# Fundamentals of Probability with Stochastic Processes 3E - Ghahramani.pdf

## Mathematics 214 with Morse at George Mason University *

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n+ m) P(X>m) = P(X > n). Therefore, P(X>n+ m) = P(X > n)P(X > m). (19) 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) i−1 ,i≤ n. (20) We will show that (18) is true for n + 1. The induction hypothesis [relation (20)] implies that P(X≤ n) = n summationdisplay i=1 P(X= i) = n summationdisplay i=1 p(1 − p) i−1 = p 1 − (1 − p) n 1 − (1 − p) = 1 − (1 − p) n . So P(X>n)= (1 − p) n and, similarly, P(X>n− 1) = (1 − p) n−1 . 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 X 1 be the number of flips until the first head, X 2 be the total number of flips until the second head, X 3 be the total number of flips until the third head, and so on. Then the length of the first character of the message and X 1 are identically distributed. The total number of the bits forming the first two characters of the message and X 2 are identically distributed. The total number of the bits forming the first three characters of the message and X 3 are identically distributed, and so on. Therefore, the total number of the bits forming the message has the same distribution as X k . 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 obvious that a geometric random variable with parameter p = parenleftbigg 1000 12 parenrightbigg slashBig parenleftbigg 1200 12 parenrightbigg = 0.1109 is 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 Nth success should occur on the (2N − M)th trial or the Nth failure should occur on the (2N − M)th trial. By symmetry, the answer is 2 · parenleftbigg 2N − M − 1 N − 1 parenrightbigg parenleftBig 1 2 parenrightBig N parenleftBig 1 2 parenrightBig N−M = parenleftbigg 2N − M − 1 N − 1 parenrightbigg parenleftBig 1 2 parenrightBig 2N−M−1 . 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 parenleftbigg 2N − 1 N parenrightbigg parenleftBig 1 2 parenrightBig N parenleftBig 1 − 1 2 parenrightBig (2N−1)−N · parenleftBig 1 − 1 2 parenrightBig 2 = parenleftbigg 2N − 1 N parenrightbigg parenleftBig 1 2 parenrightBig 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 parenleftbig Y>X parenrightbig = ∞ summationdisplay i=1 P parenleftbig Y>X| X = i parenrightbig P parenleftbig X = i parenrightbig = ∞ summationdisplay i=1 P parenleftbig Y>i parenrightbig P parenleftbig X = i parenrightbig = ∞ summationdisplay i=1 parenleftBig 1 2 parenrightBig i · parenleftBig 5 6 parenrightBig i−1 1 6 = 6 5 · 1 6 ∞ summationdisplay i=1 parenleftBig 5 12 parenrightBig i = 1 5 · 5 12 1 − 5 12 = 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 p n = parenleftbigg 50 4 parenrightbiggparenleftbigg n − 50 46 parenrightbigg parenleftbigg n 50 parenrightbigg . It is logical to find the value of n for which p n 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 p n p n−1 = (n − 50) 2 n(n − 96) . Section 5.3 Other Discrete Random Variables 105 Now p n ≥ p n−1 if and only if (n − 50) 2 ≥ n(n − 96),orn ≤ 625. Therefore, n = 625 makes p n 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 E j be the event of obtaining exactly j defective items among the first (k − 1) draws. Let A k be the event that the kth item drawn is defective. We have P(A k ) = k−1 summationdisplay j=0 P(A k | E j )P(E j ) = k−1 summationdisplay j=0 D − j N − k + 1 · parenleftbigg D j parenrightbiggparenleftbigg N − D k − 1 − j parenrightbigg parenleftbigg N k − 1 parenrightbigg . Now (D − j) parenleftbigg D j parenrightbigg = D parenleftbigg D − 1 j parenrightbigg and (N − k + 1) parenleftbigg N k − 1 parenrightbigg = N parenleftbigg N − 1 k − 1 parenrightbigg . Therefore, P(A k ) = k−1 summationdisplay j=0 D parenleftbigg D − 1 j parenrightbiggparenleftbigg N − D k − 1 − j parenrightbigg N parenleftbigg N − 1 k − 1 parenrightbigg = D N k−1 summationdisplay j=0 parenleftbigg D − 1 j parenrightbiggparenleftbigg N − D k − 1 − j parenrightbigg parenleftbigg N − 1 k − 1 parenrightbigg = D N , where k−1 summationdisplay j=0 parenleftbigg D − 1 j parenrightbiggparenleftbigg N − D k − 1 − j parenrightbigg parenleftbigg N − 1 k − 1 parenrightbigg = 1 since parenleftbigg D − 1 j parenrightbiggparenleftbigg N − D k − 1 − j parenrightbigg parenleftbigg N − 1 k − 1 parenrightbigg is the probability mass function of a hypergeometric random 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 A k be as before and let F j 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 P(B)= k−2 summationdisplay j=0 P(B | F j )P(F j ) 106 Chapter 5 Special Discrete Distributions = k−2 summationdisplay j=0 (D − j)(D − j − 1) (N − k + 2)(N − k + 1) · parenleftbigg D j parenrightbiggparenleftbigg N − D k − 2 − j parenrightbigg parenleftbigg N k − 2 parenrightbigg = k−2 summationdisplay j=0 D(D − 1) parenleftbigg D − 2 j parenrightbiggparenleftbigg N − D k − 2 − j parenrightbigg N(N − 1) parenleftbigg N − 2 k − 2 parenrightbigg = D(D − 1) N(N − 1) k−2 summationdisplay j=0 parenleftbigg D − 2 j parenrightbiggparenleftbigg N − D k − 2 − j parenrightbigg parenleftbigg N − 2 k − 2 parenrightbigg = D(D − 1) N(N − 1) . Using this, we have that the desired probability is P(A k | A k−1 ) = P(A k A k−1 ) P(A k−1 ) = P(B) P(A k−1 ) = D(D − 1) N(N − 1) D N = D − 1 N − 1 . REVIEW PROBLEMS FOR CHAPTER 5 1. 20 summationdisplay i=12 parenleftbigg 20 i parenrightbigg (0.25) i (0.75) 20−i = 0.0009. 2. N(t), the number of customers arriving at the post office at or prior to t is a Poisson process with λ = 1/3. Thus P parenleftbig N(30) ≤ 6 parenrightbig = 6 summationdisplay i=0 P parenleftbig N(30) = i parenrightbig = 6 summationdisplay i=0 e −(1/3)30 bