The average number of digits until the fifth 3 is 5/(1/10) = 50. So the average number of
digits before the fifth 3 is 49. 12. The probability that a random bridge hand has three aces is
4 3 48 10 52 13 p= = 0.0412. Therefore, the average number of bridge hands until one has three aces is 1/p = 1/0.0412 = 24.27. 13. Either the (N + 1)st success must occur on the (N + M  m + 1)st trial, or the (M + 1)st Section 5.3 Other Discrete Random Variables 101 failure must occur on the (N + M  m + 1)st trial. The answer is N +M m N 1 2
N+Mm+1 + N +M m M 1 2 N+Mm+1 . 14. We have that X + 10 is negative binomial with parameters (10, 0.15). Therefore, i 0,
P (X = i) = P (X + 10 = i + 10) = i+9 (0.15)10 (0.85)i . 9 15. Let X be the number of good diskettes in the sample. The desired probability is
10 90 1 9 100 10 90 10 10 0 100 10 P (X 9) = P (X = 9) + P (X = 10) = + 0.74. 16. We have that 560(0.35) = 196 persons make contributions. So the answer is
364 15 560 15 364 196 14 1 560 15 1  = 0.987. 17. The transmission of a message takes more than t minutes, if the first [t/2] + 1 times it is sent it will be garbled, where [t/2] is the greatest integer less than or equal to t/2. The probability of this is p[t/2]+1 . 18. The probability that the sixth coin is accepted on the nth try is
n1 (0.10)6 (0.90)n6 . 5 Therefore, the desired probability is n=50 n1 n1 (0.10)6 (0.90)n6 = 1  (0.10)6 (0.90)n6 = 0.6346. 5 5 n=6 49 19. The probability that the station will successfully transmit or retransmit a message is (1p)N1 .
This is because for the station to successfully transmit or retransmit its message, none of the other stations should transmit messages at the same instance. The number of transmissions and retransmissions of a message until the success is geometric with parameter (1  p)N1 . Therefore, on average, the number of transmissions and retransmissions is 1/(1  p)N1 . 102 Chapter 5 Special Discrete Distributions 20. If the fifth tail occurs after the 14th trial, ten or more heads have occurred. Therefore, the fifth
tail occurs before the tenth head if and only if the fifth tail occurs before or on the 14th flip. Calling tails success, X, the number of flips required to get the fifth tail is negative binomial with parameters 5 and 1/2. The desired probability is given by
14 14 P (X = n) =
n=5 n=5 n1 4 1 2 5 1 2 n5 0.91. 21. The probability of a straight is
10 45  40 = 0.003924647. 52 5 Therefore, the expected number of poker hands required until the first straight is 1/0.003924647 = 254.80. 22. (a) Since P (X = n  1) 1 = > 1, P (X = n) 1p P (X = n) is a decreasing function of n; hence its maximum is at n = 1. (b) The probability that X is even is given by P (X = 2k) =
k=1 k=1 p(1  p)2k1 = p(1  p) 1p . = 2 1  (1  p) 2p (c) We want to show the following: Let X be a discrete random variable with the set of possible values 1, 2, 3 . . . . If for all positive integers n and m, P (X > n + m  X > m) = P (X > n), (17) then X is a geometric random variable. That is, there exists a number p, 0 < p < 1, such that P (X = n) = p(1  p)n1 . To prove this, note that (17) implies that for all positive integers n and m, P (X > n + m) = P (X > n). P (X > m) Therefore, P (X > n + m) = P (X > n)P (X > m). (19) (18) Section 5.3 Other Discrete Random Variables 103 Let p = P (X = 1); using induction, we prove that (18) is valid for all positive integers n. To show (18) for n = 2, note that (19) implies that P (X > 2) = P (X > 1)P (X > 1). Since P (X > 1) = 1  P (X = 1) = 1  p, this relation gives 1  P (X = 1)  P (X = 2) = (1  p)2 , or 1  p  P (X = 2) = (1  p)2 , which yields P (X = 2) = p(1  p), so (18) is also true for n = 2. Now assume that (18) is valid for all positive integers i, i n; that is, assume that P (X = i) = p(1  p)i1 , i n. (20) We will show that (18) is true for n + 1. The induction hypothesis [relation (20)] implies that
n n P (X n) =
i=1 P (X = i) =
i=1 n p(1  p)i1 = p 1  (1  p)n = 1  (1  p)n . 1  (1  p) So P (X > n) = (1  p) and, similarly, P (X > n  1) = (1  p)n1 . Now (19) yields P (X > n + 1) = P (X > n)P (X > 1), which implies that 1  P (X n)  P (X = n + 1) = (1  p)n (1  p). Substituting P (X n) = 1  (1  p)n in this relation, we obtain P (X = n + 1) = p(1  p)n , which establishes (18) for n + 1. Therefore, we have what we wanted to show. 23. Consider a coin for which the probability of tails is 1  p and the probability of heads is p.
In successive and independent flips of the coin, let X1 be the number of flips until the first head, X2 be the total number of flips until the second head, X3 be the total number of flips until the third head, and so on. Then the length of the first character of the message and X1 are identically distributed. The total number of the bits forming the first two characters of the message and X2 are identically distributed. The total number of the bits forming the first three characters of the message and X3 are identically distributed, and so on. Therefore, the total number of the bits forming the message has the same distribution as Xk . This is negative binomial with parameters k and p. 104 Chapter 5 Special Discrete Distributions 24. Let X be the number of cartons to be opened before finding one without rotten eggs. X is not a
geometric random variable because the number of cartons is limited, and one carton not having rotten eggs is not independent of another carton not having rotten eggs. However, it should be 1000 1200 obvious that a geometric random variable with parameter p = = 0.1109 is 12 12 a good approximation for X. Therefore, we should expect approximately 1/p = 1/0.1109 = 9.015 cartons to be opened before finding one without rotten eggs. 25. Either the N th success should occur on the (2N  M)th trial or the N th failure should occur
on the (2N  M)th trial. By symmetry, the answer is 2 2N  M  1 N 1 1 2
N 1 2 NM = 2N  M  1 N 1 1 2 2NM1 . 26. The desired quantity is 2 times the probability of exactly N successes in (2N  1) trials and
failures on the (2N )th and (2N + 1)st trials: 2 2N  1 N 1 2
N 1 1 2 (2N1)N 1 1 2 2 = 2N  1 N 1 2 2N . 27. Let X be the number of rolls until Adam gets a six. Let Y be the number of rolls of the die until Andrew rolls an odd number. Since the events (X = i), 1 i < , form a partition of the sample space, by Theorem 3.4, P Y >X =
i=1 P Y >XX=i P X=i =
i=1 P Y >i P X=i 5 12 5 1 12 =
i=1 1 2 i 5 6 i1 6 1 1 = 6 5 6 i=1 5 12 i = 1 5 1 = , 7 where P (Y > i) = (1/2)i since for Y to be greater than i, Andrew must obtain an even number on each of the the first i rolls. 28. The probability of 4 tagged trout among the second 50 trout caught is
pn = 50 4 n  50 46 . n 50 It is logical to find the value of n for which pn is maximum. (In statistics this value is called the maximum likelihood estimate for the number of trout in the lake.) To do this, note that pn (n  50)2 = . pn1 n(n  96) Section 5.3 Other Discrete Random Variables 105 Now pn pn1 if and only if (n  50)2 n(n  96), or n 625. Therefore, n = 625 makes pn maximum, and hence there are approximately 625 trout in the lake. 29. (a) Intuitively, it should be clear that the answer is D/N . To prove this, let Ej be the event of
obtaining exactly j defective items among the first (k  1) draws. Let Ak be the event that the kth item drawn is defective. We have
k1 k1 P (Ak ) =
j =0 P (Ak  Ej )P (Ej ) =
j =0 Dj N k+1 D j N D k1j . N k1 Now (D  j ) and (N  k + 1) Therefore,
k1 D D1 =D j j N N 1 =N . k1 k1 D P (Ak ) =
j =0 D1 j N D k1j N 1 N k1 D1 j D = N k1 j =0 D1 j N D k1j N 1 k1 = D , N where
k1 j =0 N D k1j N 1 k1 =1 N D k1j since is the probability mass function of a hypergeometric random N 1 k1 variable with parameters N  1, D  1, and k  1. (b) Intuitively, it should be clear that the answer is (D  1)/(N  1). To prove this, let Ak be as before and let Fj be the event of exactly j defective items among the first (k  2) draws. Let B be the event that the (k  1)st and the kth items drawn are defective. We have
k2 D1 j P (B) =
j =0 P (B  Fj )P (Fj ) 106 Chapter 5 Special Discrete Distributions k2 =
j =0 (D  j )(D  j  1) (N  k + 2)(N  k + 1) D j N D k2j N k2 k2 D(D  1) =
j =0 N D k2j N 2 N (N  1) k2 D2 j N D k2j N 2 k2 D2 j D(D  1) = N (N  1) = D(D  1) . N (N  1) k2 j =0 Using this, we have that the desired probability is D(D  1) N (N  1) P (B) P (Ak Ak1 ) D1 = P (Ak  Ak1 ) = = . = P (Ak1 ) P (Ak1 ) N 1 D N REVIEW PROBLEMS FOR CHAPTER 5
20 1.
i=12 20 (0.25)i (0.75)20i = 0.0009. i 2. N (t), the number of customers arriving at the post office at or prior to t is a Poisson process
with = 1/3. Thus
6 6 P N (30) 6 =
i=0 P N (30) = i =
i=0 e(1/3)30 (1/3)30 i! i = 0.130141. 3. 4 2 8 = 1.067. 30 12 (0.30)i (0.70)12i = 0.253. i 4.
i=0 Chapter 5 Review Problems 107 5. 5 (0.18)2 (0.82)3 = 0.179. 2
1999 6.
i=2 i1 21 160 i 1 1000 2 999 1000 i2 = 0.59386. 12 7.
i=7 200 12  i 360 12 = 0.244. 8. Call a train that arrives between 10:15 A.M. and 10:28 A.M. a success. Then p, the probability
of success is p= 28  15 13 = . 60 60 Therefore, the expected value and the variance of the number of trains that arrive in the given period are 10(13/60) = 2.167 and 10(13/60)(47/60) = 1.697, respectively. 9. The number of checks returned during the next two days is Poisson with = 6. The desired
probability is
4 P (X 4) =
i=0 e6 6i = 0.285. i! 10. Suppose that 5% of the items are defective. Under this hypothesis, there are 500(0.05) = 25
defective items. The probability of two defective items among 30 items selected at random is 25 2 475 28 500 30 = 0.268. Therefore, under the above hypothesis, having two defective items among 30 items selected at random is quite probable. The shipment should not be rejected. 11. N is a geometric random variable with p = 1/2. So E(N ) = 1/p = 2, and Var(N ) =
(1  p)/p 2 = 1  (1/2) /(1/4) = 2. 5 6
5 12. 1 = 0.067. 6 13. The number of times a message is transmitted or retransmitted is geometric with parameter
1  p. Therefore, the expected value of the number of transmissions and retransmissions of a 108 Chapter 5 Special Discrete Distributions message is 1/(1  p). Hence the expected number of retransmissions of a message is 1 p 1= . 1p 1p 14. Call a customer a "success," if he or she will make a purchase using a credit card. Let E
be the event that a customer entering the store will make a purchase. Let F be the event that the customer will use a credit card. To find p, the probability of success, we use the law of multiplication: p = P (EF ) = P (E)P F  E = (0.30)(0.85) = 0.255. The random variable X is binomial with parameters 6 and 0.255. Hence P X=i = 6 i 0.255
i 1  0.255 6i , i = 0, 1, . . . , 6. Clearly, E(X) = np = 6(0.255) = 1.53 and Var(X) = np(1  p) = 6(0.255)(1  0.255) = 1.13985.
5 15.
i=3 18 i 10 5i 28 5 = 0.772. 16. By the formula for the expected value of a hypergeometric random variable, the desired quantity
is (5 6)/16 = 1.875. 17. We want to find the probability that at most 4 of the seeds do not germinate:
4 i=0 2 40 (0.06)i (0.94)40i = 0.91. i 18. 1 
i=0 20 (0.06)i (0.94)20i = 0.115. i 19. Let X be the number of requests for reservations at the end of the second day. It is reasonable
to assume that X is Poisson with parameter 3 3 2 = 18. Hence the desired probability is
23 23 P (X 24) = 1 
i=0 P (X = i) = 1 
i=0 e18 (18)i = 1  0.89889 = 0.10111. i! Chapter 5 Review Problems 109 20. Suppose that the company's claim is correct. Then the probability of 12 or less drivers using
seat belts regularly is
12 i=0 20 (0.70)i (0.30)20i 0.228. i Therefore, under the assumption that the company's claim is true, it is quite likely that out of 20 randomly selected drivers, 12 use seat belts. This is not a reasonable evidence to conclude that the insurance company's claim is false. 21. (a) (0.999)999 (0.001)1 = 0.000368. (b) 2999 (0.001)3 (0.999)2997 = 0.000224. 2 22. Let X be the number of children having the disease. We have that the desired probability is
P (X = 3) P (X = 3  X 1) = = P (X 1) 5 (0.23)3 (0.77)2 3 = 0.0989. 1  (0.77)5 23. (a) w w+b n1 b . w+b (b) w w+b n1 . 24. Let n be the desired number of seeds to be planted. Let X be the number of seeds which
will germinate. We have that X is binomial with parameters n and 0.75. We want to find the smallest n for which P (X 5) 0.90. or, equivalently, P (X < 5) 0.10. That is, we want to find the smallest n for which
4 i=0 n (0.75)i (.25)ni 0.10. i By trial and error, as the following table shows, we find that the smallest n satisfying P (X < 5) 0.10 is 9. So at least nine seeds is to be planted. n 5 6 7 8 9
4 n i=0 i (0.75)i (.25)ni 0.7627 0.4661 0.2436 0.1139 0.0489 110 Chapter 5 Special Discrete Distributions 25. Intuitively, it must be clear that the answer is k/n. To prove this, let B be the event that the ith
baby born is blonde. Let A be the event that k of the n babies are blondes. We have P (AB) = P (A) p n  1 k1 p (1  p)nk k1 = n k nk p (1  p) k n1 k1 n k k . n P (B  A) = = 26. The size of a seed is a tiny fraction of the size of the area. Let us divide the area up into many
small cells each about the size of a seed. Assume that, when the seeds are distributed, each of them will land in a single cell. Accordingly, the number of seeds distributed will equal the number of nonempty cells. Suppose that each cell has an equal chance of having a seed independent of other cells (this is only approximately true). Since is the average number of seeds per unit area, the expected number of seeds in the area, A, is A. Let us call a cell in A a "success" if it is occupied by a seed. Let n be the total number of cells in A and p be the probability that a cell will contain a seed. Then X, the number of cells in A with seeds is a binomial random variable with parameters n and p. Using the formula for the expected number of successes in a binomial distribution (= np), we see that np = A and p = A/n. As n goes to infinity, p approaches zero while np remains finite. Hence the number of seeds that fall on the area A is a Poisson random variable with parameter A and P (X = i) = eA (A)i . i! 27. Let D/N p, then by the Remark 5.2, for all n,
D x N D nx N n n x p (1  p)nx . x Now since n and nD/N , n is large and np is appreciable, thus e x n x . p (1  p)nx x! x Chapter 6 C ontinuous R andom Variables
6.1 PROBABILITY DENSITY FUNCTIONS 1. (a)
0 ce3x dx = 1 c = 3.
1/2 0 (b) P (0 < X 1/2) = 32 2. (a) f (x) = x 3 0 x4 x < 4. 3e3x dx = 1  e3/2 0.78. (b) P (X 5) = 1  (16/25) = 9/25, P (X 6) = 16/36 = 4/9, P (5 X 7) = 1  (16/49)  1  (16/25) = 0.313, P (1 X < 3.5) = 0  0 = 0.
2 3. (a)
1 c(x  1)(2  x) dx = 1
x 1 c  x 3 3x 2 +  2x 3 2 2 1 = 1 c = 6. (b) F (x) = 6(x  1)(2  x) dx, 1 x < 2. Thus 0 F (x) = 2x 3 + 9x 2  12x + 5 1 x<1 1x<2 x 2. (c) P (X < 5/4) = F (5/4) = 5/32, P (3/2 X 2) = F (2)  F (3/2) = 1  (1/2) = 1/2. 4. (a) P (X < 1.5) =
1 1.5 2 2 dx = . 2 x 3 112 Chapter 6 Continuous Random Variables
1.25 (b) P (1 < X < 1.25  X < 1.5) = 1 1.5 1 2 dx x2 2 dx x2
1 1 = 2/5 3 = . 2/3 5 1 5. (a)
1 c 1 x2 dx = 1 c arcsin x = 1 c = 1/. (b) For 1 < x < 1, F (x) = Thus
x 1 1 1  x2 dx = 1 1 arcsin x + . 2 x < 1 1 x < 1 x 1. 0 1 1 F (x) = arcsin x + 2 1 6. Since h(x) 0 and 1 f (x) dx = 1  F () 1  F () f (x) dx = 1 1  F () = 1, 1  F () h is a probability density function. 7. (a) Let F be the distribution function of X. Then X is symmetric about if and only if for all x, 1  F ( + x) = F (  x), or upon differentiation f ( + x) = f (  x). (b) f ( + x) = f (  x) if and only if (  x  3)2 = ( + x  3)2 . This is true for all x, if and only if  x  3 = ( + x  3) which gives = 3. A similar argument shows that g is symmetric about = 1.  0 1 8. (a) Since f is a probability density function,  f (x) dx = 1. But
0 1 f (x) dx = 0 1 k(2x  3x 2 ) dx = k (2x  3x 2 ) dx = k x 2  x 3 = 2k. So 2k = 1 or k = 1/2. (b) The loss is at most $500 if and only if X 1/2. Therefore, the desired probability is P X 1 = 2 1 1  (2x  3x 2 ) dx =  x 2  x 3 2 2 1/2
0 0 1/2 = 3 . 16 Section 6.2 15 Density Function of a Function of a Random Variable 113 9. P (X > 15) = 1 x/15 1 e dx = . Thus the answer is 15 e
8 i=4 8 i 1 e i 1 1 e 8i = 0.3327. 10. Since f + g 0 and  f (x) + g(x) dx =  f (x) dx +  g(x) dx = + = 1, f + g is also a probability density function. 11. Since F () = 0 and F () = 1, We have that + (/2) = 0 + (/2) = 1. Solving this system of two equations in two unknown, we obtain = 1/2 and = 1/. Thus f (x) = F (x) = 2 ,  < x < . (4 + x 2 ) 6.2 DENSITY FUNCTION OF A FUNCTION OF A RANDOM VARIABLE 1. Let G be the distribution function of Y ; for 8 < y < 8,
G(y) = P (Y y) = P (X y) = P (X 3 3 y)= 3y 2 1 1 1 dx = 3 y + . 4 4 2 Therefore, 0 1 1 3 G(y) = y+ 4 2 1 y < 8 8 y < 8 y 8. 8 < y < 8 otherwise. This gives 1 2/3 y g(y) = G (y) = 12 0 114 Chapter 6 Continuous Random Variables Let H be the distribution function of Z; for 0 z < 16, H (z) = P (X z) = P ( 4 z x 4 z ) =
4 4z  z 4 1 1 dx = 4 z. 4 2 Thus 0 14 H (z) = z 2 1 z<0 0 z < 16 z 16. This gives 1 3/4 z h(z) = H (z) = 8 0 0 < z < 16 otherwise. 2. Let G be the probability distribution function of Y and g be its probability density function.
For t > 0, G(t) = P eX t = P (X ln t) = F (ln t). For t 0, G(t) = 0. Therefore, 1 f (ln t) t > 0 g(t) = G (t) = t 0 t 0. 3. The set of possible values of X is A = (0, ). Let h : (0, ) R be defined by h(x) = x x.
The set of possible values of h is B = (0, ). The inverse of h is g, where g(y) = y 2/3 . Thus g (y) = 2/(3 3 y ) and hence
2/3 2 fY (y) = ey , y (0, ). 33y To find the probability density function of eX , let h : (0, ) R be defined by h(x) = ex ; h is an invertible function with the set of possible values B = (0, 1). The inverse of h is g(z) =  ln z. So g (z) = 1/z. Therefore, fZ (z) = e( ln z)  0, otherwise. 1 1 = z = 1, z (0, 1); z z Section 6.2 Density Function of a Function of a Random Variable 115 4. The set of possible values of X is A = (0, ). Let h : (0, ) R be defined by h(x) =
log2 x. The set of possible values of h is B = (, ). h is invertible and its inverse is g(y) = 2y , where g (y) = (ln 2)2y . Thus
y y fY (y) = 3e3 2 (ln 2)2y = (3 ln 2)2y e3(2 ) , y (, ). 5. Let G and g be the probability distribution and the probability density functions of Y , respectively. Then G(y) = P (Y y) = P =
0 y y 3 X2 y = P (X y y ) y ex dx = 1  ey , y [0, ). So g(y) = G (y) = 0, otherwise. 3 y y ye , y 0; 2 6. Let G and g be the probability distribution and density functions of X2 , respectively. For
t 0, G(t) = P (X2 t) = P ( t < X < t ) = F ( t )  F ( t ). Thus 1 1 1 g(t) = G (t) = f ( t ) + f ( t ) = f ( t ) + f ( t ) , t 0. 2 t 2 t 2 t For t < 0, g(t) = 0. 7. Let G and g be the distribution and density functions of Z, respectively. For /2 < z < /2,
G(z) = P (arctan X z) = P (X tan z) = 1 = arctan x Thus
tan z  tan z  1 dx (1 + x 2 ) 1 1 = z+ . 2  <z< 2 2 1 g(z) = 0 elsewhere. 8. Let G and g be distribution and density functions of Y , respectively. Then
G(t) = P (Y t) = P (Y t  X 1)P (X 1) + P (Y t  X > 1)P (X > 1) 1 X > 1 P (X > 1). = P (X t  X 1)P (X 1) + P X t 116 Chapter 6 Continuous Random Variables For t 1, this gives G(t) = 1 0 1 ex dx + 1 1 ex dx = 1. For 0 < t < 1, this gives G(t) = P (X t) + P X Hence 1 = t
t 0 ex dx + 1/t ex dx = 1  et + e1/t . 0 G(t) = 1  et + e1/t 1 t e + 1 e1/t t2 g(t) = G (t) = 0 t 0 0<t <1 t 1. 0<t <1 elsewhere. Therefore, 6.3 EXPECTATIONS AND VARIANCES 32/x 3 0 x4 x < 4. 1. The probability density function of X is f (x) =
(a) E(X) = Thus 32 dx = 8. x2 4 32 dx = ; so Var(X) = E(X2 )  E(X) (b) E(X2 ) = x 4
2 2 does not exist. 2. (a) E(X) = 6 3 . 2 1 2 23 9 1 23 1 ; so Var(X) =  = , and X = . (x 4 + 3x 3  2x 2 ) dx = (b) E(X 2 ) = 6 10 10 4 20 20 1 (x 3 + 3x 2  2x) dx = 3. The standardized value of the lifetime of a car muffler manufactured by company A is
(4.255)/2 = 0.375. The corresponding value for company B is (3.754)/1.5 = 0.167. Therefore, the muffler of company B has performed relatively better. 4. E eX =
0 ex (3e3x ) dx =
0 3e2x dx = 3/2. Section 6.3
1 1 Expectations and Variances 117 5. E(X) = x dx = 0, because the integrand is an odd function. 1  x2 k ek(y)/A A 0 6. Let f be the probability density function of Y . Clearly,
 < y y > . f (y) = F (y) = Therefore, E(Y ) =  A ky/A A2 ky/A k k(y)/A k ye ye dy = ek/A  2e A A k k  = A . k 7. Let H be the distribution function of C; then
P (F t) = P C t  32 t  32 =H . 1.8 1.8 Hence the probability density function of F is 1 t  32 5 t  32 d P (F t) = h = h . dt 1.8 1.8 9 1.8 The expected value of F is given by E(F ) = 1.8E(C) + 32 = 1.8
2  xh(x) dx + 32. 8. E(ln X) = 2 ln x dx. To calculate this integral, let U = ln x, dV = 1/x 2 , and use x2 1 integration by parts:
2 1 2 ln x 2 ln x dx =  2 x x 2 
1 1 2  2 dx = 1  ln 2 = 0.3069. x2 9. The expected value of the length of the other side is given by
E 81  X 2 =
2 4 81  x 2 x dx. 6 Letting u = 81  x 2 , we get du = 2x dx and E 81  X 2 = 1 12
77 65 u du 8.4. 118 Chapter 6 Continuous Random Variables 10. E(X) =  1 x xe dx = 0, because the integrand is an odd function. Now 2 E(X2 ) =  1 2 x x e dx = 2 0 x 2 ex dx since the integrand is an even function; applying integration by parts to the last integral twice, we obtain E(X2 ) = 2. Hence Var(X) = 2  02 = 2. 11. Note that
E X =  x 2 dx = 2) (1 + x 1 0 0 x dx (1 + x 2 ) x dx. 1 + x2 since the integrand is an even function. Now for 0 < < 1, 0 x dx = 1 + x2 x dx + 1 + x2 1 Clearly, the first integral in the right side is convergent. To show that the second one is also convergent, note that. x 1 x 2 = 2 . 1 + x2 x x Therefore, 1 x dx 1 + x2 1 1 1 x 2 dx = 1 (  1)x 1 1 = 1 < . 1 1 For 1, 0 x 1 + x2 x dx 1 + x2 1 x 1 dx = ln(1 + x 2 ) 2 1+x 2 = . So
0 x dx diverges. 1 + x2 12. By Remark 6.4,
E(X) =
0 P (X > t) dt =
0 (et + et ) dt = c1 + . cn 13. (a) c1 is an arbitrary positive number because c1 ,
. 1 implies that cn = n1/(n1) cn (b) E(Xn ) = dx = n n(n2)/(n1) /(n  1) cn x (c) P (Zn t) = P (ln Xn t) = P (Xn et ) = c1 dx = 1. For n > 1, x2 cn x n+1 dx = if n = 1 if n > 1.
et cn cn x n+1 dx = cn 1 1  nt , where n n cn e Section 6.3 Expectations and Variances 119 cn = n1/(n1) . Let gn be the probability density function of Zn . Then gn (t) = cn ent , t ln cn .
m+1 (d) E(Xn ) = cn cn x m+1 dx. This integral exists if and only if m  n < 1. x n+1 14. Using integration by parts twice, we obtain
E(X n+1 ) = 1 0 x n+2 sin x dx = n+1 + (n + 2) 1 n x sin x dx = n+1 + (n + 2)  (n + 1) 0 = n+1 + (n + 2)  (n + 1)E(X n1 ) . Hence E(Xn+1 ) + (n + 1)(n + 2)E(X n1 ) = n+1 . 1 0 x n+1 cos x dx 15. Since X is symmetric about , for all x (, ), f ( +x) = f ( x). Letting y = x +,
we have E(X) =   yf (y) dy =  (x + )f (x + ) dx  = xf (x + ) dx + f (x + ) dx. Now since f is symmetric about , xf (x + ) is an odd function, xf (x + ) =  xf (x + ) . Therefore,
 xf (x + ) = 0. Since  f (x + ) dx =  f (y) dy = 1, we have E(X) = 0 + 1 = . To show that the median of X is , we will show that P (X ) = P (X ). This also shows that the value of these two probabilities is 1/2. Letting u =  x, we have P (X ) = Letting u = x  , we have that P (X ) =  f (x) dx =
0 f (  u) du. f (x) dx =
0 f (u + ) du. 120 Chapter 6 Continuous Random Variables Since for all u, we have that f (  u) = f ( + u), P (X ) = P (X ) = 1/2. 16. By Theorem 6.3,
E X  y = =y Hence dE X  y = dy = Setting
y  y   x  yf (x)dx = f (x) dx 
y  y  (y  x)f (x) dx + y y (x  y)f (x) dx y y  xf (x) dx + xf (x) dx  y f (x) dx. f (x) dx + yf (y)  yf (y)  yf (y)  f (x) dx  y y f (x) dx + yf (y) f (x) dx. dE X  y = 0, we obtain that y is the solution of the following equation: dy
y  f (x) dx = y f (x) dx. By the definition of the median of a continuous random variable, the solution to this equation is y = median(X). Hence E X  y is minimum for y = median(X). 17. (a)
0 I (t) dt =
0 X I (t) dt + X I (t) dt =
0 X dt + X 0 dt = X. (Note that
0 I (t) dt is a random variable.) 0 (b) E(X) = E (c) By part (b), I (t) dt =
0 E I (t) dt =
0 P (X > t) dt =
0 1  F (t) dt. E(Xr ) =
0 P (Xr > t) dt =
0 P X> 0 r t dt =
0 1F r t dt = r y r1 1  F (y) dy, where the last equality follows by the substitution y = r t. Section 6.3 Expectations and Variances 121 18. On the interval [n, n + 1),
P X n + 1 P X > t P X n . Therefore,
n+1 n P X n + 1 dt n+1 n n+1 n P X > t dt n+1 n P X n dt, or P X n + 1 So P X > t dt P X n . P X n + 1 n=0 n=0 n+1 n P X > t dt n=0 P X > n , and hence P X n E X 1 +
n=1 n=1 P X n . 19. By Exercise 12,
E(X) = Using Exercise 16, we obtain E(X2 ) = 2
0 + . 2 2 + 2. 2 x(ex + ex ) dx = Hence Var(X) = 2 2 + 2  + 2 2 = 2  2 2  2 2 . +  2 2 20. X st Y implies that for all t,
P (X > t) P (Y > t). Taking integrals of both sides of (21) yields, 0 (21) P (X > t) dt 0 P (Y > t) dt. (22) Relation (21) also implies that 1  P (X t) 1  P (Y t), or, equivalently, P (X t) P (Y t) 122 Chapter 6 Continuous Random Variables Since this is true for all t, we have P (X t) P (Y t) Taking integrals of both sides of this inequality, we have 0 P (X t) 0 P (Y t) dt, or, equivalently, 
0 P (X t) 
0 P (Y t) dt. (23) Adding (22) and (23) yields 0 P (X > t) dt 
0 P (X t) dt 0 P (Y > t) dt 
0 P (Y t) dt By Theorem 6.2, this gives E(X) E(Y ). To show that the converse of this theorem is false, let X and Y be discrete random variables both with set of possible values {1, 2, 3}. Let the probability mass functions of X and Y be defined by pX (1) = 0.3 pY (1) = 0.5 pX (2) = 0.4 pY (2) = 0.1 pX (3) = 0.3 pY (3) = 0.4 We have that E(X) = 2 > E(Y ) = 1.9. However, since P (X > 2) = 0.3 < P (Y > 2) = 0.4, we see that X is not stochastically larger than Y . 21. First, we show that limx xP X x = 0. To do so, since x , we concentrate on
negative values of x. Letting u = t, we have xP X x = x
x  f (t) dt = x x x f (u) du =  x xf (u) du. So it suffices to show that as x , x xf (u) du 0. Now x xf (u) du x uf (u) du. Therefore, it remains to prove that  uf (u) du 0 as x . But this is true because  uf (u) du = xf (x) dx < . Chapter 6 Review Problems 123 Next, we will show that limx xP X > x = 0. To do so, note that lim xP X > x = lim x
x x x f (t) dt lim x x tf (t) dt = 0 since  tf (t) dt < . REVIEW PROBLEMS FOR CHAPTER 6 1. Let F be the distribution function of Y . Clearly, F (y) = 0 if y 1. For y > 1,
1 1 1 1 y F (y) = P y =P X = =1 . X y 10 y 1 So f (y) = F (y) = 1/y 2 0 y>1 elsewhere. 2. E(X) = 2 2 2 dx = dx =  = 2, x3 x2 x 1 1 1 2 x 2 3 dx = 2 ln x = . So Var(X) does not exist. E(X2 ) = 1 x 1 x 3. E(X) = 1 6 1 (6x 2  6x 3 ) dx = 2x 3  x 4 = , 0 4 2 0 1 1 6 6 3 , (6x 3  6x 4 ) dx = x 4  x 5 = E(X 2 ) = 0 4 5 10 0 3 1 1 1 2 Var(X) = =  , X = . 10 2 20 2 5 1 Therefore, 1 1 2 2 P  <X< + = 2 2 5 2 2 5
1 1 2+ 5 1 1 2 5 (6x  6x 2 ) dx
3
1 1 2+ 5 1 1 2 5 = 3x  2x
2 11 = . 5 5 124 Chapter 6 Continuous Random Variables 4. We have that
P (2 < X < 1) =
0 1 ex dx = ex dx + 2 2 2 2 1 1  2 = 0.748. =1 2e 2e 1 1 0 ex dx 5. For all c > 0, c dx = c ln(1 + x) = . 0 1+x 0 So, for no value of c, f (x) is a probability density function. 6. The set of possible values of X is A = [1, 2]. Let h : [1, 2] R be defined by h(x) = ex . The set of possible values of eX is B = [e, e2 ]; the inverse of h is g(y) = ln y, where g (y) = 1/y. Therefore, 4(ln y)3 4(ln y)3 fY (y) = g (y) = , y [e, e2 ]. 15 15y Applying the same procedure to Z and W , we obtain 4( z )3 1 2z fZ (z) = = , z [1, 4]. 15 2 z 15 2(1 + w )3 fW (w) = w [0, 1]. 15 w 7. The set of possible values of X is A = (0, 1). Let h : (0, 1) R be defined by h(x) = x 4 .
The set of possible values of X4 is B = (0, 1). The inverse of h(x) = x 4 is g(y) = 1 1 g (y) = y 3/4 = . We have that 4 4 y4y fY (y) = 30( 4 y )2 (1  4 y )2 1
4 4 y. So 4 y3 2 15(1  4 y ) = , y (0, 1). 24y 1 = 30 y(1  4 y )2 4 y4y 8. We have that 1 f (x) = F (x) = 1  x 2 0 E(X) =
1 1 1 < x < 1 otherwise. Therefore, x dx = 0 1  x2 since the integrand is an odd function. Chapter 6 Review Problems 125 9. Clearly n i=1 i fi 0. Since  n n i fi (x) dx =
i=1 i=1 i  n fi (x) dx =
i=1 i = 1, n i=1 i fi is a probability density function. 10. Let U = x and dV = f (x)dx. Then dU = dx and V = F (x). Since F () = 1,
E(X) =
0 xf (x) dx = xF (x) 0 0 
0 F (x) dx F (x) dx = F ()  =
0 F (x) dx =  F (x) dx =
0 0 dx 
0 1  F (x) dx. 11. Let X be the lifetime of a random light bulb. The probability that it lasts over 1000 hours is
P (X > 1000) = 1000 1 5 105 dx = 5 105  2 x3 2x 1000 1 = . 4 Thus the probability that out of six such light bulbs two last over 1000 hours is 6 2 1 4
2 3 4 4 0.3 12. Since Y 0, P (Y t) = 0 for t < 0. For t 0,
P (Y t) = P X t = P (t X t) = P (X t)  P (X < t) = P (X t)  P (X t) = F (t)  F (t). Hence G, the probability distribution function of X is given by G(t) = F (t)  F (t) if t 0 0 if t < 0; g, the probability density function of X is obtained by differentiating G: g(t) = G (t) = f (t) + f (t) if t 0 0 if t < 0. Chapter 7 Special C ontinuous Distributions
7.1 UNIFORM RANDOM VARIABLES 1. (23  20)/(27  20) = 3/7. 2. 15(1/4) = 3.75. 3. Let 2:00 P.M. be the origin, then a and b satisfy the following system of two equations in two
unknown. a + b = 0 2 (b  a)2 = 12. 12 Solving this system, we obtain a = 6 and b = 6. So the bus arrives at a random time between 1:54 P.M. and 2:06 P.M. 4. P (b2  4 0) = P (b > 2 or b < 2) = 2/6 = 1/3. 5. The probability density function of R, the radius of the sphere is 1 =1 2 f (r) = 4  2 0 E(V ) =
2 4 2<r<4 elsewhere. Thus 4 3 1 r dr = 40. 3 2 1 4 P R 3 < 36 = P (R 3 < 27) = P (R < 3) = . 3 2 6. The problem is equivalent to choosing a random number X from (0, ). The desired probability
is P X 3 +P X 2 3 = /3 +  (2 /3) = 2 . 3 Section 7.1 Uniform Random Variables 127 7. Let X be a random number from (0, ). The probability of the desired event is
P min(X,  X) =P X , 3 3 X =P X 2 3 = 2  3 3 1 = . 3 3 3 8. 180  90 3 = . 180  60 4 the segments a, X, and b  X are sides of a triangle. The probability of this is P a < X + (b  X), X < a + (b  X), b  X < a + X = P 9. Let X be a random point from (0, b). A triangular pen is possible to construct if and only if
a+b ba <X< 2 2 a+b ba  a 2 2 = = . b b 10. Let F be the probability distribution function and f be the probability density function of X.
By definition, F (x) = P (X x) = P (tan x) = P ( arctan x) arctan x   1 1 2 = = arctan x + ,  < x < . 2   2 2 Thus f (x) = F (x) = 1 , (1 + x 2 )  < x < . 11. For i = 0, 1, 2, . . . , n  1,
i+1 i X< = P [nX] = i = P (i nX < i + 1) = P n n P [nX] = i = 0, otherwise. 0, 1, 2, . . . , n  1 . G(x) = 0 if x 0. If x 0, i i+1  1 n n = . n 10 Therefore, [nX] is a random number from the set 12. (a) Let G and g be the distribution and density functions of Y , respectively. Since Y 0,
G(x) = P (Y x) = P  ln(1  X) x = P X 1  ex = (1  ex )  0 = 1  ex . 10 128 Chapter 7 Special Continuous Distributions Thus g(x) = G (x) = ex 0 x0 otherwise. (b) Let H and h be the probability distribution and probability density functions of Z, respectively. For n > 0, H (x) = P (Z x) = 0, x < 0; H (x) = P (Z x) = P (X n x ) = n x, 0 < x < 1; H (x) = 1, if x 1. Therefore, 1 n 1 x1 h(x) = H (x) = n 0 0<x<1 elsewhere. For n < 0, H (x) = P (Xn x) = 0, x < 1; H (x) = P (Xn x) = P Xn 1 1 =P X x x 1/n 1/n = P (X x ) = 1  x , x 1. 1 n 1  x 1 n h(x) = 0 if x 1 if x < 1.
1 n Therefore, 13. Cleary, E(X) = (1 + )/2. This implies that = 2E(X)  1. Now
Var(X) = E X2  E(X) Therefore, E X2  This yields, E X2 = So 3E(X2 )  2  1 = 2 . But = 2E(X)  1; so 3E(X2 )  2 2E(X)  1  1 = 2 . This implies that E(3X 2  4X + 1) = 2 . Therefore, one choice for g(X) is g(X) = 3X2  4X + 1. 1+ 2
2 2 = (1 +  0)2 . 12 = 1 + 2 + 2 . 12 2 + 2 + 1 . 3 Section 7.1 Uniform Random Variables 129 14. Let S be the sample space over which X is defined. The functions X : S R and F : R [0, 1] can be composed to obtain the random variable F (X) : S [0, 1]. Clearly, P F (X) t = 1 0 if t 1 if t 0. Let t (0, 1); it remains to prove that P F (X) t = t. To show this, note that since F is continuous, F () = 0, and F () = 1, the inverse image of t, F 1 {t} , is nonempty. We know that F is nondecreasing; since F is not necessarily strictly increasing, F 1 {t} might have more than one element. For example, if F is the constant t on some internal (a, b) (0, 1), then F (x) = t for all x (a, b), implying that (a, b) is contained in F 1 {t} . Let x0 = inf x : F (x) > t . Then F (x0 ) = t and Therefore, P F (X) t = P X x0 = F (x0 ) = t. We have shown that 0 P F (X) t = t 1 if t 0 if 0 t 1 if t 1, F (x) t if and only if x x0 . meaning that F (X) is uniform over (0, 1). 15. We are given that Y is a uniform random variable. First we show that Y is uniform over the
interval (0, 1). To do this, it suffices to show that P (Y 1) = 1 and P (Y < 0) = 0. These  are obvious implications of the fact that g is nonnegative and P (Y 1) = P P (Y < 0) = P
X  X  g(x) dx = 1: g(t) dt 1 = 1. g(t) dt < 0 = 0, The following relation shows that the probability density function of X is g. u g(t) dt  0 u d d  d = g(u), g(t) dt = P (X u) = P Y du du du 10  where the last equality follows from the fundamental theorem of calculus. 130 Chapter 7 Special Continuous Distributions 16. Let F be the distribution function of X, then F (t) = P (X t) is 0 for t < 1 and is 1 for
t 4. Let 1 t < 4; we have that F (t) = P (X t) = P (5  1 t) = P = P 0, Therefore, t +1 5 =
0 (t+1)/5 t +1 5 t +1 dx = . 5 0 t +1 F (t) = 5 1 t < 1 1 t < 4 t 4. This is the distribution function of a uniform random variable over (1, 4). 17. We have that X = n if and only if Y = 0.y1 ny3 y4 y5 , or, equivalently, if and only if, 10 Y = y1 .ny3 y4 y5 . Therefore, X = n if and only if for some k 0, 1, 2, . . . , 9 , k+ This is equivalent to n 1 k+ 100 10 Therefore, the desired probability is
9 2 n+1 n 10 Y < k + . 10 10 Y < n+1 1 k+ 100 10 2 . P
k=0 1 n k+ 100 10
9 2 Y <
2 1 n+1 k+ 100 10  n 1 k+ 100 10 2 =
k=0 9 n+1 1 k+ 100 10 2 =
k=0 20k + 2n + 1 = 0.091 + 0.002n. 10, 000 We see that this quantity increases as n does. Section 7.2 Normal Random Variables 131 7.2 NORMAL RANDOM VARIABLES 1. Since np = (0.90)(50) = 45 and np(1  p) = 2.12,
P (X 44.5) = P Z =1 44.5  45 = P (Z 0.24) 2.12 (0.24) = (0.24) = 0.5948. 3 364 = 1.73. Therefore, 365 (1.45) = 0.0735. 2. np = 1095/365 = 3 and np(1  p) =
P (X 5.5) = P Z 5.5  3 =1 1.73 3. We have that
P (Z) x) = P (x Z x) = = (x)  1  (x)  (x) (x). (x) = 2 (x)  1 =
x+ x 4. Let 1 g(x) = P (x < Z < x + ) = 2 ey 2 /2 dy. The number x that maximizes P (x < Z < x + ) is the root of g (x) = 0; that is, it is the solution of 1 2 2 e(x+) /2  ex /2 = 0, g (x) = 2 which is x = /2. 5. E(X cos X), E(sin X), and E 1 X 2 are, respectively, (x cos x)ex /2 dx, 2 1+X 2  1 x 1 2 2 (sin x)ex /2 dx, and ex /2 dx. Since these are integrals of 1 + x2 2  2  odd functions from  to , all three of them are 0. 6. (a) P (X > 35.5) = P X  35.5 35.5  35.5 > =1 4.8 4.8 (0) = 0.5. (b) The desired probability is given by P (30 < X < 40) = P = 40  35.5 30  35.5 <X< = (0.94)  (1.15) 4.8 4.8 (0.94) + (1.15)  1 = 0.8264 + 0.8749  1 = 0.701. 132 Chapter 7 Special Continuous Distributions 7. Let X be the grade of a randomly selected student;
90  67 = 1  (2.88) = 1  0.9980 = 0.002, 8 80  67 90  67 P (80 X < 90) = P Z< = (2.88)  (1.63) 8 8 P (X 90) = P Z = 0.9980  0.9484 = 0.0496. Similarly, P (70 X < 80) = 0.3004, P (60 X < 70) = 0.4586, and P (X < 60) = 0.1894. Therefore, approximately 0.2%, 4.96%, 30.04%, 45.86%, and 18.94% get A, B, C, D, and F, respectively. 8. Let X be the blood pressure of a randomly selected person;
P (89 < X < 96) = P 89  80 96  80 <Z< = P (1.29 < Z < 2.29) = 0.0875, 7 7 95  80 = 0.016. 7 P (X > 95) = P Z > Therefore, 8.75% have mild hypertension while 1.6% are hypertensive. 9. P (74.5 < X < 75.8) = P (0.5 < Z < 0.8) = (0.8)  1  (0.5) = 0.4796. 10. We must find x so that P (110  x < X < 110 + x) = 0.50, or, equivalently,
P  x X  110 x < < = 0.50. 20 20 20 Therefore, we must find the value of x which satisfies P  x/20 < Z < x/20 = 0.50 or (x/20) (x/20) = 0.50. Since (x/20) = 1 (x/20), x satisfies 2 (x/20) = 1.50 or (x/20) = 0.75. Using Table 1 of the appendix, we get x/20 = 0.67 or x = 13.4 So the desired interval is (110  13.4, 110 + 13.4) = (96.6, 123.4). 11. Let X be the amount of cereal in a box. We want to have P (X 16) 0.90. This gives
P Z 16  16.5 0.90, or (0.5/ ) 0.90. The smallest value for 0.5/ satisfying this inequality is 1.29; so the largest value for is obtained from 0.5/ = 1.29. This gives = 0.388. 12. Let X be the score of a randomly selected individual;
P (X 14) = P Z 14  12 = P (Z 0.67) = 0.2514. 3 Therefore, the probability that none of the eight individuals make a score less than 14 is (0.2514)8 = 0.000016. Section 7.2 Normal Random Variables 133 13. We want to find t so that P (X t) = 1/2. This implies that
P or t  1 X = , 2 1 t  t  = ; so = 0 which gives t = . 2 14. We have that
P (X   > k ) = P (X  > k ) + P (X  < k ) = P (Z > k) + P (Z < k) = 1 (k) + 1  (k) = 2 1  (k) . This shows that P (X   > k ) does not depend on or . 15. Let X be the lifetime of a randomly selected light bulb.
900  1000 =1 100 Hence the company's claim is false. P (X 900) = P Z (1) = (1) = 0.8413. 16. Let X be the lifetime of the light bulb manufactured by the first company. Let Y be the lifetime of the light bulb manufactured by the second company. Assuming that X and Y are independent, the desired probability, P max(X, Y ) 980 , is calculated as follows. P max(X, Y ) 980 = 1  P max(X, Y ) < 980 = 1  P (X < 980, Y < 980) = 1  P (X < 980) P (Y < 980) =1P Z < 980  900 980  1000 P Z< 100 150 (0.2) (0.53) = 1  P (Z < 0.2)P (Z < 0.53) = 1  1  = 1  (1  0.5793)(0.7019) = 0.7047. 17. Let r be the rate of return of this stock; r is a normal random variable with mean = 0.12 and standard deviation = 0.06. Let n be the number of shares Mrs. Lovotti should purchase. We want to find the smallest n for which the probability of profit in one year is at least $1000. Let X be the current price of the total shares of the stock that Mrs. Lovotti buys this year, and Y be the total price of the shares next year. We want to find the smallest n for which P (Y  X 1000). We have P (Y  X 1000) = P Y X 1000 1000 =P r X X X 1000  0.12 1000 Z 35n 0.90. =P r =P 35n 0.06 134 Chapter 7 Special Continuous Distributions Therefore, we want to find the smallest n for which 1000  0.12 35n 0.10. P Z 0.06 By Table 1 of the Appendix, this is satisfied if 1000  0.12 35n 1.29. 0.06 This gives n 670.69. Therefore, Mrs. Lovotti should buy 671 shares of the stock. 18. We have that
f (x) = 1 (x  1)2 1 (x  1)2 = . exp  exp  1/2 2(1/4) 1/2 (1/2) 2 This shows that f is the probability density function of a normal random variable with mean 1 and standard deviation 1/2 (variance 1/4). 19. Let F be the distribution function of X  . F (t) = 0 if t < 0; for t 0,
F (t) = P X   t = P (t X  t) t X t = P (  t X + t) = P  t t t t   =  1 =2 = Therefore, F (t) = This gives F (t) = Hence E X   =
0 t  1. 2 0 2 t 1 t 0 otherwise. t t 0. 2 t dt. t substituting u = t/ , we obtain E(X  ) = 2 2 2 (u) du = ueu /2 du 2 0 0 2 2 2 2 = = . =  eu /2 0 2 2 u Section 7.2 Normal Random Variables 135 20. The general form of the probability density function of a normal random variable is
f (x) = 1 2 1 (x  )2 1 2 x + 2x  = exp  . exp  2 2 2 2 2 2 2 2 Comparing this with the given probability density function, we see that 1 k= 2 2 k = 1 2 2 2k =  2 2 2 = 1. 2 Solving the first two equations for k and , we obtain k = and = 1/( 2). These and the third equation give = 1/ which satisfy the fourth equation. So k = and f is the 1 1 probability density function of N  , 2 . 2 21. Let X be the viscosity of the given brand. We must find the smallest x for which P (X x) 0.90 or P Z x = 49.9. x  37 0.90. This gives 10 x  37 0.90 or (x  37)/10 = 1.29; so 10 22. Let X be the length of the residence of a family selected at random from this town. Since
P (X 96) = P Z 96  80 = 0.298, 30 using binomial distribution, the desired probability is
2 1
i=0 12 (0.298)i (1  0.298)12i = 0.742. i 23. We have
E(eZ ) = 1 2 ex ex /2 dx 2 
2 /2 = e = e  1 1 2 1 2 e 2 +x 2 x dx 2 1 1 2 2 e 2 (x) dx = e /2 , 2 2 /2  136 Chapter 7 Special Continuous Distributions 1 1 1 1 2 2 e 2 (x) dx = 1, since e 2 (x) is the probability density function 2 2  of a normal random variable with mean and variance 1. where 24. For t 0, P (Y t) = P  t X t = P  t t Z =2 t  1. Let f be the probability density function of Y . Then d 1 f (t) = P (Y t) = 2 dt 2 t So 1 t exp  2 2 f (t) = 2 t 0 t , t 0. t 0 t 0. 25. For t 0,
P (Y t) = P eX t = P (X ln t) = P Z Let f be the probability density function of Y . We have f (t) = So f (t) = d 1 P (Y t) = dt t ln t  , t 0. t 0 otherwise. ln t  = ln t  . 1 (ln t  )2 exp  2 2 t 2 0 26. Let f be the probability density function of Y . Since for t 0,
P (Y t) = P we have that X t = P X t 2 = P  t 2 X t 2 = 2 (t 2 )  1, t 0 otherwise. 4t 1 et 4 /2 d 2 f (t) = P (Y t) = dt 0 27. Suppose that X is the number of books sold in a month. The random variable X is binomial with parameters n = (800)(30) = 24, 000 and p = 1/5001. Moreover, E(X) = np = 4.8 and X = np(1  p) = 2.19. Let k be the number of copies of the bestseller to be ordered Section 7.2 Normal Random Variables 137 every month. We want to have P (X < k) > 0.98 or P (X k  1) > 0.98. Using De MoivreLaplace theorem and making correction for continuity, this inequality is valid if P X  4.8 k  1 + 0.5  4.8 < > 0.98. 2.19 2.19 From Table 1 of the appendix, we have (k  1 + 0.5  4.8)/2.19 = 2.06, or k = 9.81. Therefore, the store should order 10 copies a month. 28. Let X be the number of light bulbs of type I. We want to calculate P (18 X 22).
Since the number of light bulbs is large and half of the light bulbs are type I, we can assume that X is approximately binomial with parameters 40 and 1/2. Note that np = 20 and np(1  p) = 10. Using De MoivreLaplace theorem and making correction for continuity, we have P (17.5 X 22.5) = P = 17.5  20 X  20 22.5  20 10 10 10 (0.79)  (0.79) = 2 (0.79)  1 = 0.5704. Remark: Using binomial distribution, the solution to this problem is
22 i=18 40 i 1 2 i 1 2 40i = 0.5704. As we see, up to at least 4 decimal places, this solution gives the same answer as obtained above. This indicates the importance of correction for continuity; if it is ignored, we obtain 0.4714, an answer which is almost 10% lower than the actual answer. 29. Let X be the number of 1's selected; X is binomial with parameters 100, 000 and 1/40. Thus np = 2500 and np(1  p) = 49.37. So P (X 3500) P Z 3499.50  2500 =1 49.37 (20.25) = 0. Hence it is fair to say that the algorithm is not accurate. 30. Note that x2 . 1/ ln a Comparing this with the probability density function of a normal random variable with pa rameters and , we see that = 0 and 2 2 = 1/ ln a. Thus = 1/(2 ln a), and hence ka x = k exp  x 2 ln a = k exp 
2 k= 1 = 2 ln a . So, for this value of k, the function f is the probability density function a normal random variable with mean 0 and standard deviation 1/(2 ln a). 138 Chapter 7 Special Continuous Distributions 31. (a) The derivation of these inequalities from the hint is straightforward.
(b) By part (a), 1 Thus 1 lim from which (b) follows.
x 1 1  (x) < < 1. 2 2 x 1/(x 2 ) ex /2 1  (x) 1, 2 1/(x 2 ) ex /2 32. By part (b) of Exercise 31,
x lim P Z > t + t t Z t = lim P Z>t+ x t P (Z t)
2 t = lim t 1 x x exp  t + t t+ 2 t 1 2 et /2 t 2 2 = lim x2 t2 exp  x  2 = ex . t t 2 + x 2t 33. Let X be the amount of soft drink in a random bottle. We are given that P (X < 15.5) = 0.07
15.5  and P (X > 16.3) = 0.10. These imply that = 0.07 and and 0.90. Using Tables 1 and 2 of the appendix, we obtain 15.5  = 1.48 16.3  = 1.28. 16.3  = Solving these two equations in two unknowns, we obtain = 15.93 and = 0.29. 34. Let X be the height of a randomly selected skeleton from group 1. Then
P (X > 185) = P Z > 185  172 = P (Z > 1.44) = 0.0749. 9 Section 7.3 Exponential Random Variables 139 Now suppose that the skeleton's of the second group belong to the family of the first group. The probability of finding three or more skeleton's with heights above 185 centimeters is
5 i=3 5 (0.0749)i (0.9251)5i = 0.0037. i Since the chance of this event is very low, it is reasonable to assume that the second group is not part of the first one. However, we must be careful that in reality, this observation is not sufficient to make a judgment. In the lack of other information, if a decision is to be made solely based on this observation, then we must reject the hypothesis that the second group is part of the first one. 35. For t (0, ), let A be the region whose points have a (positive) distance t or less from the given tree. The area of A is t 2 . Let X be the distance from the given tree to its nearest tree. We have that P (X > t) = P (no trees in A) = Now by Remark 6.4, E(X) =
0 et ( t 2 )0 2 = et . 0!
2 P (X > t) dt =
0 et dt.
2 2 t, we obtain Letting u = 1 1 E(X) = 2 0 eu 2 /2 1 1 1 = . du = 2 2 36. Note that dy = xds; so
I2 =
0 0 e(x 2 +x 2 s 2 )/2 x ds dx =
0 0 ex 2 (1+s 2 )/2 x dx ds 0 (let u = x 2 ) =
0 =
0 1 1 du ds = 2 2 0 1 ds = arctan s = . 0 1 + s2 2 eu(1+s
2 )/2 0  2 2 eu(1+s )/2 2 1+s ds 7.3 EXPONENTIAL RANDOM VARIABLES 1. Let X be the time until the next customer arrives; X is exponential with parameter = 3.
Hence P (X > x) = ex , and P (X > 3) = e9 = 0.0001234. 140 Chapter 7 Special Continuous Distributions 2. Let m be the median of an exponential random variable with rate . Then P (X > m) = 1/2;
thus em = 1/2 or m = ln 2 . 3. For  < y < ,
P (Y y) = P ( ln X y) = P X ey = ee Thus g(y), the probability density function of Y is given by g(y) =
y y d P (Y y) = ey ee = ey  e . dy y . 4. Let X be the time between the first and second heart attacks. We are given that P (X 5) =
1/2. Since exponential is memoryless, the probability that a person who had one heart attack five years ago will not have another one during the next five years is still P (X > 5) which is 1  P (X 5) = 1/2. 5. (a) Suppose that the next customer arrives in X minutes. By the memoryless property, the
desired probability is 1 = 1  e5(1/30) = 0.1535. 30 (b) Let Y be the time between the arrival times of the 10th and 11th customers; Y is exponential with = 5. So the answer is 1 P Y = 1  e5(1/30) = 0.1535. 30 P X< 6.
P X  E(X) 2X = P 2 1 2 1 2 1 +P X  =P X 1 3 +P X  =P X (3/) 3 =e + 0 = e = 0.049787. X 7. (a) P (X > t) = et .
(b) P (t X s) = 1  es  1  et = et  es . 8. The number of documents typed by the secretary on a given eighthour working day is Poisson
with parameter = 8. So the answer is e8 8 i =1 i! i=12 11 i=0 e8 8 i = 1  0.888 = 0.112. i! Section 7.3 Exponential Random Variables 141 9. The answer is
E 350  40N (12) = 350  40 1 12 = 323.33. 18 10. Mr. Jones makes his phone calls when either A or B is finished his call. At that time the
remaining phone call of A or B, whichever is not finished, and the duration of the call of Mr. Jones both have the same distribution due to the memoryless property of the exponential distribution. Hence, by symmetry, the probability that Mr. Jones finishes his call sooner than the other one is 1/2. 11. Let N (t) be the number of changeofstates occurring in [0, t]. Let X1 be the time until the machine breaks down for the first time. Let X2 be the time it will take to repair the machine, X3 be the time since the machine was fixed until it breaks down again, and so on. Clearly, X1 , X2 , . . . are the times between consecutive change of states. Since {X1 , X2 , . . . } is a sequence of independent and identically distributed exponential random variables with mean 1/, by Remark 7.2, N (t) : t 0 is a Poisson process with rate . Therefore, N (t) is a Poisson random variable with parameter t. 12. The probability mass function of L is given by
P (L = n) = (1  p)n1 p, Hence Therefore, P (T x) = P (L 1000x) = 1  P (L > 1000x) = 1  (1  p)1000x = 1  e1000x ln(1p) = 1  ex[1000 ln(1p)] , x > 0. This shows that T is exponential with parameter = 1000 ln(1  p). n = 1, 2, 3, . . . . n = 0, 1, 2, . . . . P (L > n) = (1  p)n , 13. (a) We must have
 cex dx = 1; thus c= 1  ex dx = 2
0 1 ex dx = 1 . 2 (b) E(X2n+1 ) =  1 2n+1 x e dx = 0, because the integrand is an odd function. x 2 E(X2n ) =  1 2n x dx = x e 2 0 x 2n ex dx, because the integrand is an even function. We now use induction to prove that
0 x n ex dx = n!. For n = 1, the integral is the expected value of an exponential random variable with 142 Chapter 7 Special Continuous Distributions parameter 1; so it equals to 1 = 1!. Assume that the identity is valid for n  1. Using integration by parts, we show it for n. 0 x n ex dx =   x n ex 0 +
0 nx n1 ex dx = 0 + n(n  1)! = n!. Hence E(X2n ) = (2n)!. 14. P [X] = n = P (n X < n + 1) = n+1 n ex dx = ex n+1 n = e n 1  e . This is the probability mass function of a geometric random variable with parameter p = 1  e . 15. Let that G(t) = P (X > t) = 1  F (t). By the memoryless property of X,
P (X > s + t  X > t) = P (X > s), for all s 0 and t 0. This implies that P (X > s + t) = P (X > s)P (X > t), or G(s + t) = G(s)G(t), t 0, s 0. (24) Now for arbitrary positive integers n and m, (24) gives that G G 1 1 1 1 1 2 =G + =G G = G n n n n n n
2 2 ,
3 3 2 1 2 1 1 =G + =G G = G n n n n n n . . . 1 m m = G . G n n G(1) = G 1 1 1 + + + n n n n terms 1 = G(1) n G 1 1 =G n n , Also = G 1 n n yields G Hence G(m/n) = G(1)
m/n 1/n . (25) . (26) Section 7.3 Exponential Random Variables 143 Now we show that G(1) > 0. If not, G(1) = 0 and by (25), G(1/n) = 0 for all positive integer n. This and right continuity of G imply that P (X 0) = F (0) = 1  G(0) = 1  G lim = 1  lim G
n n 1 = 1  0 = 1, n 1 n which is a contradiction to the given fact that X is a positive random variable. Thus G(1) > 0 and we can define =  ln G(1) . This gives G(1) = e , and by (26), G(m/n) = e(m/n) . Thus far, we have proved that for any positive rational t, G(t) = et . (27) To prove the same relation for a positive irrational number t, recall from calculus that for each 1 1 . Since t < tn < t + , positive integer n, there exists a rational number tn in t, t + n n limn tn exists and is t. On the other hand because F is right continuous, G = 1  F is also right continuous and so G(t) = lim G(tn ).
n But since tn is rational, (27) implies that, G(tn ) = etn . Hence G(t) = lim etn = et .
n Thus F (t) = 1  et for all t, and X is exponential. Remark: If X is memoryless, then P (X 0) = 0. To see this, note that P (X > s + t  X > t) = P (X > s) implies P (X s + t  X > t) = P (X s). Letting s = t = 0, we get P (X 0  X > 0) = P (X 0). But P (X 0  X > 0) = 0; therefore P (X 0) = 0. This shows that the memoryless property cannot be defined for random variables possessing nonpositive values with positive probability. 144 Chapter 7 Special Continuous Distributions 7.4 GAMMA DISTRIBUTIONS
. Then 1. Let f be the probability density function of a gamma random variable with parameters r and
f (x) = Therefore, f (x) = r+1 r2 x r 1 r  ex x r1 + ex (r  1)x r2 =  x e . x (r) (r) r x r1 ex . (r) This relation implies that the function f is increasing if x < (r  1)/, it is decreasing if x > (r  1)/, and f (x) = 0 if x = (r  1)/. Therefore, x = (r  1)/ is a maximum of the function f . Moreover, since f has only one root, the point x = (r  1)/ is the only maximum of f . 2. We have that
P (cX t) = P (X t/c) =
0 t/c =
0 t =
0 t (ex )(x)r1 dx (let u = cx) (r) eu/c (u/c)r1 (1/c) du (r) (/c)eu/c (u/c)r1 du. (r) This shows that cX is gamma with parameters r and /c. 3. Let N (t) be the number of babies born at or prior to t. N (t) : t 0 is a Poisson process
with = 12. Let X be the time it takes before the next three babies are born. The random variable X is gamma with parameters 3 and 12. The desired probability is P (X 7/24) = 7/24 12e12x (12x)2 dx = 864 (3) 7/24 x 2 e12x dx. Applying integration by parts twice, we get x 2 e12x dx =  Thus 7 1 1 1 12x = 864  x 2 e12x  xe12x  e = 0.3208. 7/24 24 12 72 864 Remark: A simpler way to do this problem is to avoid gamma random variables and use the properties of Poisson processes: P X 7 P N 2 = 24
2 i=0 1 2 12x 1 1 12x x e e  xe12x  + c. 12 72 864 7 P N =i = 24 2 i=0 e(7/24)12 (7/24)12 i! i = 0.3208. Section 7.4 Gamma Distributions 145 4.  f (x) dx =
0 r ex (x)r1 dx = (r) (r) 0 ex x r1 dx. Let t = x; then dt = dx, so  f (x) dx = r (r) 1 = (r) t r1 1 dx r1 0 1 et t r1 dt = (r) = 1. (r) 0 et 5. Let X be the time until the restaurant starts to make profit; X is a gamma random variable with
parameters 31 and 12. Thus E(X) = 31/12; that is, two hours and 35 minutes. 6. By the method of Example 5.17, the number of defective light bulbs produced is a Poisson
process at the rate of (200)(0.015) = 3 per hour. Therefore, X, the time until 25 defective light bulbs are produced is gamma with parameters = 3 and r = 25. Hence r 25 = = 8.33. 3 E(X) = That is, it will take, on average, 8 hours and 20 minutes to fill up the can. 7.
1 = 2 Making the substitution t = y 2 /2, we get 2 y 2 /2 1 y 2 /2 e dy = e dy = 2 2 2  0 1 2 = ey /2 dy = . 2  0 t 1/2 et dt. 146 Chapter 7 Special Continuous Distributions Hence 1 3 = 2 2 3 5 = 2 2 5 7 = 2 2 . . . n+ 1 = 2 = 2n + 1 2n  1 2n  3 7 5 3 1 = 2 2 2 2 2 2 2 (2n)! (2n) 6 4 2 1 1 = , 2 2 3 3 1 = , 2 2 2 5 5 3 1 = , 2 2 2 2 22n (2n)! (2n)! = n n = n . 2 2 n! 4 n! 8. (a) Let F be the probability distribution function of Y . For t 0, F (t) = P (Z 2 t) = 0.
For t > 0, F (t) = P (Y t) = P Z 2 t = P  = t   t = t Z t t =2 t  1. t  1 Let f be the probability density function of Y . For t 0, f (t) = 0. For t > 0, 1 t/2 1 1/2 e t 1 t/2 1 1 2 et/2 = 2 , t = e = (1/2) t 2 2 t where by the previous exercise, = (1/2). This shows that Y is gamma with parameters = 1/2 and r = 1/2. 1 f (t) = F (t) = 2 2 t (b) Since (X  )/ is standard normal, by part (a), W is gamma with parameters = 1/2 and r = 1/2. 9. The following solution is an intuitive one. A rigorous mathematical solution would have to
consider the sum of two random variables, each being the minimum of n exponential random Section 7.5 Beta Distributions 147 variables; so it would require material from joint distributions. However, the intuitive solution has its own merits and it is important for students to understand it. Let the time Howard enters the bank be the origin and let N (t) be the number of customers served by time t. As long as all of the servers are busy, due to the memoryless property of the exponential distribution, N (t) : t 0 is a Poisson process with rate n. This follows because if one server serves at the rate , n servers will serve at the rate n. For the Poisson process N (t) : t 0 , every time a customer is served and leaves, an "event" has occurred. Therefore, again because of the memoryless property, the service time of the person ahead of Howard begins when the first "event" occurs and Howard's service time begins when the second "event" occurs. Therefore, Howard's waiting time in the queue is the time of the second event of the Poisson process N (t), t 0 . This period, as we know, has a gamma distribution with parameters 2 and n. 10. Since the lengths of the characters are independent of each other and identically distributed,
for any two intervals 1 and 2 with the same length, the probability that n characters are emitted during 1 is equal to the probability that n characters are emitted in 2 . Moreover, for s > 0, the number of characters being emitted during (t, t + s] is independent of the number of characters that have been emitted in [0, t]. Clearly, characters are not emitted simultaneously. Therefore, N (t) : t 0 is stationary, possesses independent increments, and is orderly. So it is a Poisson process. By Exercise 11, Section 7.3, the time until the first character is emitted is exponential with parameter = 1000 ln(1  p). Thus N (t) : t 0 is a Poisson process with parameter = 1000 ln(1  p). Knowing this, we have that the time until the message is emitted, that is, the time until the kth character is emitted is gamma with parameters k and = 1000 ln(1  p). 7.5 BETA DISTRIBUTIONS
1 4! = = 12. We have B(2, 3) 1! 2! E(X) = 2 , 5 1. Yes, it is a probability density function of a beta random variable with parameters = 2 and = 3. Note that VarX = 6 1 = . 2) 6(5 25 2. No, it is not because, for = 3 and = 5, we have
7! 1 = = 105 = 120. B(3, 5) 2! 4! 3. Let = 5 and = 6. Then f is the probability density function of a beta random variable
with parameters 5 and 6 for c= 1 10! = = 1260. B(5, 6) 4! 5! 148 Chapter 7 Special Continuous Distributions For this value of c, E(X) = 5 , 11 VarX = 30 5 = . 2) 12(11 242 4. The answer is
P (p 0.60) = =
1 0.60 1 x 19 (1  x)12 dx B(20, 13)
1 0.60 32! 19! 12! x 19 (1  x)12 dx = 0.538. 5. Let X be the proportion of resistors the procurement office purchases from this vendor. We
know that X is beta. Let and be the parameters of the density function of X. Then =1 + 3 1 = . 2 ( + + 1)( + ) 18 Solving this system of 2 equations in 2 unknowns, we obtain = 1 and = 2. The desired probability is P (X 7/12) =
1 7/12 1 x 11 (1  x)21 dx = 2 B(1, 2) 1 7/12 (1  x) dx = 50 0.17. 288 6. Let X be the median of the fractions for the 13 sections of the course; X is a beta random variable with parameters 7 and 7. Let Y be a binomial random variable with parameters 13 and 0.40. By Theorem 7.2, P (X 0.40) = P (Y 7). Therefore,
6 P (X 0.40) = P (Y 6) =
i=0 13 (0.40)i (0.60)13i = 0.771156. i 7. Let Y be a binomial random variable with parameters 25 and 0.25; by Theorem 7.2,
P (X 0.25) = P (Y 5). Therefore,
4 P (X 0.25) = P (Y < 5) =
i=0 25 (0.25)i (0.75)25i = 0.214. i Section 7.5 Beta Distributions 149 8. (a) Clearly,
E(Y ) = a + (b  a)E(X) = a + (b  a) Var(X) = (b  a)2 Var(X) = (b) , + (b  a)2 . ( + + 1)( + )2 Note that 0 < X < 1 implies that a < Y < b. Let a < t < b; then P (Y t) = P a + (b  a)X t = P X =
0 (ta)/(ba) t a ba 1 x 1 (1  x)1 dx. B(, ) ya ba Let y = (b  a)x + a; we have P (Y t) = =
t a t a 1 ya B(, ) b  a 1 1 1 1 dy ba dy. 1 ya 1 b  a B(, ) b  a
1 1 by ba
1 1 This shows that the probability density function of Y is f (y) = (c) 1 1 ya b  a B(, ) b  a by ba , a < y < b. Note that a = 2, b = 6. Hence P (Y < 3) =
2 3 1 4! y  2 4 1! 2! 4
3 2 6y 4 2 dy = 3 64 (y  2)(6  y)2 dy = 67 3 67 = 0.26. 64 12 256 9. Suppose that
f (x) = 1 x 1 (1  x)1 , B(, ) 0 < x < 1, is symmetric about a point a. Then f (a  x) = f (a + x). That is, for 0 < x < min(a, 1  a), (a  x)1 (1  a + x)1 = (a + x)1 (1  a  x)1 . (28) Since and are not necessarily integers, for (a x)1 and (1a x)1 to be welldefined, we need to restrict ourselves to the range 0 < x < min(a, 1  a). Now, if a < 1  a, then, by continuity, (28) is valid for x = a. Substituting a for x in (28), we obtain (2a)1 (1  2a)1 = 0. 150 Chapter 7 Special Continuous Distributions Since a = 0, this implies that a = 1/2. If 1  a < a, then, by continuity, (28) is valid for x = 1  a. Substituting 1  a for x in (28), we obtain (2a  1)1 (2  2a)1 = 0. Since a = 1, this implies that a = 1/2. Therefore, in either case a = 1/2. In (28), substituting a = 1/2, and taking x = 1/4, say, we get (1/4)1 (3/4)1 = (3/4)1 (1/4)1 . This gives 3 = 0, which can only hold for = . Therefore, only beta density functions with = are symmetric, and they are symmetric about a = 1/2. 10. t = 0 gives x = 0; t = gives x = 1. Since dx =
B(, ) =
0 2t dt, we have (1 + t 2 )2 0 t2 1 + t2 1 1 1 + t2 1 2t dt = 2 (1 + t 2 )2 t 21 (1 + t 2 )(+) dt. 11. We have that
B(, ) =
0 1 x 1 (1  x)1 dx. Let x = cos to obtain
2 B(, ) = 2
0 /2 (cos )21 (sin )21 d. 0 Now () = Use the substitution t = y 2 to obtain () = 2
0 t 1 et dt. y 21 ey dy.
2 This implies that () () = 4
0 0 x 21 y 21 e(x 2 +y 2 ) dxdy. Now we evaluate this double integral by means of a change of variables to polar coordinates: y = r sin , x = r cos ; we obtain () () = 4
0 0 /2 r 2(+)1 (cos )21 (sin )21 er ddr
2 = 2B(, )
0 r 2(+)1 er dr = B(, )
2 0 u+1 eu du (let u = r 2 ) = B(, ) ( + ). Section 7.5 Beta Distributions 151 Thus B(, ) = () () . ( + ) 12. We will show that E(X2 ) = n/(n  2). Since E(X2 ) < , by Remark 6.6, E(X) < .
Since E(X) exists and xf (x) is an odd function, we have E(X) = Consequently, Var(X) = E(X2 )  E(X)
2  xf (x) dx = 0. = n . n2 Therefore, all we need to find is E(X2 ). By Theorem 6.3, n+1 x2 2 x2 1 + E(X 2 ) = n n  n 2 Substituting x = ( n )t in this integral yields (n+1)/2 dx. n+1 2 E(X2 ) = (nt 2 )(1 + t 2 )(n+1)/2 n dt n  n 2 n+1 = 2 n 2n t 2 (1 + t 2 )(n+1)/2 dt. 0 2 By the previous two exercises, 3 n2 3 n2 2 2 . = t 2 (1 + t 2 )(n+1)/2 dt = B , 2 n+1 2 2 0 2 Therefore, 3 3 n2 n2 n+1 n 2 2 2 2 2 . = E(X2 ) = n n n n+1 2 2 2 By the solution to Exercise 7, Section 7.4, (1/2) = . Using the identity (r +1) = r (r), we have 3 1 1 = = ; 2 2 2 2 n n2 n2 n2 = +1 = . 2 2 2 2 152 Chapter 7 Special Continuous Distributions Consequently, n2 n 2 2 E(X2 ) = n2 n2 2 2 = n . n2 7.6 SURVIVAL ANALYSIS AND HAZARD FUNCTIONS 1. Let X be the lifetime of the electrical component, F be its probability distribution function,
and (t) be its failure rate. For some constants and , we are given that (t) = t + . Since (48) = 0.10 and (72) = 0.15, 48 + = 0.10 72 + = 0.15. Solving this system of two equations in two unknowns gives = 1/480 and = 0. Hence (t) = t/480. By (7.6), for t > 0, P (X > t) = F (t) = exp 
0 t u 2 du = et /960 . 480 Let f be the probability density function of X. This also gives f (t) =  The answer to part (a) is P (X > 30) = e900/960 = e0.9375 = 0.392. The exact value for part (b) is P (X < 31  X > 30) = = P (30 < X < 31) P (X > 30) 1 0.392
31 30 d t t 2 /960 F (t) = e . dt 480 (t/480)et 2 /960 dt = 0.02411 = 0.0615. 0.392 Note that for small t , (t) t is approximately the probability that the component fails within t hours after t, given that it has not yet failed by time t. Letting t = 1, for t = 30, (t) t 0.0625 which is relatively close to the exact value of 0.0615. This is interesting because t = 1 is not that small, and one may not expect close approximations anyway. Chapter 7 Review Problems 153 2. Let F be the survival function of a Weibull random variable. We have F (t) = t x 1 ex dx. Letting u = x , we have du = x 1 dx. Thus F (t) = Therefore, t eu du = eu t = et . t 1 et = t 1 et (t) = 1, for = 1; so the Weibull in this case is exponential with parameter 1. Clearly, for < 1, (t) < 0; so (t) is decreasing. For > 1, (t) > 0; so (t) is increasing. Note that for = 2, the failure rate is the straight line (t) = 2t. (t) = REVIEW PROBLEMS FOR CHAPTER 7 1.
5 30  25 = . 37  25 12 is 170  130 P (X > 170) 20 = P (X > 170  X > 140) = P (X > 140) 140  130 P Z> 20 P Z> = 1  (2) 1  0.9772 P (Z > 2) = = = 0.074. P (Z > 0.5) 1  (0.5) 1  0.6915 90 = 9.49. Using normal approximation 2. Let X be the weight of a randomly selected women from this community. The desired quantity 3. Let X be the number of times the digit 5 is generated; X is binomial with parameters n = 1000 and p = 1/10. Thus np = 100 and np(1  p) = and making correction for continuity, P (X 93.5) = P Z 93.5  100 = P (Z 0.68) = 1  9.49 (0.68) = 0.248. 4. The given relation implies that
1  e2 = 2 (1  e3 )  (1  e2 ) . 154 Chapter 7 Special Continuous Distributions This is equivalent to 3e2  2e3  1 = 0, or, equivalently, e  1
2 2e + 1 = 0. The only root of this equation is = 0 which is not acceptable. Therefore, it is not possible that X satisfy the given relation. 5. Let X be the lifetime of a random light bulb. Then
P (X < 1700) = 1  e(1/1700)1700 = 1  e1 . The desired probability is 1  P (none fails)  P (one fails) 20 20 =1 (1  e1 )0 (e1 )20  0 1 1  e1 e1 19 = 0.999999927. 6. Note that limx0 x ln x = 0; so
E( ln X) =
0 1 ( ln x) dx = x  x ln x 1 0 = 1. 7. Let X be the diameter of the randomly chosen disk in inches. We are given that X N (4, 1).
We want to find the distribution function of 2.5X; we have 1 P (2.5X x) = P (X x/2.5) = 2
x/2.5  e(t4) 2 /2 dt. 8. If < 0, then + < ; therefore,
P ( X + ) = P (0 X + ) P (0 X ). If > 0, then e < 1. Thus P ( X + ) = 1  e(+)  1  e = e 1  e < 1  e = P (0 X ). 9. We are given that 1/ = 1.25; so = 0.8. Let X be the time it takes for a random student to
complete the test. Since P (X > 1) = e(0.8)1 = e0.8 , the desired probability is 1  e0.8
10 = 1  e8 = 0.99966. Chapter 7 Review Problems 155 10. Note that f (x) = ke[x(3/2)] 2 +17/4 = ke17/4 e[x(3/2)] .
2 Comparing this with the probability density function of a normal random variable with mean 3/2, we see that 2 = 1/2 and ke17/4 = 1/( 2 ). Therefore, k= 1 1 e17/4 = e17/4 . 2 90  72 =1 7 11. Let X be the grade of a randomly selected student.
P (X 90) = P Z Similarly, P (80 X < 90) = P (1.14 Z < 2.57) = 0.122, P (70 X < 80) = P (0.29 Z < 1.14) = 0.487, P (60 X < 70) = P (1.71 Z < 0.29) = 0.3423, P (X < 60) = P (Z < 1.71) = 0.0436. Therefore, approximately 0.51% will get A, 12.2% will get B, 48.7% will get C, 34.23% D, and 4.36% F. (2.57) = 0.0051. 12. Since E(X) = 1/,
P X > E(X) = e(1/) = e1 = 0.36788. 13. Round off error to the nearest integer is uniform over (0.5, 0.5); round off error to the nearest 1st decimal place is uniform over (0.05, 0.05); round off error to the nearest 2nd decimal place is uniform over (0.005, 0.005), and so on. In general, round off error to the nearest k decimal places is uniform over (5/10k+1 , 5/10k+1 ). 14. We want to find the smallest a for which P (X a) 0.90. This implies
P Z a  175 0.90. 22 Using Table 1 of the appendix, we see that (a  175)/22 = 1.29 or a = 203.38. 15. Let X be the breaking strength of the yarn under consideration. Clearly,
P (X 100) = P Z So the desired probability is 1 10 10 (0.33)1 (0.67)9 = 0.89. (0.33)0 (0.67)10  1 0 100  95 =1 11 (0.45) = 0.33. 156 Chapter 7 Special Continuous Distributions 16. Let X be the time until the 91st call is received. X is a gamma random variable with parameters
r = 91 and = 23. The desired probability is P (X 4) =
4 23e23x (23x)911 dx (91)
4 0 =1 =1 23e23x (23x)911 dx 90!
4 0 2391 90! x 90 e23x dx = 1  0.55542 = 0.44458. 17. Clearly,
E(X) = Var(X) = Now E X2  E(X) implies that
2 (1  ) + (1 + ) = 1, 2 2 (1 +  1 + )2 = . 12 3 = 2 3 2 + 1, 3 which yeilds 3E(X2 )  1 = 2 , or, equivalently, E(3X2  1) = 2 . Therefore, one choice for g(X) is g(X) = 3X2  1. E X2 = 18. Let and be the parameters of the density function of X/ . Solving the following two
equations in two unknowns, E(X/ ) = Var(X/ ) = 3 = , + 7 3 = , 2 ( + + 1)( + ) 98 we obtain = 3 and = 4. Therefore, X/ is beta with parameters 3 and 4. The desired probability is P ( /7 < X < /3) = P (1/7 < X/ < 1/3) = = 60
1/3 1/7 1/3 1/7 1 x 2 (1  x)3 dx B(3, 4) x 2 (1  x)3 dx = 0.278. Chapter 8 Bivariate Distributions
8.1 JOINT DISTRIBUTIONS OF TWO RANDOM VARIABLES
2 x=1 2 y=1 1. (a) k(x/y) = 1 implies that k = 2/9.
2 y=1 2 x=1 (b) pX (x) = pY (y) = (2x)/(9y) = x/3, x = 1, 2. y = 1, 2. (2x)/(9y) = 2/(3y), p(2, 1) 2 4/9 = . = pY (1) 2/3 3 (c) P (X > 1  Y = 1) =
2 2 (d) E(X) =
y=1 x=1 3 x=1 2 y=1 x 2 x 5 = ; 9 y 3 2 2 E(Y ) =
y=1 x=1 y 2 x 4 = . 9 y 3 2. (a) c(x + y) = 1 implies that c = 1/21.
2 y=1 (1/21)(x 3 x=1 (1/21)(x (b) pX (x) = pY (y) = + y) = (2x + 3)/21. x = 1, 2, 3. + y) = (6 + 3y)/21. y = 1, 2. (c) P (X 2  Y = 1) =
3 2 p(2, 1) + p(3, 1) 7/21 7 = = . pY (1) 9/21 9
3 2 (d) E(X) =
x=1 y=1 46 1 x(x + y) = ; 21 21 E(Y ) =
x=1 y=1 11 1 y(x + y) = . 21 7 3. (a) k(1 + 1 + 1 + 9 + 4 + 9) = 1 implies that k = 1/25.
(b) pX (1) = p(1, 1) + p(1, 3) = 12/25, pY (1) = p(1, 1) = 2/25, pX (2) = p(2, 3) = 13/25; pY (3) = p(1, 3) + p(2, 3) = 23/25. 158 Chapter 8 Bivariate Distributions Therefore, pX (x) = 12/25 13/25 if x = 1 if x = 2, pY (y) = 2/25 23/25 if y = 1 if y = 3. (c) E(X) = 1 13 38 12 +2 = ; 25 25 25 E(Y ) = 1 2 23 71 +3 = . 25 25 25 4. P (X > Y ) = p(1, 0) + p(2, 0) + p(2, 1) = 2/5, P (X + Y 2) = p(1, 0) + p(1, 1) + p(2, 0) = 7/25, P (X + Y = 2) = p(1, 1) + p(2, 0) = 6/25. 5. Let X be the number of sheep stolen; let Y be the number of goats stolen. Let p(x, y) be the
7 x 8 y 5 4xy ; 20 4 joint probability mass function of X and Y . Then, for 0 x 4, 0 y 4, 0 x + y 4, p(x, y) = p(x, y) = 0, for other values of x and y. 6. The following table gives p(x, y), the joint probability mass function of X and Y ; pX (x), the marginal probability mass function of X; and pY (y), the marginal probability mass function of Y . x 2 3 4 5 6 7 8 9 10 11 12 pY (y) y 0 1/36 0 1/36 0 1/36 0 1/36 0 1/36 0 1/36 6/36 1 0 2/36 0 2/36 0 2/36 0 2/36 0 2/36 0 10/36 2 0 0 2/36 0 2/36 0 2/36 0 2/36 0 0 8/36 3 0 0 0 2/36 0 2/36 0 2/36 0 0 0 6/36 4 0 0 0 0 2/36 0 2/36 0 0 0 0 4/36 5 0 0 0 0 0 2/36 0 0 0 0 0 2/36 pX (x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 7. p(1, 1) = 0, p(1, 0) = 0.30, p(0, 1) = 0.50, p(0, 0) = 0.20. Section 8.1 Joint Distributions of Two Random Variables 159 8. (a) For 0 x 7, 0 y 7, 0 x + y 7,
p(x, y) = 13 x 13 y 26 7xy . 52 7 For all other values of x and y, p(x, y) = 0. (b) P (X Y ) =
3 y=0 7y x=y p(x, y) = 0.61107. 0 x 1;
1 0 9. (a) fX (x) =
0 x 2 dy = 2x, fY (y) = 1 y 2 dx = 2(1  y), 0 y 1. (b) E(X) =
0 1 xfX (x) dx = yfY (y) dy =
0 1/2 0 1 0 x x/2 x(2x) dx = 2/3; 2y(1  y) dy = 1/3.
1/2 0 E(Y ) =
0 1 1 (c) P X < 1 = 2 fX (x) dx = 2 dy dx = 1 2x dx = , 4 P (X < 2Y ) = P (X = Y ) = 0. 1 , 2 10. (a) fX (x) =
0 x 8xy dy = 4x 3 , 0 x 1, fY (y) = 1 y 8xy dx = 4y(1  y 2 ), 0 y 1.
1 0 (b) E(X) =
0 1 xfX (x) dx = yfY (y) dy =
0 x 4x 3 dx = 4/5; y 4y(1  y 2 ) dy = 8/15. fY (y) =
0 E(Y ) =
0 1 1 11. fX (x) =
0 2 1 x ye dy = ex , x > 0; 2 1 x 1 ye dx = y, 0 < y < 2. 2 2 12. Let R = (x, y) : 0 x 1, 0 y 1 . Since area(R) = 1, P (X + Y 1/2) is the area of the region (x, y) R : x + y 1/2 which is 1/8. Similarly, P (X  Y 1/2) is the 160 Chapter 8 Bivariate Distributions area of the region (x, y) R : x  y 1/2 which is 7/8. P (X 2 + Y 2 1) is the area of the region (x, y) R : x 2 + y 2 1 which is /4. P (XY 1/4) is the sum of the area of the region (x, y) : 0 x 1/4, 0 y 1 which is 1/4 and the area of the region under the curve y = 1/(4x) from 1/4 to 1. (Draw a figure.) Therefore, 1 + 4
1 1/4 P (XY 1/4) = 1 dx 0.597. 4x 13. (a) The area of R is
0 1 1 (x  x 2 ) dx = ; so 6 if (x, y) R elsewhere. f (x, y) = 6 0 (b) fX (x) = fY (y) = (c) E(X) = x x2 y 1 0 f (x, y) dy =
y x x2 6 dy = 6x(1  x), y 0 < x < 1; 0 < y < 1. f (x, y) dx =
1 0 y 6 dx = 6( y  y), xfX (x) dx = yfY (y) dy =
0 6x 2 (1  x) dx = 1/2; 6y( y  y) dy = 2/5. E(Y ) =
0 1 1 14. Let X and Y be the minutes past 11:30 A.M. that the man and his fiance arrive at the lobby, respectively. We have that X and Y are uniformly distributed over (0, 30). Let S = (x, y) : 0 x 30, 0 y 30 , and R = (x, y) S : y x  12 or y x + 12 . The desired probability is the area of R divided by the area of S: 324/900 = 0.36. (Draw a figure.) 15. Let X and Y be two randomly selected points from the interval (0, ). We are interested in
E X  Y  . Since the joint probability density function of X and Y is 1 0 f (x, y) = 2 0<x< , 0<y< elsewhere, Section 8.1 Joint Distributions of Two Random Variables 161 E X  Y  =
0 0 x  y 1
2 0 0 y 1
2 dx dy 1
2 0 y = = (y  x) dx dy + (x  y) dx dy 6 + 6 = . 3 16. The problem is equivalent to the following: Two random numbers X and Y are selected at
random and independently from (0, ). What is the probability that X  Y  < X? Let S = (x, y) : 0 < x < , 0 < y < and R = (x, y) S : x  y < x = (x, y) S : y < 2x . The desired probability is the area of R which is 3 2 /4 divided by (Draw a figure.)
2 . So the answer is 3/4. 17. Let S = (x, y) : 0 < x < 1, 0 < y < 1 and R = (x, y) S : y x and x 2 + y 2 1 . The desired probability is the area of R which is /8 divided by the area of S which is 1. So the answer is /8. probability density function f . For discrete random variables the proof is similar. The relation P (X Y ) = 1, implies that f (x, y) = 0 if x > y. Hence by Theorem 8.2, E(X) = = =
x      y  y   18. We prove this for the case in which X and Y are continuous random variables with joint xf (x, y) dx dy xf (x, y) dx dy yf (x, y) dx dy yf (x, y) dx dy = E(Y ). 19. Let H be the distribution function of a random variable with probability density function h.
That is, let H (x) =
 h(y) dy. Then  x  P (X Y ) = h(x)h(y) dy dx =  2 h(x) = x  h(y) dy dx 1 = h(x)H (x) dx = H (x) 2   1 2 1 (1  02 ) = . 2 2 20. Since 0 2G(x)  1 1, 0 2H (y)  1 1, and 1 1, we have that
1 2G(x)  1 2H (y)  1 1. 162 Chapter 8 Bivariate Distributions So 0 1 + 2G(x)  1 2H (y)  1 2. This and g(x) 0, h(y) 0 imply that f (x, y) 0. To prove that f is a joint probability density function, it remains to show that
    f (x, y) dx dy = 1. f (x, y) dx dy  =  g(x)h(y) dx dy +  =1+ =1+  h(y) 2H (y)  1 dy  g(x)h(y) 2G(x)  1 2H (y)  1 dx dy g(x) 2G(x)  1 dx
2   1 2 1 2H (y)  1 2G(x)  1  4 4 Now we calculate the marginals. fX (x) = =   = 1 + 0 0 = 1. g(x)h(y) 1 + 2G(x)  1 2H (y)  1 g(x)h(y) dy +   dy g(x)h(y) 2G(x)  1 2H (y)  1 dy  2 = g(x) h(y) dy + g(x) 2G(x)  1 h(y) 2H (y)  1 dy 1 2H (y)  1  4 = g(x) + g(x) 2G(x)  1 0 = g(x) + 0 = g(x). = g(x) + g(x) 2G(x)  1 Similarly, fY (y) = h(y). 21. Orient the circle counterclockwise and let X be the length of the arc N M and Y be length of the arc N L. Let R be the radius of the circle; clearly, 0 X 2 R and 0 Y 2 R. The angle MN L is acute if and only if Y  X < R. Therefore, the sample space of this experiment is S = (x, y) : 0 x 2 R, 0 y 2 R and the desired event is E = (x, y) S : y  x < R . The probability that MN L is acute is the area of E which is 3 2 R 2 divided by the area of S which is 4 2 R 2 ; that is, 3/4. 22. Let
S = (x, y) R2 : 0 x 1, 0 y 1 , B = (x, y) S : 0.5 < x + y < 1.5 , A = (x, y) S : 0 < x + y < 0.5 , C = (x, y) S : x + y > 1.5 . Section 8.1 Joint Distributions of Two Random Variables 163 The probability that the integer nearest to x + y is 0 is integer nearest to x + y is 1 is x + y is 2 is area (C) 1 = . area(S) 8 area(B) 3 = , and the probability that the nearest integer to area(S) 4 1 area(A) = , The probability that the area (S) 8 23. Let X be a random number from (0, a) and Y be a random number from (0, b). In 4 =4 3 ways we can select three of X, a  X, Y , and b  Y . If X, a  X, and Y are selected, a triangular pen is possible to make if and only if X < (a  X) + Y , a  X < X + Y , and Y < X + (a  X). The probability of this event is the area of (x, y) R2 : 0 < x < a, 0 < y < b, 2x  y < a, 2x + y > a, y < a which is a 2 /2 divided by the area of S = (x, y) R2 : 0 < x < a, 0 < y < b which is ab: (a 2 /2)/ab = a/(2b). Similarly, for each of the other three 3combinations of X, a  x, Y , and b  Y also the probability that the three segments can be used to form a triangular pen is a/(2b). Thus the desired probability is 1 a 1 a 1 a a 1 a + + + = . 4 2b 4 2b 4 2b 4 2b 2b 24. Let X and Y be the two points that are placed on the segment. Let E be the event that the length of none of the three parts exceeds the given value . Clearly, P (E  X < Y ) = P (E  Y < X) and P (X < Y ) = P (Y < X) = 1/2. Therefore, P (E) = P (E  X < Y )P (X < Y ) + P (E  Y < X)P (Y < X) 1 1 = P (E  X < Y ) + P (E  X < Y ) = P (E  X < Y ). 2 2 This shows that for calculation of P (E), we may reduce the sample space to the case where X < Y . The reduced sample space is S = (x, y) : x < y, 0 < x < , 0 < y < The desired probability is the area of R = (x, y) S : x < , y  x < , y >  divided by area(S) =
2 . (3  )2 2 area(R) = 2 2 3 1 2 2 /2. But if
2 3 2 2 if . 164 Chapter 8 Bivariate Distributions Hence the desired probability is 3 1 2 if
2 P (E) = 3 2 2 1  3 1  if . 25. R is the square bounded by the lines x + y = 1, x + y = 1, x  y = 1, and x  y = 1; its area is 2. To find the probability density function of X, the xcoordinate of the point selected at random from R, first we calculate P (X t), t. For 1 t < 0, P (X t) is the area of the triangle bound by the lines x + y = 1, x  y = 1, and x = t which is (1 + t)2 divided by area(R) = 2. (Draw a figure.) For 0 t < 1, P (X t) is the area inside R to the left of the line x = t which is 2  (1  t)2 divided by area(R) = 2. Therefore, 0 (1 + t)2 2 P (X t) = 2  (1  t)2 2 1 and hence 1 + t d P (X t) = 1  t dt 0 t < 1 1 t < 0 0t <1 t 1, 1 t < 0 0t <1 otherwise. This shows that fX (t), the probability density function of X is given by fX (t) = 1  t, 1 t 1; 0, elsewhere. 26. Clearly,
P (Z z) =
{(x,y) : y/xz} f (x, y) dx dy. Now for x > 0, y/x z if and only if y xz; for x < 0, y/x z if and only if y xz. Therefore, integration region is (x, y) : x < 0, y xz (x, y) : x > 0, y xz . Thus P (Z z) =
0  xz 0 xz  f (x, y) dy dx + f (x, y) dy dx. Section 8.1 Joint Distributions of Two Random Variables 165 Using the substitution y = tx, we get P (Z z) = = = =
0  0  0   z  xf (x, tx) dt dx +
0 z  z xf (x, tx) dt dx xf (x, tx) dt dx xf (x, tx) du dx xf (x, tx) dx dt. z  z  z  xf (x, tx) dt dx +
0 xf (x, tx) dt dx +
0  z   xf (x, tx) dt dx =  z  Differentiating with respect to z, Fundamental Theorem of Calculus implies that, fZ (z) = d P (Z z) = dz xf (x, xz) dx. 27. Note that there are exactly n such closed semicircular disks because the probability that the diameter through Pi contains any other point Pj is 0. (Draw a figure.) Let E be the event that all the points are contained in a closed semicircular disk. Let Ei be the event that the points are all in Di . Clearly, E = n Ei . Since there is at most one Di , 1 i n, that contains all i=1 the Pi 's, the events E1 , E2 , . . . , En are mutually exclusive. Hence
n n n P (E) = P
i=1 Ei =
i=1 P (Ei ) =
i=1 1 2 n1 =n 1 2 n1 , where the nexttothelast equality follows because P (Ei ) is the probability that P1 , P2 , . . . , Pi1 , Pi+1 , . . . , Pn fall inside Di . The probability that any of these falls inside Di is (area of Di )/(area of the disk) = 1/2 independently of the others. Hence the probability that all of them fall inside Di is (1/2)n1 . 28. We have that
fX (x) = ( + + ) 1x 1 1 x y (1  x  y) 1 dy () () ( ) 0 1x 1 x 1 y 1 (1  x  y) 1 dy. = B(, + )B(, ) 0
1 0 Let z = y/(1  x); then dy = (1  x) dz, and
1x 0 y 1 (1  x  y) 1 dy = (1  x)+ 1 z1 (1  z) 1 dz = (1  x)+ 1 B(, ). So fX (x) = 1 x 1 (1  x)+ 1 B(, ) B(, + )B(, ) 1 x 1 (1  x)+ 1 . = B(, + ) 166 Chapter 8 Bivariate Distributions This shows that X is beta with parameters (, + ). A similar argument shows that Y is beta with parameters (, + ). 29. It is straightforward to check that f (x, y) 0, f is continuous and   f (x, y) dx dy = 1. F Therefore, f is a continuous probability density function. We will show that does not x F exist at (0, 0). Similarly, one can show that does not exist at any point on the yaxis. Note x that for small x > 0, F ( x, 0)  F (0, 0) = P (X x , Y 0)  P (X 0 , Y 0) x , Y 0) =
0  0 x = P (0 X f (x, y) dx dy. Now, from the definition of f (x, y), we must have y > ln(2 x). Thus, for small x > 0, F ( x, 0)  F (0, 0) = This implies that
x0+ 0 ln(2 x) 0 x x < (1/2)ey or, equivalently, x . 2 (1  2xey ) dx dy = ( x)2  ( x) ln(2 x) + lim F ( x, 0)  F (0, 0) = lim x0+ x x  ln(2 x)  1 = , 2 showing that F does not exist at (0, 0). x 8.2 INDEPENDENT RANDOM VARIABLES 1. Note that pX (x) = (1/25)(3x 2 + 5), pY (y) = (1/25)(2y 2 + 5). Now pX (1) = 8/25,
pY (0) = 5/25, and p(1, 0) = 1/25. Since p(1, 0) = pX (1)pY (0), X and Y are dependent. 2. Note that
1 p(1, 1) = , 7 pX (1) = p(1, 1) + p(1, 2) = 1 2 3 + = , 7 7 7 1 5 6 pY (1) = p(1, 1) + p(2, 1) = + = . 7 7 7 Since p(1, 1) = pX (1)pY (1), X and Y are dependent. Section 8.2 Independent Random Variables 167 3. By the independence of X and Y ,
P (X = 1, Y = 3) = P (X = 1)P (Y = 3) = 1 2 1 2 2 3 2 3
3 = 4 . 81 P (X + Y = 3) = P (X = 1, Y = 2) + P (X = 2, Y = 1) = 1 2 1 2 2 3 2 3
2 + 1 2 2 3 2 1 2 4 = . 2 3 27 4. No, they are not independent because, for example, P (X = 0  Y = 8) = 1 but
39 8 52 8 P (X = 0) = = 0.08175 = 1, showing that P (X = 0  Y = 8) = P (X = 0). 5. The answer is 7 2 1 2 2 1 2 5 8 2 1 2 2 1 2 6 = 0.0179. 6. We have that
P max(X, Y ) t = P (X t, Y t) = P (X t)P (Y t) = F (t)G(t). P min(X, Y ) t = 1  P min(X, Y ) > t = 1  P (X > t, Y > t) = 1  P (X > t)P (Y > t) = 1  1  F (t) 1  G(t) = F (t) + G(t)  F (t)G(t). 7. Let X and Y be the number of heads obtained by Adam and Andrew, respectively. The desired
probability is
n n P (X = i, Y = i) =
i=0 i=0 n P (X = i)P (Y = i) n i
2n n i=0 =
i=0 1 2 i 1 2
2 ni 1 2 n i
2n 1 2 i 1 2 ni 1 = 2 n i = 2n , n where the last equality follows by Example 2.28. 168 Chapter 8 Bivariate Distributions An Intuitive Solution: Let Z be the number of tails obtained by Andrew. The desired probability is
n n n P (X = i, Y = i) =
i=0 i=0 P (X = i, Z = i) =
i=0 P (X = i, Y = n  i) = P (Adam and Andrew get a total of n heads) 1 2n 2n = P ( n heads in 2n flips of a fair coin) = . 2 n 8. For i, j 0, 1, 2, 3 , the sum of the numbers in the ith row is pX (i) and the sum of the
numbers in the j th row is pY (j ). We have that pX (0) = 0.41, pY (0) = 0.41, pX (1) = 0.44, pY (1) = 0.44, pX (2) = 0.14, pY (2) = 0.14, pX (3) = 0.01; pY (3) = 0.01. Since for all x, y 0, 1, 2, 3 , p(x, y) = pX (x)pY (y), X and Y are independent. 9. They are not independent because
fX (x) =
0 x 2 dy = 2x, 0 x 1; 0 y 1; fY (y) = and so f (x, y) = fX (x)fY (y). 1 y 2 dx = 2(1  y), 10. Let X and Y be the amount of cholesterol in the first and in the second sandwiches, respectively.
Since X and Y are continuous random variables, P (X = Y ) = 0 regardless of what the probability density functions of X and Y are. 11. We have that
fX (x) =
0 x 2 ex(y+1) dy = xex , x 2 ex(y+1) dx = x 0; y 0, fY (y) =
0 2 , (y + 1)3 where the second integral is calculated by applying integration by parts twice. Now since f (x, y) = fX (x)fY (y), X and Y are not independent. Section 8.2 Independent Random Variables 169 12. Clearly,
E(XY ) =
0 1 x 1 0 1 1 (xy)(8xy) dy dx =
0 1 x 1 4 8y 2 dy x 2 dx = , 9 E(X) = E(Y ) =
0 1 8 x(8xy) dy dx = , 15 x 1 4 y(8xy) dy dx = . 5 x So E(XY ) = E(X)E(Y ). 13. Since
f (x, y) = ex 2e2y = fX (x)fY (y), X and Y are independent exponential random variables with parameters 1 and 2, respectively. Therefore, 1 E(X 2 Y ) = E(X2 )E(Y ) = 2 = 1. 2 14. The joint probability density function of X and Y is given by
f (x, y) = e(x+y) 0 x > 0, y > 0 elsewhere. Let G be the probability distribution function, and g be the probability density function of X/Y . For t > 0, G(t) = P =
0 X t = P (X tY ) Y 0 ty e(x+y) dx dy = t . 1+t Therefore, for t > 0, g(t) = G (t) = 1 . (1 + t)2 Note that G (t) = 0 for t < 0; G (0) does not exist. 15. Let F and f be the probability distribution and probability density functions of max(X, Y ),
respectively. Clearly, F (t) = P max(X, Y ) t = P (X t, Y t) = (1  et )2 , t 0. Thus f (t) = F (t) = 2et (1  et ) = 2et  2e2t . 170 Chapter 8 Bivariate Distributions Hence E max(X, Y ) = 2
0 tet dt 
0 2te2t dt = 2  1 3 = . 2 2 Note that
0 tet dt is the expected value of an exponential random variable with parameter 0 1, thus it is 1. Also, 2te2t dt is the expected value of an exponential random variable with parameter 2, thus it is 1/2. 16. Let F and f be the probability distribution and probability density functions of max(X, Y ).
For 1 < t < 1, F (t) = P max(X, Y ) t = P (X t, Y t) = P (X t)P (Y t) = Thus f (t) = F (t) = Therefore, E(X) =
1 1 t +1 2 2 . t +1 , 2 t 1 < t < 1. 1 t +1 dt = . 2 3 17. Let F and f be the probability distribution and probability density functions of XY , respectively. Clearly, for t 0, F (t) = 0 and for t 1, F (t) = 1. For 0 < t < 1, F (t) = P (XY t) = 1  P (XY > t) = 1  Hence f (t) = F (t) =  ln t 0
1 t 1 t/x dy dx = t  t ln t. 0<t <1 elsewhere. 18. The joint probability density function of X and Y is given by
f (x, y) = Now fX (x) = 0 1 1 = area (R) if (x, y) R otherwise.
1x 2  1x 2 1 2 dy = 2 1 dx = 1  x2, fY (y) =
 1y 2 1y 2 1  y2. Since f (x, y) = fX (x)fY (y), the random variables X and Y are not independent. Section 8.2 Independent Random Variables 171 19. Let X be the number of adults and Y be the number of children who get sick. The desired
probability is
5 6 5 6 P (Y = i, X = j ) =
i=0 j =i+1 5 6 i=0 j =i+1 P (Y = i)P (X = j ) =
i=0 j =i+1 6 6 (0.30)i (0.70)6i (0.2)j (0.8)6j = 0.22638565. i j 20. Let X be the lifetime of the muffler Elizabeth buys from company A and Y be the lifetime of
the muffler she buys from company B. The joint probability density function of X and Y is h(x, y) = f (x)g(y), x > 0, y > 0. So the desired probability is P (Y > X) =
0 x 2 (2y)/11 1 11 dy ex/6 dx = . e 11 6 23 21. If IA and IB are independent, then
P (IA = 1, IB = 1) = P (IA = 1)P (IB = 1). This is equivalent to P (AB) = P (A)P (B) which shows that A and B are independent. On the other hand, if {A, B} is an independent set, so are the following: A, B c , Ac , B , and Ac , B c . Therefore, P (AB) = P (A)P (B), P (Ac B) = P (Ac )P (B), These relations, respectively, imply that P (IA = 1, IB = 1) = P (IA = 1)P (IB = 1), P (IA = 1, IB = 0) = P (IA = 1)P (IB = 0), P (IA = 0, IB = 1) = P (IA = 0)P (IB = 1), P (IA = 0, IB = 0) = P (IA = 0)P (IB = 0). These four relations show that IA and IB are independent random variables. P (AB c ) = P (A)P (B c ), P (Ac B c ) = P (Ac )P (B c ). 22. The joint probability density function of B and C is 2 2 9b c 676 f (b, c) = 0 1 < b < 3, 1 < c < 3 otherwise. For X2 +BX+C to have two real roots we must have B 2 4C > 0, or, equivalently, B 2 > 4C. Let E = (b, c) : 1 < b < 3, 1 < c < 3, b2 > 4c ; 172 Chapter 8 Bivariate Distributions the desired probability is 9b2 c2 db dc = 676
3 2 1 b2 /4 E 9b2 c2 dc db 0.12. 676 (Draw a figure to verify the region of integration.) 23. Note that
fX (x) = fY (y) = Now fX (x)fY (y) = g(x)h(y) = f (x, y) = f (x, y)      g(x)h(y) dy = g(x) g(x)h(y) dx = h(y)   h(y) dy, g(x) dx. h(y) dy    g(x) dx h(y)g(x) dy dx f (x, y) dy dx = f (x, y). This relation shows that X and Y are independent. 24. Let G and g be the probability distribution and probability density functions of
max(X, Y ) min(X, Y ). Then G(t) = 0 if t < 1. For t 1, G(t) = P max(X, Y ) t = P max(X, Y ) t min(X, Y ) min(X, Y ) = P X t min(X, Y ), Y t min(X, Y ) = P min(X, Y ) =P X =P Y X Y , min(X, Y ) t t X X Y Y , Y , X , Y t t t t X Y , X t t =P X Y tX . t x y tx t This quantity is the area of the region (x, y) : 0 < x < 1, 0 < y < 1, Section 8.2 Independent Random Variables 173 which is equal to (t  1)/t. Hence G(t) = 0 t  1 t t <1 t 1, and therefore, 1 2 g(t) = G (t) = t 0 t 1 elsewhere. 25. Let F be the distribution function of X/(X + Y ). Since X/(X + Y ) (0, 1), we have that
F (t) = For 0 t < 1, P X 1t t =P Y X = 2 X+Y t =
0 0 [(1t)x]/t 0 0 1 t <0 t 1. ex ey dy dx ex e[(1t)x]/t dx = ex/t dt = t. Therefore, 0 F (t) = t 1 t <0 0t <1 t 1. This shows that X/(X + Y ) is uniform over (0, 1). 26. The fact that if X and Y are both normal with mean 0 and equal variance implies that f (x, y) is circularly symmetrical is straightforward. We prove the converse; suppose that f is circularly symmetrical, then there exists a function so that fX (x)fY (y) = x2 + y2 . Differentiating this relation with respect to x and using fY (y) = yields x2 + y2 x2 + y2 x2 + y2 = fX (x) . xfX (x) fX (x)fY (y) = fX (x) x 2 + y 2 /fX (x) 174 Chapter 8 Bivariate Distributions Now the right side of this equation is a function of x while its left side is a function of x 2 + y 2 . This implies that fX (x)/ xfX (x) is constant. To prove this, we show that for any given x1 and x2 , fX (x1 ) f (x2 ) = X . x1 fX (x1 ) x2 fX (x2 )
2 2 2 2 Let y1 = x2 and y2 = x1 ; then x1 + y1 = x2 + y2 and we have fX (x1 ) = x1 fX (x1 ) 2 2 x1 + y 1 2 2 x1 + y1 2 2 x1 + y 1 = 2 2 x2 + y2 2 2 x2 + y2 2 2 x2 + y2 = fX (x2 ) . x2 fX (x2 ) We have shown that for some constant k, fX (x) = k. xfX (x) Therefore, fX (x) 1 = kx and hence ln fX (x) = kx 2 + c, or fX (x) 2 fX (x) = e(1/2)kx where = ec . Now since then fX (x) = ex
2 /(2 2 ) 2 +c = e(1/2)kx ,
2  e(1/2)kx dx = 1, we have that k < 0. Let =
2 1/k;  and ex 2 /(2 2 ) dx = 1 implies that = 1/( 2 ). So 1 2 2 fX (x) = ex /(2 ) , showing that X N (0, 2 ). The fact that Y N (0, 2 ) is 2 proved similarly. 8.3 CONDITIONAL DISTRIBUTIONS
2 1. pY (y) =
x=1 p(x, y) = 1 (2y 2 + 5). Thus 25 pXY (xy) = p(x, y) (1/25)(x 2 + y 2 ) x2 + y2 = = 2 x = 1, 2, y = 0, 1, 2, pY (y) (1/25)(2y 2 + 5) 2y + 5 P (X = 2  Y = 1) = pXY (21) = 5/7,
2 2 E(XY = 1) =
x=1 xpXY (x1) =
x=1 x x2 + 1 12 = . 7 7 Section 8.3 Conditional Distributions 175 2. Since
fY (y) =
0 y 2 dx = 2y, 0 < y < 1, we have that fXY (xy) = 2 1 f (x, y) = = , fY (y) 2y y 0 < x < y, 0 < y < 1. 3. Let X be the number of flips of the coin until the sixth head is obtained. Let Y be the number of flips of the coin until the third head is obtained. Let Z be the number of additional flips of the coin after the third head occurs until the sixth head occurs; Z is a negative binomial random variable with parameters 3 and 1/2. By the independence of the trials, pXY (x5) = P (Z = x  5) = = x6 2 1 2
x5 x6 2 1 2 3 1 2 x8 , x = 8, 9, 10, . . . . 4. Note that
fXY x Therefore, P 3 x 2 + (9/16) 3 1 = = (48x 2 + 27). 4 (27/16) + 1 43
1/2 1/4 1 1 3 <X< Y = = 4 2 4 1 17 (48x 2 + 27) dx = . 43 86 5. In the discrete case, let p(x, y) be the joint probability mass function of X and Y , and let A
be the set of possible values of X. Then E(X  Y = y) =
xA x p(x, y) = pY (y) xA xpX (x)pY (y) = pY (y) xpX (x) = E(X).
xA In the continuous case, letting f (x, y) be the joint probability density function of X and Y , we get E(X  Y = y) = =   x f (x, y) dx = fY (y)  xfX (x)fY (y) dx fY (y) xfX (x) dx = E(X). 6. Since
fY (y) =  f (x, y) dx =
0 1 (x + y) dx = 1 + y, 2 176 Chapter 8 Bivariate Distributions the desired quantity is given by x+y fXY (xy) = (1/2) + y 0 0 x 1, 0 y 1 elsewhere. 7. Clearly,
fY (y) =
0 ex(y+1) dx = 1 , 0 y e  1. y+1 0 Therefore, E(X  Y = y) = =
0  xfXY (xy) dx = xf (x, y) dx fY (y) 1 xex(y+1) dx = . 1/(y + 1) y+1 Note that, the last integral, 0 x(y + 1)ex(y+1) dx is 1/(y + 1) because it is the expected value of an exponential random variable with parameter y + 1. 8. Let f (x, y) be the joint probability density function of X and Y . Clearly,
f (x, y) = fXY (xy)fY (y). Thus fX (x) = Now fY (y) = and fXY (xy) = Therefore, for 0 < x < 1, fX (x) =
0 x  fXY (xy)fY (y) dy. 1 0 0<y<1 elsewhere, 0 < y < 1, y < x < 1 elsewhere. 0 1 1y 1 dy =  ln(1  x), 1y and hence fX (x) =  ln(1  x) 0 < x < 1 0 elsewhere. Section 8.3 Conditional Distributions 177 9. f (x, y), the joint probability density function of X and Y is given by 1 f (x, y) = 0 Thus fY Now fXY x Therefore, P 0X x x if x 2 + y 2 1 otherwise. 4 = 5 1(16/25)  1(16/25) 6 1 dx = . 5 4 f x, 4 5 = 4 5 fY 5 5 = , 6 3 3  x . 5 5 4 4 y= = 11 5 4/11 0 10 5 dx = . 6 33 10. (a)
0 cex dy dx = 1 implies that c = 1/2. f (x, y) = fY (y) (1/2)ex (1/2)ex dy y (b) fXY (xy) = (1/2)ex (1/2)e
x = ex+y , dx x > y, fY X (yx) = x = 1 , x < y < x. 2x x (c) By part (b), given X = x, Y is a uniform random variable over (x, x). Therefore, E(Y X = x) = 0 and 2 x  (x) x2 Var(Y X = x) = = . 3 12 11. Let f (x, y) be the joint probability density function of X and Y . Since
fXY (xy) = 3 1 = 2y 20 + (2y)/3  20 20 < x < 20 + otherwise, 1/30 0 0 < y < 30 elsewhere, 2y 3 0 and fY (y) = 178 Chapter 8 Bivariate Distributions we have that 1 f (x, y) = fXY (xy)fY (y) = 20y 0 20 < x < 20 + elsewhere. 2y , 0 < y < 30 3 12. Let X be the first arrival time. Clearly,
P X x  N (t) = 1 = For 0 x < t, P X x  N (t) = 1 = P X x, N (t) = 1 P N (t) = 1 = P N (x) = 1, N (t  x) = 0 P N (t) = 1
0 0 1 if x < 0 if x t. ex (x)1 e(tx) (t  x) P N (x) = 1 P N (t  x) = 0 1! 0! = = t 1 P N (t) = 1 e (t) 1! = x , t where the third equality follows from the independence of the random variables N (x) and N (t  x) (recall that Poisson processes possess independent increments). We have shown that 0 if x < 0 P X x  N (t) = 1 = x/t if 0 x < t 1 if x t. This shows that the conditional distribution of X given N (t) = 1 is uniform on (0, 1). 13. For x y, the fact that the conditional distribution of X given Y = y is hypergeometric
follows from the following: P (X = x  Y = y) = P (X = x, Y = y) P (X = x)P (Y  X = y  x) = P (Y = y) P (Y = y) m x nm yx . n y = n  m yx m x p (1  p)(nm)(yx) p (1  p)mx yx x = n y ny p (1  p) y Section 8.3 Conditional Distributions 179 It must be clear that the conditional distribution of Y given that X = x is binomial with parameters n  m and p. That is, P (Y = y  X = x) = n  m yx p (1  p)nmy+x , y = x, x + 1, . . . , n  m + x. yx 14. Let f (x, y) be the joint probability density function of X and Y . By the solution to Exercise 25,
Section 8.1, f (x, y) = and fY (y) = 1  y, 1 y 1. Hence fXY (xy) = 1 1/2 = , 1 + y x 1  y, 1 y 1. 1  y 2 1  y 1/2 0 x + y 1 elsewhere, 15. Let be the parameter of N (t) : t 0 . The fact that for s < t, the conditional distribution of N (s) given N (t) = n is binomial with parameters n and p = s/t, follows from the following relations for i n. P N (s) = i  N (t) = n = P N (s) = i, N (t) = n P N (t) = n P N (s) = i P N (t)  N (s) = n  i P N (t) = n
ni = P N (s) = i, N (t)  N (s) = n  i P N (t) = n = = P N (s) = i P N (t  s) = n  i P N (t) = n = es (s)i e(ts) (t  s) i! (n  i)! et (t)n n! = n i s t i 1 s t ni , where the third equality follows since Poisson processes possess independent increments and the fourth equality follows since Poisson processes are stationary. 180 Chapter 8 Bivariate Distributions For i k, P N (t) = i  N (s) = k = P N (t)  N (s) = i  k  N (s) = k = P N (t)  N (s) = i  k = P N (t  s) = i  k e(ts) (t  s) = (i  k)!
ik shows that the conditional distribution of N (t) given N (s) = k is Poisson with parameter (t  s). 16. Let p(x, y) be the joint probability mass function of X and Y . Clearly,
pY (5) = and 11 13 p(x, 5) = 0 11 13 12 13 1 13
4 1 , 13
4x x1 12 13 12 13 1 13 x<5 x=5 4 1 13 x6 1 13 x > 5. Using these, we have that E(X  Y = 5) =
x=1 4 xpXY (x5) =
x=1 x p(x, 5) pY (5) 11 12 4 =
x=1 11 1 x 11 12 x +
x=6 4 x 1 13 1 13
4 1 13 12 13 12 13
y y x6 11 = 0.72932 + 12 = 0.72932 + 11 12 (y + 6)
y=0 4 y
y=0 12 13 +6
y=0 12 13 y 11 = 0.702932 + 12 1 13 12/13 1 = 13.412. +6 2 (1/13) 1  (12/13) Remark: In successive draws of cards from an ordinary deck of 52 cards, one at a time, randomly, and with replacement, the expected value of the number of draws until the first ace is 1/(1/13) = 13. This exercise shows that knowing the first king occurred on the fifth trial will increase, on the average, the number of trials until the first ace 0.412 draws. Section 8.3 Conditional Distributions 181 17. Let X be the number of blue chips in the first 9 draws and Y be the number of blue chips drawn
altogether. We have that
9 E(X  Y = 10) =
x=0 9 x p(x, 10) pY (10) 9 x 12 22
x =
x=1 x 10 22 9x 18 10 9 x 9 10  x 18 10 = 9 12 10  x 22 12 10 10 8 22 22 10x 10 22 x1 9 =
x=1 x 9 10 = 5, 18 where the last sum is (9 10)/18 because it is the expected value of a hypergeometric random variable with N = 18, D = 9, and n = 10. 18. Clearly,
fX (x) = Thus fY X (yx) = Therefore, E(Y  X = x) = But
1 x 1 x 1 x n(n  1)(y  x)n2 dy = n(1  x)n1 . f (x, y) n(n  1)(y  x)n2 (n  1)(y  x)n2 = = . fX (x) n(1  x)n1 (1  x)n1 (n  1)(y  x)n2 n1 dy = n1 (1  x) (1  x)n1
1 x y y(y  x)n2 dy. y(y  x)n2 dy = 1 x 1 x (y  x + x)(y  x)n2 dy = (y  x)n1 dy + 1 x x(y  x)n2 dy = Thus E(Y  X = x) = (1  x)n x(1  x)n1 + . n n1 n1 n1 1 (1  x) + x = + x. n n n 182 Chapter 8 Bivariate Distributions 19. (a) The area of the triangle is 1/2. So
f (x, y) =
1y 0 2 0 if x 0, y 0, x + y 1 elsewhere. (b) fY (y) = 2 dx = 2(1  y), 0 < y < 1. Therefore, 1 2 = , 0 x 1  y, 0 y < 1. 2(1  y) 1y fXY (xy) = (c) By part (b), given that Y = y, X is a uniform random variable over (0, 1  y). Thus E(X  Y = y) = (1  y)/2, 0 < y < 1. 20. Clearly,
x pX (x) =
y=0 1 1 = 2 2 y! (x  y)! e e x! x y=0 e2 x! = y! (x  y)! x! x y=0 x y = e2 2x , x! where the last equality follows since x is the number of subsets of a set with x y elements and hence is equal to 2x . Therefore, pX (x) is Poisson with parameter 2 and so
x y=0 pY X (yx) = This yields
x p(x, y) x x = 2 . pX (x) y E(Y  X = x) =
y=0 y x x 2 = y x y
y=0 x y 1 2 y 1 2 xy = x , 2 where the last equality follows because the last sum is the expected value of a binomial random variable with parameters x and 1/2. 21. Let X be the lifetime of the dead battery. We want to calculate E(X  X < s). Since X is a
continuous random variable, this is the same as E(X  X s). To find this quantity, let FXXs (t) = P (X t  X s), and fXXs (t) = FXXs (t). Then E(X  X s) =
0 tfXXs (t) dt. Section 8.4 Transformations of Two Random Variables 183 Now P (X t, X s) FXXs (t) = P (X t  X s) = P (X s) P (X t) if t < s = P (X s) 1 if t s. Differentiating FXXs (t) with respect to t, we obtain f (t) if t < s fXXs (t) = F (s) 0 otherwise. This yields E(X  X s) = 1 F (s)
0 s tf (t) dt. 8.4 TRANSFORMATIONS OF TWO RANDOM VARIABLES 1. Let f be the joint probability density function of X and Y . Clearly,
f (x, y) = 1 0 0 < x < 1, 0 < y < 1 elswhere. 2 ln x = u 2 ln y = v defines a onetoone transformation of R = (x, y) : 0 < x < 1, 0 < y < 1 onto the region Q = (u, v) : u > 0, v > 0 . It has the unique solution x = eu/2 , y = ev/2 . Hence 1  eu/2 2 0 0 1 = e(u+v)/2 = 0. 4 The system of two equations in two unknowns 1  ev/2 2 By Theorem 8.8, g(u, v), the joint probability density function of U and V is g(u, v) = f eu/2 , ev/2 1 1 (u+v)/2 e = e(u+v)/2 , u > 0, v > 0. 4 4 J= 184 Chapter 8 Bivariate Distributions 2. Let f (x, y) be the joint probability density function of X and Y . Clearly,
f (x, y) = f1 (x)f2 (y), x > 0, y > 0. Let V = X and g(u, v) be the joint probability density functions of U and V . The probability density function of U is gU (u), its marginal density function. The system of two equations in two unknowns x/y = u x=v defines a onetoone transformation of R = (x, y) : x > 0, y > 0 onto the region Q = (u, v) : u > 0, v > 0 . It has the unique solution x = v, y = v/u. Hence 0 J=  By Theorem 8.8, g(u, v) = f v, Therefore, gU (u) =
0 1 1 u = v u2 v = 0. u2 v u v v v v v = 2 f v, = 2 f1 (v)f2 2 u u u u u u > 0, v > 0. v v f (v)f2 dv, 2 1 u u u > 0. 3. Let g(r, ) be the joint probability density function of R and . We will show that g(r, ) =
gR (r)g ( ). This proves the surprising result that R and are independent. Let f (x, y) be the joint probability density function of X and Y . Clearly, f (x, y) = 1 (x 2 +y 2 )/2 , e 2  < x < ,  < y < . Let R be the entire xyplane excluded the set of points on the xaxis with x 0. This causes no problems since P (Y = 0, X 0) = P (Y = 0)P (X 0) = 0. The system of two equations in two unknowns x2 + y2 = r arctan y = x Section 8.4 Transformations of Two Random Variables 185 defines a onetoone transformation of R onto the region Q = (r, ) : r > 0, 0 < < 2 . It has the unique solution x = r cos y = r sin . Hence J= cos sin r sin r cos = r = 0. By Therorem 8.8, g(r, ) is given by g(r, ) = f (r cos , r sin )r = Now gR (r) =
0 2 1 r 2 /2 re 2 0 < < 2, r > 0. 1 r 2 /2 2 d = rer /2 , re 2 r > 0, 1 1 r 2 /2 dr = re , 0 < < 2. 2 2 0 Therefore, g(r, ) = gR (r)g ( ), showing that R and are independent random variables. The formula for g ( ) indicates that is a uniform random variable over the interval (0, 2). The probability density function obtained for R is called Rayleigh. g ( ) = and 4. Method 1: By the convolution theorem (Theorem 8.9), g, the probability density function of
the sum of X and Y , the two random points selected from (0, 1) is given by g(t) =  f1 (x)f2 (t  x) dx, where f1 and f2 are, respectively, the probability density functions of X and Y . Since f1 (x) = f2 (x) = 1 0 x (0, 1) elsewhere, the integrand, f1 (x)f2 (t  x) is nonzero if 0 < x < 1 and t  1 < x < t. This shows that for t < 0 and t 2, g(t) = 0. For 0 t < 1, t  1 < 0; thus g(t) =
0 t dx = t. For 1 t < 2, 0 < t  1 < 1; therefore, g(t) =
1 t1 dx = 1  (t  1) = 2  t. 186 Chapter 8 Bivariate Distributions So t g(t) = 2  t 0 if 0 t < 1 if 1 t < 2 otherwise. Method 2: Note that the sample space of the experiment of choosing two random numbers from (0, 1) is S = (x, y) R2 : 0 < x < 1, 0 < y < 1 . So, for 0 t < 1, P (X + Y t) is the area of the region (x, y) S : 0 < x t, 0 < y t, x + y t divided by the area of S: t 2 /2. For 1 t < 2, P (X + Y t) is the area of S  (x, y) S : t  1 x < 1, t  1 y < 1, x + y > t (2  t)2 . (Draw figures to verify these regions.) Let G be the divided by the area of S: 1  2 probability distribution function of X + Y . We have shown that 0 t <0 2 t 0t <1 2 G(t) = 2 1  (2  t) 1t <2 2 1 t 2. Therefore, t g(t) = G (t) = 2  t 0 0t <1 1t <2 otherwise. 5. (a) Clearly, pX (x) = 1/3 for x = 1, 0, 1 and pY (y) = 1/3 for y = 1, 0, 1. Since 1/9 z = 2, +2 P (X + Y = z) = 2/9 z = 1, +1 3/9 z = 0, P (X + Y = z) =
x the relation pX (x)pY (z  x) is easily seen to be true. Section 8.4 Transformations of Two Random Variables 187 (b) p(x, y) = pX (x)pY (y) for all possible values x and y of X and Y if and only if (1/9)+c = 1/9 and (1/9)  c = 1/9; that is, if and only if c = 0. 6. Let h(x, y) be the joint probability density function of X and Y . Then 1 2 2 x y 0 x 1, y 1 elsewhere. h(x, y) = Consider the system of two equations in two unknowns x/y = u xy = v. This system has the unique solution uv y = v/u. x= We have that x 1 y 1 uv 1 u 1 , v (30) (29) v/u 1 v u. 1 Clearly, x 1, y 1 imply that v = xy 1, so > 0. Therefore, the system of equations v (29) defines a onetoone transformation of R = (x, y) : x 1, y 1 onto the region Q = (u, v) : 0 < By (30), J= 1 2 v u 1 2 u v = 1 = 0. 2u 1 uv . v v  2u u 1 2 uv Hence, by Theorem 8.8, g(u, v), the joint probability density function of U and V is given by g(u, v) = h uv, v 1 , J = u 2uv 2 0< 1 u v. v 188 Chapter 8 Bivariate Distributions 7. Let h be the joint probability density function of X and Y . Clearly,
h(x, y) = e(x+y) 0 x > 0, y > 0 elsewhere. Consider the system of two equations in two unknowns x+y =u ex = v. This system has the unique solution x = ln v y = u  ln v. We have that x > 0 ln v > 0 v > 1, (32) (31) y > 0 u  ln v > 0 eu > v. Therefore, the system of equations (31) defines a onetoone transformation of R = (x, y) : x > 0, y > 0 onto the region By (32), 0 J= 1 v = 1 = 0. v Q = (u, v) : u > 0, 1 < v < eu . 1 v Hence, by Theorem 8.8, g(u, v), the joint probability density function of U and V is given by 1  g(u, v) = h(ln v, u  ln v)J = 1 u e , v u > 0, 1 < v < eu . 8. Let U = X + Y and V = X  Y . Let g(u, v) be the joint probability density function of U and V . We will show that g(u, v) = gU (u)gV (v). To do so, let f (x, y) be the joint probability density function of X and Y . Then f (x, y) = 1 (x 2 +y 2 )/2 e , 2  < x < ,  < y < . The system of two equations in two unkowns x+y =u xy =v Section 8.4 Transformations of Two Random Variables 189 defines a onetoone correspondence from the entire xyplane onto the entire uvplane. It has the unique solution x = u+v 2 uv y = . 2 Hence 1/2 1/2 1 J= =  = 0. 2 1/2 1/2 By Theorem 8.8, g(u, v) = f u+v uv , J 2 2 uv u+v 2 + 1 2 2 = exp  4 2 2 = 1 e(u2 +v2 )/4 , 4  < u, v < . This gives gU (u) = 1 1 u2 /4 v2 /4 2 2 e(u +v )/4 dv = e e dv 4  4  1 1 1 2 2 2 = eu /4 ev /4 dv = eu /4 ,  < u < , 2 2  2 1 2 ev /4 is the probability density function of 2 a normal random variable with mean 0 and variance 2. Thus its integral over the interval (, ) is 1. Similarly, where the last equality follows because 1 2 gV (v) = ev /2 , 2  < v < . Since g(u, v) = gU (u)gV (v), U and V are independent normal random variables each with mean 0 and variance 2. 9. Let f be the joint probability density function of X and Y . Clearly,
f (x, y) = r1 +r2 x r1 1 y r2 1 e(x+y) , (r1 ) (r2 ) x > 0, y > 0. Consider the system of two equations in two unknowns x +y = u x = v. x+y (33) 190 Chapter 8 Bivariate Distributions Clearly, (33) implies that u > 0 and v > 0. This system has the unique solution x = uv y = u  uv. We have that x > 0 uv > 0 u > 0 and v > 0, v < 1. (34) y > 0 u  uv > 0 R = (x, y) : x > 0, y > 0 onto the region Therefore, the system of equations (33) defines a onetoone transformation of Q = (u, v) : u > 0, 0 < v < 1 . By (34), J= v u 1  v u = u = 0. Hence by Thereom 8.8, the joint probability density function of U and V is given by g(u, v) = f (uv, u  uv)J = Note that g(u, v) = = This shows that g(u, v) = gU (u)gV (v). That is, U and V are independent. Furthermore, it shows that gU (u) is the probability density function of a gamma random variable with parameter r1 + r2 and ; gV (v) is the probability density function of a beta random variable with parameters r1 and r2 . eu (u)r1 +r2 1 (r1 + r2 ) (r1 + r2 ) r1 1 v (1  v)r2 1 (r1 ) (r2 ) u > 0, 0 < v < 1. r1 +r2 ur1 +r2 1 eu v r1 1 (1  v)r2 1 (r1 ) (r2 ) u > 0, 0 < v < 1. eu (u)r1 +r2 1 1 v r1 1 (1  v)r2 1 , (r1 + r2 ) B(r1 , r2 ) 10. Let f be the joint probability density function of X and Y . Clearly,
f (x, y) = 2 e(x+y) , The system of two equations in two unknowns x+y =u x/y = v x > 0, y > 0. Chapter 8 Review Problems 191 defines a onetoone transformation of R = (x, y) : x > 0, y > 0 onto the region Q = (u, v) : u > 0, v > 0 . It has he unique solution x = uv/(1 + v), y = u/(1 + v). Hence v u 1+v (1 + v)2 u J= = 0. = (1 + v)2 u 1  1+v (1 + v)2 By Theorem 8.8, g(u, v), the joint probability density function of U and V is g(u, v) = f uv u 2 u eu , , J = 1+v 1+v (1 + v)2 gU (u) = 2 ueu , and gV (v) = u > 0, u > 0, v > 0. This shows that g(u, v) = gU (u)gV (v), where 1 , v > 0. (1 + v)2 Therefore, U = X + Y and V = X/Y are independent random variables. REVIEW PROBLEMS FOR CHAPTER 8 1. (a) We have that
P (XY 6) = p(1, 2) + p(1, 4) + p(1, 6) + p(2, 2) + p(3, 2) = 0.05 + 0.14 + 0.10 + 0.25 + 0.15 = 0.69. (b) First we calculate pX (x) and pY (y), the marginal probability mass functions of X and Y . They are given by the following table. x y 2 4 6 pX (x) 1 0.05 0.14 0.10 0.29 2 0.25 0.10 0.02 0.37 3 0.15 0.17 0.02 0.34 pY (y) 0.45 0.41 0.14 192 Chapter 8 Bivariate Distributions Therefore, E(X) = 1(0.29) + 2(0.37) + 3(0.34) = 2.05; E(Y ) = 2(0.45) + 4(0.41) + 6(0.14) = 3.38. 2. (a) and (b) p(x, y), the joint probability mass function of X and Y , and pX (x) and pY (y), the
marginal probability mass functions of X and Y are given by the following table. x 2 3 4 5 6 7 8 9 10 11 12 pY (y) (c) E(X) =
15 x=2 y 1 1/36 0 0 0 0 0 0 0 0 0 0 1/36 2 0 2/36 1/36 0 0 0 0 0 0 0 0 3/36 3 0 0 2/36 2/36 1/36 0 0 0 0 0 0 5/36 4 0 0 0 2/36 2/36 2/36 1/36 0 0 0 0 7/36
6 y=1 5 0 0 0 0 2/36 2/36 2/36 2/36 1/36 0 0 9/36 6 0 0 0 0 0 2/36 2/36 2/36 2/36 2/36 1/36 11/36 pX (x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 xpX (x) = 7; E(Y ) = ypY (y) = 161/36 4.47. 3. Let X be the number of spades and Y be the number of hearts in the random bridge hand. The
desired probability mass function is 13 x pXY (x4) = p(x, 4) = pY (4) 13 26 4 9x 52 13 13 4 52 13 39 9 = 13 x 26 9x , 39 9 0 x 9. 4. The set of possible values of X and Y , both, is 0, 1, 2, 3 . Let p(x, y) be their joint probability
mass function; then p(x, y) = 13 x 13 y 26 3xy , 0 x, y, x + y 3. 52 3 Chapter 8 Review Problems 193 5. Reducing the sample space, the answer is 13 x 13 6x , 26 6 0 x 6. 2 x 0 6. (a)
0 c dy dx = 1 x
x 0 c = 1/2. 0<x<
2 y (b) fX (x) = fY (y) = 1 1 dy = , 2x 2 1 , 2 0 < y < 2. 2 y 1 1 dx = ln x 2x 2 = 1 2 ln , 2 y 7. Note that f (x, y) = y 1 3 2 1 1 1 3 x + , where y, 0 < y < 2 and x 2 + , 0 < x < 1 are 2 2 2 2 2 2 probability density functions. Therefore, fY (y) = 1 y, 0 < y < 2, 2 3 2 1 x + , 0 < x < 1. 2 2 fX (x) = We observe that f (x, y) = fX (x)fY (y). This shows that X and Y are independent random variables and hence E(XY ) = E(X)E(Y ). This relation can also be verified directly: E(XY ) =
0 1 0 2 3 3 2 1 2 5 x y + xy dy dx = , 4 4 6 5 1 3 3 x y + xy dy dx = , 4 4 8 3 2 2 1 2 4 x y + y dy dx = . 4 4 3 E(X) =
0 1 0 2 E(Y ) =
0 1 0 2 Hence E(XY ) = 5 5 4 = = E(X)E(Y ). 6 8 3 8. A distribution function is 0 at  and 1 at , so it cannot be constant everywhere. F (x, y)
is not a joint probability distribution function because assuming it is, we get that FX (x) is constant everywhere: FX (x) = F (x, ) = 1, x. 194 Chapter 8 Bivariate Distributions
2 2 r2  r3 r2  r2 = 2 2 3. 2 r1 r1 9. The answer is 10. Let Y be the total number of heads obtained. Let X be the total number of heads in the first
10 flips. For 2 x 10, 10 x 1 2
10 p(x, 12) = pXY (x  12) = pY (12) 10 12  x 20 1 20 12 2 1 2 10 = 10 x 10 12  x . 20 12 This is the probability mass function of a hypergeometric random variable with parameters 12 10 nD = = 6, as expected. N = 20, D = 10, and n = 12. Its expected value is N 20 11. f (x, y), the joint probability density function of X and Y is given by
f (x, y) = Therefore, by symmetry, P (X > 2Y ) + P (Y > 2X) = 2P (X > 2Y ) = 2
0 2y 2 2 2 F (x, y) = 4xyex ey , x > 0, y > 0. x y 4xyex ey dx dy =
2 2 2 . 5 12. We have that
fX (x) =
0 1x 3 3 3(x + y) dy =  x 2 + , 0 < x < 1, 2 2 By symmetry, 3 3 fY (y) =  y 2 + , 0 < y < 1. 2 2 Therefore, P (X + Y > 1/2) =
0 1/2 1x (1/2)x 1 1/2 0 1x 3(x + y) dy dx + 3(x + y) dy dx = 9 5 29 + = . 64 16 64 f (x, y) = fY (y) ey
1 y 0 e 1 0 13. Since
fXY (xy) = we have that dx = 1, 0 < x < 1, y > 0, 1 , n+1 E(Xn  Y = y) = x n 1 dx = n 1. Chapter 8 Review Problems 195 14. Let p(x, y) be the joint probability mass function of X and Y . We have that
p(x, y) = 10 x 10 x 1 4 15 y
x 3 4 1 4 10x 15 y 1 4 y 3 4 15y =
1 1 x x+y 3 4 25xy , 0 x 10, 0 y 15. 15.
0 cx(1  x) dy dx = 1 fX (x) =
1 x y 0 c = 12. Clearly, 12x(1  x) dy = 12x(1  x)2 , 0 < x < 1, fY (y) = 12x(1  x) dx = 6y 2  4y 3 , 0 < y < 1. Since f (x, y) = fX (x)fY (y), X and Y are not independent. 16. The area of the region bounded by y = x 2  1 and y = 1  x 2 is
1 1 1x 2 x 2 1 8 dy dx = . 3 Therefore f (x, y), the joint probability density function of X and Y is given by f (x, y) = Clearly, fX (x) =
1x 2 x 2 1 3/8 0 x 2  1 < y < 1  x 2 , 1 < x < 1 elsewhere. 3 3 dy = (1  x 2 ), 1 < x < 1. 8 4 To find fY (y), note that for 1 < y < 0, fY (y) = and, for 0 y < 1, fY (y) =
1+y  1+y 3 3 1+y dx = 8 4 1y  1y 3 3 dx = 1  y. 8 4 196 Chapter 8 Bivariate Distributions So 3 1 + y 4 fY (y) = 3 4 1  y 0 1 < y < 0 0y<1 otherwise. Since f (x, y) = fX (x)fY (y), X and Y are not independent. 17. Let f (x, y) be the joint probability density function of X and Y , G be the probability distribution function of X/Y , and g be the probability density function of X/Y . We have that f (x, y) = 1/2 0 0 < x < 1, 0 < y < 2 otherwise. Clearly, P (X/Y t) = 0 if t < 0. For 0 t < 1/2, P For t 1/2, P X t = Y
1 0 2 0 0 ty 1 dx dy = t. 2 X t = Y 2 x/t 1 1 dy dx = 1  . 2 4t (Draw appropriate figures to verify the limits of these integrals.) Therefore, 0 t <0 1 0t < G(t) = t 2 1 1 1  t . 4t 2 This gives 0 g(t) = G (t) = 1 1 4t 2 t <0 0t < t 1 . 2 1 2 18. No, because G(, ) = F () + F () = 2 = 1. 19. The problem is equivalent to the following: Two points X and Y are selected independently
and at random from the interval (0, ). What is the probability that the length of at least one Chapter 8 Review Problems 197 interval is less than /20? The solution to this problem is as follows: P min(X, Y  X,  Y) < 20 X < Y P (X < Y ) X > Y P (X > Y ) X < Y P (X < Y ) X<Y X<Y X<Y 1 2 + P min(Y, X  Y, = 2P min(X, Y  X, = 2P min(X, Y  X,  X) <  Y) <  Y) < 20 20 20 = 1  P min(X, Y  X, =1P X =1P X Now P X 20 , Y X 20  Y) , 20 20 20 , Y X , Y X 19 20 20 20 Y 19 20 20 ,Y X<Y . ,Y X < Y is the area of the region 20 , yx 20 , y 19 20 (x, y) R2 : 0 < x < , 0 < y < , x divided by the area of the triangle (x, y) R2 : 0 < x < , 0 < y < , y > x ; that is, 17 17 2 20 20 = 0.7225. 2 2 Therefore, the desired probability is 1  0.7225 = 0.2775. 20. Let p(x, y) be the joint probability mass function of X and Y .
p(x, y) = P (X = x, Y = y) = (0.90)x1 (0.10)(0.90)y1 (0.10) = (0.90)x+y2 (0.10)2 . 21. We have that
fX (x) = x x 1 y 1 y dy = 2x, 0 < x < 1, dx = 1 + y dx = 1  y 1 < y < 0 0 < y < 1, fY (y) = 198 Chapter 8 Bivariate Distributions fXY (xy) = 1 1 + y 1 1  y y < x < 1, 1 < y < 0 y < x < 1, 0 < y < 1, and fY X (yx) = Thus E(Y  X = x) = and E(X  Y = y) = 1 y 1 y 1 , x < y < x. 2x
x x y dy = 0 = 0 x + 0, 2x x 1y dx = , 1 < y < 0 1+y 2 1+y x dx = , 1y 2 0 < y < 1. 22. We present the solution given by Merryfield, Viet, and Watson, in the AugustSeptember 1997
issue of the American Mathematical Monthly. Let f be the joint probability density function of X and Y . E(WA ) = E(WB ) =
b b a b a a a b WA (x, y)f (x, y) dxdy, WB (x, y)f (x, y) dxdy. Let U = Y , V = X, h1 (x, y) = y and h2 (x, y) = x. Then the system of equations y=u x=v has the unique solution x = v, y = u, and J= 0 1 = 1 = 0. 1 0 Applying the change of variables formula for multiple intergrals, we obtain E(WA ) = =
b a a b WA (x, y)f (x, y) dxdy = WA (v, u)f (v, u) dudv. b a a b WA (v, u)f (v, u)J dudv b b a a Chapter 8 Review Problems 199 Since the distribution of the money in each player's wallet is the same, the joint distributions of (X, Y ) and (Y, X) have the same probability density function f satisfying f (x, y) = f (y, x). Observing that WA (Y, X) = WB (X, Y ), we have that WA (v, u) = WB (u, v). This and f (v, u) = f (u, v) imply that E(WA ) =
b a a b WB (u, v)f (u, v) dudv = E(WB ). On the other hand, WA (X, Y ) = WB (X, Y ) implies that E(WA ) = E(WB ). Thus E(WA ) = E(WA ), implying that E(WA ) = E(WB ) = 0. Chapter 9 Multivariate Distributions
9.1 JOINT DISTRIBUTIONS OF n > 2 RANDOM VARIABLES
clubs, and spades selected. We have 13 h 13 d 52 13 13 c 13 s 1. Let p(h, d, c, s) be the joint probability mass function of the number of hearts, diamonds, p(h, d, c, s) = , h + d + c + s = 13, 0 h, d, c, s 13. 2. Let p(a, h, n, w) be the joint probability mass function of A, H , N , and W . Clearly,
3 20 n w p(a, h, n, w) = , 38 12 a + h + n + w = 12, 0 a 8, 0 h 7, 0 n 3, 0 w 12. The marginal probability mass function of A is given by pA (a) = 8 a 30 12  a , 0 a 8. 38 12 8 a 7 h 3. (a) The desired joint marginal probability mass functions are given by
2 pX,Y (x, y) =
z=1 5 xy xyz = , x = 4, 5, y = 1, 2, 3. 162 54 yz xyz = , y = 1, 2, 3, z = 1, 2. 162 18 xyz xz = , x = 4, 5, z = 1, 2. 162 27 pY,Z (y, z) =
x=4 3 pX,Z (x, z) =
y=1 Section 9.1
3 2 3 Joint Distributions of n > 2 Random Variables
2 201 (b) E(Y Z) =
y=1 z=1 yzpY,Z (y, z) =
y=1 z=1 (yz)2 35 = . 18 9 4. (a) The desired marginal joint probability mass functions are given by
fX,Y (x, y) = fX,Z (x, z) = fY,Z (y, z) =
0 y z x y 6exyz dz = 6ex2y , 0 < x < y < . 6exyz dy = 6exz (ex  ez ), 0 < x < z < . 6exyz dx = 6eyz (1  ey ), 0 < y < z < . 0 x (b) E(X) = 1/3.
0 x xfX,Y (x, y) dy dx = 6xex2y dy dx =
0 3xe3x dx = 5. They are not independent because P (X1 = 1, X2 = 1, X3 = 0) = 1/4, whereas
P (X1 = 1)P (X2 = 1)P (X3 = 0) = 1/8. 6. Note that
fX (x) =
0 0 x 2 ex(1+y+z) dy dz exz
0 = x 2 ex
0 0 exy dy dz = ex , x > 0, fY (y) = and similarly, x 2 ex(1+y+z) dz dx = 0 1 , y > 0, (1 + y)2 fZ (z) = Also fX,Y (x, y) =
0 1 , z > 0. (1 + z)2 x 2 ex(1+y+z) dz = xex(1+y) , y > 0. Since f (x, y, z) = fX (x)fY (y)fZ (z), X, Y , and Z are not independent. Since fX,Y (x, y) = fX (x)fY (y), X, Y , and Z are not pairwise independent either. 202 Chapter 9 Multivariate Distributions 7. (a) The marginal probability distribution functions of X, Y , and Z are, respectively, given by
FX (x) = F (x, , ) = 1  e1 x , x > 0, FY (y) = F (, y, ) = 1  e2 y , y > 0, FZ (z) = F (, , z) = 1  e3 z , z > 0. Since F (x, y, z) = FX (x)FY (y)FZ (z), the random variables X, Y , and Z are independent. (b) From part (a) it is clear that X, Y , and Z are independent exponential random variables with parameters 1 , 2 , and 3 , respectively. Hence their joint probability density functions is given by f (x, y, z) = 1 2 3 e1 x2 y3 z . (c) The desired probability is calculated as follows: P (X < Y < Z) =
0 x y 0 f (x, y, z) dz dy dx x = 1 2 3 = e1 x e2 y y e3 z dz dy dx 1 2 . (2 + 3 )(1 + 2 + 3 ) 8. (a) Clearly f (x, y, z) 0 for the given domain. Since
1 0 0 x 0 y  ln x dz xy dy dx = 1, f is a joint probability density function. (b) fX,Y (x, y) =
0 y  fY (y) = 1 y ln x ln x dz =  , 0 y x 1. xy x y 1 ln x dz dx = (ln y)2 , 0 y 1.  xy 2 0 9. For 1 i n, let Xi be the distance of the ith point selected at random from the origin. For
r < R, the desired probability is P (X1 r, X2 r, . . . , Xn r) = P (X1 r)P (X2 r) P (Xn r) = For r R, the desired probability is 0. R2  r 2 R2
n = 1 r2 R2 n . 10. The sphere inscribed in the cube has radius a and is centered at the origin. Hence the desired
probability is (4/3) a 3 /(8a 3 ) = /6. Section 9.1 Joint Distributions of n > 2 Random Variables 203 11. Yes, it is because f 0 and 0 x1 x2 0 x1 0 xn1 x2 exn dxn dxn1 dx1 xn2 = exn1 dxn1 dx1 0 = = x1 ex2 dx2 dx1 = ex1 dx1 = 1. 12. Let f (x1 , x2 , x3 ) be the joint probability density function of X1 , X2 , and X3 , the lifetimes of
the original, the second, and the third transistors, respectively. We have that 1 1 1 1 (x1 +x2 +x3 )/5 e . f (x1 , x2 , x3 ) = ex1 /5 ex2 /5 ex3 /5 = 5 5 5 125 Now P (X1 + X2 + X3 < 15) =
0 15 0 15 0 0 15 0 15x1 15x1 0 15x1 x2 1 (x1 +x2 +x3 )/5 dx3 dx2 dx1 e 125 = = 1 1 (x1 +x2 )/5  e3 dx2 dx1 e 25 25 dx1 1 1 x1 /5 4 3  e + e3 x1 e 5 5 25 17 3 e = 0.5768. 2 =1 Therefore, the desired probability is P (X1 + X2 + X3 15) = 1  0.5768 = 0.4232. 13. Let F be the distribution function of X. We have that
F (t) = P (X t) = 1  P (X > t) = 1  P (X1 > t, X2 > t, . . . , Xn > t) = 1  P (X1 > t)P (X2 > t) P (Xn > t) = 1  e1 t e2 t en t = 1  e(1 +2 ++n )t , t > 0. Thus X is exponential with parameter 1 + 2 + + n . 14. Let Y be the number of functioning components of the system. The random variable Y is
binomial with parameters n and p. The reliability of this system is given by
n r = P (X = 1) = P (Y k) =
i=k n i p (1  p)ni . i 204 Chapter 9 Multivariate Distributions 15. Let Xi be the lifetime of the ith part. The time until the item fails is the random variable min(X1 , X2 , . . . , Xn ) which by the solution to Exercise 13 is exponentially distributed with parameter n. Thus the average life of the item is 1/(n). 16. Let X1 , X2 , . . . be the lifetimes of the transistors selected at random. Clearly,
N = min n : Xn > s . Note that P XN t  N = n = P Xn t  X1 s, X2 s, . . . , Xn1 s, Xn > s). This shows that for s t, P XN t  N = n = 0. For s < t, P XN t  N = n = P (s < Xn t, X1 s, X2 s, . . . , Xn1 s) P (X1 s, X2 s, . . . , Xn1 s, Xn > s) P (s < Xn t)P (X1 s)P (X2 s) P (Xn1 s) P (X1 s)P (X2 s) P (Xn1 s)P (Xn > s) P (s < Xn t) F (t)  F (s) = . P (Xn > s) 1  F (s) = = This relation shows that the probability distribution function of XN given N = n does not depend on n. Therefore, XN and N are independent. 17. Clearly,
X = X1 1  (1  X2 )(1  X3 ) 1  (1  X4 )(1  X5 X6 ) X7 = X1 X7 X2 X4 + X3 X4  X2 X3 X4 + X2 X5 X6 + X3 X5 X6  X2 X3 X5 X6  X2 X4 X5 X6  X3 X4 X5 X6 + X2 X3 X4 X5 X6 . The reliability of this system is r = p1 p7 p2 p4 + p3 p4  p2 p3 p4 + p2 p5 p6 + p3 p5 p6  p2 p3 p5 p6  p2 p4 p5 p6  p3 p4 p5 p6 + p2 p3 p4 p5 p6 . 18. Let G and F be the distribution functions of max1in Xi and min1in Xi , respectively. Let
g and f be their probability density functions, respectively. For 0 t < 1, G(t) = P (X1 t, X2 t, . . . , Xn t) = P (X1 t)P (X2 t) P (Xn t) = t n . Section 9.1 Joint Distributions of n > 2 Random Variables 205 So 0 G(t) = t n 1 t <0 0t <1 t 1. nt n1 0
1 Therefore, g(t) = G (t) = This gives E max Xi =
1in 0 0<t <1 elsewhere. n . n+1 nt n dt = Similarly, for 0 t < 1, F (t) = P
1in min Xi t = 1  P 1in min Xi > t = 1  P (X1 > t)P (X2 > t) P (Xn > t) = 1  (1  t)n , 0 t < 1. Hence 0 F (t) = 1  (1  t)n 1 f (t) = So E min Xi =
1in 0 t <0 0t <1 t 1, 0<t <1 otherwise. 1 . n+1 and n(1  t)n1 0
1 nt (1  t)n1 dt = 19. We have that
P max(X1 , X2 , . . . , Xn ) t = P (X1 t, X2 t, . . . , Xn t) = P (X1 t)P (X2 t) P (Xn t) = F (t) , and P min(X1 , X2 , . . . , Xn ) t = 1  P min(X1 , X2 , . . . , Xn ) > t = 1  P (X1 > t, X2 > t, . . . , Xn > t) = 1  P (X1 > t)P (X2 > t) P (Xn > t) = 1  1  F (t) .
n n 206 Chapter 9 Multivariate Distributions 20. We have that
P (Yn > x) = P min(X1 , X2 , . . . , Xn ) > = P X1 > x n x x x , X2 > , . . . , Xn > n n n x x x = P X1 > P X2 > P Xn > n n n x n = 1 . n Thus
n lim P (Yn > x) = lim 1 
n x n n = ex , x > 0. 21. We have that
P (X < Y < Z) = = = =    x y h(x)h(y)h(z) dz dy dx x h(x)h(y) 1  H (y) dy dx 1 1  H (y) 2
2 x h(x)  dx 1 2 h(x) 1  H (x) dx 2  1 1 1 3 =  1  H (x) = . 2 3 6  22. Noting that Xi2 = Xi , 1 i 5, we have
X = max{X2 X5 , X2 X3 X4 , X1 X4 , X1 X3 X5 } = 1  (1  X2 X5 )(1  X2 X3 X4 )(1  X1 X4 )(1  X1 X3 X5 ) = X2 X5 + X1 X4 + X1 X3 X5 + X2 X3 X4  X1 X2 X3 X4  X1 X2 X3 X5  X1 X2 X4 X5  X1 X3 X4 X5  X2 X3 X4 X5 + 2X1 X2 X3 X4 X5 . Therefore, whenever the system is turned on for water to flow from A to B, water reaches B with probability r given by, r = P (X = 1) = E(X) = p2 p5 + p1 p4 + p1 p3 p5 + p2 p3 p4  p1 p2 p3 p4  p1 p2 p3 p5  p1 p2 p4 p5  p1 p3 p4 p5  p2 p3 p4 p5 + 2p1 p2 p3 p4 p5 . 23. Clearly, B = (1 1)/2 and h = 1. So the volume of the pyramid is (1/3)Bh = 1/6.
Therefore, the joint probability density function of X, Y , and Z is f (x, y, z) = 6 0 (x, y, z) V otherwise. Section 9.1 Joint Distributions of n > 2 Random Variables 207 Thus fX (x) =
0 1x 0 1xy 6 dz dy = 3(1  x)2 , 0 < x < 1. Similarly, fY (y) = 3(1  y)2 , 0 < y < 1, and fZ (z) = 3(1  z)2 , 0 < z < 1. Since f (x, y, z) = fX (x)fY (y)fZ (z), X, Y , and Z are not independent. 24. The probability that Ax 2 +Bx+C = 0 has real roots is equal to the probability that B 2 4AC 0. To calculate this quantity, we will first evaluate the distribution functions of B 2 and 4AC and then use the convolution theorem to find the distribution function of B 2  4AC. 0 if t < 0 2 FB 2 (t) = P (B t) = t if 0 t < 1 1 if t 1, 1 if 0 < t < 1 fB 2 (t) = F 2 (t) = 2 t B 0 otherwise, and F4AC (t) = P (4AC t) = 0 if t < 4 if 4 t < 0 if t 0. t P AC  4 1 Now A and C are random numbers from (0, 1); hence (A, C) is a random point from the square (0, 1) (0, 1) in the acplane. Therefore, P (AC t/4) = P C t/(4A) is the t area of the shaded region bounded by a = 1, c = 1, c =  of Figure 1. 4a c
1 t/4 0 t/4 1 a Figure 1 The shaded region of Exercise 24. 208 Chapter 9 Multivariate Distributions Thus, for 4 t < 0, F4AC (t) = Therefore, 0 t t t F4AC (t) = P (4AC t) = 1 +  ln  4 4 4 1 Applying convolution theorem, we obtain P B 2  4AC 0 = 1  P B 2  4AC < 0 =1 =1 Letting y =  1 0 1 t/4 1 t/(4a) dc da = 1 + t t t  ln  . 4 4 4 if t < 4 if 4 t < 0 if t > 0. F4AC (0  x)fB 2 (x)dx 1 x x 1 x + ln dx. 4 4 4 2 x 1 x/2, we get dy = dx. So 4 x P B 2  4AC 0 = 1 
0 1/2 (1  y 2 + y 2 ln y 2 )2dy 2dy + 2
0 1/2 =1
0 1/2 (y 2  y 2 ln y 2 )dy =2
0 1/2 (y 2  y 2 ln y 2 )dy. Now by integration by parts (u = ln y 2 , dv = y 2 dy), 1 2 y 2 ln y 2 dy = y 3 ln y 2  y 3 . 3 9 Thus P B 2  4AC 0 = 10 3 2 3 y  y ln y 2 9 3
1/2 0 = 1 5 + ln 2 0.25. 36 6 25. The following solution by Scott Harrington, Duke University, Durham, NC, was given in The
College Mathematics Journal, September 1993.
Let V be the set of points (A, B, C) [0, 1]3 such that f (x) = x 3 +Ax 2 +Bx +C = 0 has all real roots. The probability that all of the roots are real is the volume of V . Section 9.1 Joint Distributions of n > 2 Random Variables 209 The function is cubic, so it either has one real root and two complex roots or three real roots. Since the coefficient of x 3 is positive, limx f (x) =  and limx+ f (x) = +. The number of real roots of the graph of f (x) depends on the nature of the critical points of the function f . f (x) = 3x 2 + 2Ax + B = 0, with roots 1 1 x = A A2  3B. 3 3 1 1 Let D = A2  3B, x1 =  (A + D), and x2 =  (A  D). If A2 < 3B then the 3 3 critical points are imaginary, so the graph of f (x) is strictly increasing and there must be exactly one real root. Thus we may assume A2 3B. In order for there to be three real roots, counting multiplicities, the local maximum x1 , f (x1 ) and local minimum x2 , f (x2 ) must satisfy f (x1 ) 0 and f (x2 ) 0; that is, f (x1 ) =  1 3 (A + 3A2 D + 3AD 2 + D 3 ) 27 1 1 + A(A2 + 2AD + D 2 )  B(A + D) + C 0, 9 3 f (x2 ) =  1 3 (A  3A2 D + 3AD 2  D 3 ) 27 1 1 + A(A2  2AD + D 2 )  B(A  D) + C 0. 9 3 Simplifying produces two halfspaces: C C 1  2A3 + 9AB  2(A2  3B)3/2 , 27 1  2A3 + 9AB + 2(A2  3B)3/2 , 27 (constraint surface 1); (constraint surface 2). 1 These two surfaces intersect at the curve given parametrically by A = t, B = t 2 3 1 3 and C = t . Note that all points in the intersection of these two halfspaces 27 1 satisfy B A2 . Surface 2 intersects the plane C = 0 at the Aaxis, but surface 1 3 1 intersects the plane C = 0 at the curve B = A2 , which is a quadratic curve in the 4 plane C = 0 located between the Aaxis and the upper limit B = 1 A2 . Therefore, V 3 is the region above the plane C = 0 and constraint surface 1, and below constraint surface 2. The volume of V is the volume V2 under surface 2 minus the volume V1 under surface 1. Now V1 =
1 (1/3)a 2 a=0 b=(1/4)a 2 1  2a 3 + 9ab  2(a 2  3b)3/2 db da 27 210 Chapter 9 Multivariate Distributions
1 0 (1/3)a 2 = =
0 1 9 4  2a 3 b + ab2 + (a 2  3b)5/2 27 2 15 1 7 7 5 a da = , 27 160 25, 920
(1/3)a 2 da
b=(1/4)a 2 1 and V2 = = 1 a=0 b=0 1 0 1  2a 3 + 9ab + 2(a 2  3b)3/2 db da 27
(1/3)a 2 1 9 4  2a 3 b + ab2  (a 2  3b)5/2 27 2 15 V = V2  V1 = da =
b=0 0 1 1 5 1 a da = . 270 1620 Thus 1 7 1  = . 1, 620 25, 920 2, 880 9.2 ORDER STATISTICS 1. By Theorem 9.5, we have that
f3 (x) = where f (x) = and 4! f (x) F (x) 2! 1! 1 0
2 1  F (x) , 0<x<1 otherwise, x<0 0x<1 x 1. 0 < x < 1. 0 F (x) = x 1 Therefore, f3 (x) = 12x 2 (1  x), Hence the desired probability is
1/2 1/4 12x 2 (1  x) dx = 67 = 0.26172. 256 2. Let X1 and X2 be the points selected at random. By Theorem 9.6, the joint probability density
function of X(1) and X(2) is given by f12 (x, y) = 2! x 11 (y  x)211 , (1  1)! (2  1  1)! (2  2)! 0 < x < y < 1. Section 9.2 Order Statistics 211 So f12 (x, y) = 2, 0 < x < y < 1. We have that, the desired probability is given by P X(2) 3X(1) =
0 1 0 y/3 1 2 dx dy = . 3 3. By Theorem 9.5, f4 (x), the probability density function of X(4) is given by
f4 (x) = 4! ex 1  ex 3! 0! 3 3 ex 44 = 4ex 1  ex .
3 The desired probability is 4ex 1  ex
3 dx = 1  1  e3 2 4 . 4. By Remark 6.4,
E X(n) =
0 P X(n) > x) dx. Now P X(n) > x = 1  P X(n) x = 1  P (X1 x, X2 x, . . . , Xn x) = 1  F (x) . So E X(n) =
0 n 1  F (x) n dx. 5. To find P X(i) = k , 0 k n, note that
P X(i) = k = 1  P X(i) < k  P X(i) > k . Let N be the number of Xj 's that are less than k. Then N is a binomial random variable with parameters m and
k1 p1 =
l=0 n l p (1  p)nl . l (35) Let L be the number of Xj 's that are greater than k. Then L is a binomial random variable with parameters m and
n p2 =
l=k+1 n l p (1  p)nl . l (36) 212 Chapter 9 Multivariate Distributions Clearly, P X(i) < k = P (N i) = m j =i m j p1 (1  p1 )mj , j
m and P X(i) > k = P (L m  i + 1) = j =mi+1 m j p2 (1  p2 )mj . j
m Thus, for 0 k n,
m P X(i) = k = 1 
j =i m j m j p1 (1  p1 )mj  p2 (1  p2 )mj , j j j =mi+1 where p1 and p2 are given by (35) and (36). 6. By Theorem 9.6, the joint probability density function of X(1) and X(n) is given by
f1n (x, y) = n(n  1)f (x)f (y) F (y)  F (x) Therefore, G(t) = P =
t  t  x n2 , x < y. X(1) + X(n) t 2
2tx = P X(1) + X(n) 2t
n2 n(n  1)f (x)f (y) F (y)  F (x)
n1 dy dx =n F (2t  x)  F (x) f (x) dx. 7. By Theorem 9.5, f1 (x), the probability density function of X(1) is given by
f1 (x) = 2! ex 1  ex (1  1)! (2  1)!
11 ex 21 = 2e2x , x 0. By Theorem 9.6, f12 (x, y), the joint probability density function of X(1) and X(2) is given by f12 (x, y) = 2! ex ey (1  1)! (2  1  1)! (2  2)!
11 = 1  ex ex  ey 211 = 22 e(x+y) , 0 x < y < . Let U = X(1) and V = X(2)  X(1) . We will show that g(u, v), the joint probability density function of U and V satisfy g(u, v) = gU (u)gV (v). This proves that U and V are independent. To find g(u, v), note that the system of two equations in two unknowns x=u yx =v Section 9.2 Order Statistics 213 defines a onetoone transformation of R = (x, y) : 0 x < y < onto the region Q = (u, v) : u 0, v > 0 . It has the unique solution x = u, y = u + v. Hence 1 0 J= 1 1 By Thereom 8.8, g(u, v) = f12 (u, u + v)J = 22 e(u+2v) , Since g(u, v) = gU (u)gV (v), where and gU (u) = 2e2u , gV (v) = ev , u 0, v > 0, u 0, v > 0. = 1 = 0. we have that U and V are independent. Furthermore, U is exponential with parameter 2 and V is exponential with parameter . 8. Let f12 (x, y) be the joint probability density function of X(1) and X(2). By Theorem 9.6,
1 1 2 2 2 2 f12 (x, y) = 2! f (x)f (y) = 2 ex /2 ey /2 2 2 1 x 2 /2 2 y 2 /2 2 = 2 e e ,  < x < y < . Therefore, E X(1) = = =  y     x 1 x 2 /2 2 y 2 /2 2 e dx dy e 2
2 /2 2 1 2 1 2 1 ey ey y  xex 2 /2 2 dx dy dy 2 /2 2 ( 2 )ey 2 /2 2 = ey 2 / 2 dy 214 Chapter 9 Multivariate Distributions = 1 1 2 2 1 =  1 =  .  y2 2 e 2(/ 2) dy  9. (a) By Theorem 9.6, the joint probability density function of X(1) and X(n) is given by
n(n  1)f (x)f (y) F (y)  F (x) f1n (x, y) = 0
n2 x<y elsewhere. We will use this to find g(r, v), the joint probability density function of R = X(n)  X(1) and V = X(n) . The probability density function of the sample range, R, is then the marginal probability density function of R. That is, gR (r) =  g(r, v)dv. To find g(r, v), we will use Theorem 8.8. The system of two equations in two unknowns yx =r y=v defines a onetoone transformation of (x, y) :  < x < y < onto the region (r, v) :  < v < , r > 0 . It has the unique solution x = v  r, y = v. Hence J= By Theorem 8.8, g(u, v) is given by g(r, v) = f1n (v  r, v)J = n(n  1)f (v  r)f (v) F (v)  F (v  r) This implies gR (r) =  n2 1 1 0 1 = 1 = 0. ,  < v < , r > 0. n(n  1)f (v  r)f (v) F (v)  F (v  r) n2 dv, r > 0. (37) Section 9.3 Multinomial Distributions 215 (b) The probability density function of n random numbers from (0, 1) is obtained by letting 1 0<v<1 f (v) = 0 otherwise, and F (v)  F (v  r) = v  (v  r) = r in (37). Note that the integrand of the integral in (37) is nonzero if 0 < v < 1 and if 0 < v  r < 1; that is, if 0 < r < v < 1. Therefore, gR (r) =
1 r n(n  1)r n2 dv = n(n  1)r n2 (1  r), 0 < r < 1. 10. Let f and F be the probability density and distribution functions of Xi , 1 i n, respectively.
We have that f (x) = and 1/ 0 0<x< elsewhere x<0 0x< x . n(n  1)r n2 (  r) 0 < r < . n 0 F (x) = x/ 1 r Let g(r) be the probability density function of R = X(n)  X(1) . By part (a) of Exercise 9, g(r) = n(n  1) v vr  0 n2 dv = (Note that 0 < v < and 0 < v  r < imply that r < v < .) Therefore, E(R) = r n(n  1)r n2 n1 (  r) dr = . n n+1 9.3 MULTINOMIAL DISTRIBUTIONS 1. The desired probability is
8! 3! 2! 3! 150 800
3 400 800 2 250 800 3 = 0.028. 2. We have that
P (B = i, R = j, G = 20  i  j ) 20! (0.2)i (0.3)j (0.5)20i , = i! j ! (20  i  j )! 0 i, j 20, i + j 20. 216 Chapter 9 Multivariate Distributions 3. Let U , D, and S be the number of days among the next six days that the stock market moves
up, moves down, and remains the same, respectively. The desired probability is P (U = 0, D = 0, S = 6) + P (U = 1, D = 1, S = 4) + P (U = 2, D = 2, S = 2) + P (U = 3, D = 3, S = 0) = 6! 0! 0! 6! + 6! 2! 2! 2! 1 4
0 5 12
2 0 1 3
2 6 + 1 3
2 6! 1! 1! 4! + 6! 3! 3! 0! 1 4 1 5 12
3 1 1 3
3 4 1 4 5 12 1 4 5 12 1 3 0 = 0.171. 4. Let A, B, C, D, and F be the number of students who get A, B, C, D, and F, respectively. The
desired probability is given by P (A = 2, B = 5, C = 5, D = 2, F = 1) + P (A = 3, B = 5, C = 5, D = 2, F = 0) = 15! (0.16)2 (0.34)5 (0.34)5 (0.14)2 (0.02)1 2! 5! 5! 2! 1! 15! (0.16)3 (0.34)5 (0.34)5 (0.14)2 (0.02)0 + 3! 5! 5! 2! 0! = 0.0172. 5. Let L, M, and S be the number of large, medium, and small watermelons among the five
watermelons Joanna buys, respectively. (a) We have that P (L 2) = 1  P (L = 0)  P (L = 1) 5 5 (0.50)1 (0.50)4 = 0.8125. =1 (0.50)0 (0.50)5  1 0 5! (0.5)2 (0.3)2 (0.2)1 = 0.135. (b) P (L = 2, M = 2, S = 1) = 2! 2! 1! (c) Using parts (a) and (b) and P (L = 3, M = 2, S = 0) = we have that P (M = 2  L 2) = = = P (M = 2, L 2) P (L 2) P (L = 2, M = 2, S = 1) + P (L = 3, M = 2, S = 0) P (L 2) 0.135 + 0.1125 = 0.3046. 0.8125 5! (0.5)3 (0.3)2 (0.2)0 = 0.1125, 3! 2! 0! Section 9.3 Multinomial Distributions 217 6. Let X be the number of faculty members who are below 40 and Y be the number of those who
are above 50 in the committee. The desired probability mass function is 10! (0.5)x (0.3)2 (0.2)8x x! 2! (8  x)! 10 (0.3)2 (0.7)8 2
x 8x pXY (x2) = = 8 x 5 7 2 7 , 0 x 8. 7. The probability is 1/4 that the blood type of a child of this man and woman is AB. The
probability is 1/4 that it is A, and the probability is 1/2 that it is B. The desired probability is equal to 6! 1 3 1 2 1 1 15 = 0.117. = 3! 2! 1! 2 4 4 128 8. The probability of two AA's, two Aa's, and two aa's is
g(p) = 6! (p2 )2 2p(1  p) 2! 2! 2!
2 (1  p)2 2 = 360p6 (1  p)6 . To find the maximum of this function, set g (p) = 0 to obtain p = 1/2. 9. Let N (t) be the number of customers who arrive at the store by time t. We are given that N(t) : t 0 is a Poisson process with = 3. Let X, Y , and Z be the number of customers who use charge cards, write personal checks, and pay cash in five operating minutes, respectively. Then P (X = 5, Y = 2, Z = 3) =
n=10 P X = 5, Y = 2, Z = 3  N (5) = n P N (5) = n n! e15 15n (0.40)5 (0.10)2 (0.20)3 (0.30)n10 5! 2! 3! (n  10)! n! n=10 (0.40)5 (0.10)2 (0.20)3 e15 1510 3! 5! 2! = = (0.30)n10 (15)n10 (n  10)! n=10 = (0.00010035) (4.5)n10 = (0.00010035)e4.5 = 0.009033. (n  10)! n=10 218 Chapter 9 Multivariate Distributions REVIEW PROBLEMS FOR CHAPTER 9 1. Let p(b, r, g) be the joint probability mass function of B, R, and G. Then
20 b 30 r 100 20 50 g p(b, r, g) = , b + r + g = 20, 0 b, r, g 20. 2. Let F be the distribution function of X. Let X1 , X2 , . . . , Xn be the outcomes of the first,
second, . . . , and the nth rolls, respectively. Then X = min(X1 , X2 , . . . , Xn ). Therefore, F (t) = P (X t) = 1  P (X > t) = 1  P (X1 > t, X2 > t, . . . , Xn > t) t <1 0 1  5 n 1 t < 2 6 1  4 n 2 t < 3 6 n 3 n = 1  P (X1 > t) = 1  6 3t <4 n 1  2 4t <5 6 1  1 n 5 t < 6 6 1 t 6. The probability mass function of X is p(x) = P (X = x) = 7x 6
n  6x 6 n , x = 1, 2, 3, 4, 5, 6. 3. Let D1 , D2 , . . . , Dn be the distances of the points selected from the origin. Let D =
min(D1 , D2 , . . . , Dn ). The desired probability is P (D r) = P (D1 r, D2 r, . . . , Dn r) = P (D1 r) = 1
1 0 0 1 0 1 n = 1  P (D1 < r) n (4/3) r 3 8a 3 n = 1 6 r a 3 n . 4. (a) c (x + y + 2z) dz dy dx = 1 c = 1/2. Chapter 9 Review Problems 219 (b) We have that 1 1 1 P X< ,Y < ,Z< 1 1 1 3 2 4 Y < ,Z< = P X< 1 1 3 2 4 P Y < ,Z< 2 4
1/3 1/2 0 1 0 0 1/2 0 0 1/4 1/4 = 0 1 (x + y + 2z) dz dy dx 2 1 (x + y + 2z) dz dy dx 2 = 1/36 2 = . 1/8 9 5. The joint probability mass function of the number of times each face appears is multinomial.
18! Hence the desired probability is (3!)6 1 6
18 = 0.00135. 6. Using the multinomial distribution, the answer is
7! (0.4)3 (0.35)2 (0.25)2 = 0.1029. 3! 2! 2! 7. For 1 i n, let Xi be the lifetime of the ith component. Then (t) be the survival function of min(X1 , X2 , . . . , Xn ) is the lifetime of the system. Let F the system. By the independence of the lifetimes of the components, for all t > 0, F (t) = P min(X1 , X2 , . . . , Xn ) > t = P (X1 > t, X2 > t, . . . , Xn > t) = P (X1 > t)P (X2 > t) P (Xn > t) = F1 (t)F2 (t) Fn (t). 8. For 1 i n, let Xi be the lifetime of the ith component. Then max(X1 , X2 , . . . , Xn ) is the lifetime of the system. Let F (t) be the survival function of the system. By the independence of the lifetimes of the components, for all t > 0, F (t) = P max(X1 , X2 , . . . , Xn ) > t = 1  P max(X1 , X2 , . . . , Xn ) t = 1  P (X1 t, X2 t, . . . , Xn t) = 1  P (X1 t)P (X2 t) P (Xn t) = 1  F1 (t)F2 (t) Fn (t). 9. The problem is equivalent to the following: Two points X and Y are selected independently and at random from the interval (0, ). What is the probability that the length of at least one 220 Chapter 9 Multivariate Distributions interval is less than /20? The solution to this problem is as follows: P min(X, Y  X,  Y) < 20 X < Y P (X < Y ) X > Y P (X > Y ) X < Y P (X < Y ) X<Y X<Y X<Y 1 2 + P min(Y, X  Y, = 2P min(X, Y  X, = 2P min(X, Y  X,  X) <  Y) <  Y) < 20 20 20 = 1  P min(X, Y  X, =1P X =1P X Now P X , Y X  Y) , 20 20 20 , Y X , Y X 19 20 20 20 Y 19 20 20 ,Y X<Y . 20 20 ,Y X < Y is the area of the region , yx , y 19 20 (x, y) R2 : 0 < x < , 0 < y < , x divided by the area of the triangle 20 20 (x, y) R2 : 0 < x < , 0 < y < , y > x ; that is, 17 17 2 20 20 = 0.7225. 2 2 Therefore, the desired probability is 1  0.7225 = 0.2775. 10. Let f13 (x, y) be the joint probability density function of X(1) and X(3) . By Theorem 9.6,
f13 (x, y) = 6(y  x), 0 < x < y < 1. X(1) + X(3) and V = X(1) . Using Theorem 8.8, we will find g(u, v), the joint Let U = 2 probability density function of U and V . The probability density function of the midrange of these three random variables is gU (u). The system of two equations in two unknowns x+y =u 2 x = v Chapter 9 Review Problems 221 defines a onetoone transformation of R = (x, y) : 0 < x < y < 1 onto the region Q = (u, v) : 0 < v < u < that has the unique solution x=v y = 2u  v. Hence 0 J= therefore, g(u, v) = f13 (v, 2u  v)J = 24(u  v), To find gU (u), draw the region Q to see that u 0 24(u  v) dv gU (u) = u 24(u  v) dv
2u1 v+1 <1 2 1 = 2 = 0; 2 1 0<v<u< v+1 < 1. 2 0 < u < 1/2 1/2 u < 1. Therefore, gU (u) = 12u2 12(u  1)2 0 < u < 1/2 1/2 u < 1. The expected value of U is given by E(U ) =
0 1/2 12u3 du + 1 1/2 12u(u  1)2 du = 3 5 1 + = . 16 16 2 Chapter 10 More E xpectations and Variances
10.1 EXPECTED VALUES OF SUMS OF RANDOM VARIABLES 1. Since
E(X) =
0 1 x(1  x) dx +
1 2 x(x  1) dx = 2 , 3 3 , 2 and E(X2 ) =
0 1 x 2 (1  x) dx +
1 2 x 2 (x  1) dx = we have that E(X2 + X) = 13 3 2 + = . 2 3 6 2. By Example 10.7, the answer is 5 = 12.5. 2/5
2 3. We have that E(X2 ) = Var(X) + E(X) = 1. Similarly, E(Y 2 ) = E(Z 2 ) = 1. Thus
E X2 (Y + 5Z)2 = E(X2 )E (Y + 5Z)2 = E(Y 2 + 25Z 2 + 10Y Z) = E(Y 2 ) + 25E(Z 2 ) + 10E(Y )E(Z) = 26. 4. Since f (x, y) = ex 2e2y , X and Y are independent exponential random variables with
parameters 1 and 2, respectively. Thus E(X) = 1, E(Y ) = 1/2, E(X 2 ) = Var(X) + E(X) and E(Y 2 ) = Var(Y ) + E(Y ) Therefore, E(X2 + Y 2 ) = 2 + 5 1 = . 2 2
2 2 = 1 + 1 = 2, 1 1 1 + = . 4 4 2 = Section 10.1 Expected Values of Sums of Random Variables 223 5. let X1 , X2 , X3 , X4 , and X5 be geometric random variables with parameters 1, 4/5, 3/5, 2/5,
and 1/5, respectively. The desired quantity is E(X1 + X2 + X3 + X4 + X5 ) = E(X1 ) + E(X2 ) + E(X3 ) + E(X4 ) + E(X5 ) 5 5 5 = 1 + + + + 5 = 11.42. 4 3 2 6. Clearly,
E(Xi ) = 1 Thus E(X1 + X2 + + Xn ) = n 1 1 1 +0 1 = . n n n 1 = 1 is the desired quantity. n 7. Let X1 , X2 , X3 , and X4 be the cost of a band to play music, the amount the caterer will charge,
the rent of a hall to give the party, and other expenses, respectively. Let N be the number of people who participate. We have that E(X1 ) = 1550, E(X2 ) = 1900, E(X3 ) = 1000, E(X4 ) = 550, and
200 E(N ) =
i=151 i 1 1 = 50 50 200 150 i
i=1 i=1 i = 1 200 201 150 151  = 175.50. 50 2 2 To have no loss on average, let x be the amount (in dollars) that the society should charge each participant. We must have E(X1 + X2 + X3 + X4 ) E(xN ) = xE(N ). This gives x 1550 + 1900 + 1000 + 550 E(X1 ) + E(X2 ) + E(X3 ) + E(X4 ) = = 28.49. 175.50 175.50 So to have no loss on the average, the society should charge each participant $28.49. 8. (a) E( 007) = E(007 007) = 1, 000.
(b) E( 156156) = E( 156) + E(156 156156) = E(156 156) + E(156156 156156) = 1, 000 + 1, 000, 000 = 1, 001, 000. (c) E( 575757) = E( 57) + E(57 5757) + E(5757 575757) = E(57 57) + E(5757 5757) + E(575757 575757) = 100 + 10, 000 + 1, 000, 000 = 1, 010, 100. 224 Chapter 10 More Expectations and Variances 9. Let X be the number of students standing at the front of the room after k, 1 k < n names
have been called. The k students whose names have been called are not standing. Let A1 , A2 , . . . , Ank be the students whose names have not been called. Let Xi = Clearly, X = X1 + X2 + + Xnk . k . n This is because Ai is standing if and only if his or her original seat was among the first k. Hence k (n  k)k . E(X) = E(X1 ) + E(X2 ) + + E(Xnk ) = (n  k) = n n E(Xi ) = P Ai is standing = For i, 1 i n  k, 1 0 if Ai is standing otherwise. 10. By Theorem 10.2, E min(X1 , X2 , . . . , Xn ) =
k=1 P min(X1 , X2 , . . . , Xn ) k P (X1 k, X2 k, . . . Xn k)
k=1 = =
k=1 P (X1 k)P (X2 k) P (Xn k) P (X1 k)
k=1 n = =
k=1 i=k pi n =
k=1 hn . k 11. Let E1 be the event that the first three outcomes are heads and the fourth outcome is tails.
For 2 i n  3, let Ei be as defined in the hint. Let En2 be the event that the outcome (n  3) is tails and the last three outcomes are heads. The expected number of exactly three consecutive heads is
n3 n3 E X1 +
i=2 Xi + Xn2 = E(X1 ) +
i=2 n3 E(Xi ) + E(Xn2 ) P (Ei ) + P (En2 )
i=2 = P (E1 ) + = = 1 2 1 2
4 n3 +
i=2 1 2 5 +
5 1 2 = 4 3 + (n  4) 1 2 n . 32 Section 10.1 Expected Values of Sums of Random Variables 225 12. Let
Xi = 1 0 if the ith box is empty otherwise; The expected number of the empty boxes is E(X1 + X2 + + X40 ) = 40E(Xi ) = 40P (Xi = 1) = 40 39 40
80 5.28. 13. The expected number of birthdays that belong to one student is
E(X1 + X2 + + X25 ) = 25E(Xi ) = 25P (Xi = 1) = 25 364 365
24 = 23.41. 14. Let Xi = 1, if the birthdays of at least two students are on the ith day of the year, and Xi = 0,
otherwise. The desired quantity is
365 E
i=1 Xi = 365E(Xi ) = 365P (Xi = 1) = 365 1  364 365
25  25 1 1 365 364 365 24 = 0.788. 15. Let u1 , u2 , . . . , u39 be an enumeration of the nonheart cards. Let
Xi = 1 0 if no heart is drawn before ui is drawn otherwise.
39 i=1 Let N be the number of cards drawn until a heart is drawn. Clearly, N = 1 + the result of Exercise 9, Section 3.2,
39 39 Xi . By E(N ) = 1 +
i=1 39 E(Xi ) = 1 +
i=1 P (Xi = 1) =1+
i=1 1 1 = 1 + 39 = 3.786. 14 14 Note that if the experiment was performed with replacement, then E(N ) = 4. 16. We have that
E( THTHTTHTHT) = E( T) + E(T THT) + E(THT THTHT) + E(THTHT THTHTTHTHT) = E( T) + E(THT THT) + E(THTHT THTHT) + E(THTHTTHTHT THTHTTHTHT) = 2 + 8 + 32 + 1, 024 = 1, 066. 226 Chapter 10 0 More Expectations and Variances 17. (a)
0 I (x, y) dx dy is the area of the rectangle (x, y) R2 : 0 x < X, 0 y < Y ; therefore it is equal to XY . (b) Part (a) implies that E(XY ) =
0 0 E I (x, y) dx dy =
0 0 P (X > x, Y > y) dx dy. 18. Clearly N > i if and only if
X1 X2 X3 Xi . Hence for i 2, P (N > i) = P X1 X2 X3 Xi1 Xi = 1 i! because Xi 's are independent and identically distributed. So, by Theorem 10.2, E(N ) =
i=1 P (N i) = P (N > i) = P (N > 0) + P (N > 1) +
i=0 i=0 i=2 1 i! =1+1+
i=2 1 = i! 1 = e. i! 19. If the first red chip is drawn on or before the 10th draw, let N be the number of chips before
the first red chip. Otherwise, let N = 10. Clearly, P (N = i) = The desired quantity is
9 1 2 i 1 1 = 2 2 i+1 , 0 i 9; P (N = 10) = 1 2 10 . E(10  N ) =
i=0 (10  i) 1 2 i+1 + (10  10) 1 2 10 9.001. 20. Clearly, if for some R, X = Y , CauchySchwarz's inequality becomes equality. We
show that the converse of this is also true. Suppose that for random variables X and Y , E(XY ) = Then 4 E(XY )
2 E(X2 )E(Y 2 ).  4E(X2 )E(Y 2 ) = 0. Section 10.2 Covariance 227 Now the left side of this equation is the discriminant of the quadratic equation E(Y 2 )2  2 E(XY ) + E(X2 ) = 0. Hence this quadratic equation has exactly one root. On the other hand, E(Y 2 )2  2 E(XY ) + E(X2 ) = E (X  Y )2 . So the equation E (X  Y )2 = 0 has a unique solution. That is, there exists a unique number 1 R such that E (X  1 Y )2 = 0. Since the expected value of a positive random variable is positive, this implies that with probability 1, X  1 Y = 0 or X = 1 Y. 10.2 COVARIANCE 1. Since X and Y are independent random variables, Cov(X, Y ) = 0.
3 4 2. E(X) =
x=1 y=3 17 1 2 x (x + y) = ; 70 7
4 1 y=3 70 E(Y ) =
3 3 x=1 4 xy(x + y) = 124 ; 35 E(XY ) =
x=1 y=3 1 2 43 x y(x + y) = . 70 5 Therefore, Cov(X, Y ) = E(XY )  E(X)E(Y ) = 1 43 17 124  = . 5 7 35 245 3. Intuitively, E(X) is the average of 1, 2, . . . , 6 which is 7/2; E(Y ) is (7/2)(1/2) = 7/4. To
show these, note that
6 6 E(X) =
x=1 xpX (x) =
x=1 x(1/6) = 7/2. 228 Chapter 10 More Expectations and Variances By the table constructed for p(x, y) in Example 8.2, E(Y ) = 0 63 120 99 64 29 8 1 7 +1 +2 +3 +4 +5 +6 = . 384 384 384 384 384 384 384 4
6 6 By the same table, E(XY ) =
x=1 y=0 xyp(x, y) = 91/12. Therefore, Cov(X, Y ) = E(XY )  E(X)E(Y ) = 91 7 7 35  = > 0. 12 2 4 24 This shows that X and Y are positively correlated. The higher the outcome from rolling the die, the higher the number of tails obtaineda fact consistent with our intuition. 4. Let X be the number of sheep stolen; let Y be the number of goats stolen. Let p(x, y) be the
7 x 8 y 5 4xy ; 20 4 joint probability mass function of X and Y . Then, for 0 x 4, 0 y 4, 0 x + y 4, p(x, y) = p(x, y) = 0, for other values of x and y. Clearly, X is a hypergeometric random variable with parameters n = 4, D = 7, and N = 20. Therefore, E(X) = nD 28 7 = = . N 20 5 Y is a hypergeometric random variable with parameters n = 4, D = 8, and N = 20. Therefore, E(Y ) = Since
4 4x nD 32 8 = = . N 20 5 168 , 95 E(XY ) =
x=0 y=0 xyp(x, y) = we have Cov(X, Y ) = E(XY )  E(X)E(Y ) = 168 7 8 224  = < 0. 95 5 5 475 Therefore, X and Y are negatively correlated as expected. Section 10.2 Covariance 229 5. Since Y = n  X,
E(XY ) = E(nX  X2 ) = nE(X)  E(X2 ) = nE(X)  Var(X) + E(X)2 = n np  np(1  p) + n2 p2 = n(n  1)p(1  p), and Cov(X, Y ) = E(XY )  E(X)E(Y ) = n(n  1)p(1  p)  np n(1  p) = np(1  p). This confirms the (obvious) fact that X and Y are negatively correlated. 6. Both (a) and (b) are straightforward results of relation (10.6). 7. Since Cov(X, Y ) = 0, we have
Cov(X, Y + Z) = Cov(X, Y ) + Cov(X, Z) = Cov(X, Z). 8. By relation (10.6),
Cov(X + Y, X  Y ) = E(X2  Y 2 )  E(X + Y )E(X  Y ) = E(X2 )  E(Y 2 )  E(X)
2 + E(Y ) 2 = Var(X)  Var(Y ). 9. In Theorem 10.4, let a = 1 and b = 1. 10. (a) This is an immediate result of Exercise 8 above.
(b) By relation (10.6), Cov(X, XY ) = E(X2 Y )  E(X)E(XY ) = E(X2 )E(Y )  E(X) E(Y ) = E(Y )Var(X).
2 11. The probability density function of is given by 1 if [0, 2 ] f ( ) = 2 0 otherwise. Therefore, E(XY ) =
0 2 sin cos 1 d = 0, 2 E(X) =
0 2 sin 1 d = 0, 2 E(Y ) =
0 2 cos 1 d = 0. 2 Thus Cov(X, Y ) = E(XY )  E(X)E(Y ) = 0. 230 Chapter 10 More Expectations and Variances 12. The joint probability density function of X and Y is given by 1 f (x, y) = 0 x2 + y2 1 elsewhere. X and Y are dependent because, for example, P 0<X< while, 1 P 0<X< =2 2
1/2 0 0 1x 2 1 1 Y =0 = 2 4 2 1 dy dx = 1/2 0 1  x 2 dx 1 3 1 =P 0<X< Y =0 . = + 6 4 2 X and Y are uncorrelated because E(X) =
x 2 +y 2 1 x 1 1 dx dy = 1 1 dx dy = 1 0 1 0 0 0 2 r 2 cos d dr = 0, E(Y ) =
x 2 +y 2 1 y 2 r 2 sin d dr = 0, and E(XY ) =
x 2 +y 2 1 xy 1 1 dx dy = 1 0 0 2 r 3 cos sin d dr = 0, implying that Cov(X, Y ) = E(XY )  E(X)E(Y ) = 0. 13. We have that
E(X) =
2 1/2 8 2 x dx = 1.4, 15 6 3/2 y dy = 1.396, 13 E X2 = 2 1/2 8 3 x dx = 2.125, 15 6 5/2 y dy = 2.252. 13 E(Y ) = 9/4 1/4 E Y2 = 9/4 1/4 These give Var(X) = 2.125  1.42 = 0.165, and Var(Y ) = 2.252  1.3962 = 0.303. Hence E(X + Y ) = 1.4 + 1.396 = 2.796, and by independence of X and Y , Var(X + Y ) = Var(X) + Var(Y ) = 0.165 + 0.303 = 0.468. Therefore, the expected value and variance of the total raise Mr. Jones will get next year are $2796 and $468, respectively. Section 10.2 Covariance 231 14. We have that
Var(XY ) = E(X2 Y 2 )  E(X)E(Y )
2 = E(X2 )E(Y 2 )  2 2 1 2 2 2 2 2 2 2 = (2 + 1 )(2 + 2 )  2 2 = 1 2 + 2 2 + 2 1 . 1 2 1 2 1 2 15. (a) Let U1 and U2 be the measurements obtained using the voltmeter for V1 and V2 , respec tively. Then V1 = U1 + X1 and V2 = U2 + X2 , where X1 and X2 , the measurement errors, are independent random variables with mean 0 and variance 2 . So the error variance in the estimation of V1 and V2 using the first method is 2 . (b) Let U3 and U4 be the measurements obtained,using the voltmeter, for V and W , respectively. Then V = U3 +X3 and W = U4 +X4 , where X3 and X4 , the measurement errors, are independent random variables with mean 0 and variance 2 . Since (U3 + U4 )/2 is used to estimate V1 , and (U3  U4 )/2 is used to estimate V2 , V1 = and U3  U4 X3  X4 V W = + , 2 2 2 we have that, for part (b), (X3 + X4 )/2 and (X3  X4 )/2 are the measurement errors in measuring V1 and V2 , respectively. The independence of X3 and X4 yields V2 = Var and 1 2 1 X3  X4 = Var(X3 ) + Var(X4 ) = ( 2 + 2 ) = . 2 4 4 2 Therefore, the error variances in the estimation of V1 and V2 , using the second method, is 2 /2, showing that the second method is preferable. Var 1 X3 + X4 1 2 = Var(X3 ) + Var(X4 ) = ( 2 + 2 ) = , 2 4 4 2 V +W U3 + U4 X3 + X4 = + , 2 2 2 16. Let r be the annual rate of return for Mr. Ingham's total investment. We have
Var(r) = Var(0.18r1 + 0.40r2 + 0.42r3 ) = (0.18)2 Var(r1 ) + (0.40)2 Var(r2 ) + (0.42)2 Var(r3 ) + 2(0.18)(0.40)Cov(r1 , r2 ) + 2(0.18)(0.42)Cov(r1 , r3 ) + 2(0.40)(0.42)Cov(r2 , r3 ) = (0.18)2 (0.064) + (0.40)2 (0.0144) + (0.42)2 (0.01) + 2(0.18)(0.40)(0.03) + 2(0.18)(0.42)(0.015) + 2(0.40)(0.42)(0.021) = 0.01979. 232 Chapter 10 More Expectations and Variances Hence the standard deviation of the annual rate of return for Mr. Ingham's total investment is 0.01979 = 0.14. 17. Let r1 , r2 , and r3 be the annual rates of return for Mr. Kowalski's investments in financial
assets 1, 2, and 3, respectively. Let r be the annual rate of return for his total investment. Then, by Example 4.25, r = 0.25r1 + 0.40r2 + 0.35r3 . Since the assets are uncorrelated, we have E(r) = (0.25)(0.12) + (0.40)(0.15) + (0.35)(0.18) = 0.153, Var(r) = (0.25)2 (0.08)2 + (0.40)2 (0.12)2 + (0.35)2 (0.15)2 = 0.00546, r = Var(r) = 0.074. Hence r N (0.153, 0.00546). Let X be the total investment of Mr. Kowalski. We are given that X = 50, 000. Let Y be the total return of Mr. Kowalski's investment next year. The desired probability is P (Y  X 10, 000) = P Y X 10, 000 X 50, 000 0.2  0.153 0.074 (0.64) = 1  0.7389 = 0.2611. = P (r 0.2) = P Z = P (Z 0.64) = 1  18. (a) We have that
E(X) =
0 1 x 1 0 x 1 0 x 1 1 1 8x 2 y dy dx = 8 , 15 E(Y ) =
0 1 x 1 0 x 1 4 8xy 2 dy dx = , 5 2 , 3 E(X 2 ) = 1 8x 3 y dy dx = , 3 4 8x 2 y 2 dy dx = , 9 E(Y 2 ) = 1 8xy 3 dy dx = E(XY ) = Cov(X, Y ) = E(XY )  E(X)E(Y ) = Var(X) = Therefore, Var(X + Y ) = 1 8  3 15
2 = 11 , 225 8 4  9 15 2 Var(Y ) =  3 4 4 = , 5 225 4 2 2 = . 5 75 11 2 4 1 + +2 = . 225 75 225 9 Section 10.2 Covariance 233 (b) Since Cov(X, Y ) = 0, X and Y are not independent. This does not contradict Exercise 23 of Section 8.2 because although f (x, y) is the product of a function of x and a function of y, its domain is not of the form (x, y) : a x b, c y d . In the domain of f , x and y are related by x y. 19. For 1 i n, let Xi be the ith random number selected; we have
n n n Var
i=1 Xi =
i=1 Var(Xi ) =
i=1 n (1  0)2 = . 12 12 20. By the hint,
E(X) =
0 0 0 0 1 4 (y+1)x x e dx dy = 2 1 3 x(y+1) dx dy = x ye 2 0 4! 1 dy = 3, 2 (y + 1)5 3! 1 1 y dy = , 2 (y + 1)4 2 E(Y ) = 0 and E(XY ) =
0 0 1 4 (y+1)x dx dy = x ye 2 0 4! 1 y dy = 1. 3 (y + 1)5 Since Cov(X, Y ) = 1  1 3 =  < 0,, X and Y are negatively correlated. 2 2 21. Note that
E (X  t)2 = E (X  +  t)2 = E (X  )2 + 2(  t)E(X  ) + (  t)2 = E (X  )2 + (  t)2 . This relation shows that E (X  t)2 is minimum if (  t)2 = 0; that is, if t = . For this value, E (X  t)2 = Var(X). 22. Clearly,
Cov(IA , IB ) = E(IA IB )  E(IA )E(IB ) = P (AB)  P (A)P (B). This shows that Cov(IA , IB ) > 0 P (AB) > P (A)P (B) P (AB) > P (A), P (B) P (A  B) > P (A). The proof that IA and IB are positively correlated if and only if P (BA) > P (B) follows by symmetry. 234 Chapter 10 More Expectations and Variances 23. By Exercise 6,
Cov(aX + bY, cZ + dW ) = a Cov(X, cZ + dW ) + b Cov(Y, cZ + dW ) = ac Cov(X, Z) + ad Cov(X, W ) + bc Cov(Y, Z) + bd Cov(Y, W ). 24. By Exercise 6 and an induction on n,
n m n m Cov
i=1 ai Xi ,
j =1 bj Yj =
i=1 ai Cov Xi ,
j =1 bj Yj . By Exercise 6 and an induction on m,
m m Cov Xi ,
j =1 bj Yj =
j =1 bj Cov(Xi , Yj ). The desired identity follows from these two identities. 25. For 1 i n, let Xi = 1 if the outcome of the ith throw is 1; let Xi = 0, otherwise. For 1 j n, let Yj = 1 if the outcome of the j th throw is 6; let Yj = 0, otherwise. Clearly, Cov(Xi , Yj ) = 0 if i = j . By Exercise 24,
n n n n n Cov
i n Xi ,
j =1 Yj =
j =1 i=1 Cov(Xi , Yj ) =
i=1 n Cov(Xi , Yi ) n 1 1 = . 6 6 36 =
i=1 E(Xi Yi )  E(Xi )E(Yi ) =
i=1 0 As expected, in n throws of a fair die, the number of ones and the number of sixes are negatively correlated. 26. Let Sn = n i=1 ai Xi , i = E(Xi ); then
n n E(Sn ) =
i=1 ai i , Sn  E(Sn ) =
i=1 ai (Xi  i ). Thus
n Var(Sn ) = E
i=1 n ai (Xi  i ) 2 =
i=1 n ai2 E (Xi  i )2 + 2 ai2Var(Xi ) + 2
i=1 ai aj E (Xi  i )(Xj  j ) i<j ai aj Cov(Xi , Xj ). = i<j Section 10.2 Covariance 235 27. To find Var(X), we use the following identity:
n n Var
i=1 Xi =
i=1 Var(Xi ) + 2 i<j Cov(Xi , Xj ). (38) Now for 1 i n, E(Xi ) = P (Ai ) = Thus Var(Xi ) = E(Xi2 )  E(Xi ) Also for i < j, Xi Xj = Therefore, E(Xi Xj ) = P (Ai Aj ) = P (Aj  Ai )P (Ai ) = and Cov(Xi , Xj ) = E(Xi Xj )  E(Xi )E(Xj ) D D D(N  D) (D  1)D  . = = (N  1)N N N (N  1)N 2 Substituting the values of Var(Xi )'s and Cov(Xi , Xj ) back into (38), we get Var(X) = n n D(N  D) D(N  D) +2 N2 (N  1)N 2 2 nD(N  D) n1 . = 1 2 N N 1 and
i<j 2 D , N = E(Xi2 ) = P (Ai ) = D D  N N
2 D . N = D(N  D) . N2 1 0 if Ai Aj occurs otherwise. (D  1)D D1 D = , N 1 N (N  1)N This follows since in (38), have n and n n(n  1) = equal terms, respectively. 2 2
n i=1 28. Let Xi = 1, if the ith couple is left intact; 0, otherwise. We are interested in Var(
where Var
i=1 n n Xi ), Xi =
i=1 Var(Xi ) + 2 i<j Cov(Xi , Xj ). To find Var(Xi ), note that since Xi2 = Xi , Var(Xi ) = E Xi2  E(Xi )
2 = E(Xi ) E(Xi ) . 2 236 Chapter 10 More Expectations and Variances By Example 10.3, E(Xi ) = So Var(Xi ) = (2n  m)(2n  m  1) . 2n(2n  1) (2n  m)(2n  m  1) (2n  m)(2n  m  1) 1 . 2n(2n  1) 2n(2n  1) To find Cov(Xi , Xj ), note that Xi Xj = 1 if the ith and j th couples are left intact; 0, otherwise. Now Cov(Xi , Xj ) = E(Xi Xj )  E(Xi )E(Xj ) = P (Xi Xj = 1)  E(Xi )E(Xj ) 2n  4 (2n  m)(2n  m  1) 2 m  . = 2n 2n(2n  1) m Therefore, Cov(Xi , Xj ) = (2n  m)(2n  m  1)(2n  m  2)(2n  m  3) 2n(2n  1)(2n  2)(2n  3) (2n  m)(2n  m  1) 2 .  2n(2n  1) So
n Var
i=1 Xi =n (2n  m)(2n  m  1) (2n  m)(2n  m  1) 1 2n(2n  1) 2n(2n  1) +2 n(n  1) (2n  m)(2n  m  1)(2n  m  2)(2n  m  3) 2 2n(2n  1)(2n  2)(2n  3)  (2n  m)2 (2n  m  1)2 4n2 (2n  1)2 = (2n  m)(2n  m  1) (2n  m)(2n  m  1) 1 2(2n  1) 2n(2n  1) + (n  1) (2n  m)(2n  m  1)(2n  m  2)(2n  m  3) 2(2n  1)(2n  2)(2n  3)  (2n  m)2 (2n  m  1)2 4n(2n  1)2 Section 10.3 Correlation 237 = (2n  m)(2n  m  1) (2n  m)(2n  m  1) 1 2(2n  1) 2n(2n  1) + (n  1)(2n  m  2)(2n  m  3) (2n  2)(2n  3)  (n  1)(2n  m)(2n  m  1) 2n(2n  1) = (n  1)(2n  m  2)(2n  m  3) (2n  m)(2n  m  1) 1+ 2(2n  1) (2n  2)(2n  3) (2n  m)(2n  m  1) .  2(2n  1) 10.3 CORRELATION 1. We have that Cov(X, Y ) = (X, Y )X Y = 3; thus
Var(2X  4Y + 3) = Var(2X  4Y ) = 4Var(X) + 16Var(Y )  16Cov(X, Y ) = 4(4) + 16(9)  16(3) = 112. 2. By Exercise 23 of Section 8.2, X and Y are independent random variables. This can also be shown directly by verifying the relation f (x, y) = fX (x)fY (y). Hence Cov(X, Y ) = 0, and therefore (X, Y ) = 0. 3. Let X and Y be the lengths of the pieces obtained. Since Y = 1 X, by Theorem  10.5,
(X, Y ) = 1. Since X and Y are uniform over (0, 1), X = 1/ 12 and Y = 1/ 12. Therefore, 1 1 1 Cov(X, Y ) = (X, Y )X Y = (1) = . 12 12 12 4. If 1 1 = 0, both sides of the relation are 0 and the equality holds. If 1 1 = 0, then
(1 X + 2 , 1 Y + 2 ) = Cov(1 X + 2 , 1 Y + 2 ) 1 X+2 1 Y +2 Cov(1 X, 1 Y ) 1 1 Cov(X, Y ) = = sgn(1 1 )(X, Y ). 1 X 1 Y 1 1 X Y = 5. No, because for all random variables X and Y , 1 (X, Y ) 1. 238 Chapter 10 More Expectations and Variances 6. By Exercise 6 of Section 10.2,
Cov(X + Y, X  Y ) = Var(X)  Var(Y ). Since Cov(X, Y ) = 0, X+Y XY = = Var(X + Y ) Var(X  Y ) Var(X) + Var(Y ) Var(X) + Var(Y ) = Var(X) + Var(Y ). Therefore, (X + Y, X  Y ) = Var(X)  Var(Y ) Cov(X + Y, X  Y ) = . X+Y XY Var(X) + Var(Y ) 7. Using integration by parts, we obtain
1 2 1 E(X 2 ) = 2 E(X) = Hence Var(X) = By symmetry, E(Y ) =
/2 0 /2 0 , 4 0 /2 2 x 2 sin(x + y) dx dy = +  2. 8 2 0 x sin(x + y) dx dy = 2 2 2 + 2  = 2+ . 8 2 16 2 16 /2 2 and Var(Y ) =  2 + . Since 4 2 16 1 2
/2 0 0 /2 E(XY ) = Cov(X, Y ) = xy sin(x + y) dx dy =  1, 2 2 1  . Therefore, 2 16 (/2)  1  ( 2 /16) Cov(X, Y ) = = 0.245. (X, Y ) = (/2)  2 + ( 2 /16) Var(X) Var(Y ) Since (X, Y ) = 1, there is no linear relation between X and Y . Section 10.4 Conditioning on Random Variables 239 10.4 CONDITIONING ON RANDOM VARIABLES 1. Let N be the number of tosses required; then
E(N ) = E E(N X) = E(N  X = 0)P (X = 0) + E(N  X = 1)P (X = 1) 1 1 1 1 = 1 + E(N ) + 1 + 2 + E(N ) . 2 2 2 2 Solving this equation for E(N ), we obtain E(N ) = 5. 2. We have that
E Y (t) = E E Y (t)X = E Y (t)  X < t P (X < t) + E Y (t)  X t P (X t) a = E aX  (t  X) P (X < t) + E(at)P (X t) 3 at 4a X P (X < t) + atP (X t) =E 3 3 at t  4 7t 4a 11  + at = 3 2 3 74 3 1 1 = a(22  t)(t  4) + at (7  t). 9 3 d 1 To find the value of t that maximizes E Y (t) , we solve E Y (t) = a(8t + 47) = 0 for dt 9 t. We get t = 47/8 = 5.875. 3. (a) Clearly,
E(Xn  Xn1 = x) = x This implies that E(Xn Xn1 ) = 1 + 1  Therefore, E(Xn ) = E E(Xn Xn1 ) = 1 + 1  Now we use induction to prove that E(Xn ) = b  d 1  For n = 1, (40) holds since E(X1 ) = (b  d) d bd 1 + (b  d + 1) = b  d 1  . b b b 1 b
n x bx 1 + (x + 1) =1+ 1 x. b b b 1 Xn1 . b 1 E(Xn1 ). b (39) . (40) 240 Chapter 10 More Expectations and Variances Suppose that (40) is valid for n, we show that it is valid for n + 1 as well. By (39), E(Xn+1 ) = 1 + 1  1 n 1 1 E(Xn ) = 1 + 1  bd 1 b b b 1 n+1 1 n+1 1 d 1 =bd 1 . =1+b 1 b b b This shows that (40) holds for n + 1, and hence for all n. (b) We have that
b b P (En ) =
x=bd b P (En  Xn1 = x)P (Xn1 = x) =
x=bd x P (Xn1 = x) b
n1 = 1 d 1 1 xP (Xn1 = x) = E(Xn1 ) = 1  1 b x=bd b b b . 4. Let V be a random variable defined by
V = Then X= Therefore, E(X) = E E(X  V ) = E(X  V = 1)P (V = 1) + E(X  V = 0)P (V = 0) = E(Y )p + E(Z)(1  p). 1 0 with probability p with probability 1  p. Y Z if V = 1 if V = 0. 5. The probability that a page should be retyped is
p =1 e3/2 (3/2)0 e3/2 (3/2)1 e3/2 (3/2)2   = 0.1912. 0! 1! 2! Thus E(X1 ) = 200(0.1912) and
200 E(X2 ) = E E(X2 X1 ) =
x=0 200 E(X2  X1 = x)P (X1 = x) =
x=0 (0.1912)xP (X1 = x) = (0.1912)E(X1 ) = (0.1912)2 (200). Similarly, E(X3 ) = E E(X3 X2 ) = (0.1912)3 (200) Section 10.4 Conditioning on Random Variables 241 and, in general, E(Xn ) = (0.1912)n (200). Therefore, by (10.2), E
i=1 Xi =
i=1 E(Xi ) =
i=1 (0.1912)i (200) = 200 0.1912 = 47.28. 1  0.1912 6. For i 1, let Xi be the length of the ith character of the message. Since the total number of
the bits of the message is K Xi , and since it will take (1/1000)th of a second to emit a bit, i=1 we have that T = (1/1000) K Xi . By Wald's equation and Theorem 10.8, i=1 E(T ) = Var(T ) = = 1 1 1 E(K)E(X1 ) = = 1000 1000 p 1000p 1 1000 1 1000
2 E(K)Var(X1 ) + E(X1 ) Var(K) 1 1p (1  p) + 2 + 22 = . p2 p 1, 000, 000p2 2 2 7. We have that
E(Xn ) = E E(Xn Y ) = E(Xn  Y = 1)P (Y = 1) + E(Xn  Y = 0)P (Y = 0) 39  n . = 0 P (Y = 1) + 1 + E(Xn+1 ) 52  n This recursive relation and E(X39 ) = 0 imply that E(X38 ) = 1/14, E(X37 ) = 2/14, E(X36 ) = 3/14, and, in general, E(Xi ) = (39  i)/14. The answer is 1 + E(X0 ) = 1 + 39 53 = = 3.786. 14 14 8. Let F be the distribution function of X. We have
P (X < Y ) = =   P (X < Y  Y = y)g(y) dy P (X < y) g(y) dy =  F (y) g(y) dy. 242 Chapter 10 More Expectations and Variances 9. Let f be the probability density function of ; then
P (N = i) =
0 P (N = i  = x)f (x) dx ex x i x e dx = i!
i 0 i+1 0 0 =
0 e2x x i dx i! = = 1 1 i! 2 1 1 i! 2 0 e2x (2x)i dx eu ui du = 1 2
i+1 . In these calculations, we have used the substitution u = 2x and the relation eu ui du = i!. 10. Suppose that player A carries x dollars in his wallet. Then player A wins if and only if player B carries y dollars, y (x, 1] in his wallet. Thus player A wins y dollars with probability 1  x. In such a case, the expected amount player A wins is (1 + x)/2. Player A loses x dollars with probability x. Therefore, 1 3 1+x (1  x) + (x) x =  x 2 . 2 2 2 Let fX be the probability density function of X, then E(WA  X = x) = fX (x) = 1 0 if 0 x 1 otherwise. Therefore, E(WA ) = E E(WA  X) =
0 1 E(WA  X = x)fX (x) dx
1 0 =
0 1 1 1 3 2 1  x dx = x  x 3 2 2 2 2 = 0. The solution above was presented by Kent G. Merryfield, Ngo Viet, and Saleem Watson in their joint paper "The Wallet Paradox" published in the AugustSeptember 1977 issue of the American Mathematical Monthly. Note the following observations by the authors.
It is interesting to consider special cases of this formula for the conditional expectation. Since E(WA  X = 1) = 1 and E(WA  X = 0) = 1/2, we see that a player carrying one dollar in his wallet should expect to lose it, whereas a player carrying nothing in his wallet should expect to gain half a dollar (the mean). Interestingly, if a player is carrying half a dollar (the mean) in his wallet, then E(WA  X = 1/2) = 1/8; that is, his expectation of winning is positive. Section 10.4 Conditioning on Random Variables 243 11. (a) To derive the relation
1 1 E(Kn  Kn1 = i) = (i + 1) + i + 1 + E(Kn ) 2 2 1 = (i + 1) + E(Kn ), 2 we noted the following. It took i tosses of the coin to obtain n  1 consecutive heads. If the result of the next toss is heads, we have the desired n consecutive heads. This occurs with probability 1/2. However, if the result of the next toss is tails, then, on the average, we need an additional E(Kn ) tosses [a total of i + 1 + E(Kn ) tosses] to obtain n consecutive heads. This also happens with probability 1/2. (b) From (a) it should be clear that 1 E(Kn  Kn1 ) = (Kn1 + 1) + E(Kn ). 2 (c) Finding the expected values of both sides of (b) yields 1 E(Kn ) = E(Kn1 ) + 1 + E(Kn ). 2 Solving this for E(Kn ), we obtain E(Kn ) = 2 + 2E(Kn1 ). (d) Note that K1 is a geometric random variable with parameter 1/2. Thus E(K1 ) = 2. Solving E(Kn ) = 2 + 2E(Kn1 ) recursively, we get E(Kn ) = 2 + 22 + 23 + + 2n = 2(1 + 2 + + 2n1 ) 2n  1 = 2(2n  1). =2 21 12. Suppose that the last tour left at time 0. Let X be the time from 0 until the next guided tour begins. Let S10 be the time from 0 until 10 new tourists arrive. The random variable S10 is gamma with parameters = 1/5 and n = 10. Let F and f be the probability distribution and density functions of S10 . Then, for t 0, (t/5)9 1 . f (t) = et/5 5 9! To find E(X), note that E(X) = E(X  S10 < 60)P (S10 < 60) + E(X  S10 60)P (S10 60) = E(S10  S10 < 60)P (S10 < 60) + 60P (S10 60). 244 Chapter 10 More Expectations and Variances Now P (S10 < 60) =
0 60 1 t/5 (t/5)9 e dt = 0.7576, 5 9! and, by Remark 8.1, E(S10  S10 < 60) = 1 F (60) 1 = 0.7576
60 0 60 0 tf (t) dt 1 t/5 (t/5)9 te dt = 43.0815. 5 9! Therefore, E(X) = (43.0815)(0.7576) + 60(1  0.7576) = 47.18. This shows that the expected length of time between two consecutive tours is approximately 47 minutes and 10 seconds. 13. Let X1 be the time until the first application arrives. Let X2 be the time between the first and
second applications, and so forth. Then Xi 's are independent exponential random variables with mean 1/ = 1/5 of a day. Let N be the first integer for which X1 2, X2 2, . . . , XN 2, XN+1 > 2. The time that the admissions office has to wait before doubling its student recruitment efforts is SN+1 = X1 + X2 + + XN+1 . Therefore, E(SN+1 ) = E E(SN+1  N ) =
i=0 E(SN+1  N = i)P (N = i). Now, for i 0,
i+1 E(SN+1  N = i) = E(X1 + X2 + + Xi+1  N = i) =
j =1 i E(Xj  N = i) =
j =1 E(Xj  Xj 2) + E(Xi+1  Xi+1 > 2), where by Remark 8.1, E(Xj  Xj 2) = E(Xi+1  Xi+1 > 2) = 1 F (2)
2 0 2 tf (t) dt, tf (t) dt, 1 1  F (2) Section 10.4 Conditioning on Random Variables 245 F and f being the probability distribution and density functions of Xi 's, respectively. That is, for t 0, F (t) = 1  e5t , f (t) = 5e5t . Thus, for 1 j i, E(Xj  Xj 2) = 1 1  e10
2 0 5t e5t dt = (1.0000454) t  1 5t e 5 2 0 = (1.0000454)(0.19999) = 0.1999092 and, for j = i + 1, E(Xi+1  Xi+1 > 2) = Thus, for i 0, 1 e10
2 5t e5t dt = e10 t  1 5t e 5 2 = 2.2. E(SN+1  N = i) = (0.1999092)i + 2.2. To find P (N = i), note that for i 0, P (N = i) = P (X1 2, X2 2, . . . , Xi 2, Xi+1 > 2) = F (2)
i 1  F (2) = (0.9999546)i (0.0000454). Putting all these together, we obtain E(SN+1 ) =
i=0 E(SN+1  N = i)P (N = i) (0.1999092)i + 2.2 (0.9999546)i (0.0000454) =
i=0 = (0.00000908)
i=0 i(0.9999546) + (0.00009988)
i=0 i (0.9999546)i 0.9999546 1 = (0.00000908) + (0.00009988) 2 (1  0.9999546) 1  0.9999546 = 4407.286, i i = r/(1  r)2 , and = where the next to last equality follows from i=1 ir i=0 r 1/(1  r), r < 1. Since an academic year is 9 months long, and contains approximately 180 business days, the admission officers should not be concerned about this rule at all. It will take 4,407.286 business days, on average, until there is a lapse of two days between two consecutive applications. 14. Let Xi be the number of calls until Steven has not missed Adam in exactly i consecutive calls.
We have that E Xi  Xi1 = with probability p Xi1 + 1 Xi1 + 1 + E(Xi ) with probability 1  p. 246 Chapter 10 More Expectations and Variances Therefore, E(Xi ) = E E(Xi  Xi1 ) = E(Xi1 ) + 1 p + E(Xi1 ) + 1 + E(Xi ) (1  p). Solving this equation for E(Xi ), we obtain E(Xi ) = 1 1 + E(Xi1 ) . p Now X1 is a geometric random variable with parameter p. So E(X1 ) = 1/p. Thus E(X2 ) = E(X3 ) = . . . E(Xk ) = 1 1 1  pk 1 (1/p k )  1 1 1 1 + + 2 + + k1 = = k . p p p p p (1/p)  1 p (1  p) 1 1 1 1 + E(X1 ) = 1+ , p p p 1 1 1 1 1 + E(X2 ) = 1+ + 2 , p p p p 15. Let N be the number of games to be played until Emily wins two of the most recent three games. Let X be the number of games to be played until Emily wins a game for the first time. The random variable X is geometric with parameter 0.35. Hence E(X) = 1/0.35. First, we find the random variable E(N  X) in terms of X. Then we obtain E(N ) by calculating the expected value of E(N  X). Let W be the event that Emily wins the (X + 1)st game as well. Let LW be the event that Emily loses the (X + 1)st game but wins the (X + 2)nd game. Let LL be the event that Emily loses both the (X + 1)st and the (X + 2)nd games. Given X = x, we have E(N  X = x) = (x + 1)P (W ) + (x + 2)P (LW ) + (x + 2) + E(N ) P (LL). So E(N  X = x) = (x + 1)(0.35) + (x + 2)(0.65)(0.35) + (x + 2) + E(N ) (0.65)2 . This gives E(N  X = x) = x + (0.4225)E(N ) + 1.65. Therefore, E(N  X) = X + (0.4225)E(N ) + 1.65. Hence E(N ) = E E(N  X) = E(X) + (0.4225)E(N ) + 1.65 = Solving this for E(N ) gives E(N ) = 7.805. 1 + (0.4225)E(N ) + 1.65. 0.35 Section 10.4 Conditioning on Random Variables 247 16. Since hemophilia is a sexlinked disease, and John is phenotypically normal, John is H .
Therefore, no matter what Kim's genotype is, none of the daughters has hemophilia. Whether a boy has hemophilia or not depends solely on the genotype of Kim. Let X be the number of the boys who have hemophilia. To find, E(X), the expected number of the boys who have hemophilia, let 0 if Kim is hh Z = 1 if Kim is H h 2 if Kim is H H . Then E(X) = E E(X  Z) = E(X  Z = 0)P (Z = 0) + E(X  Z = 1)P (Z = 1) + E(X  Z = 2)P (Z = 2) = 4(0.02)(0.02) + 4(1/2) 2(0.98)(0.02) + 0 0.98)(0.98) = 0.08. Therefore, on average, 0.08 of the boys and hence 0.08 of the children are expected to have hemophilia. 17. Let X be the number of bags inspected until an unacceptable bag is found. Let Kn be the number of consequent bags inspected until n consecutive acceptable bags are found. The number of bags inspected in one inspection cycle is X + Km . We are interested in E(X + Km ) = E(X) + E(Km ). Clearly, X is a geometric random variable with parameter (1  p). So E(X) = 1/ (1  p) . To find E(Km ), note that n, E(Kn ) = E E(Kn  Kn1 ) . Now E(Kn  Kn1 = i) = (i + 1)p + i + 1 + E(Kn ) (1  p) = (i + 1) + (1  p)E(Kn ). (41) To derive this relation, we noted the following. It took i inspections to find n  1 consecutive acceptable bags. If the next bag inspected is also acceptable, we have the n consecutive acceptable bags required in i + 1 inspections. This occurs with probability p. However, if the next bag inspected is unacceptable, then, on the average, we need an additional E(Kn ) inspections a total of i + 1 + E(Kn ) inspections until we get n consecutive acceptable bags of cinnamon. This happens with probability 1  p. From (41), we have E(Kn  Kn1 ) = (Kn1 + 1) + (1  p)E(Kn ). Finding the expected values of both sides of this relation gives E(Kn ) = E(Kn1 ) + 1 + (1  p)E(Kn ). 248 Chapter 10 More Expectations and Variances Solving for E(Kn ), we obtain E(Kn ) = E(Kn1 ) 1 + . p p Noting that E(K1 ) = 1/p and solving recursively, we find that E(Kn ) = Therefore, the desired quantity is E(X + Km ) = E(X) + E(Km ) = 1 1 1 1 + 1 + + + m1 (1  p) p p p 1 1 1 + + + n. p p2 p 1 m 1 1 1 (1  )pm + p = + . = 1 (1  p) p pm (1  p) 1 p 18. For 0 < t 1, let N (t) be the number of batteries changed by time t. Let X be the lifetime
of the initial battery used; X is a uniform random variable over the interval (0, 1). Therefore, fX , the probability density function of X, is given by fX (x) = 1 0 if 0 < x < 1 otherwise. We are interested in K(t) = E N(t) . Clearly, E N (t) = E E N(t)  X =
0 t =
0 E N (t)  X = x fX (x) dx
t 0 1 + E N (t  x)
t 0 dx = t + E N (t  x) dx =t+ K(u) du, where the last equality follows from the substitution u = t  x. Differentiating both sides of t K(t) = t + 0 K(u) du with respect to t, we obtain K (t) = 1 + K(t) which is equivalent to K (t) = 1. 1 + K(t) Thus, for some constant c, ln 1 + K(t) = t + c, Section 10.4 Conditioning on Random Variables 249 or, 1 + K(t) = et+c . The initial condition K(0) = E N (0) = 0 yields ec = 1; so K(t) = et  1. On average, after 950 hours of operation, K(0.95) = 1.586 batteries are used. 19. Since E(XY ) is a function of Y , by Example 10.23,
E(XZ) = E E(XZY ) = E E XE(XY )Y = E E(XY )E(XY ) = E(Z 2 ). Therefore, E X  E(XY )
2 = E (X  Z)2 = E(X2  2ZX + Z 2 ) = E(X2 )  2E(Z 2 ) + E(Z 2 ) = E(X2 )  E(Z 2 ) = E(X2 )  E E(XY )2 . 20. Let Z = E(XY ); then
Var(XY ) = E (X  Z)2 Y = E(X2  2XZ + Z 2 Y ) = E(X2 Y )  2E(XZY ) + E(Z 2 Y ). Since E(XY ) is a function of Y , by Example 10.23, E(XZY ) = E XE(XY )Y = E(XY )E(XY ) = Z 2 . Also E(Z 2 Y ) = E E(XY )2 Y = E(XY )2 = Z 2 since, in general, E f (Y )Y = f (Y ): if Y = y, then E f (Y )Y is defined to be E f (Y )Y = y = E f (y)Y = y = f (y). Therefore, Var(XY ) = E(X2 Y )  2Z 2 + Z 2 = E(X2 Y )  E(XY )2 . 21. By the definition of variance,
N N Var
i=1 Xi = E
i=1 Xi 2 N  E
i=1 Xi 2 , (42) 250 Chapter 10 More Expectations and Variances where by Wald's equation,
N E
i=1 Xi 2 = E(X)E(N ) 2 = E(N ) 2 E(X) . 2 (43) Now since N is independent of {X1 , X2 , . . . },
N E
i=1 Xi 2 N =E E
i=1 Xi
N 2 N
2 =
n=1 E
i=1 n Xi N = n P (N = n) =
n=1 E
i=1 n Xi
2 2 N = n P (N = n) =
n=1 E
i=1 Xi P (N = n). Thus
N E
i=1 Xi 2 n =
n=1 E
i=1 Xi2 + 2
i<j Xi Xj P (N = n) E(Xi )E(Xj ) P (N = n)
i<j =
n=1 nE(X2 ) + 2 = E(X 2 )
n=1 nP (N = n) +
2 n=1 n=1 2 n E(X)E(X)P (N = n) = 2 = E(X2 )E(N ) + E(X)
2 n(n  1)P (N = n) = E(X )E(N ) + E(X) E N (N  1) = E(X2 )E(N ) + E(X) E(N 2 )  E(X) E(N ). Putting this and (43) in (42), we obtain
N 2 2 2 Var
i=1 Xi = E(X2 )E(N ) + E(X) E(N 2 )  E(X) E(N )  E(N ) = E(N ) E(X2 )  E(X)
2 2 2 2 E(X) 2 + E(X) 2 E(N 2 )  E(N ) 2 . Section 10.5 Bivariate Normal Distribution 251 Therefore, Var N Xi = E(N )Var(X) + [E(X)]2Var(N ).
i=1 10.5 BIVARIATE NORMAL DISTRIBUTION 1. The conditional probability density function of Y , given that X = 70 is normal with mean
E(Y  X = x) = Y + and standard deviation
2 Y X=x = 2 (1  2 )Y = 2.7 1  (0.45)2 = 2.411. Y 2.7 (x  X ) = 60 + (0.45) (70  71) = 59.595, X 3 Therefore, the desired probability is P (Y 59  X = 70) = P Y  59.595 59  59.595 X = 70 2.411 2.411 = 1  (0.25) = (0.25) = 0.5987. 2. By (10.24),
1 1 exp x2 + y2 . 162 162 (a) Since = 0, X and Y are independent normal random variables with mean 0 and standard deviation 9. Therefore, f (x, y) = P X 6, Y 12 = P (X 6)P (Y 12) = P = X0 6 Y 0 12 P 9 9 12 9 (0.67) (1.23) = (0.7486)(0.9082) = 0.68. (b) To find P X2 + Y 2 36 , we use polar coordinates. P X2 + Y 2 36 = = 1 162
x 2 +y 2 36 exp 
2 6 0 1 x2 + y2 162 dy dx 1 2 exp  0 1 2 2r r dr d. 162 162 Now let u = r 2 /162; du = (2r/162)dr and we get P X2 + Y 2 36 = 1 2
2 0 0 2/9 eu du d = 1  e2/9 = 0.8. 252 Chapter 10 More Expectations and Variances 3. Note that
2 2 Var(X + Y ) = 2 X + Y + 2(X, Y )X Y . Setting d Y Var(X + Y ) = 0, we get = (X, Y ) . d X x  X and X 4. By (10.24), f (x, y) is maximum if and only if Q(x, y) is minimum. Let z1 =
z2 = y  Y . Then  1 implies that Y
2 2 2 2 Q(x, y) = z1  2z1 z2 + z2 z1  2z1 z2  + z2 2 2 z1  2z1 z2  + z2 = z1   z2  2 0. This inequality shows that Q is minimum if Q(x, y) = 0. This happens at x = X and y = Y . Therefore, (X , Y ) is the point at which the maximum of f is obtained. 5. We have that
fX (x) =
0 x 2 dy = 2x, 0 < x < 1, fY (y) = 1 y 2 dx = 2(1  y), 0 < y < 1, fXY (xy) = 1 2 = , 2(1  y) 1y 2 1 = , 2x x y < x < 1. fY X (yx) = Therefore, E(X  Y = y) =
1 y x 0 0 < y < x. xfXY (xy) dx = 1 y x 0 x 1 1+y dx = , 1y 2 1 x dy = , x 2 0 < y < 1, E(Y  X = x) = yfY X (yx) dy = y 0 < x < 1. Now since E(Y  X = x) is a linear function of x and E(X  Y = y) is a linear function of y, by Lemma 10.3, Y x Y + (x  X ) = X 2 Section 10.5 Bivariate Normal Distribution 253 and X + These relations imply that Hence > 0 and 2 = X 1+y . (y  Y ) = Y 2 and X 1 = . Y 2 Y 1 = X 2 Y X 1 = . Therefore = 1/2. X Y 4 6. We use Theorem 8.8 to find the joint probability density function of X and Y . The joint
probability density function of Z and W is given by f (z, w) = 1 1 exp  z2 + w2 . 2 2 Let h1 (z, w) = 1 z + 1 and h2 (z, w) = 2 z + 1  2 w + 2 . The system of equations 1 z + 1 = x z + 2 1  2 w + 2 = y defines a onetoone transformation of R2 in the zwplane onto R2 in the xyplane. It has a unique solution z= w= x  1 , 1 1 12 y  2 (x  1 )  2 1 for z and w in terms of x and y. Moreover, 1 1  1 1  2 0 1 2 1  2 = 1 1 2 1  2 = 0. J= Hence, by Theorem 8.8, the joint probability density function of X and Y is given by 1 1 2 1  2 f x  1 , 1 1 1 2 x  1 y  2  2 1 . 1 1 Noting that f (z, w) = exp  z2 + w2 . Straightforward calculations will result in 2 2 (10.24), showing that the joint probability density function of X and Y is bivariate normal. 254 Chapter 10 More Expectations and Variances 7. Using Theorem 8.8, it is straightforward to show that the joint probability density function of
X + Y and X  Y is bivariate normal. Since (X + Y, X  Y ) = Var(X)  Var(Y ) Cov(X + Y, X  Y ) = = 0, X+Y XY X+Y XY X + Y and X  Y are uncorrelated. But for bivariate normal, uncorrelated and independence are equivalent. So X + Y and X  Y are independent. REVIEW PROBLEMS FOR CHAPTER 10 1. Number the last 10 graduates who will walk on the stage 1 through 10. Let Xi = 1 if the ith
graduate receives his or her own diploma; 0, otherwise. The number of graduates who will receive their own diploma is X = X1 + X2 + + Xn . Since E(Xi ) = 1 we have E(X) = E(X1 ) + E(X2 ) + + E(Xn ) = n 1 = 1. n 1 1 1 +0 1 = , n n n 2. Since
E(X) =
1 2 5 (2x 2  2x) dx = , 3 49 , 10 and E(X 3 ) =
1 2 (2x 4  2x 3 ) dx = we have that E(X3 + 2X  7) = 37 49 10 + 7= . 10 3 30 1 , 2 94 , 135 3. Since
E(X2 ) = and E(XY ) = we have that E(X 2 + 2XY ) = 1 3
1 0 0 2 1 3 1 0 0 2 (3x 5 + x 3 y) dy dx = (3x 4 y + x 2 y 2 ) dy dx = 1 188 511 + = . 2 135 270 Chapter 10 Review Problems 255 4. Let X1 , X2 , . . . , Xn be geometric random variables with parameters 1, (n  1)/n, (n  2)/n,
. . . , 1/n, respectively. The desired quantity is n n + + + n E(X1 + X2 + + Xn ) = 1 + n1 n2 1 1 1 + + + + 1 = 1 + nan1 . =1+n n1 n2 2 the first nonsix outcome is obtained. We have 5. Let X be the number of tosses until 4 consecutive sixes. Let Y be the number of tosses until
E(X) = E E(XY ) =
i=1 4 E(X  Y = i)P (Y = i) E(X  Y = i)P (Y = i)
i=5 =
i=1 4 E(X  Y = i)P (Y = i) + i + E(X)
i=1 = This equation reduces to E(X) = 1 + E(X) 1 6 i1 5 + 6 4
i=5 1 6 i1 5 . 6 5 1 5 + 2 + E(X) + 3 + E(X) 6 6 6 + 4 + E(X) 1 6 1 6 2 3 5 6 5 (1/6)4 5 +4 . 6 6 1  (1/6) Solving this equation for E(X), we obtain E(X) = 1554. 6. f (x, y, z) = (2x)(2y)(2z), 0 < x < 1, 0 < y < 1, 0 < z < 1. Since 2x, 0 < x < 1
is a probability density function, 2y, 0 < y < 1 is a probability density function, and 2z, 0 < z < 1 is also a probability density function, these three functions are fX (x), fY (y), and fZ (z), respectively. Therefore, f (x, y, z) = fX (x)fY (y)fZ (z) showing that X, Y , and Z are independent. Thus (X, Y ) = (Y, Z) = (X, Z) = 0. 7. Since Cov(X, Y ) = X Y (X, Y ) = 2,
Var(3X  5Y + 7) = Var(3X  5Y ) = 9Var(X) + 25Var(Y )  15Cov(X, Y ) = 9 + 225  30 = 204. 8. Clearly,
pX (1) = p(1, 1) + p(1, 3) = 12/25, pY (1) = p(1, 1) = 2/25, pX (2) = p(2, 3) = 13/25; pY (3) = p(1, 3) + p(2, 3) = 23/25. 256 Chapter 10 More Expectations and Variances Therefore, pX (x) = These yield 12 13 38 +2 = ; 25 25 25 2 23 71 E(Y ) = 1 +3 = ; 25 25 25 1 1 1 22 E(XY ) = (1)(1) (12 + 12 ) + (1)(3) (12 + 32 ) + (2)(3) (22 + 32 ) = . 25 25 25 5 E(X) = 1 Thus Cov(X, Y ) = E(XY )  E(X)E(Y ) = 22 38 71 52  = . 5 25 25 625 12/25 13/25 if x = 1 if x = 2, pY (y) = 2/25 23/25 if y = 1 if y = 3. 9. In Exercise 6, Section 8.1, we calculated p(x, y), pX (x), and pY (y). The results of that
exercise yield
12 E(X) =
x=2 5 xpX (x) = 7; E(Y ) =
y=0 12 ypY (y) = 35/18;
5 E(XY ) =
x=2 y=0 xyp(x, y) = 245/18. Therefore, Cov(X, Y ) = E(XY )  E(X)E(Y ) = (245/18)  7(35/18) = 0. This shows that X and Y are uncorrelated. Note that X and Y are not independent as the following shows. 1/36 = p(2, 0) = pX (2)pY (0) = (1/36)(6/36) = 1/216. 10. Let p be the probability mass function of X Y , q be the probability mass function of X +Y ,
and r be the probability mass function of X2  Y 2 . We have x p(x) 0 726/1296 1 520/1296 2 50/1296, Chapter 10 Review Problems 257 x q(x) 0 625/1296 x r(x) 1 500/1296 2 150/1296 3 20/1296 4 1/1296, 0 726/1296 1 500/1296 3 20/1296 4 50/1296. Using these we obtain E X2  Y 2  = E X  Y 2 X+Y Therefore, X  Y , X + Y  = = = Cov X  Y , X + Y XY  X+Y E X2  Y 2   E X  Y  E(X + Y ) XY  X+Y (760/1296)  (620/1296)(864/1296) = 0.563. (0.831)(0.572) 760 , 1296 720 = , 1296 = 0.831. E X  Y  = 620 , 1296 E(X + Y ) = 864 , 1296 E (X + Y )2 = 1, XY  = 0.572, 11. One way to solve this problem is to note that the desired probability is the area of the region
under the curve y = sin x from x = 0 to x = /2 divided by the area of the rectangle [0, /2] [0, 1]. Hence it is
/2 sin x dx = 0 /2 2 . A second way to find this probability is to note that (X, Y ) lies below the curve y = sin x if and only if Y < sin X. Noting that f , the probability density function of X is given by 2 if 0 < x < 2 f (x) = 0 otherwise, and conditioning on X, we obtain P (Y < sin X) =
0 /2 P (Y < sin X  X = x)f (x)dx =
0 /2 0 /2 sin x  0 2 dx 10 = 2 cos x = 2 . 258 Chapter 10 More Expectations and Variances 12. (a) Clearly,
fX (x) =
0 y x ex dy = xex , 0 < x < , fY (y) = (b) we have that E(X) =
0 ex dx = ey , 0 < y < . x 2 ex = 2, x 3 ex dx = 6, E(Y ) =
0 yey = 1, y 2 ey = 2. E X2 =
0 E Y2 =
0 Therefore, Var(X) = 2 and Var(Y ) = 1. Also E(XY ) =
0 y ex dx dy = 3. Thus (X, Y ) = 1 E(XY )  E(X)E(Y ) 32 = . = X Y 21 2 13. Let h(, ) = E (Y   X)2 . Then
h(, ) = E Y 2 + 2 + 2 E X2  2E(Y )  2E(XY ) + 2E(X). Setting h h = 0 and = 0, we obtain + E(X) = E(Y ) E(X) + E X2 = E(XY ). Solving this system of two equations in two unknowns, we obtain = Cov(X, Y ) X Y Y = = , 2 2 X X X Y = Y  X . X Therefore, Y = Y + Y (X  X ). X 0 14. We have that
E(X) =
0 0 xyey(1+x) dy dx = x 1+x 0 (1 + x)yey(1+x) dy dx. Chapter 10 Review Problems 259 Now 0 (1 + x)yey(1+x) dy is the expected value of an exponential random variable with parameter 1 + x, so it is 1/(1 + x). Letting u = 1 + x, we have E(X) =
0 x u1 dx = du 2 (1 + x) u2 1 1 1 du  du = ln u  1 = . u u2 1 1 =
1 (b) To find E(XY ), note that E(X  Y = y) =
0 xfXY (xy) dx =
0 0 x f (x, y) dx, fY (y) where fY (y) =
0 yey(1+x) dx = ey yeyx dx = ey . Note that 0 yeyx dx = 1 because yeyx is the probability density function of an exponential random variable with parameter 1. So E(X  Y = y) =
0 x yey eyx dx = ey 0 xyexy dx = 1 , y where the last equality holds because the last integral is the expected value of an exponential random variable with parameter y. Since y > 0, E(X  Y = y) = 1/y, E(XY ) = 1/Y. 15. Let X and Y denote the number of minutes past 10:00 A.M. that bus A and bus B arrive at the station, respectively. X is uniformly distributed over (0, 30). Given that X = x, Y is uniformly distributed over (0, x). Let f (x, y) be the joint probability density function of X and Y . We calculate E(Y ) by conditioning on X: E(Y ) = E E(Y X) =  E(Y  X = x)fX (x) dx =
0 30 x 1 30 dx = . 2 30 4 Thus the expected arrival time of bus B is 7.5 minutes past 10:00 A.M. 16. To find the distribution function of
N N i=1 Xi , note that
N P
i=1 Xi t =
n=1 P
i=1 n Xi t N = n P (N = n) Xi t N = n P (N = n)
i=1 n =
n=1 P P
n=1 i=1 = Xi t P (N = n), 260 Chapter 10 More Expectations and Variances
n i=1 where the last inequality follows since N is independent of X1 , X2 , X3 , . . . . Now is a gamma random variable with parameters n and . Thus
N Xi P
i=1 Xi t =
n=1 0 t 0 t ex
x n=1 (x)n1 dx (1  p)n1 p (n  1)! (1  p)x (n  1)! (1  p)x (n  1)!
n1 =
n=1 pe
x dx
n1 =
0 t pe dx =
0 t pex e(1p)x dx pepx dx = 1  ept . =
0 t This shows that N i=1 Xi is exponential with parameter p. 17. Let X1 , X2 , . . . , Xi , . . . , X20 be geometric random variables with parameters 1, 19/20, . . . ,
20  (i  1) /20, . . . , 1/20. The desired quantity is
20 20 20 E
i=1 Xi =
i=1 E(Xi ) =
i=1 20 = 71.9548. 20  (i  1) Chapter 11 S ums of I ndependent R andom Variables and L imit Theorems
11.1 MOMENTGENERATING FUNCTIONS
5 1. MX (t) = E etX =
x=1 etx p(x) = 1 t e + e2t + e3t + e4t + e5t . 5 2. (a) For t = 0,
MX (t) = E etX = whereas for t = 0, MX (0) = 1. Thus
3 1 1 tx 1 e3t  et e dx = , 4 4 t 1 e3t  et t MX (t) = 4 1 if t = 0 if t = 0.
2 3  (1) 1 + 3 Since X is uniform over (1, 3), E(X) = = 1 and Var(X) = 2 12 (b) By the definition of derivative, E(X) = MX (0) = lim MX (h)  MX (0) 1 e3h  eh = lim 1 h0 h0 h h 4h 4 = . 3 = lim e3h  eh  4h 3e3h + eh  4 9e3h  eh = lim = lim = 1, h0 h0 h0 4h2 8h 8 where the fifth and sixth equalities follow from L'Hpital's rule. 262 Chapter 11 Sums of Independent Random Variables and Limit Theorems 3. Note that
MX (t) = E e
tX 1 e 2 = 3 x=1
tx x =2
x=1 e e tx x ln 3 =2
x=1 ex(tln 3) . Restricting the domain of MX (t) to the set t : t < ln 3 and using the geometric series theorem, we get MX (t) = 2 etln 3 2et . = 3  et 1  etln 3 (Note that e ln 3 = 1/3.) Differentiating MX (t), we obtain MX (t) = which gives E(X) = MX (0) = 3/2. 6et 3  et
2 , 4. For t = 0, MX (0) = 1. For t = 0, using integration by parts, we obtain
MX (t) =
0 1 2xetx dx = 2et 2et 2  2 + 2. t t t 5. (a) For t = 0, MX (0) = 1. For t = 0,
MX (t) =
0 1 etx 6x(1  x) dx = 6
0 1 xetx dx  6
0 1 x 2 etx dx =6 et et 1 2et et 2et 2 12(1  et ) 6(1 + et )  2 + 2 6  2 + 3  3 = + . t t t t t t t t3 t2 (b) By the definition of derivative, 12(1  et ) 6(1 + et ) + 1 MX (t)  MX (0) t3 t2 = lim E(X) = MX (0) = lim t0 t0 t t = lim 12(1  et ) + 6t (1 + et )  t 3 1 = , 4 t0 t 2 where the last equality is calculated by applying L'Hpital's rule four times. 6. Let A be the set of possible values of X. Clearly, MX (t) = xA etx p(x), where p(x) is the Section 11.1 MomentGenerating Functions 263 probability mass function of X. Therefore, MX (t) =
xA xetx p(x), x 2 etx p(x),
xA MX (t) = . . .
(n) MX (t) = xA x n etx p(x). Therefore, (n) MX (0) = xA x n p(x) = E(Xn ). 7. (a) By definition,
MX (t) = E e (b) From and
tX =
x=0 e tx e  x (et )x = e = e exp(et ) = exp (et  1) . x! x! x=0 MX (t) = et exp (et  1) MX (t) = et
2 exp (et  1) + et exp (et  1) , we obtain E(X) = MX (0) = and E(X2 ) = MX (0) = 2 + . Therefore, Var(X) = (2 + )  2 = . 8. The probability density function of X is given by
f (x) = Therefore, for t = 0, MX (t) = E etX =
b a 0 1 ba if a < x < b otherwise. 1 tx 1 etb  eta e dx = , ba ba t whereas for t = 0, MX (0) = 1. Thus etb  eta 1 MX (t) = b  a t 1 if t = 0 if t = 0. 264 Chapter 11 Sums of Independent Random Variables and Limit Theorems 9. The probability mass function of a geometric random variable X, p(x) with parameter p is
given by Thus MX (t) =
x=1 x=1 t x p(x) = pq x1 , q = 1  p, pq x1 etx = x = 1, 2, 3, . . . . p q x=1 qet . x qe converges to qet / 1  qet if qet < 1 Now by the geometric series theorem, or, equivalently, if t <  ln q. Restricting the domain of MX (t) to the set {t : t <  ln q}, we obtain qet p p pet x qet = = . MX (t) = q x=1 q 1  qet 1  qet Now MX (t) = Therefore, E(X) = MX (0) = and E(X2 ) = MX (0) = Thus Var(X) = E(X2 )  E(X)
2 pet (1  qet )2 and MX (t) = pet + pqe2t . (1  qet )3 p 1 = . 2 (1  q) p p(1 + q) 1+q = . 3 (1  q) p2 = 1+q 1 q  2 = 2. p2 p p 10. Let X be a discrete random variable with the probability mass function p(x) = x/21, x =
1, 2, 3, 4, 5, 6. The momentgenerating function of X is the given function. function x p(x) 1 5/15 3 4/15 4 2/15 5 4/15. et , 1  2t 11. X is a discrete random variable with the set of possible values {1, 3, 4, 5} and probability mass 12. We have that
M2X+1 (t) = E e(2X+1)t = et E e2tX = et MX (2t) = t< 1 . 2 13. Note that
MX (t) = Therefore, E(X) = MX (0) = 24 , (2  t)4 MX (t) = 96 . (2  t)5 96 = 3, 32 24 3 = , 16 2 and hence Var(X) = 3  (9/4) = 3/4. E(X2 ) = MX (0) = Section 11.1 MomentGenerating Functions 265 (r) (r) 14. Since for odd r's, MX (t) = (et  et )/6 and for even r's, MX (t) = (et + et )/6, we have that E(X r ) = 0 if r is odd and E(Xr ) = 1/3 if r is even. momentgenerating function. 15. For a random variable X, we must have MX (0) = 1. Since t/(1  t) is 0 at 0, it cannot be a 16. (a) The distribution of X is binomial with parameters 7 and 1/4.
(b) The distribution of X is geometric with parameter 1/2. (c) The distribution of X is gamma with parameters r and 2. (d) The distribution of X is Poisson with parameter = 3. MX (t) = 17. Since 1 t 2 4 e + , 3 3 X is a binomial random variable with parameters 4 and 1/3; therefore,
2 P (X 2) =
i=0 4 i 1 3 n=0 i 2 3 4i 8 = . 9 18. By relation (11.2),
MX (t) = n=0 2n n t = n! (2t)n = e2t . n! This shows that X = 2 with probability 1. 19. We know that for t = 0,
MX (t) = Therefore, for t = 0, et  1 et  1 = . t (1  0) t MaX+b (t) = E et (aX+b) = ebt E eatX = ebt MX (at) = ebt eat  1 e(a+b)t  ebt , = at (a + b)  b t which is the momentgenerating function of a uniform random variable over (b, a + b). 20. Let n = E(Z n ); then MX (t) =
n=0 n MX (0) n t = n! n=0 n n t . n! (44) Now et = n n=0 (t /n!). Therefore, et 2 /2 =
n=0 (t 2 /2)n = n! n=0 t 2n = 2n n! n=0 (2n)! t 2n . 2n n! (2n)! 266 Chapter 11 Sums of Independent Random Variables and Limit Theorems (2n)! comparing this relation with (44), we obtain E(Z 2n+1 ) = 0, n 0 and E(Z 2n ) = n , 2 n! n 1. 21. By definition,
MX (t) = r (r) 0 etx x r1 ex dx = r (r) 0 e(t)x x r1 dx. This integral converges if t < . Therefore, if we restrict the range of MX (t) to t < , by the substitution u = (  t)x, we obtain MX (t) = r (r) 0 r (r) eu ur1 du = = (  t)r (r) (  t)r t r . Now MX (t) = rr (  t)r1 ; thus E(X) = MX (0) = r/. Also MX (t) = r(r + 1)r (  t)r2 ; therefore, E(X2 ) = MX (0) = r(r + 1) /2 , and hence Var(X) = r r(r + 1)  2 2 = r . 2 22. (a) Let F be the distribution function of X. We have that
P (X t) = P (X t) = t f (x) dx. Letting u = x and noting that f (u) = f (u), we obtain P (X t) =
 t f (u) ( du) = t  f (u) du = F (t). This shows that the distribution function of X is also F . (b) Clearly, MX (t) = Letting u = x, we get MX (t) =   etx f (x) dx. etu f (u) du =  etu f (u) du = MX (t). A second way to explain this is to note that MX (t) is the momentgenerating function of X. Since X and X are identically distributed, we must have that MX (t) = MX (t). Section 11.1 MomentGenerating Functions 267 23. Note that
MX (t) = E etX = Now by the ratio test, x=1 6 tx 6 e = 2 2x2 x=1 etx . x2 et (x+1) /(x + 1)2 x2 = lim 2 et = et x x x + 2x + 1 etx /x 2 lim which is > 1 for t (0, ). Therefore, of the form (, ), > 0, MX (t) exists. x=1 etx diverges on (0, ) and thus on no interval x2 24. For t < 1/2, (11.2) implies that MX (t) =
n=0 E(Xn ) n t = n! n=0 n=0 (n + 1)(2t)n = 1 d 2 dt
2 n=0 1 2 n=0 d (2t)n+1 dt 1 d 1 1 2 dt 1  2t = 1 d 2 dt (2t)n+1 = (2t)n  1 = 1/2 1 = = 2 (1  2t) (1/2)  t . We see that for t < 1/2, MX (t) exists; furthermore, it is the momentgenerating function of a gamma random variable with parameters r = 2 and = 1/2. 25. (a) At the end of the first period, with probability 1, the investment will grow to
A+A X X =A 1+ ; k k
2 at the end of the second period, with probability 1, it will grow to A 1+ X X X X +A 1+ =A 1+ k k k k ; and, in general, at the end of the nth period, with probability 1, it will grow to A 1+ (b) X n . k Dividing a year into k equal periods allows the banks to compound interest quarterly, monthly, or daily. If we increase k, we can compound interest every minute, second, or even fraction of a second. For an infinitesimal > 0, suppose that the interest is compounded at the end of each period of length . If 0, then the interest is compounded continuously. Since a year is 1/ periods, each of length , the interest rate per period of length is the random variable X/(1/) = X. Suppose that at time t, the investment has grown to A(t). Then at t + , with probability 1, the investment will be A(t + ) = A(t) + A(t) X. 268 Chapter 11 Sums of Independent Random Variables and Limit Theorems This implies that P Letting 0, yields P lim A(t + )  A(t) = XA(t) = 1 0 A(t + )  A(t) = XA(t) = 1. or, equivalently, with probability 1, A (t) = XA(t). (c) Part (b) implies that, with probability 1, A (t) = X. A(t) Integrating both sides of this equation, we obtain that, with probability 1, ln[A(t)] = tX + C, or A(t) = etX+c . Considering the fact that A(0) = A, this equation yields A = ec . Therefore, with probability 1, A(t) = etX ec = AetX . This shows that if the interest rate is compounded continuously, then an initial investment of A dollars will grow, in t years, with probability 1, to the random variable AetX , whose expected value is E(AetX ) = AE(etX ) = AMX (t). We have shown the following: If money is invested in a bank at an annual rate X, where X is a random variable, and if the bank compounds interest continuously, then, on average, the money will grow by a factor of MX (t), the momentgenerating function of the interest rate. 26. Since Xi and Xj are binomial with parameters (n, pi ) and (n, pj ),
E(Xi ) = npi , Xi = npi (1  pi ), E(Xj ) = npj , Xj = npj (1  pj ). Section 11.2 Sums of Independent Random Variables 269 To find E(Xi Xj ), note that M(t1 , t2 ) = E et1 Xi +t2 Xj
n nxi =
xi =0 xj =0 n nxi et1 xi +t2 xj P (Xi = xi , Xj = xj ) et1 xi +t2 xj n! x pixi pj j (1  pi  pj )nxi xj xi ! xj ! (n  xi  xj )!
xi =
xi =0 xj =0 n nxi =
xi =0 xj n! et1 pi xi ! xj ! (n  xi  xj )! =0
n et2 pj xj (1  pi  pj )nxi xj = pi et1 + pj et2 + 1  pi  pj , where the last equality follows from multinomial expansion (Theorem 2.6). Therefore, 2M (t1 , t2 ) = n(n  1)pi pj et1 et2 pi et1 + pj et2 + 1  pi  pj t1 t2 and so E(Xi Xj ) = Thus 2M (0, 0) = n(n  1)pi pj . t1 t2
n2 , n(n  1)pi pj  (npi )(npj ) pi pj (Xi , Xj ) = . = (1  pi )(1  pj ) npi (1  pi ) npj (1  pj ) 11.2 SUMS OF INDEPENDENT RANDOM VARIABLES 1. MX (t) = E etX = MX (t) = exp t + (1/2) 2 2 t 2 . 2. Since
MX1 +X2 ++Xn (t) = MX1 (t)MX2 (t) MXn (t) = pet 1  (1  p)et
n , X1 + X2 + + Xn is negative binomial with parameters (n, p). 3. Since
MX1 +X2 ++Xn (t) = MX1 (t)MX2 (t) MXn (t) = X1 + X2 + + Xn is gamma with parameters n and . t n , 270 Chapter 11 Sums of Independent Random Variables and Limit Theorems 4. For 1 i n, let Xi be negative binomial with parameters ri and p. We have that
M X1 +X2 ++Xn (t) = MX1 (t)MX2 (t) MXn (t) = pet 1  (1  p)et pet 1  (1  p)et
r1 pet 1  (1  p)et . r2 pet 1  (1  p)et rn = r1 +r2 ++rn Thus X1 + X2 + + Xr is negative binomial with parameters r1 + r2 + + rn and p. 5. Since
MX1 +X2 ++Xn (t) = MX1 (t)MX2 (t) MXn (t) r1 r2 = t t t r1 +r2 ++rn = , t rn X1 + X2 + + Xn is gamma with parameters r1 + r2 + + rn and . 6. By Theorem 11.4, the total number of underfilled bottles is binomial with parameters 180 and
0.15. Therefore, the desired probability is 180 (0.15)27 (0.85)153 = 0.083. 27 7. For j < i, P (X = i  X + Y = j ) = 0. For j i,
P (X = i  X + Y = j ) = P (X = i)P (Y = j  i) P (X = i, Y = j  i) = P (X + Y = j ) P (X + Y = j ) n i m j i . n+m j = m n i p j i (1  p)m(j i) p (1  p)ni j i i = n+m j p (1  p)n+mj j Interpretation: Given that in n + m trials exactly j successes have occurred, the probability mass function of the number of successes in the first n trials is hypergeometric. This should be intuitively clear. Section 11.2 Sums of Independent Random Variables 271 8. Since X + Y + Z is Poisson with parameter 1 + 2 + 3 and X + Z is Poisson with parameter
1 + 3 , we have that P (Y = y  X + Y + Z = t) = P (Y = y, X + Z = t  y) P (X + Y + Z = t) e2 2 e(1 +3 ) (1 + 3 )ty y! (t  y)! e(1 +2 +3 ) (1 + 2 + 3 )t t! t y 2 1 + 2 + 3
y y = = 1 + 3 1 + 2 + 3 ty . 9. Let X be the remaining calling time of the person in the booth. Let Y be the calling time of the person ahead of Mr. Watkins. By the memoryless property of exponential, X is exponential with parameter 1/8. Since Y is also exponential with parameter 1/8, assuming that X and Y are independent, the waiting time of Mr. Watkins, X + Y , is gamma with parameters 2 and 1/8. Therefore, P (X + Y 12) = 12 1 x/8 5 xe dx = e3/2 = 0.558. 64 2 10. By Theorem 11.7, X + Y N (5, 9), X  Y N (3, 9), and 3X + 4Y N (19, 130). Thus
P (X + Y > 0) = P 05 X+Y 5 > =1 3 3 (1.67) = (1.67) = 0.9525, P (X  Y < 2) = P and P (3X + 4Y > 20) = P 2+3 XY +3 < = 3 3 (1.67) = 0.9525, 20  19 3X + 4Y  19 > =1 130 130 (0.9) = 0.4641. 11. Theorem 11.7 implies that X N (110, 1.6), where X is the average of the IQ's of the randomly selected students. Therefore, P ( X 112) = P 112  110 X  110 1.6 1.6 =1 (1.58) = 0.0571. 12. Let X1 be the average of the accounts selected at store 1 and X2 be the average of the accounts selected at store 2. We have that 900 = N (90, 90) X1 N 90, 10 and 500 2500 X2 N 100, = N 100, . 15 3 272 Chapter 11 Sums of Independent Random Variables and Limit Theorems Therefore, X1  X2 N  10, 770 and so 3 X1  X2 + 10 0 + 10 > 770/3 770/3 P ( X1 > X2 ) = P ( X1  X2 > 0) = P =1 (0.62) = 0.2676. 13. By Exercise 6, Section 10.5, X and Y are sums of independent standard normal random
variables. Hence X + Y is a linear combination of independent standard normal random variables. Thus, by Theorem 11.7, X + Y is normal. 14. By Exercise 13, X  Y is normal; its mean is 71  60 = 11, its variance is
Var(X  Y ) = Var(X) + Var(Y )  2Cov(X, Y ) = Var(X) + Var(Y )  2(X, Y )X Y = 9 + (2.7)2  2(0.45)(3)(2.7) = 9. Therefore, P (X  Y 8) = P 8  11 X  Y  11 =1 3 3 (1) = (1) = 0.8413. 15. Let X be the average of the weights of the 12 randomly selected athletes. Let X1 , X2 , . . . , X12 be the weights of these athletes. Since 25 X N 225, 12 we have that 2700 = P ( X 225) P (X1 + X2 + + X12 2700) = P X 12 225  225 X  225 = =P 625/12 625/12 Dr. Olwell teaches, respectively. We have that 418 448 X1 N 65, = N(65, 19) and X2 N 72, = N (72, 16). 22 28 Therefore, X1  X2 N (7, 35) and hence the desired probability is P X1  X2  2 = P ( X1  X2 2) + P ( X1  X2 2) =P =1 2+7 X1  X2 + 7 35 35 (1.52) + +P 2 + 7 X1  X2 + 7 35 35
2 = N 225, 625 , 12 (0) =
1 . 2 16. Let X1 and X2 be the averages of the final grades of the probability and calculus courses (0.85) = 1  0.9352 + 0.8023 = 0.8671. Section 11.2 Sums of Independent Random Variables 273 17. Let X and Y be the lifetimes of the mufflers of the first and second cars, respectively.
(a) To calculate the desired probability, P (X  Y  1.5), note that by symmetry, P X  Y  1.5 = 2P (X  Y 1.5). Now X  Y N (0, 2), hence P X  Y  1.5 = 2P 1.5  0 XY 0 2 2 =2 1 (1.06) = 0.289. (b) Let Z be the lifetime of the first muffler the family buys. By symmetry, the desired probability is 2P (Y > X + Z) = 2P (Y  X  Z > 0). Now Y  X  Z N (3, 3). Hence 2P (Y  X  Z > 0) = 2P Y XZ+3 0+3 > 3 3 =2 1 (1.73) = 0.0836. 18. Let n be the maximum number of passengers who can use the elevator and X1 , X2 , . . . , Xn
be the weights of n random passengers. We must have P (X1 + X2 + Xn > 3000) < 0.0003 or, equivalently, P (X1 + X2 + + Xn 3000) > 0.9997. Let X be the mean of the weights of the n random passengers. We must have 3000 P X n 625 , we must have Since X N 155, n P or (3000/n)  155 X  155 > 0.9997, 25/ n 25/ n 155 n 3000 > 0.9997.  25 25 n 3000 155 n 3.49  25 25 n or, equivalently, 155n + 87.25 n  3000 0. > 0.9997. Using Table 2 of the Appendix, this gives 274 Chapter 11 Sums of Independent Random Variables and Limit Theorems Since the roots of the quadratic equation 155n + 87.25 n  3000 = 0 are (approximately) n = 4.127 and n = 4.69, the inequality is valid if and only if n + 4.69 n  4.127 0. But n + 4.69 > 0, so the inequality is valid if and only if n  4.127 0 or n 17.032. Therefore the answer is n = 17. 19. By Remark 9.3, the marginal joint probability mass function of X1 , X2 , . . . , Xk is multinomial
with parameters n and (p1 , p2 , . . . , pk , 1  p1  p2   pk ). Thus, letting p = p1 + p2 + + pk and x = x1 + x2 + + xk , we have that p(x1 , x2 , . . . , xk ) = This gives P (X1 + X2 + + Xk = i) =
x1 +x2 ++xk =i n! x px1 px2 pk k (1  p)nx . x1 ! x2 ! xk ! (n  x)! 1 2 n! x x x p11 p22 pk k (1  p)ni x1 ! x2 ! xk ! (n  i)! i! x p x1 px2 pk k x1 ! x2 ! xk ! 1 2 = n! (1  p)ni i! (n  i)! x 1 +x2 ++xk =i n = (1  p)ni (p1 + p2 + + pk )i i n i = p (1  p)ni . i This shows that X1 + X2 + + Xk is binomial with parameters n and p = p1 + p2 + + pk . 20. First note that if Y1 and Y2 are two exponential random variables each with rate , min(Y1 , Y2 ) is exponential with rate 2. Now let A1 , A2 , . . . , A11 be the customers in the line ahead of Kim. Due to the memoryless property of exponential random variables, X1 , the time until A1 's turn to make a call is exponential with rate 2(1/3) = 2/3. The time until A2 's turn to call is X1 + X2 , where X2 is exponential with rate 2(1/3) = 2/3. Continuing this argument and considering the fact that Kim is the 12th person waiting in the line, we have that the time until Kim's turn to make a phone call is X1 + X2 + + X12 , where {X1 , X2 , . . . , X12 } is an independent and identically distributed sequence of exponential random variables each with rate 2/3. Hence the distribution of the waiting time of Kim is gamma with parameters (12, 2/3). Her expected waiting time is 12(2/3) = 18. 11.3 MARKOV AND CHEBYSHEV INEQUALITIES
E(X) = 22. By Markov inequality, 1. Let X be the lifetime (in months) of a randomly selected dollar bill. We are given that Section 11.3 Markov and Chebyshev Inequalities 275 22 = 0.37. 60 This shows that at most 37% of the onedollar bills last 60 or more months; that is, at least five years. P (X 60) 2. We have that P (X 2) = 2/5. Hence, by Markov's inequality,
E(X) 2 = P (X 2) . 5 2 This gives E(X) 4/5. E(X) 5 = = 0.4545. 11 11 42  25 2 = = 0.472. 36 36 3. (a) P (X 11) (b) P (X 11) = P (X  5 6) P X  5 6 4. Let X be the lifetime of the randomly selected light bulb; we have
P (X 700) P X  800 100 2500 = 0.25. 10, 000 5. Let X be the number of accidents that will occur tomorrow. Then
(a) P (X 5) 2 = 0.4. 5
4 (b) P (X 5) = 1 
i=0 e2 2i = 0.053. i! 2 = 0.222 9 (c) P (X 5) = P (X  2 3) P X  2 3 6. Let X be the IQ of a randomly selected student from this campus; we have
15 = 0.017. 900 Therefore, less than 1.7% of these students have an IQ above 140. P (X > 140) P X  110 > 30 7. Let X be the waiting period from the time Helen orders the book until she receives it. We want
to find a so that P (X < a) 0.95 or, equivalently, P (X a) 0.05. But P (X a) = P (X  7 a  7) P X  7 a  7 4 . (a  7)2 So we should determine the value of a for which 4/(a  7)2 0.05; it is easily seen that a 15.9 or a = 16. Therefore, Helen should order the book 16 days earlier. 276 Chapter 11 Sums of Independent Random Variables and Limit Theorems 8. By Markov's inequality, P (X 2) 1 = . 2 2 1 = . 2 9. P (X > 2) = P (X  > ) P X   10. We have that P (38 < X < 46) = P (4 < X  42 < 4) = P X  42 < 4 = 1  P X  42 4 . By (11.3), P X  42 4 Hence P (38 < X < 46) 1  3 60 = . 16(25) 20 17 3 = = 0.85. 20 20 11. For i = 1, 2, . . . , n, let Xi be the IQ of the ith student selected at random. We want to find n,
so that P 3< or, equivalently, X1 + X2 + + Xn  < 3 0.92 n P (X   3) 0.08. Since E(Xi ) = and Var(Xi ) = 150, by (11.3), P (X   3) 150 . 32 n Therefore, all we need to do is to find n for which 150/(9n) 0.08. This gives n 150/[9(0.08)] = 208.33. Thus the psychologist should choose a sample of size 209. 12. Let X1 , X2 , . . . , Xn be the random sample, be the expected value of the distribution, and 2
be the variance of the distribution. We want to find n so that P (X   < 2 ) 0.98 or, equivalently, By (11.3), P (X   2 ) < 0.02. P (X   2 ) 1 2 = . 2n (2 ) 4n Therefore, all we need to do is to make sure that 1/(4n) 0.02. This gives n 12.5. So a sample of size 13 gives a mean which is within 2 standard deviations from the expected value with a probability of at least 0.98. Section 11.3 Markov and Chebyshev Inequalities 277 13. Call a random observation success, if the operator is busy. Call it failure, if he is free. In
(11.5), let = 0.05 and = 0.04; we have n 1 = 2500. 4(0.05)2 (0.04) Therefore, at least 2500 independent observations should be made to ensure that (1/n) n i=1 estimates p, the proportion of time that the airline operator is busy, with a maximum error of 0.05 with probability 0.96 or higher. 14. By (11.5),
1 = 1666.67. 4(0.05)2 (0.06) Therefore, it suffices to flip the coin n = 1667 times independently. n 15. P X   = P X   2n 2n E (X  )2n . 2n E ekX . ekt 16. By Markov's inequality, P (X > t) = P ekX > ekt 17. By the Corollary of CauchySchwarz Inequality (Theorem 10.3),
E(X  Y )
2 E (X  Y )2 = 0. This gives that E(X  Y ) = 0. Therefore, Var(X  Y ) = E (X  Y )2  E(X  Y )
2 = 0. We have shown that X Y is a random variable with mean 0 and variance 0; by Example 11.16, P (X  Y = 0) = 1. So with probability 1, X = Y . 18. If Y = X with probability 1, Theorem 10.5 implies that (X, Y ) = 1. Suppose that (X, Y ) = 1; we show that X=Y with probability 1. Note that E(X) = E(Y ) = (n + 1)/2, Var(X) = Var(Y ) = (n2  1)/12, and X = Y = (n2  1)/12. These and 1 = (X, Y ) = E(XY )  E(X)E(Y ) X Y imply that E(XY ) = (2n2 + 3n + 1)/6. Therefore, E (X  Y )2 = E(X2  2XY + Y 2 ) = E(X2 ) + E(Y 2 )  2E(XY ) = Var(X) + E(X) = n+1 n 1 + 12 2
2 2 2 + Var(Y ) + E(Y )
2  2E(XY )
2 + n+1 n 1 + 12 2
2  2n2 + 3n + 1 = 0. 3 E (X  Y )2 = 0 implies that with probability 1, X=Y (see Exercise 17 above). 278 Chapter 11 Sums of Independent Random Variables and Limit Theorems 19. By Markov's inequality,
P X E etX 1 1 ln = P (tX ln ) = P etX = MX (t). t 0 0 20. Using gamma function introduced in Section 7.4,
E(X) = 1 n! 1 n! x n+1 ex dx = (n + 1)! (n + 2) = = n + 1, n! n! (n + 2)! (n + 3) = = (n + 1)(n + 2). n! n! E(X2 ) = x n+2 ex dx = 2 Hence X = (n + 1)(n + 2)  (n + 1)2 = n + 1. Now P (0 < X < 2n + 2) = 1  P (X 2n + 2), and by Chebyshev's inequality, P (X 2n + 2) = P X  (n + 1) n + 1 P X  (n + 1) n + 1 n+1 1 = . (n + 1)2 n+1 Therefore, P (0 < X < 2n + 1) 1  1 n = . n+1 n+1 11.4 LAWS OF LARGE NUMBERS
1 0 1. Since
E(Xi ) = by the strong law of large numbers, P 1 x 4x(1  x) dx = , 3 X1 + X2 + + Xn 1 = = 1. n n 3 lim 2. If X1 > M with probability 1, then X2 > M with probability 1 since X1 and X2 are identically
distributed. Therefore, X1 + X2 > 2M > M with probability 1. This argument shows that {X1 > M} {X1 + X2 > M} {X1 + X2 + X3 > M} . Therefore, by the continuity of probability function (Theorem 1.8),
n lim P (X1 + X2 + + Xn > M) = P lim X1 + X2 + + Xn > M .
n Section 11.4 Laws of Large Numbers 279 By this relation, it suffices to show that M > 0,
n lim X1 + X2 + + Xn > M (45) with probability 1. Let S be the sample space over which Xi 's are defined. Let = E(Xi ); we are given that > 0. By the central limit theorem, P Therefore, letting V = S : lim X1 () + X2 () + Xn () = , n n
n lim X1 + X2 + Xn = = 1. n we have that P (V ) = 1. To establish (45), it is sufficient to show that V ,
n lim X1 () + X2 () + Xn () = . (46) To do so, applying the definition of limit to
n lim X1 () + X2 () + Xn () = , n we have that for = /2, there exists a positive integer N (depending on ) such that n > N, X1 () + X2 () + Xn ()  < = n 2 or, equivalently,  This yields X1 () + X2 () + Xn () < < . 2 n 2 X1 () + X2 () + Xn () > . n 2 X1 () + X2 () + Xn () > which establishes (46). n , 2 Thus, for all n > N, 3. For 0 < < 1,
P Yn  0 > = 1  P Yn  0 = 1  P (X n) = 1 
0 n f (x) dx. Therefore,
n lim P Yn  0 > = 1 
0 f (x) dx = 1  1 = 0, showing that Yn converges to 0 in probability. 280 Chapter 11 Sums of Independent Random Variables and Limit Theorems 4. By the strong law of large numbers, Sn /n converges to almost surely. Therefore, Sn /n
converges to in probability and hence
n lim P n(  ) Sn n( + ) = lim P  n = lim P Sn  n n Sn  > = 1  0 = 1. = 1  lim P n n Sn + n 5. Suppose that the bank will never be empty of customers again. We will show a contradiction.
Let Un = T1 + T2 + + Tn . Then Un is the time the nth new customer arrives. Let Wi be the service time of the ith new customer served. Clearly, W1 , W2 , W3 , . . . are independent and identically distributed random variables with E(Wi ) = 1/. Let Zn = T1 +W1 +W2 + +Wn . Since the bank will never be empty of customers, Zn is the departure time of the nth new customer served. By the strong law of large numbers,
n lim 1 Un = n and
n lim T1 W1 + W2 + + Wn Zn = lim + n n n n T1 W1 + W2 + + Wn 1 1 + lim =0+ = . = lim n n n n Clearly, the bank will never remain empty of customers again if and only if n, Un+1 < Zn . This implies that Un+1 Zn < n n Zn n + 1 Un+1 < . n n+1 n n + 1 Un+1 Zn lim n n n + 1 n n lim Since lim
n or, equivalently, Thus (47) n+1 Un+1 Zn 1 1 = 1, and with probability 1, lim = and lim = , (47) n n + 1 n n n 1 1 implies that or . This is a contradiction to the fact that < . Hence, with probability 1, eventually, for some period, the bank will be empty of customers again. Section 11.4 Laws of Large Numbers 281 6. Suppose that the bank will never be empty of customers again. We will show a contradiction.
Let Un = T1 + T2 + + Tn . Then Un is the time the nth new customer arrives. Let R be the sum of the remaining service time of the customer being served and the sums of the service times of the m customers present in the queue at t = 0. Let Zn = R + S1 + S2 + + Sn . Since the bank will never be empty of customers, and customers are served on a firstcome, firstserved basis, we have that U1 < R and hence Zn is the departure time of the nth new customer. By the strong law of large numbers,
n lim 1 Un = n and R S1 + S2 + + Sn Zn = lim + n n n n n 1 1 R S1 + S2 + + Sn + lim =0+ = . = lim n n n n lim Clearly, the bank will never remain empty of customers if and only if n, Un+1 < Zn . This implies that Un+1 Zn < n n n + 1 Un+1 Zn < . n n+1 n n + 1 Un+1 Zn lim n n n + 1 n n lim Since lim
n or, equivalently, Thus (48) n+1 Un+1 Zn 1 1 = 1, and with probability 1, lim = and lim = , (48) n n + 1 n n n 1 1 implies that or . This is a contradiction to the fact that < . Hence, with probability 1, eventually, for some period, the bank will be empty of customers. 7. Xn converges to 0 in probability because for every > 0, P Xn  0 is the probability i i+1 that the random point selected from [0, 1] is in k , k . Now n implies that 2k 2 2 i i+1 and the length of the interval k , k 0, Therefore, limn P Xn  0 = 0. 2 2 However, Xn does not converge at any point because for all positive natural number N, there are always m > N and n > N, such that Xm = 0 and Xn = 1 making it impossible for Xn  Xm  to be less than a given 0 < < 1. 282 Chapter 11 Sums of Independent Random Variables and Limit Theorems 11.5 CENTRAL LIMIT THEOREM 1. Let X1 , X2 , . . . , X150 be the random points selected from the interval (0, 1). For 1 i 150, Xi is uniform over (0, 1). Therefore, E(Xi ) = = 0.5 and Xi = 1/ 12. We have P 0.48 < X1 + X2 + + X150 < 0.52 = P (72 < X1 + X2 + + X150 < 78) 150 =P 72  (150)(0.5) X1 + X2 + + X150  (150)(0.5) 78  (150)(0.5) < < 150 1/ 12 150 1/ 12 150 1/ 12 (0.85)  (0.85) = 2 (0.85)  1 = 2(0.8023)  1 = 0.6046. 2. For 1 i 35, let Xi be the score of the ith student selected at random. By the central limit
theorem P (460 < X < 540) = P 460 < X1 + X2 + + X35 < 540 35 = P (16100 < X1 + X2 + + X35 < 18900) =P X1 + X2 + + X35  35(500) 18900  35(500) 16100  35(500) < < 100 35 100 35 100 35 X1 + X2 + + X35  35(500) < 2.37 100 35 (2.37) = 0.9911  0.0089 = 0.9822. = P  2.37 < = (2.37)  3. We have that
=
1 3 5 56 1 x x+ dx = = 2.07, 9 2 27 125 1 2 5 x x+ dx = , 9 2 27 E(X2 ) =
1 3 X = The desired probability is (125/27)  (56/27)2 = 0.57. X1 + X2 + + X24 P (2 < X < 2.15) = P 2 < < 2.15 24 = P (48 < X1 + X2 + + X24 < 51.6) Section 11.5 Central Limit Theorem 283 =P X1 + X2 + + X24  24(2.07) 51.6  24(2.07) 48  24(2.07) < < 0.57 24 0.57 24 0.57 24 (0.69)  (0.60) = 0.7549  0.2743 = 0.4806. 4. Let X1 , X2 , . . . , Xn be the sample. Since f is an even function, for 1 i n,
E(Xi ) = E(Xi2 ) = Xi By the central limit theorem, X1 + X2 + + Xn >0 P (X > 0) = P n =P X1 + X2 + + Xn  n(0) >0 =1 2 n (0) = 0.5.  1 x xe dx = 0 2 0 1 2 x x e dx = 2  = 2  0 = 2. x 2 ex dx = 2 5. Let = E(Xi ) and = Xi . Clearly, E(Sn ) = n and Sn = n; thus, by the central limit
theorem, P E(Sn )  Sn Sn E(Sn ) + Sn = P n  n Sn n + n Sn  n =P 1 1 (1)  (1) = 2 (1)  1 = 0.6826. n 6. For 1 i 300, let Xi be the amount of the ith expenditure minus Jim's ith record; Xi is approximately uniform over (1/2, 1/2). Hence E(Xi ) = 0 and Xi = 1/(2 3). The desired probability is P (10 < X1 + X2 + + X300 < 10) 10  300(0) X1 + X2 + + X300  300(0) 10  300(0) =P < < 300 1/(2 3) 300 1/(2 3) 300 1/(2 3) (2)  (2) = 0.9772  0.0228 = 0.9544. (1/2)  (1/2) /12 =
2 7. Note that actual value is a nebulous concept. In this exercise, like everywhere else, we are
using it to mean the average of a very large number of measurements. Let Xi be the error in 284 Chapter 11 Sums of Independent Random Variables and Limit Theorems the ith measurement; = E(Xi ) = 0, = Xi = 1/ 3. Hence P  0.25 < X1 + X2 + + X50 < 0.25 50 = P (12.5 < X1 + X2 + + X50 < 12.5) =P 12.5 X1 + X2 + + X50 12.5 < < 1/ 3 50 1/ 3 50 1/ 3 50 (3.06)  (3.06) = 2 (3.06)  1 = 0.9778. 8. For 1 i 300, let Xi = 2, if the ith employee attends with his or her spouse; let Xi = 1, if the ith employee attends alone; let Xi = 0, if the ith employee does not attend. To find the desired quantity, the probability of the event 300 Xi 320, note that i=1 = E(Xi ) = 2 E(Xi2 ) = 4 2 Xi = 1 1 1 + 1 + 0 = 1, 3 3 3 1 1 1 5 +1 +0 = , 3 3 3 3 Xi = 2 . 3 5 2 1= , 3 3 Thus
300 P
i=1 Xi 320 = P 300 i=1 Xi  300 320  300 2/3 300 2/3 300 1 (1.41) = 0.0793. 9. Direct calculations show that
=
4 6 xf (x) dx = 2/ ln(3/2) = 4.93, x 2 f (x) dx = 10/ ln(3/2) E(X2 ) =
4 6 X = We want to find n so that or, equivalently, 4 10  = 0.577. ln(3/2) [ln(3/2)]2 P X   0.07 0.98 P (0.07 X  0.07) 0.98. Section 11.5 Central Limit Theorem 285 Since P  0.07 X1 + X2 + + Xn  < 0.07 n 0.07n X1 + X2 + + Xn  n 0.07n 0.577 n 0.577 n 0.577 n  0.12 n = 2 0.12 n  1, 0.12 n  = P (0.07n X1 + X2 + + Xn  n 0.07n) =P all we need to do is to find n so that 2 0.12 n  1 0.98, or 0.12 n 0.99. By Table 2 of the appendix, this is satisfied if 0.12 n 2.33, or n 377.007. Therefore, for all sample sizes of 378 or larger, the sample mean is within 0.07 of the . 10. Let
Xi = 0.125 with probability 1/2 0.125 with probability 1/2. The change in the stock price, per share, after 60 days is X1 + X2 + + X60 . Clearly, E(Xi ) = 0 and Xi = 0.125. To find the distribution of X1 + X2 + + X60 , note that for all t, 60 60 t t i=1 Xi  60(0) P . Xi t = P 0.968 0.125 60 0.125 60 i=1 This relation implies that P X1 + X2 + + X60 t 0.968 (t). So (X1 + X2 + + X60 )/0.968 is approximately standard normal and hence X1 + X2 + + X60 N (0, 0.9682 ). Since the most likely value of a normal random variable with mean 0 is 0, the change in the stock price after 60 days is most likely 0 and hence the most likely value of the holdings of this investor after 60 days is 50,000. 11. Let X1 be the number of tosses until the first tails. Let X2 be the number of additional tosses until the second tails; X3 be the number of tosses after the second tails until the third tails, and so on. Clearly, Xi 's are independent geometric random variables, each with parameter 286 Chapter 11 Sums of Independent Random Variables and Limit Theorems 1/2. To the desired probability, P (X1 + X2 + + X50 75), note that E(Xi ) = 2 and find 1  (1/2) = 2 1/2. Therefore, Xi = 1/2 P (X1 + X2 + + X50 75) X1 + X2 + + X50  50(2) 75  50(2) 50 2 1/2 50 2 1/2 1  (2.5) = (2.5) = 0.9938. =P 12. By Exercise 8, Section 7.4, for each i, i 1, the random variable Xi2 is gamma with parameters = 1/2 and r = 1/2. Therefore, = E(Xi2 ) = and 2 = Var(Xi2 ) = Therefore, by central limit theorem,
n r =1 r = 2. 2 lim P Sn n + Sn  n 1 2n = lim P n 2n Sn  n = lim P 1 = n n (1) = 0.8413. 13. Let Yn = n i=1 Xi ; Yn is Poisson with rate n. On the one hand,
n P (Yn n) =
k=0 en nk 1 = n k! e n k=0 nk , k! and on the other hand,
n n lim P (Yn n) = lim P
n Xi n
i=1 n i=1 = lim P
n Xi  n nn n n nk 1 = . k! 2 = (0) = 1 . 2 So 1 lim n en k=0 Chapter 11 Review Problems 287 REVIEW PROBLEMS FOR CHAPTER 11 1. X, the average wage of a sample of 10 employees is normal with mean $27000 and standard deviation $4900/ 10 = $1549.52. Therefore, the desired probability is 30, 000  27000 X  27000 = 1  (1.94) = 0.0262. P ( X 30, 000) = P 1549.52 1549.52 2 1 20 = and 3 3 9
10 2. MX (t) is the momentgenerating function of a binomial random variable with parameters 10
and 2/3. Therefore, Var(X) = 10 P (X 8) =
i=8 10 i 2 3 i 1 3 10i = 0.299. 3. MX (t) is the momentgenerating function of a discrete random variable X with P (X = 1) =
1/6, P (X = 2) = 1/3, and P (X = 3) = 1/2. Therefore, F , the distribution function of X is given by t <1 0 1/6 1 t < 2 F (x) = 1/2 2 t < 3 1 t 3. 4. MX (t) is the momentgenerating function of a normal random variable with mean 1 and
variance 4. 5. X is a uniform random variable over the interval (1/2, 1/2). 6. X is a Poisson random variable with parameter = 1/2. Therefore,
P (X > 0) = 1  P (X = 0) = 1  e1/2 = 0.393. 7. Note that
(n) MX (t) = (1)n+1 (n + 1)! . (1  t)n+2 (n) Therefore, E(Xn ) = MX (0) = (1)n+1 (n + 1)!. 8. Let X be the average of the heights of 10 randomly selected men and Y be the average 40 heights of 6 randomly selected women. Theorem 10.7 implies that X N 173, and 10 20 22 Y N 160, ; thus X  Y N 13, . Therefore, 6 3 5  13 X  Y  13 = (2.95) = 0.9984. P ( X  Y 5) = P 22/3 22/3 288 Chapter 11 Sums of Independent Random Variables and Limit Theorems 9. By definition,
E etX =  1 x tx e e dx = 2
0 0 1 = 2 1 e(1+t)x dx + 2   0 1 x tx e e dx + 2 ex(t1) dx. 0 1 x tx e e dx 2 Now for these integrals to exist, we must restrict the domain of the momentgenerating function of X to {t R :  1 < t < 1}. In this domain, MX (t) = E etX =
0 1 1 + e(1+t)x ex(t1)  0 2(1 + t) 2(t  1) 1 1 1 + = . = 2(1 + t) 2(1  t) 1  t2 10. (a) By the law of total probability (Theorem 3.4),
n P (X + Y = n) =
i=0 n P (X + Y = n  X = i)P (X = i)
n =
i=0 n P (X + Y = n, X = i) =
i=0 P (Y = n  i, X = i) =
i=0 P (X = i)P (Y = n  i). (b) By part (a),
n P (X + Y = n) =
i=0 1 e i e ni = e(+) i! (n  i)! n! n i=0 n i ni i e(+) ( + )n , = n! where the last equality follows from the binomial expansion (Theorem 2.5). 11. We have
P 0.95 < X1 + X2 + + X28 < 1.05 = P (26.6 < X1 + X2 + + X28 < 29.4) 28 X1 + X2 + + X28  28(1) 29.4  28 26.6  28 < < 2 28 2 28 2 28 (0.13)  (0.13) = 0.5517  0.4483 = 0.1034. =P Chapter 11 Review Problems 289 12. In (11.5), let = 0.01 and = 0.06; we have
n 1 = 41, 666.67. 4(0.01)2 (0.06) Therefore, at least 41667 patients should participate in the trial. 13. By (11.4),
P p  p < 0.05 1  p(1  p) 1 1 = 0.98, 2 5000 (0.05) 4(0.05)2 5000 since p(1  p) 1/4 implies that p(1  p) 1/4. 14. For i = 1, 2, 3, . . . , n, let Xi be the IQ of the ith student of the sample. We want to determine
n so that X1 + X2 + + Xn  < .2 0.98. n Since E(Xi ) = and Var(Xi ) = 170, by the central limit theorem, P  0.2 < P  0.2 <
n i=1 Xi n n  < 0.2 = P  (0.2)n <
i=1 n i=1 Xi  n < (0.2)n =P (0.2)n < 170n (0.2)n  170n Xi  n 170n (0.2)n < 170n (0.2)n 0.2 n  1 0.98. =2 170n 170 Therefore, we should determine n so that (0.2 n/ 170) 0.98. From Table 2 of the Appendix, we find (0.2) n/ 170 = 2.33, which implies that n = 23072.8250; therefore, the psychologist should choose a sample of size 23073. 15. Let Xi be the amount chopped off on the ith charge in dollars. Let X be the actual amount
Ed has charged to his credit card this month minus the amount his record shows. Clearly, X = X1 + X2 + + X20 , and for 1 i 20, Xi is uniform over (0, 1). Thus E(Xi ) = 1/2 and Var(Xi ) = 1/12 and hence E(X) = 20/2 = 10 and Var(X) = 20/12 = 5/3. Therefore, by Chebyshev's inequality, P (X > 15) = P (X  10 > 5) P X  10 > 5 5/3 = P X  E(X) > 5 = 0.0667. 25 16. P (X 45) P X  0 45 152 /452 = 1/9. 290 Chapter 11 Sums of Independent Random Variables and Limit Theorems 17. Suppose that the ith randomly selected book is Xi centimeters thick. The desired probability
is P (X1 + X2 + + X31 87) = P X1 + X2 + + X31  3(31) 87  3(31) 1 31 1 31 87  93 = (1.08) = 1  0.8599 = 0.1401. 31 18. For 1 i 20, let Xi denote the outcome of the ith roll. We have
6 E(Xi ) =
i=1 i 7 1 = , 6 2 6 E(Xi2 ) =
i=1 i2 91 1 = . 6 6 Thus Var(Xi ) =
20 35 91 49  = , and hence 6 4 12 Xi 75 = P 20 Xi  70 65  70 75  70 i=1 35/12 20 35/12 20 35/12 20 P 65 i=1 (0.65)  (0.65) = 2 (0.65)  1 = 0.4844. 19. By Markov's inequality, P (X n) 20. Let X =
26 i=1 1 = . So nP (X n) 1. n n Xi. We have that E(Xi2 ) = E(Xi ) = 0.5098, 26 25 = 0.2601, 51 49 E(Xi ) = 26/51 = 0.5098, Var(Xi ) = 0.5098  (0.5098)2 = 0.2499, E(Xi Xj ) = P (Xi = 1, Xj = 1) = P (Xi = 1)P (Xj = 1  Xi = 1) = and Cov(Xi , Xj ) = E(Xi Xj )  E(Xi )E(Xj ) = 0.2601  (0.5098)2 = 0.0002. Thus E(X) = 26(0.5098) = 13.2548 and
26 Var(X) = Cov(Xi , Xj ) i<j 26 = 26(0.2499) + 2 (0.0002) = 6.6274. 2
i=1 Var(Xi ) + 2
Therefore, by Chebyshev's inequality, P (X 10) P X  13.2548 3.2548 6.6274 = 0.6256. (3.2548)2 Chapter 12 Stochastic Processes
12.2 MORE ON POISSON PROCESSES 1. We know that E N (t) = Var N (t) = t. Hence E N (t)/t = and Var N (t)/t = /t.
Applying Chebyshev's inequality to N (t)/t, we have P N (t)  2. t t As t , the result follows from this relation. 2. By Wald's equation,
E Y (52) = E N (52) E(Xi ) = 52(2.3) (1.2) = 143.52. By Theorem 10.8, Var Y (52) = E N (52) Var(Xi ) + E(Xi ) Var N (52) Y (52) = 52(2.3) (0.7)2 + (1.2)2 52(2.3) = 230.828, = 230.828 = 15.193.
2 3. Let X1 be the time between Linda's arrival at the point and the first car passing by her. Let
X2 be the time between the first and second cars passing Linda, and so forth. The Xi 's are independent exponential random variables with mean 1/ = 7. Let N be the first integer for which X1 15, X2 15, . . . , XN 15, XN+1 > 15. The time Linda has to wait before being able to cross the street is 0 if N = 0 (i.e., X1 > 15), and is SN = X1 + X2 + + XN , otherwise. Therefore, E(SN ) = E E(SN  N ) =
i=0 E(SN  N = i)P (N = i) =
i=1 E(SN  N = i)P (N = i), 292 Chapter 12 Stochastic Processes where the last equality follows since for N = 0, we have that SN = 0. Now
i E(SN  N = i) = E(X1 + X2 + + Xi  N = i) =
j =1 i E(Xj  N = i) =
j =1 E(Xj  Xj 15), where by Remark 8.1, E(Xj  Xj 15) = 1 F (15)
15 0 tf (t) dt; F and f being the probability distribution and density functions of Xi 's, respectively. That is, for t 0, F (t) = 1  et/7 , f (t) = (1/7)et/7 . Thus E(Xj  Xj 15) = 1 1  e15/7
15 0 t t/7 dt = (1.1329)  (t + 7)et/7 e 7 15 0 = (1.1329)(4.41898) = 5.00631. This gives E(SN  N = i) = 5.00631i. To find P (N = i), note that for i 1, P (N = i) = P (X1 15, X2 15, . . . , Xi 15, Xi+1 > 15) = F (15) i 1  F (15) = (0.8827)i (0.1173). Putting all these together, we obtain E(SN ) =
i=1 E(SN  N = i)P (N = i) =
i=1 (5.00631i)(0.8827)i (0.1173) 0.8827 = 37.6707, (1  0.8827)2 = (0.5872)
i=1 i(0.8827)i = (0.5872) where the next to last equality follows from ir i = r/(1  r)2 , r < 1. Therefore, on i=1 average, Linda has to wait approximately 38 seconds before she can cross the street. 4. Label the time point 9:00 A.M. as t = 0. Then t = 4 corresponds to 1:00 P.M. Let N (t) be the number of fish caught at or prior to t; N (t) : t 0 is a Poisson process with rate 2. Let X1 , X2 , . . . , X6 be six uniformly distributed independent random variables over [0, 4]. By theorem 12.4, given that N(4) = 6, the time that the fisherman caught the first fish is Y = min(X1 , X2 , . . . , X6 ). Therefore, the desired probability is P (Y < 1) = 1  P (Y 1) = 1  P min(X1 , X2 , . . . , X6 ) 1 = 1  P (X1 1, X2 1, . . . , X6 1) = 1  P (X1 1)P (X2 1) P (X6 1) = 1  3 4
6 = 0.822. Section 12.2 More on Poisson Processes 293 5. Let S1 , S2 , and S3 be the number of meters of wire manufactured, after the inspector left,
until the first, second, and third fractures appeared, respectively. By Theorem 12.4, given that N (200) = 3, the joint probability density function of S1 , S2 , and S3 is fS1 ,S2 ,S3 N(200) (t1 , t2 , t3  3) = 3! , 8, 000, 000 0 < t1 < t2 < t3 < 200. Using this, the probability we are interested in, is given by the following triple integral: P (S1 + 60 < S2 , S2 + 60 < S3 ) =
0 80 140 t1 +60 200 t2 +60 80 0 80 0 3! dt3 dt2 dt1 8, 000, 000
140 t1 +60 = = = = 3! 8, 000, 000 6 8, 000, 000 (140  t2 ) dt2 dt1 1 2 3200  80t1 + t1 dt1 2
80 0 1 3 6 2 t  40t1 + 3200t1 8, 000, 000 6 1 8 = 0.064. 125 6. By (12.8), the conditional probability density function of Sk , given that N (t) = n, is
fSk N(t) (xn) = Therefore, E Sk  N (t) = n =
0 t n! 1 x (n  k)! (k  1)! t t k1 1 x t nk , 0 x t. n! 1 x x (n  k)! (k  1)! t t n! t (n  k)! (k  1)!
1 0 k1 1 x t nk dx. Letting x/t = u, we have (1/t) dx = du. Thus E Sk  N (t) = n = uk (1  u)nk du. What we want to show follows from the following relations discussed in Section 7.5:
1 0 uk (1  u)nk du = B(k + 1, n  k + 1) = (k + 1) (n  k + 1) k! (n  k)! = . (n + 2) (n + 1)! 7. Let T be the time until the next arrival, and let S be the time until the next departure. By the memoryless property of exponential random variables, T and S are exponential random variables with parameters and , respectively. They are independent by the definition of an M/M/1 queue. Thus P (A) = P (T > t and S > T ) = P (T > t)P (S > t) = et et = e(+)t , 294 Chapter 12 Stochastic Processes 0 P (B) = P (S > T ) = =
0 P (S > T  T = u)eu du 0 P (S > u  T = u)eu du = P (S > u)eu du =
0 eu eu du = . + t A similar calculation shows that P (AB) = P (S > T > t) = = t P (S > T  T = u)eu du e(+)t = P (A)P (B). + eu eu du = 8. (a) Let X be the number of customers arriving to the queue during a service period S. Then
P (X = n) =
0 P (X = n  S = t)et dt =
0 0 et (t)n t e dt n! = n! n t n e(+)t dt = n! ( + ) n t n ( + )e(+)t dt. 0 Note that ( + )e(+)t is the probability density function of an exponential random variable Z with parameter + . Hence P (X = n) = By Example 11.4, E(Z n ) = Therefore, P (X = n) = n = 1 n+1 ( + ) +
n n E(Z n ). n! ( + ) n! . ( + )n , + n 0. This is the probability mass function of a geometric random variable with parameter /( + ). (b) Due to the memoryless property of exponential random variables, the remaining service time of the customer being served is also exponential with parameter . Hence we want to find the number of new customers arriving during a period, which is the sum of n + 1 independent exponential random variables. Since during each of these service times the number of new arrivals is geometric with parameter /( + ), during the entire period under consideration, the distribution of the total number of new customers arriving is the sum of n + 1 independent geometric random variables each with parameter /( + ), which is negative binomial with parameters n + 1 and /( + ). Section 12.2 More on Poisson Processes 295 9. It is straightforward to check that M(t) is stationary, orderly, and possesses independent
increments. Clearly, M(0) = 0. Thus M(t) : t 0 is a Poisson process. To find its rate, note that, for 0 k < , P M(t) = k =
n=k P M(t) = k  N (t) = n P N (t) = n n k et (t)n p (1  p)nk k n! n=k =
n=k = et pk k! (1  p)k t (1  p) (n  k)!
k n et pk = t (1  p) k! (1  p)k = n=k t (1  p) (n  k)! nk et pk (pt)k pt . (t)k et (1p) = e k! k! This shows that the parameter of M(t) : t 0 is p. 10. Note that P Vi = min(V1 , V2 , . . . , Vk ) is the probability that the first shock occurring to the system is of type i. Suppose that the first shock occurs to the system at time u. If we label the time point u as t = 0, then from that point on, by stationarity and the independentincrements property, probabilistically, the behavior of these Poisson processes is identical to the system considered prior to u. So the probability that the second shock is of type i is identical to the probability that the first shock is of type i, and so on. Hence they are all equal to P Vi = min(V1 , V2 , . . . , Vk ) . To find this probability, note that, for 1 j k, Vj 's, are independent exponential random variables, and the probability density function of Vj is j ej t . Thus P (Vj > u) = ej u . By conditioning on Vi , we have P Vi = min(V1 , . . . , Vk ) =
0 P min(V1 , . . . , Vk ) = Vi  Vi = u i ei u du = i
0 P min(V1 , . . . , Vk ) = u  Vi = u ei u du P (V1 u, . . . , Vi1 u, Vi+1 u, . . . , Vk u  Vi = u)ei u du P (V1 u, . . . , Vi1 u, Vi+1 u, . . . , Vk u)ei u du P (V1 u) P (Vi1 u)P (Vi+1 u) P (Vk u)ei u du = i
0 = i
0 = i
0 296 Chapter 12 Stochastic Processes 0 = i = i
0 e1 u ei1 u ei+1 u ek u ei u du e(1 ++k )u du = i
0 eu du = i . 12.3 MARKOV CHAINS 1. {Xn : n = 1, 2, . . . } is not a Markov chain. For example, P (X4 = 1) depends on all the values of X1 , X2 , and X3 , and not just X3 . That is, whether or not the fourth person selected is female depends on the genders of all three persons selected prior to the fourth and not only on the gender of the third person selected. 2. For j 0, P (Xn = j ) =
i=0 P (Xn = j  X0 = i)P (X0 = i) =
i=0 n pij p(i), n where pij is the ij th entry of the matrix P n . 3. The transition probability matrix of this Markov chain is 0 1/2 0 0 0 1/2 1/2 0 1/2 0 0 0 0 1/2 0 1/2 0 0 . P = 0 0 1/2 0 1/2 0 0 0 0 1/2 0 1/2 1/2 0 0 0 1/2 0 By calculating P 4 and P 5 , we will find that, (a) the probability that in 4 transitions the Markov 4 chain returns to 1 is P11 = 3/8; (b) the probability that, in 5 transitions, the Markov chain enters 2 or 6 is 11 11 11 5 5 p12 + p16 = + = . 32 32 16 4. Solution 1: Starting at 0, the process eventually enters 1 or 2 with equal probabilities. Since 2 is absorbing, "never entering 1" is equivalent to eventually entering 2 directly from 0. The probability of that is 1/2. Solution 2: Let Z be the number of transitions until the first visit to 1. Note that state 2 is absorbing. If the process enters 2, it will always remain there. Hence Z = n if and only if the Section 12.3 Markov Chains 297 first n  1 transitions are from 0 to 0, and the nth transition is from 0 to 1, implying that P (Z = n) = 1 2 n1 1 , 4 n = 1, 2, . . . . The probability that the process ever enters 1 is P (Z < ) =
n=1 1 2 n1 1/4 1 1 = = . 4 1  (1/2) 2 Therefore, the probability that the process never enters 1 is 1  (1/2) = 1/2. 5. (a) By the Markovian property, given the present, the future is independent of the past.
Thus the probability that tomorrow Emmett will not take the train to work is, simply, p21 + p23 = 1/2 + 1/6 = 2/3. (b) The desired probability is p21 p11 + p21 p13 + p23 p31 + p23 p33 = 1/4. 6. Let Xn denote the number of balls in urn I after n transfers. The stochastic process {Xn : n =
0, 1, . . . } is a Markov chain with state space {0, 1, . . . , 5} and transition probability matrix 0 1 0 0 0 0 1/5 0 4/5 0 0 0 0 2/5 0 3/5 0 0 . P = 0 3/5 0 2/5 0 0 0 0 0 4/5 0 1/5 0 0 0 0 1 0 Direct calculations show that 241 3125 0 1022 15625 P (6) = P 6 = 0 168 3125 0 168 3125 0 . 1022 15625 0 241 3125 0 0 5293 15625 0 4746 15625 0 168 625 2044 3125 0 9857 15625 0 9492 15625 0 0 9492 15625 0 9857 15625 0 2044 3125 168 625 0 4746 15625 0 5293 15625 0 298 Chapter 12 Stochastic Processes Hence, by Theorem 12.5, P (X6 = 4) = 0 168 1 2 4746 3 4 5293 5 + 0+ + 0+ + 0 = 0.1308. 625 15 15 15625 15 15 15625 15 7. By drawing a transition graph, it is readily seen that this Markov chain consists of the recurrent
classes {0, 3} and {2, 4} and the transient class {1}. 8. Let Zn be the outcome of the nth toss. Then
Xn+1 = max(Xn , Zn+1 ) shows that {Xn : n = 1, 2, . . . } is a Markov chain. Its state space is {1, 2, . . . , 6}, and its transition probability matrix is given by 1/6 1/6 1/6 1/6 1/6 1/6 0 2/6 1/6 1/6 1/6 1/6 0 0 3/6 1/6 1/6 1/6 . P = 0 0 0 4/6 1/6 1/6 0 0 0 0 5/6 1/6 0 0 0 0 0 1 It is readily seen that no two states communicate with each other. Therefore, we have six classes of which {1}, {2}, {3}, {4}, {5}, are transient, and {6} is recurrent (in fact, absorbing). 9. This can be achieved more easily by drawing a transition graph. An example of a desired
matrix is as follows: 0 1 0 0 0 0 0 0 0 1/2 0 1/2 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1/3 2/3 0 0 0 0 . 0 0 0 0 2/5 0 3/5 0 0 0 1/2 0 1/2 0 0 0 0 0 3/5 0 2/5 0 0 0 1/3 0 2/3 0 10. For 1 i 7, starting from state i, let xi be the probability that the Markov chain will
eventually be absorbed into state 4. We are interested in x6 . Applying the law of total Section 12.3 Markov Chains 299 probability repeatedly, we obtain the following system of linear equations: x1 = (0.3)x1 + (0.7)x2 x = (0.3)x + (0.2)x + (0.5)x 2 1 2 3 x3 = (0.6)x4 + (0.4)x5 x =1 4 x5 = x3 x6 = (0.1)x1 + (0.3)x2 + (0.1)x3 + (0.2)x5 + (0.2)x6 + (0.1)x7 x = 0.
7 Solving this system of equations, we obtain x1 = x2 = x3 = x4 = x5 = 1 x = 0.875 6 x7 = 0. Therefore, the probability is 0.875 that, starting from state 6, the Markov chain will eventually be absorbed into state 4. 11. Let 1 , 2 , and 3 be the longrun probabilities that the sportsman devotes to horseback riding,
sailing, and scuba diving, respectively. Then, by Theorem 12.7, 1 , 2 , and 3 are obtained from solving the system of equations. 1 0.20 0.32 0.60 1 2 = 0.30 0.15 0.13 2 0.50 0.53 0.27 3 3 along with 1 + 2 + 3 = 1. The matrix equation above gives us the following system of equations 1 = 0.201 + 0.322 + 0.603 = 0.301 + 0.152 + 0.133 2 3 = 0.501 + 0.532 + 0.273 . By choosing any two of these equations along with 1 + 2 + 3 = 1, we obtain a system of three equations in three unknowns. Solving that system yields 1 = 0.38856, 2 = 0.200056, and 3 = 0.411383. Hence the longrun probability that on a randomly selected vacation day the sportsman sails is approximately 0.20. 12. For n 1, let
Xn = 1 0 if the nth fish caught is trout if the nth fish caught is not trout. 300 Chapter 12 Stochastic Processes Then {Xn : n = 1, 2, . . . } is a Markov chain with state space {0, 1} and transition probability matrix 10/11 1/11 8/9 1/9 Let 0 be the fraction of fish in the lake that are not trout, and 1 be the fraction of fish in the lake that are trout. Then, by Theorem 12.7, 0 and 1 satisfy 0 1 = 10/11 8/9 1/11 1/9 0 , 1 which gives us the following system of equations 0 = (10/11)0 + (8/9)1 = (1/11) + (1/9) . 1 0 1 By choosing any one of these equations along with the relation 0 + 1 = 1, we obtain a system of two equations in two unknown. Solving that system yields 0 = 88/97 0.907 and 1 = 9/97 0.093. Therefore, approximately 9.3% of the fish in the lake are trout. 13. Let 1 Xn = 2 3 if the nth card is drawn by player I if the nth card is drawn by player II if the nth card is drawn by player III. {Xn : n = 1, 2, . . . } is a Markov chain with probability transition matrix 48/52 4/52 0 39/52 13/52 . P = 0 12/52 0 40/52 Let 1 , 2 , and 3 be the proportion of cards drawn by players I, II, and III, respectively. 1 , 2 , and 3 are obtained from 1 12/13 0 3/13 1 2 = 1/13 3/4 2 0 3 3 0 1/4 10/13 and 1 + 2 + 3 = 1, which gives 1 = 39/64 0.61, 2 = 12/64 0.19, and 3 = 13/64 0.20. 14. For 1 i 9, let i be the probability that the mouse is in cell i, 1 i 9, at a random time Section 12.3 Markov Chains 301 in the future. Then i 's satisfy 1 0 1/3 0 1/3 0 0 0 0 0 1 2 2 1/2 0 1/2 0 1/4 0 0 0 0 3 0 1/3 0 0 0 1/3 0 0 0 3 4 1/2 0 0 0 1/4 0 1/2 0 0 4 5 = 0 1/3 0 1/3 0 1/3 0 1/3 0 5 . 6 0 0 1/2 0 1/4 0 0 0 1/2 6 7 0 0 0 1/3 0 0 0 1/3 0 7 8 0 0 0 0 1/4 0 1/2 0 1/2 8 0 0 0 0 0 1/3 0 1/3 0 9 9 Solving this system of equations along with
9 i=1 1 , we obtain 1 = 3 = 7 = 9 = 1/12, 2 = 4 = 6 = 8 = 1/8, 5 = 1/6. 15. Let Xn denote the number of balls in urn I after n transfers. The stochastic process {Xn : n =
0, 1, . . . } is a Markov chain with state space {0, 1, . . . , 5} and transition probability matrix 0 1 0 0 0 0 1/5 0 4/5 0 0 0 0 2/5 0 3/5 0 0 . P = 0 0 3/5 0 2/5 0 0 0 0 4/5 0 1/5 0 0 0 0 1 0 Clearly, {Xn : n = 0, 1, . . . } is an irreducible recurrent Markov chain; since it is finitestate, it is positive recurrent. However, {Xn : n = 0, 1, . . . } is not aperiodic, and the period of each state is 2. Hence the limiting probabilities do not exist. For 0 i 5, let i be the fraction of time urn I contains i balls. Then with this interpretation, i 's satisfy the following equations 0 0 1/5 0 0 0 0 0 1 1 0 2/5 0 0 0 1 2 0 4/5 0 3/5 0 0 2 = 3 0 0 3/5 0 4/5 0 3 , 4 0 0 0 2/5 0 1 4 5 0 0 0 0 1/5 0 5
5 i=0 i = 1. Solving these equations, we obtain 0 = 5 = 1/31, 1 = 4 = 5/31, 2 = 3 = 10/31. 302 Chapter 12 Stochastic Processes Therefore, the fraction of time an urn is empty is 0 + 5 = 2/31. Hence the expected number of balls transferred between two consecutive times that an urn becomes empty is 31/2 = 15.5.
16. Solution 1: Let Xn be the number of balls in urn I immediately before the nth game begins.
Then {Xn : n = 1, 2, . . . } is a Markov chain with state space {0, 1, . . . , 7} and transition probability matrix 3/4 1/4 0 0 0 0 0 0 1/4 1/2 1/4 0 0 0 0 0 0 1/4 1/2 1/4 0 0 0 0 0 0 1/4 1/2 1/4 0 0 0 . P = 0 0 1/4 1/2 1/4 0 0 0 0 0 0 0 1/4 1/2 1/4 0 0 0 0 0 0 1/4 1/2 1/4 0 0 0 0 0 0 1/4 3/4 Since the transition probability matrix is doubly stochastic; that is, the sum of each column is also 1, for i = 0, 1, . . . , 7, i , the longrun probability that the number of balls in urn I immediately before a game begins is 1/8 (see Example 12.35). This implies that the longrun probability mass function of the number of balls in urn I or II is 1/8 for i = 0, 1, . . . , 7. Solution 2: Let Xn be the number of balls in the urn selected at step 1 of the nth game. Then {Xn : n = 1, 2, . . . } is a Markov chain with state space {0, 1, . . . , 7} and transition probability matrix 1/2 0 0 0 0 0 0 1/2 1/4 1/4 0 0 0 0 1/4 1/4 0 1/4 1/4 0 0 1/4 1/4 0 0 0 1/4 1/4 1/4 1/4 0 0 . P = 0 0 1/2 1/2 0 0 0 0 0 0 1/4 1/4 1/4 1/4 0 0 0 1/4 1/4 0 0 1/4 1/4 0 1/4 1/4 0 0 0 0 1/4 1/4 Since the transition probability matrix is doubly stochastic; that is, the sum of each column is also 1, for i = 0, 1, . . . , 7, i , the longrun probability that the number of balls in the urn selected at step 1 of a game is 1/8 (see Example 12.35). This implies that the longrun probability mass function of the number of balls in urn I or II is 1/8 for i = 0, 1, . . . , 7. 17. For i 0, state i is directly accessible from 0. On the other hand, i is accessible from i + 1.
These two facts make it possible for all states to communicate with each other. Therefore, the Markov chain has only one class. Since 0 is recurrent and aperiodic (note that p00 > 0 makes 0 aperiodic), all states are recurrent and aperiodic. Let k be the longrun probability that a Section 12.3 Markov Chains 303 computer selected at the end of a semester will last at least k additional semesters. Solving 0 p1 1 0 0 . . . 0 1 p2 0 1 0 . . . 1 2 = p3 0 0 1 . . . 2 . . . . . . . . . along with i=0 i = 1, we obtain 0 = k = 1+ 1+ i=1 (1 1 ,  p1  p2   pi ) k 1. 1  p1  p2   pk , i=1 (1  p1  p2   pi ) 18. Let DN denote the state at which the last movie Mr. Gorfin watched was not a drama, but
the one before that was a drama. Define DD, N D, and N N similarly, and label the states DD, DN , N D, and N N by 0, 1, 2, and 3, respectively. Let Xn = 0 if the nth and (n  1)st movies Mr. Gorfin watched were both dramas. Define Xn = 1, 2, and 3 similarly. Then {Xn : n = 1, 2, . . . } is a Markov chain with state space {0, 1, 2, 3} and transition probability matrix 7/8 1/8 0 0 0 0 1/2 1/2 . P = 1/2 1/2 0 0 0 0 1/8 7/8 (a) If the first two movies Mr. Gorfin watched last weekend were dramas, the probability 2 2 that the fourth one is a drama is p00 + p02 . Since 49/64 7/64 1/16 1/16 1/4 1/4 1/16 7/16 , P2 = 7/16 1/16 1/4 1/4 1/16 1/16 7/64 49/64 the desired probability is (49/64) + (1/16) = 53/64. Let 0 denote the longrun probability that Mr. Gorfin watches two dramas in a row. Define 1 , 2 , and 3 similarly. We have that, 0 7/8 0 1/2 0 0 1 1/8 0 1/2 0 1 = 2 0 1/2 0 1/8 2 . 0 1/2 0 7/8 3 3 Solving this system along with 0 + 1 + 2 + 3 = 1, we obtain 0 = 2/5, 1 = 1/10, 2 = 1/10, and 3 = 2/5. Hence the probability that Mr. Gorfin watches two dramas in a row is 2/5. (b) 304 Chapter 12 Stochastic Processes 19. Clearly,
Xn+1 = 0 1 + Xn if the (n + 1)st outcome is 6 otherwise. This relation shows that {Xn : n = 1, 2, . . . } is a Markov chain. Its transition probability matrix is given by 1/6 5/6 0 0 0 ... 1/6 0 5/6 0 0 . . . 1/6 0 0 5/6 0 . . . . P = 1/6 0 0 0 5/6 . . . . . . It is readily seen that all states communicate with 0. Therefore, by transitivity of the communication property, all states communicate with each other. Therefore, the Markov chain is irreducible. Clearly, 0 is recurrent. Since p00 > 0, it is aperiodic as well. Hence all states are recurrent and aperiodic. On the other hand, starting at 0, the expected number of transitions until the process returns to 0 is 6. This is because the number of tosses until the next 6 obtained is a geometric random variable with probability of success p = 1/6, and hence expected value 1/p = 6. Therefore, 0, and hence all other states are positive recurrent. Next, a simple probabilistic argument shows that, i = 5 6
i 1 , 6 i = 0, 1, 2, . . . . This can also be shown by solving the following system of equations: 0 1/6 1/6 1/6 1/6 . . . 0 1 5/6 0 0 0 . . . 1 2 = 0 5/6 0 0 . . . 2 0 5/6 0 . . . 3 3 0 . . . . . . . . . 0 + 1 + 2 + = 1. 20. (a) Let
Xn = 1 0 if Alberto wins the nth game if Alberto loses the nth game. Then {Xn : n = 1, 2, . . . } is a Markov chain with state space {0, 1}. Its transition 1p p probability matrix is P = p 1  p . Using induction, we will now show that 1 1 1 1 n n 2 + 2 (1  2p) 2  2 (1  2p) . P (n) = P n = 1 1 1 1  (1  2p)n + (1  2p)n 2 2 2 2 Section 12.3 Markov Chains 305 Clearly, for n = 1, P (1) = P . Suppose that 1 1 n 2 + 2 (1  2p) = 1 1  (1  2p)n 2 2 1 1  (1  2p)n 2 2 . 1 1 n + (1  2p) 2 2 1 1  (1  2p)n+1 2 2 . 1 1 n+1 + (1  2p) 2 2 P (n) We will show that 1 1 n+1 2 + 2 (1  2p) = 1 1  (1  2p)n+1 2 2 P n+1 To do so, note that P (n+1) = Thus
n+1 n n p11 = p10 p01 + p11 p11 = p p00 p01 p10 p11 n n n n n n p00 p01 p00 p00 + p01 p10 p00 p01 + p01 p11 = . n n n n n n p10 p11 p10 p00 + p11 p10 p10 p01 + p11 p11 1 1 1 1  (1  2p)n + (1  p) + (1  2p)n 2 2 2 2 1 1 1 1 = p + (1  p) + (1  2p)n  p + (1  p) = + (1  2p)n+1 . 2 2 2 2 1 1 n+1 This establishes what we wanted to show. The proof that p00 = + (1  2p)n+1 is 2 2 identical to what we just showed. We have
n+1 n+1 P01 = 1  P00 = 1  1 1 1 1 + (1  2p)n =  (1  2p)n . 2 2 2 2 1 1  (1  2p)n . 2 2 Similarly,
n+1 n+1 p10 = 1  p11 = (b) Let 0 and 1 be the longrun probabilities that Alberto loses and wins a game, respectively. Then 1p p 0 0 = , p 1p 1 1 and 0 + 1 = 1 imply that 0 = 1 = 1/2. Therefore, the expected number of games Alberto will play between two consecutive wins is 1/1 = 2. 306 Chapter 12 Stochastic Processes n 21. For each j 0, limn pij exists and is independent of i if the following system of equations, in 0 , 1 , . . . , have a unique solution. 0 1p 1p 0 0 0 1 0 1p 0 0 p 2 = 0 p 0 1p 0 0 p 0 1p 3 0 . . . . . . 0 + 1 + 2 + = 1. From the matrix equation, we obtain i =
i p 0 , 1p i=0 0 0 0 0 ... 0 1 . . . . . . 2 . . . 3 . . . i = 0, 1, . . . . i p to 1p i=0 converge. Hence we must have p < 1  p, or p < 1/2. Therefore, for p < 1/2, this irreducible, aperiodic Markov chain which is positively recurrent has limiting probabilities. Note that, for p < 1/2, i p 0 =1 1p i=0 For these quantities to satisfy i = 1, we need the geometric series yields 0 = 1  p . Thus the limiting probabilities are 1p i = p 1p
i 1 p , 1p i = 0, 1, 2, . . . . 22. Let Yn be Carl's fortune after the nth game. Let Xn be Stan's fortune after the nth game. Let
Zn = Yn  Xn . The {Zn : n = 0, 1, . . . } is a random walk with state space {0, 2, 4, . . . }. We have that Z0 = 0, and at each step either the process moves two units to the right with probability 0.46 or two units to the left with probability 0.54. Let A be the event that, starting at 0, the random walk will eventually enter 2; P (A) is the desired quantity. By the law of total probability, P (A) = P (A  Z1 = 2)P (Z1 = 2) + P (A  Z1 = 2)P (Z1 = 2) = 1 (0.46) + P (A)
2 (0.54). To show that P (A  Z1 = 2) = P (A) , let E be the event of, starting from 2, eventually entering 0. It should be clear that P (E) = P (A). By independence of E and A, we have P (A  Z = 2) = P (EA) = P (E)P (A) = P (A) .
2 2 Section 12.3 Markov Chains 307 We have shown that P (A), the quantity we are interested in, satisfies (0.54) P (A)
2  P (A) + 0.46 = 0. This is a quadratic equation in P (A). Solving it gives P (A) = 23/27 0.85. 23. We will use induction on m. For m = 1, the relation is, simply, the Markovian property, which
is true. Suppose that the relation is valid for m  1. We will show that it is also valid for m. We have
P (Xn+m = j  X0 = i0 , X1 = i1 , . . . , Xn = in ) =
iS P (Xn+m = j  X0 = i0 , . . . , Xn = in , Xn+m1 = i) P (Xn+m1 = i  X0 = i0 , . . . , Xn = in ) =
iS P (Xn+m = j  Xn+m1 = i)P (Xn+m1 = i  Xn = in ) P (Xn+m = j  Xn+m1 = i, Xn = in )P (Xn+m1 = i  Xn = in )
iS = = P (Xn+m = j  Xn = in ), where the following relations are valid from the definition of Markov chain: given the present state, the process is independent of the past.
P (Xn+m = j  X0 = i0 , . . . , Xn = in , Xn+m1 = i) = P (Xn+m = j  Xn+m1 = i), P (Xn+m = j  Xn+m1 = i) = P (Xn+m = j  Xn+m1 = i, Xn = in ).
2n+1 24. Let (0, 0), the origin, be denoted by O. It should be clear that, for all n 0, POO = 0. Now, for n 1, let Z1 , Z2 , Z3 , and Z4 be the number of transitions to the right, left, up, and down, respectively. The joint probability mass function of Z1 , Z2 , Z3 , and Z4 is multinomial. We have
n 2n POO =
i=0 n P (Z1 = i, Z2 = i, Z3 = n  i, Z4 = n  i) 1 (2n)! i! i! (n  i)! (n  i)! 4
i =
i=0 n 1 4 i 1 4 ni 1 4 ni =
i=0 (2n)! n! n! 1 n! n! i! (n  i)! i! (n  i)! 4
2n 2n 1 = 4 2n n n i=0 n i 2 . 308 Chapter 12 Stochastic Processes
n By Example 2.28, (Stirling's formula), 1 4 2n i=0 2 n i 2 = 2n n . Thus POO = n (2n)! n! n! 2n n
2 2 1 4 2n 2n n 2 . Now, by Theorem 2.7 2n n 1 = 4 n=1 2n 4 n (2n)2n e2n 1 2n 4 ( 2 n nn en )2 2 = 1 . n Therefore, 1 Since n=1 n=1 n POO = 1 4 2n is convergent if and only if
n=1 1 is convergent. n 1 is divergent, n n=1 n POO is divergent, showing that the state (0, 0) is recurrent. 25. Clearly, P (Xn+1 = 1  Xn = 0) = 1. For i 1, given Xn = i, either Xn+1 = i + 1 in which case we say that a transition to the right has occurred, or Xn+1 = i  1 in which case we say that a transition to the left has occurred. For i 1, given Xn = i, when the nth transition occurs, let S be the remaining service time of the customer being served or the service time of a new customer, whichever applies. Let T be the time from the nth transition until the next arrival. By the memoryless property of exponential random variables, S and T are exponential random variables with parameters and , respectively. For i 1, P (Xn+1 = i + 1  Xn = i) = P (T < S) =
0 P (S > T  T = t)et dt 0 =
0 P (S > t)et dt = et et dt = . + Therefore, P (Xn+1 = i  1  Xn = i) = P (T > S) = 1  = . + + These calculations show that knowing Xn , the next transition does not depend on the values of Xj for j < n. Therefore, {Xn : n = 1, 2, . . . } is a Markov chain, and its transition probability matrix is given by 0 1 0 0 0 ... 0 0 0 . . . + + 0 0 . . . . 0 P = + + 0 0 . . . 0 + + . . . Since all states are accessible from each other, this Markov chain is irreducible. Starting from 0, for the Markov chain to return to 0, it needs to make as many transitions to the left as it Section 12.3 Markov Chains 309 n makes to the right. Therefore, P00 > 0 only for positive even integers. Since the greatest common divisor of such integers is 2, the period of 0, and hence the period of all other states is 2. 26. The ij th element of P Q is the product of the ith row of P with the jth column of Q. Thus
it is pi q j . To show that the sum of each row of P Q is 1, we will now calculate the sum of the elements of the ith row of P Q, which is pi q j =
j j pi q j = pi
j qj = pi = 1. Note that
j q j = 1 and pi = 1 since the sum of the elements of the th row of Q and the sum of the elements of the ith row of P are 1. 27. If state j is accessible from state i, there is a path
i = i1 , i2 , i3 , . . . , in = j from i to j . If n K, we are done. If n > K, by the pigeonhole principle, there must exist k and (k < ) so that ik = i . Now the path i = i1 , i2 , . . . , ik , ik+1 , . . . , i , i can be reduced to i = i1 , i2 , . . . , ik , i
+1 , +1 , . . . , in = j . . . , in = j which is still a path from i to j but in fewer steps. Repeating this procedure, we can eliminate all of the states that appear more than once from the path and yet reach from i to j with a positive probability. After all such eliminations are made, we obtain a path i = i1 , im1 , im2 , . . . , in = j in which the states i1 , im1 , im2 , . . . , in are distinct states. Since there are K states altogether, this path has at most K states.
n n 28. Let I = {n 1 : pii > 0} and J = {n 1 : pjj > 0}. Then d(i), the period of i, is the greatest common divisor of the elements of I , and d(j ), the period of j , is the greatest common divisor of the elements of J . If d(i) = d(j ), then one of d(i) and d(j ) is smaller than the other one. We will prove the theorem for the case in which d(j ) < d(i). The proof for the case in which d(i) < d(j ) follows by symmetry. Suppose that for positive integers n and m, m k n pij > 0 and pj i > 0. Let k J ; then pjj > 0. We have
n+m n m pii pij pj i > 0, 310 Chapter 12 Stochastic Processes and
n+k+m n k m pij pjj pj i > 0. pii By these inequalities, we have that d(i) divides n + m and n + k + m. Hence it divides (n + k + m)  (n + m) = k. We have shown that, if k J , then d(i) divides k. This means that d(i) divides all members of J . It contradicts the facts that d(j ) is the greatest common divisor of J and d(j ) < d(i). Therefore, we must have d(i) = d(j ). 29. The stochastic process {Xn : n = 1, 2, . . . } is a Markov chain with state space {0, 1, . . . , k1}. For 0 i k  2, a transition is only possible from state i to 0 or i + 1. The only transition from k  1 is to 0. Let Z be the number of weeks it takes Liz to play again with Bob from the time they last played. The event Z > i occurs if and only if Liz has not played with Bob since i Sundays ago, and the earliest she will play with him is next Sunday. Now the probability is i/k that Liz will play with Bob if last time they played was i Sundays ago; hence i P (Z > i) = 1  , k i = 1, 2, . . . , k  1. Using this fact, for 0 i k  2, we obtain pi(i+1) = P (Xn+1 = i + 1  Xn = i) = P (Z > i + 1) = P (Z > i) 1 P (Xn = i, Xn+1 = i + 1) P (Xn = i) = i+1 k = k  i  1, i ki 1 k ki1 1 = , ki ki pi0 = P (Xn+1 = 0  Xn = i) = 1  p(k1)0 = P (Xn+1 = 0  Xn = k  1) = 1. Hence the transition probability matrix of {Xn : n = 1, 2, . . . } is given by Section 12.3 Markov Chains 311 1 k 1 k  1 1 k  2 P = 1 k  3 . . . 1 2 1 1 0 0 0 1 k 1 0 1 k1 0 0 1 0 0 1 k2 0 1 0 0 0 1 k3 ... ... ... ... 0 0 0 0 0 0 0 . 0 1 2 0 0 0 0 0 0 0 0 0 ... ... 0 0 It should be clear that the Markov chain under consideration is irreducible, aperiodic, and positively recurrent. For 0 i k  1, let i be the longrun probability that Liz says no to Bob for i consecutive weeks. 0 , 1 , . . . , k1 are obtained from solving the following matrix equation along with k1 i = 1. i=0 1 k 0 1 1  k 1 2 0 3 = 0 . . . 0 k2 . . . k1 0 1 k1 0 1 1 k1 0 0 1 1 k2 0 0 1 k2 0 1 1 k3 0 0 0 1 k3 ... ... ... ... ...
1 2 1 0 0 0 0 0 1 2 3 . . . . k2 k1 0 0 0 0 0 0 0 ... 1 2 0 The matrix equation gives i = ki 0 , k i = 1, 2, . . . , k  1. 312 Chapter 12 Stochastic Processes Using k1 i=0 i = 1, we obtain 0 k1 i=0 ki =1 k
k1 or, equivalently, 0 k This implies that
k1 k
i=0 i=0 i = 1. 0 2 (k  1)k k  = 1, k 2 which gives 0 = 2/(k + 1). Hence i = 2(k  i) , k(k + 1) i = 0, 1, 2, . . . , k  1. 30. Let Xi be the amount of money player A has after i games. Clearly, X0 = a and {Xn : n = 0, 1, . . . } is a Markov chain with state space {0, 1, . . . , a, a + 1, . . . , a + b}. For 0 i a + b, let mi = E(T  X0 = i). Let F be the event that A wins the first game. Then, for 1 i a + b  1, E(T  X0 = i) = E(T  X0 = i, F )P (F  X0 = i) + E(T  X0 = i, F c )P (F c  X0 = i). This gives 1 1 mi = (1 + mi+1 ) + (1 + mi1 ) , 2 2 or, equivalently, 2mi = 2 + mi+1 + mi1 , Now rewrite this relation as mi+1  mi = 2 + mi  mi1 , and, for 1 i a + b, let Then yi+1 = 2 + yi , and, for 1 i a + b, 1 i a + b  1, 1 i a + b  1, 1 i a + b  1. 1 i a + b  1, yi = mi  mi1 . mi = y1 + y2 + + yi . Clearly, m0 = 0, ma+b = 0, y1 = m1 , and y2 = 2 + y1 = 2 + m1 , y3 = 2 + y2 = 2 + (2 + m1 ) = 4 + m1 . . . yi = 2(i  1) + m1 , 1 i a + b. Section 12.3 Markov Chains 313 Hence, for 1 i a + b, mi = y1 + y2 + + yi = im1  2 1 + 2 + + (i  1) = im1  i(i  1) = i(m1  i + 1). This and ma+b = 0 imply that (a + b)(m1  a  b + 1) = 0, or m1 = a + b  1. Therefore, mi = i(a + b  i), and hence the desired quantity is E(T  X0 = a) = ma = ab. 31. Let q be a positive solution of the equation x = show that n 0, P (Xn = 0) q. This implies that
n i=0 i x i . Then q = i=0 i q i . We will p = lim P (Xn = 0) q. To establish that P (Xn = 0) q, we use induction. For n = 0, P (X0 = 0) = 0 q is trivially true. Suppose that P (Xn = 0) q. We have P (Xn+1 = 0) =
i=0 P (Xn+1 = 0  X1 = i)P (X1 = i). It should be clear that P (Xn+1 = 0  X1 = i) = P (Xn = 0  X0 = 1) . However, since P (X0 = 1) = 1, P (Xn = 0  X0 = 1) = P (Xn = 0). Therefore, Thus P (Xn+1 = 0) =
i=0 i P (Xn+1 = 0  X1 = i) = P (Xn = 0) . i P (Xn = 0) P (X1 = i) i=0 i q i i = q. This establishes the theorem. 314 Chapter 12 Stochastic Processes 32. Multiplying P successively, we obtain
p12 =
2 p12 3 p12 1 13 9 = 13 9 = 13 2 1 1 + , 13 13 9 1 + 13 13 1 1 + , 13 13 and in general,
n p12 = 1 13 1 13 9 13 1 n1 +
n 9 13 = n2 + + 1
n = 9 13 9 1 13 9 1 1 4 13 . n Hence the desired probability is limn p12 = 1/4. 33. We will use induction. Let n = 1; then, for 1 + j  i to be nonnegative, we must have
1+j i i  1 j . For the inequality 1 to be valid, we must have j i + 1. Therefore, 2 i  1 j i + 1. But, for j = i, 1 + j  i is not even. Therefore, if 1 + j  i is an even 1+j i nonnegative integer satisfying 1, we must have j = i  1 or j = i + 1. For 2 j = i  1, 1+i1i n+j i = =0 2 2 Hence P (X1 = i  1  X0 = i) = 1  p = showing that the relation is valid. For j = i + 1, 1+i+1i n+j i = =1 2 2 Hence P (X1 = i + 1  X0 = i) = p = and nj +i 1i1+i = = 0. 2 2 1 1 p (1  p)0 , 1 and nj +i 1i+1+i = = 1. 2 2 1 0 p (1  p)1 , 0 showing that the relation is valid in this case as well. Since, for a simple random walk, the only possible transitions from i are to states i + 1 and i  1, in all other cases P (X1 = j  X0 = i) = 0. Section 12.4 ContinuousTime Markov Chains 315 We have established the theorem for n = 1. Now suppose that it is true for n. We will show it for n + 1 by conditioning on Xn : P (Xn+1 = j  X0 = i) = P (Xn+1 = j  X0 = i, Xn = j  1)P (Xn = j  1  X0 = i) + P (Xn+1 = j  X0 = i, Xn = j + 1)P (Xn = j + 1  X0 = i) = P (Xn+1 = j  Xn = j  1)P (Xn = j  1  X0 = i) + P (Xn+1 = j  Xn = j + 1)P (Xn = j + 1  X0 = i) n (n+j 1i)/2 (1  p)(nj +1+i)/2 n+j 1i p 2 n + (1  p) n + j + 1  i p (n+j +1i)/2 (1  p)(nj 1+i)/2 2 n n (n+1+j i)/2 = (1  p)(n+1j +i)/2 n1+j i + n+1+j i p 2 2 n+1 = n + 1 + j  i p (n+1+j i)/2 (1  p)(n+1j +i)/2 . 2 =p 12.4 CONTINUOUSTIME MARKOV CHAINS 1. By ChapmanKolmogorov equations, pij (t + h)  pij (t) =
k=0 pik (h)pkj (t)  pij (t) pik (h)pkj (t) + pii (h)pij (t)  pij (t)
k=i = =
k=i pik (h)pkj (t) + pij (t) pii (h)  1 . Thus pij (t + h)  pij (t) = h k=i pik (h) 1  pii (h) pkj (t)  pij (t) . h h Letting h 0, by (12.13) and (12.14), we have pij (t) =
k=i qik pkj (t)  i pij (t). 316 Chapter 12 Stochastic Processes 2. Clearly, X(t) : t is a continuoustime Markov chain. Its balance equations are as follows:
State f 0 1 2 3 Solving these equations along with f + 0 + 1 + 2 + 3 = 1 we obtain f = 1 = 3 = 2 , ( + ) , ( + )2 +
3 Input rate to 0 f + 1 + 2 + 3 0 1 2 = = = = = = Output rate from f 0 + 0 1 + 1 2 + 2 3 . 0 = 2 = , + 2 , ( + )3 . 3. The fact that X(t) : t 0 is a continuoustime Markov chain should be clear. The balance
equations are State (0, 0) (n, 0) (0, m) Input rate to (1,0) + (0,1) (n+1,0) + (n1,0) (0,m+1) + (0,m1) = = = = Output rate from (0,0) + (0,0) (n,0) + (n,0) , (0,m) + (0,m) n1 m 1. 4. Let X(t) be the number of customers in the system at time t. Then the process X(t) : t 0 is a birth and death process with n = , n 0, and n = n, n 1. To find 0 , the probability that the system is empty, we will first calculate the sum in (12.18). We have n=1 0 1 n1 = 1 2 n n=1 n = n! n n=1 1 n! n = 1 +
n=0 1 n! n = 1 + e/ . Hence, by (12.18), 0 = 1 = e/ . / 11+e Section 12.4 ContinuousTime Markov Chains 317 By (12.17), n = n 0 (/)n e/ , = n!n n! n = 0, 1, 2, . . . . This shows that the longrun number of customers in such an M/M/ queueing system is Poisson with parameter /. The average number of customers in the system is, therefore, /. 5. Let X(t) be the number of operators busy serving customers at time t. Clearly, X(t) : t 0 is a finitestate birth and death process with state space {0, 1, . . . , c}, birth rates n = , n = 0, 1, . . . , c, and death rates n = n, n = 0, 1, . . . , c. Let 0 be the proportion of time that all operators are free. Let c be the proportion of time all of them are busy serving customers. (a) c is the desired quantity. By (12.22), 0 = 1+
n=1 1
c n! n n = 1
c n=0 1 n! n . By (12.21), c = 1 (/)c c! . 1 c n (/) n=0 n! This formula is called Erlang's loss formula. (b) We want to find the smallest c for which 1/c!
c n=0 (1/n!) 0.004. For c = 5, the left side is 0.00306748. For c = 4, it is 0.01538462. Therefore, the airline must hire at least five operators to reduce the probability of losing a call to a number less than 0.004. 6. No, it is not because it is possible for the process to enter state 0 directly from state 2. In a
birth and death process, from a state i, transitions are only possible to the states i  1 and i + 1. 7. For n 0, let Hn be the time, starting from n, until the process enters state n + 1 for the first
time. Clearly, E(H0 ) = 1/ and, by Lemma 12.2, E(Hn ) = 1 + E(Hn1 ), n 1. 318 Chapter 12 Stochastic Processes Hence E(H0 ) = 1 , 1 E(H1 ) = + 1 E(H2 ) = + 1 2 = , 2 3 = . Continuing this process, we obtain, E(Hn ) = The desired quantity is
j 1 j 1 n+1 , n 0. E(Hn ) =
n=i n=i 1 n+1 = (i + 1) + (i + 2) + + j = = 1 (1 + 2 + + j )  (1 + 2 + + i) 1 j (j + 1) i(i + 1) j (j + 1)  i(i + 1)  = . 2 2 2 8. Suppose that a birth occurs each time that an outoforder machine is repaired and begins to
operate, and a death occurs each time that a machine breaks down. The fact that X(t) : t 0 is a birth and death process with state space {0, 1, . . . , m} should be clear. The birth and death rates are n = n = k n = 0, 1, . . . , m  k (m  n) n = m  k + 1, m  k + 2, . . . , m, n n = 0, 1, . . . , m. 0 = n = n , The death rates are 0 = 0 n = + (n  1) , n 1. 9. The Birth rates are
n 1. 10. Let X(t) be the population size at time t. Then X(t) : t 0 is a birth and death process
with birth rates n = n + , n 0, and death rates n = n, n 1. For i 0, let Hi Section 12.4 ContinuousTime Markov Chains 319 be the time, starting from i, until the population size reaches i + 1 for the first time. We are interested in E(H0 ) + E(H1 ) + E(H2 ). Note that, by Lemma 12.2, E(Hi ) = Since E(H0 ) = 1/ , E(H1 ) = and E(H2 ) = 1 2 + ( + ) + 2( + ) + = . 2 + 2 + ( + ) ( + )(2 + ) ( + )(2 + ) + ( + )(2 + 2 + ) + ( + ) . ( + )(2 + ) 1 1 + + = , + + ( + ) 1 i + E(Hi1 ), i i i 1. Thus the desired quantity is E(H0 ) + E(H1 ) + E(H2 ) = 11. Let X(t) be the number of deaths in the time interval [0, t]. Since there are no births, by Remark 7.2, it should be clear that X(t) : t 0 is a Poisson process with rate as long as the population is not extinct. Therefore, for 0 < j i, pij (t) = et (t)ij . (i  j )! Clearly, p00 (t) = 1. For i > 0, j = 0, we have
i i pi0 (t) = 1 
j =1 pij (t) = 1 
j =1 et (t)ij =1 (i  j )! 1 j =i et (t)ij . (i  j )! Letting k = i  j yields
i1 pi0 (t) = 1 
k=0 et (t)k = k! k=i et (t)k . k! 12. Suppose that a birth occurs whenever a physician takes a break, and a death occurs whenever
he or she becomes available to answer patients' calls. Let X(t) be the number of physicians on break at time t. Then X(t) : t 0 is a birth and death process with state space {0, 1, 2}. Clearly, X(t) = 0 if at t both of the physicians are available to answer patients' calls, X(t) = 1 if at t only one of the physicians is available to answer patients' calls, and X(t) = 2 if at t none of the physicians is available to answer patients' calls. We have that 0 = 2, 1 = , 2 = 0, 320 Chapter 12 Stochastic Processes 0 = 0, Therefore, 0 = 2, Also, p01 = p21 = 1, Therefore, q01 = 0 p01 = 2, q12 = 1 p12 = , q02 = q20 = 0. 1 = , 1 = + , 2 = 2. 2 = 2. , + p12 = . + p02 = p20 = 0, p10 = q10 = 1 p10 = , q21 = 2 p21 = 2, Substituting these quantities in the Kolmogorov backward equations pij (t) =
k=i qik pkj (t)  i pij (t), we obtain p00 (t) = 2p10 (t)  2p00 (t) p01 (t) = 2p11 (t)  2p01 (t) p02 (t) = 2p12 (t)  2p02 (t) p10 (t) = p20 (t) + p00 (t)  ( + )p10 (t) p11 (t) = p21 (t) + p01 (t)  ( + )p11 (t) p12 (t) = p22 (t) + p02 (t)  ( + )p12 (t) p20 (t) = 2p10 (t)  2p20 (t) p21 (t) = 2p11 (t)  2p21 (t) p22 (t) = 2p12 (t)  2p22 (t). 13. Let X(t) be the number of customers in the system at time t. Then X(t) : n 0 is a birth
and death process with n = , for n 0, and n = By (12.21), for n = 1, 2, . . . c, n = for n > c, n = n n cc 0 = 0 = c! c (c)nc c! cnc n c! c
n n c n = 0, 1, . . . , c n > c. n 1 = n 0 n! n! n 0 ; 0 = cc n 0 . c! Section 12.4 ContinuousTime Markov Chains 321 Noting that c n=0 n + n=c+1 n = 1, we have 1 n! n c 0
n=0 n=c+1 + 0 cc c! n=c+1 n = 1. Since < 1, we have 0 = n = c+1 . Therefore, 1 1
c n=0 1 n! n c + c! c = n c! (1  ) c! (1  )
c n=0 n=c+1 1 n! n . + cc c+1 14. Let s, t > 0. If j < i, then pij (s + t) = 0, and i1 pik (s)pkj (t) =
k=0 k=0 pik (s)pkj (t) +
k=i pik (s)pkj (t) = 0, since pik (s) = 0 if k < i, and pkj (t) = 0 if k i > j. Therefore, for j < i, the ChapmanKolmogorov equations are valid. Now suppose that j > i. Then j pik (s)pkj (t) =
k=0 k=i j pik (s)pkj (t) es (s)ki et (t)j k (k  i)! (j  k)!
j k=i j i =0 =
k=i e(t+s) = (j  i)! = e(t+s) (j  i)! e (j  i)!
(t+s) (j  i)! (s)ki (t)j k k  i)! (j  k)! (j  i)! (s) (t)(j i) ! (j  i  )! j i (s) (t)(j i) = = (t+s) j i =0 e (s + t)j i (j  i)! where the last equality follows by Theorem 2.5, the binomial expansion. Since e(t+s) (t + s) (j  i)!
j i = pij (s + t), we have shown that the ChapmanKolmogorov equations are satisfied. 322 Chapter 12 Stochastic Processes 15. Let X(t) be the number of particles in the shower t units of time after the cosmic particle enters the earth's atmosphere. Clearly, X(t) : t 0 is a continuoustime Markov chain with state space {1, 2, . . . } and i = i, i 1. In fact, X(t) : t 0 is a pure birth process, but that fact will not help us solve this exercise. Clearly, for i 1, j 1, pij = Hence qij = 1 0 i 0 if j = i + 1 if j = i + 1. if j = i + 1 if j = i + 1. We are interested in finding p1n (t). This is the desired probability. For n = 1, p11 (t) is the probability that the cosmic particle does not collide with any air particles during the first t units of time in the earth's atmosphere. Since the time it takes the particle to collide with another particle is exponential with parameter , we have p11 (t) = et . For n 2, by the Kolmogorov's forward equation, p1n (t) =
k=n qkn p1k (t)  n p1n (t) = q(n1)n p1(n1) (t)  n p1n (t) = n1 p1(n1) (t)  n p1n (t). Therefore, p1n (t) = (n  1)p1(n1) (t)  np1n (t). For n = 2, this gives or, equivalently, p12 (t) = p11 (t)  2p12 (t) p12 (t) = et  2p12 (t). (49) Solving this first order linear differential equation with boundary condition p12 (0) = 0, we obtain p12 (t) = et (1  et ). For n = 3, by (49), or, equivalently, p13 (t) = 2p12 (t)  3p13 (t) p13 (t) = 2et (1  et )  3p13 (t). p13 (t) = et (1  et )2 . Continuing this process, and using induction, we obtain that p1n (t) = et (1  et )n1 n 1. Solving this first order linear differential equation with boundary condition p13 (0) = 0 yields Section 12.4 ContinuousTime Markov Chains 323 16. It is straightforward to see that
(i,j ) = 1
i 1 1 2 j 1 , 2 i, j 0, satisfy the following balance equations for the tandem queueing system under consideration. Hence, by Example 12.43, (i,j ) is the product of an M/M/1 system having i customers in the system, and another M/M/1 queueing system having j customers in the system. This establishes what we wanted to show.
State (0, 0) (i, 0), i 1 (0, j ), j 1 (i, j ), i, j 1 Input rate to 2 (0,1) 2 (i,1) + (i1,0) 2 (0,j +1) + 1 (1,j 1) 2 (i,j +1) + 1 (i+1,j 1) + (i1,j ) = = = = = Output rate from (0,0) (i,0) + 1 (i,0) (0,j ) + 2 (0,j ) (i,j ) + 1 (i,j ) + 2 (i,j ) . 17. Clearly, X(t) : t 0 is a birth and death process with birth rates i = i, i 0, and death rates i = i + , i > 0; 0 = 0. For some m 1, suppose that X(t) = m. Then, for infinitesimal values of h, by (12.5), the population at t +h is m+1 with probability mh+o(h), it is m  1 with probability (m + )h + o(h), and it is still m with probability 1  mh  o(h)  (m + )h  o(h) = 1  (m + m + )h + o(h). Therefore, E X(t + h)  X(t) = m = (m + 1) mh + o(h) + (m  1) (m + )h + o(h) + m 1  (m + m + )h + o(h) = m + m(  )  h + o(h). This relation implies that E X(t + h)  X(t) = X(t) + (  )X(t)  h + o(h). Equating the expected values of both sides, and noting that E E X(t + h)  X(t) we obtain E X(t + h) = E X(t) + h(  )E X(t)  h + o(h). For simplicity, let g(t) = E X(t) . We have shown that g(t + h) = g(t) + h(  )g(t)  h + o(h) = E X(t + h) , 324 Chapter 12 Stochastic Processes or, equivalently, g(t + h)  g(t) o(h) = (  )g(t)  + . h h g (t) = (  )g(t)  . As h 0, this gives If = , then g (t) =  . So g(t) =  t + c. Since g(0) = n, we must have c = n, or g(t) =  t + n. If = , to solve the first order linear differential equation, g (t) = (  )g(t)  , let f (t) = (  )g(t)  . Then 1 f (t) = f (t),  or f (t) =  . f (t) ln f (t) = (  )t + c, or f (t) = e()t+c = Ke()t , where K = ec . Thus g(t) = K ()t e + .   1  e()t .  This yields Now g(0) = n implies that K = n(  )  . Thus g(t) = E X(t) = ne()t + 18. For n 0, let En be the event that, starting from state n, eventually extinction will occur. Let
n = P (En ). Clearly, 0 = 1. We will show that n = 1, for all n. For n 1, starting from n, let Zn be the state to which the process will move. Then Zn is a discrete random variable with set of possible values {n  1, n + 1}. Conditioning on Zn yields P (En ) = P (En  Zn = n  1)P (Zn = n  1) + P (En  Zn = n + 1)P (Zn = n + 1). Hence n = n1 or, equivalently, n (n+1  n ) = n (n  n1 ), n 1. n n + n+1 , n + n n + n n 1, Section 12.4 ContinuousTime Markov Chains 325 For n 0, let yn = n+1  n . We have n yn = n yn1 , or yn = Therefore, 1 y0 1 2 1 2 y2 = y1 = y0 2 1 2 y1 = . . . yn = 1 2 n y0 . 1 2 n n 1. n yn1 , n n 1, n 1. On the other hand, by yn = n+1  n , n 0, 1 = 0 + y0 = 1 + y0 2 = 1 + y1 = 1 + y0 + y1 . . . n+1 = 1 + y0 + y1 + + yn . Hence
n n+1 = 1 + y0 +
k=1 yk
n = 1 + y0 + y0
k=1 n 1 2 k 1 2 k 1 2 k 1 2 k
n = 1 + y0 1 +
k=1 = 1 + (1  1) 1 +
k=1 n 1 2 k . 1 2 k Since 1 2 k 1 2 k = , the sequence increases without bound. For 1 2 k 1 2 k k=1 k=1 n 's to exist, this requires that 1 = 1, which in turn implies that n+1 = 1, for n 1. 326 Chapter 12 Stochastic Processes 12.5 BROWNIAN MOTION 1. (a) By the independentincrements property of Brownian motions, the desired probability
is P  1/2 < Z(10) < 1/2  Z(5) = 0 = P  1/2 < Z(10)  Z(5) < 1/2  Z(5) = 0 = P  1/2 < Z(10)  Z(5) < 1/2 . Since Z(10)  Z(5) is normal with mean 0 and variance (10  5) 2 = 45, letting Z N (0, 1), we have P  1/2 < Z(10)  Z(5) < 1/2 =P 0.5  0 0.5  0 <Z< 45 45 (0.07)  (0.07) = 0.056. P (0.07 < Z < 0.07) = (b) In Theorem 12.9, let t1 = 5, t2 = 7, z1 = 0, z2 = 1. We have E Z(6)  Z(5) = 0 and Z(7) = 1 = 0 + 1  0 (6  5) = 0.5, 75 (7  6)(6  5) = 4.5. 75 Var Z(6)  Z(5) = 0 and Z(7) = 1 = 9 2. In the subsection of 12.5, The Maximum of a Brownian Motion, we have shown that
P max X(s) u =
0st 2 0 u 1 t u0 u < 0. We will show that X(t) has the same probability distribution function. To do so, note that X(t) N (0, 2 t) and X(t)/( t) is standard normal. Thus, for u 0, P X(t) u = P  u X(t) u = P X(t) u  P X(t) < u u u =P Z P Z < t t u u u =2 =  1  1. t t t For u < 0, P X(t) u = 0. Hence max X(s) and X(t) are identically distributed.
0st Section 12.5 Brownian Motion 327 3. Let Z N (0, 1). Since X(t) N (0, 2 t), we have
P X(t) > = P X(t) > t t = P X(t) > t + P X(t) < t t t t =P Z> = 1  t/ = 1  t/ =P Z> This implies that
t0 t +P Z < t t +P Z < +  t/ + 1  t/ = 2  2 t/ . lim P X(t) > = 2  1 = 1. t X(t) > = 2  2 = 0, t whereas
t lim P 4. Let F be the probability distribution function of 1/Y 2 . Let Z N (0, 1). We have F (t) = P 1/Y 2 t = P Y 2 1/t = P Y 1/ t + P Y 1/ t =P Z +P Z  t t  =2 1 =1 + , t t t which, by (12.35), is also the distribution function of T . 5. Clearly, P (T < x) = 0 if x t. For x > t, by Theorem 12.10,
P (T < x) = P at least one zero in (t, x) = 2 arccos t . x Let F be the distribution function of T . We have shown that 0 xt F (x) = 2 arccos t x t. x 6. Rewrite X(t1 ) + X(t2 ) as X(t1 ) + X(t2 ) = 2X(t1 ) + X(t2 )  X(t1 ). Now 2X(t1 ) and X(t2 ) 
X(t1 ) are independent random variables. By Theorem 11.7, 2X(t1 ) N (0, 4 2 t1 ). Since X(t2 )  X(t1 ) N 0, 2 (t2  t1 ) , applying Theorem 11.7 once more implies that 2X(t1 ) + X(t2 )  X(t1 ) N 0, 4 2 t1 + 2 (t2  t1 ) . 328 Chapter 12 Stochastic Processes Hence X(t1 ) + X(t2 ) N (0, 3 2 t1 + 2 t2 ). 7. Let f (x, y) be the joint probability density function of X(t) and X(t +u). Let fX(t+u)X(t) (ya) be the conditional probability density function of X(t + u) given that X(t) = a. Let fX(t) (x) be the probability density function of X(t). We know that X(t) is normal with mean 0 and variance 2 t. The formula for f (x, y) is given by (12.28). Using these, we obtain 1 1 a 2 (y  a)2 + exp  2 2 t u 2 2 tu 1 a2 exp  2 2 t 2 t fX(t+u)X(t) (ya) = f (a, y) = fX(t) (a) = 1 1 (y  a)2 . exp  2 2 u 2 u This shows that the conditional probability density function of X(t + u) given that X(t) = a is normal with mean a and variance 2 u. Hence E X(t + u)  X(t) = a = a. This implies that E X(t + u)  X(t) = X(t). 8. By Example 10.23,
E X(t)X(t + u)  X(t) = X(t)E X(t + u)  X(t) . By Exercise 7 above, E X(t + u)  X(t) = X(t). Hence E X(t)X(t + u) = E E X(t)X(t + u)  X(t) = E X(t)E X(t + u)  X(t) = E X(t) X(t) = E X(t)2 = Var X(t) + E X(t)
2 = 2 t + 0 = 2 t. 9. For t > 0, the probability density function of Z(t) is
t (x) = 1 x2 . exp  2 2 t 2 t Section 12.5 Brownian Motion 329 Therefore, E V (t) = E Z(t) = =2
0  xt (x) dx 0 xt (x) dx = 2 x yields t ueu
2 /2 x 2 2 ex /(2 t) dx. 2 t Making the change of variable u = 2t 0 E V (t) = du = 2t 2  eu /2 0 = 2t . Var V (t) = E V (t)2  E V (t) = 2t  since E Z(t)2 = Var Z(t) + E Z(t) 2 = E Z(t)2  2 2 t 2 2 t 2 = 2t 1  , 2 = 2 t + 0 = 2 t. To find P V (t) z  V (0) = z0 , note that, by (12.27), P V (t) z  V (0) = z0 = P Z(t) z  V (0) = z0 = P  z Z(t) z  V (0) = z0 =
z z 1 2 2 e(uz0 ) /(2 t) du. 2 t Letting U N (z0 , 2 t) and Z N (0, 1), this implies that P V (t) z  V (0) = z0 = P (z U z) =P = = z  z0 z  z0 z t t z  z0  t z + z0 + t z  z0 t z  z0  1. t 10. Clearly, D(t) = X(t)2 + Y (t)2 + Z(t)2 . Since X(t), Y (t), and Z(t) are independent and 330 Chapter 12 Stochastic Processes identically distributed normal random variables with mean 0 and variance 2 t, we have E D(t) =    1 1 2 2 2 2 x 2 + y 2 + z2 ex /(2 t) ey /(2 t) 2 t 2 t 1 2 2 ez /(2 t) dx dy dz 2 t = 1 3 t 2 t 2    x 2 + y 2 + z2 e(x 2 +y 2 +z2 )/(2 2 t) dx dy dz. We now make a change of variables to spherical coordinates: x = sin cos , y = sin sin , z = cos , 2 = x 2 + y 2 + z2 , dx dy dz = 2 sin d d d, 0 < , 0 , and 0 2 . We obtain E D(t) = = = = 1 3 t 2 t 2 1 3 t 2 t 2 1 3 t 2 t 2 2 0 2 0 2 0 0 2 0 0 0 0 0 0 e 2 /(2 2 t) 2 sin d d, d d sin d d
2 /(2 2 t) 3 e 2 /(2 2 t)  2 t ( 2 + 2 2 t)e 0 sin d d 1 2 4 t 2 3 t 2 t 2 sin d d = 2 2t . 11. Noting that 5.29 = 2.3, we have V (t) = 95e2t+2.3W (t) , where W (t) : t 0 is a standard Brownian motion. Hence W (t) N (0, t). The desired probability is P V (0.75) < 80 = P 95e2(0.75)+2.3W (0.75) < 80 = P e2.3W (0.75) < 3.774 = P W (0.75) < 0.577 =P W (0.75)  0 0.577 < = P (Z < 0.67) = 0.75 0.75 (0.67) = 0.7486. Chapter 12 Review Problems 331 REVIEW PROBLEMS FOR CHAPTER 12 1. Label the time point 10:00 as t = 0. We are given that N (180) = 10 and are interested in P S10 160  N (180) = 10 . Let X1 , X2 , . . . , X10 be 10 independent random variables uniformly distributed over the interval [0, 180]. Let Y = max(X1 , . . . , X10 ). By Theorem 12.4, P S10 > 160  N (180) = 10 = P (Y > 160) = 1  P (Y 160) = 1  P max(X1 , . . . , X10 ) 160 = 1  P (X1 160)P (X2 160) P (X10 160) =1 160 180
10 = 0.692. 2. For all positive integer n, we have that
P 2n = 1 0 0 1 and P 2n+1 = 0 1 . 1 0 Therefore, {Xn : n = 0, 1, . . . } is not regular. 3. By drawing a transition graph, it can be readily seen that, if states 0, 1, 2, 3, and 4 are renamed
0, 4, 2, 1, and 3, respectively, then the transition probability matrix P 1 will change to P 2 . 4. Let Z be the number of transitions until the first visit to 1. Clearly, Z is a geometric random
variable with parameter p = 3/5. Hence its expected value is 1/p = 5/3. classes {3, 5} and {4}, and two transient classes {1} and {2}. 5. By drawing a transition graph, it is readily seen that this Markov chain consists of two recurrent 6. We have that
Xn+1 = Xn 1 + Xn if the (n + 1)st outcome is not 6 if the (n + 1)st outcome is 6. This shows that {Xn : n = 1, 2, . . . } is a Markov chain with state space {0, 1, 2, . . . }. Its transition probability matrix is given by 5/6 1/6 0 0 0 ... 0 5/6 1/6 0 0 . . . 0 5/6 1/6 0 . . . . P = 0 0 0 0 5/6 1/6 . . . . . . All states are transient; no two states communicate with each other. Therefore, we have infinitely many classes; namely, {0}, {1}, {2}, . . . , and each one of them is transient. 332 Chapter 12 Stochastic Processes 7. The desired probability is
p11 p11 + p11 p12 + p12 p22 + p12 p21 + p21 p11 + p21 p12 + p22 p21 + p22 p22 = (0.20)2 + (0.20)(0.30) + (0.30)(0.15) + (0.30)(0.32) + (0.32)(0.20) + (0.32)(0.30) + (0.15)(0.32) + (0.15)2 = 0.4715. 8. The following is an example of such a transition probability matrix: 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1/2 0 0 1/2 0 0 0 . P = 0 0 0 0 1/3 2/3 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 9. For n 1, let
Xn = 1 0 if the nth golfball produced is defective if the nth golfball produced is good. Then {Xn : n = 1, 2, . . . } is a Markov chain with state space {0, 1} and transition probability 15/18 3/18 matrix 11/12 1/12 . Let 0 be the fraction of golfballs produced that are good, and 1 be the fraction of the balls produced that are defective. Then, by Theorem 12.7, 0 and 1 satisfy 0 1 = 15/18 11/12 3/18 1/12 0 , 1 which gives us the following system of equations 0 = (15/18)0 + (11/12)1 = (3/18) + (1/12) . 1 0 1 By choosing any one of these equations along with the relation 0 + 1 = 1, we obtain a system of two equations in two unknowns. Solving that system yields 0 = 11 0.85 13 and 1 = 2 0.15. 13 Therefore, approximately 15% of the golfballs produced have no logos. 10. Let 1 Xn = 2 3 if the nth ball is drawn by Carmela if the nth ball is drawn by Daniela if the nth ball is drawn by Lucrezia. Chapter 12 Review Problems 333 The process {Xn : n = 1, 2, . . . } is an irreducible, aperiodic, positive recurrent Markov chain with transition probability matrix 7/31 11/31 13/31 P = 7/31 11/31 13/31 . 7/31 11/31 13/31 Let 1 , 2 , and 3 be the longrun proportion of balls drawn by Carmela, Daniela, and Lucrezia, respectively. Intuitively, it should be clear that these quantities are 7/31, 11/31, and 13/31, respectively. However, that can be seen also by solving the following matrix equation along with 0 + 1 + 3 = 1. 1 7/31 7/31 7/31 1 2 = 11/31 11/31 11/31 2 . 13/31 13/31 13/31 3 3 11. Let 1 and 2 be the longrun probabilities that Francesco devotes to playing golf and playing tennis, respectively. Then, by Theorem 12.7, 1 and 2 are obtained from solving the system of equations 1 0.30 0.58 1 = 0.70 0.42 2 2 1 = 0.301 + 0.582 2 = 0.701 + 0.422 . along with 1 + 2 = 1. The matrix equation above gives the following system of equations: By choosing any one of these equations along with the relation 1 + 2 = 1, we obtain a system of two equations in two unknowns. Solving that system yields 1 = 0.453125 and 2 = 0.546875. Therefore, the longrun probability that, on a randomly selected day, Francesco plays tennis is approximately 0.55. 12. Suppose that a train leaves the station at t = 0. Let X1 be the time until the first passenger arrives at the station after t = 0. Let X2 be the additional time it will take until a train arrives at the station, X3 be the time after that until a passenger arrives, and so on. Clearly, X1 , X2 , . . . are the times between consecutive change of states. By the memoryless property of exponential random variables, {X1 , X2 , . . . } is a sequence of independent and identically distributed exponential random variables with mean 1/. Hence, by Remark 7.2, N (t) : t 0 is a Poisson process with rate . Therefore, N (t) is a Poisson random variable with parameter t. continuoustime Markov chain with state space {0, 1, 2}. Let 0 , 1 , and 2 be the longrun proportion of time the process is in states 0, 1, and 2, respectively. The balance equations for X(t) : t 0 are as follows: 13. Let X(t) be the number of components working at time t. Clearly, X(t) : t 0 is a 334 Chapter 12 Stochastic Processes State 0 1 2 Input rate to 1 22 + 0 1 = = = = Output rate from 0 1 + 1 22 From these equations, we obtain 1 = 2 0 and 2 = 2 0 . Using 0 + 1 + 2 = 1 yields 2 2 2 0 = 2 . 2 + 2 + 2 (2 + ) . 22 + 2 + 2 Hence the desired probability is 1  0 = 14. Suppose that every time an outoforder machine is repaired and is ready to operate a birth
occurs. Suppose that a death occurs every time that a machine breaks down. The fact that X(t) : t 0 is a birth and death process should be clear. The birth and death rates are k n = 0, 1, . . . , m + s  k n = (m + s  n) n = m + s  k + 1, m + s  k + 2, . . . , m + s 0 n m + s; n n = 0, 1, . . . , m n = m n = m + 1, m + 2, . . . , m + s 0 n > m + s. 15. Let X(t) be the number of machines operating at time t. For 0 i m, let i be the longrun
proportion of time that there are exactly i machines operating. Suppose that a birth occurs each time that an outoforder machine is repaired and begins to operate, and a death occurs each time that a machine breaks down. Then X(t) : t 0 is a birth and death process with state space {0, 1, . . . , m}, and birth and death rates, respectively, given by i = (m  i) and i = i for i = 0, 1, . . . , m. To find 0 , first we will calculate the following sum:
m i=1 0 1 i1 = 1 2 i = m i=1 m i=1 (m) (m  1) (m  2) (m  i + 1) (2)(3) (i)
m Pi i = i! i
m m i=1 m i i = 1 +
i=0 m i i mi 1 = 1 + 1 + m , Chapter 12 Review Problems 335 where m Pi is the number of ielement permutations of a set containing m objects. Hence, by (12.22), m m + m 0 = 1 + = = . + By (12.21), i = = =
i 0 1 i1 m Pi 0 = 0 1 2 i i! i m i m i i +
i m = m i mi i + i + mi + 1 + , 0 i m. Therefore, in steadystate, the number of machines that are operating is binomial with parameters m and /( + ). 16. Let X(t) be the number of cars at the center, either being inspected or waiting to be inspected,
at time t. Clearly, X(t) : t 0 is a birth and death process with rates n = /(n + 1), n 0, and n = , n 1. Since n=1 0 1 n1 = 1 2 n n=1 1 2 3 n = 1 + n n! n=0 n = e/  1. By (12.18), 0 = e/ . Hence, by (12.17), 2 3 n / (/)n e/ , e = n n! n = n 0. Therefore, the longrun probability that there are n cars at the center for inspection is Poisson with rate /. 17. Let X(t) be the population size at time t. Then X(t) : t 0 is a birth and death process with birth rates n = n, n 1, and death rates n = n, n 0. For i 0, let Hi be the time, starting from i, until the population size reaches i + 1 for the first time. We are interested in 4 i=1 E(Hi ). Note that, by Lemma 12.2, E(Hi ) = Since E(H0 ) = 1/, E(H1 ) = 1 1 1 + = + 2, 1 i + E(Hi1 ), i i i 1. 336 Chapter 12 Stochastic Processes E(H2 ) = E(H3 ) = E(H4 ) = 1 2 2 1 1 + + 2 = + 2 + 3, 2 2 2 3 1 1 2 2 3 1 + + 2+ 3 = + 2 + 3 + 4, 3 3 2 3 2 1 4 1 2 3 1 2 3 4 + + 2+ 3 + 4 = + 2 + 3 + 4 + 5. 4 4 3 2 4 3 2 Therefore, the answer is
4 E(Hi ) =
i=1 254 + 343 + 302 2 + 243 + 124 . 125 18. Let X(t) be the population size at time t. Then X(t) : t 0 is a birth and death process with rates n = , n 0, and n = n, n 1. To find i 's, we will first calculate the sum in the relation (12.18): n=1 0 1 n1 = 1 2 n n=1 n 1 = 1 + n n! n! n=0 n = 1 + e / . Thus, by (12.18), 0 = e / and, by (12.17), for i 1, i = n  / ( /)n e / e = . n! n n! Hence the steadystate probability mass function of the population size is Poisson with parameter /. 19. By applying Theorem 12.9 to Y (t) : t 0 with t1 = 0, t2 = t, y1 = 0, y2 = y, and t = s,
we have E Y (s)  Y (t) = y = 0 + and Var Y (s)  Y (t) = y = 2 s y0 (s  0) = y, t 0 t (t  s)(s  0) s = 2 (t  s) . t 0 t 20. First, suppose that s < t. By Example 10.23,
E X(s)X(t)  X(s) = X(s)E X(t)  X(s) . Now, by Exercise 7, Section 12.5, E X(t)  X(s) = X(s). Chapter 12 Review Problems 337 Hence E X(s)X(t) = E E X(s)X(t)  X(s) = E X(s)E X(t)  X(s) = E X(s)X(s) = E X(s)2 = Var X(s) + E X(s) = 2 s + 0 = 2 s. For t < s, by symmetry, Therefore, E X(s)X(t) = 2 min(s, t). E X(s)X(t) = 2 t.
2 21. By Theorem 12.10,
P (U < x and T > y) = P no zeros in (x, y) = 1  2 arccos x . y 22. Let the current price of the stock, per share, be v0 . Noting that
V (t) = v0 e3t+5.2W (t) , 27.04 = 5.2, we have where W (t) : t 0 is a standard Brownian motion. Hence W (t) N (0, t). The desired probability is calculated as follows: P V (2) 2v0 = P v0 e6+5.2W (2) 2v0 = P 6 + 5.2W (2) ln 2 = P W (2) 1.02 =P W (2)  0 0.72 2 (0.72) = 0.7642. = P (Z 0.72) = 1  P (Z < 0.72) =1 ...
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This homework help was uploaded on 04/20/2008 for the course MATH 3355 taught by Professor Britt during the Spring '08 term at LSU.
 Spring '08
 Britt
 Math, Probability

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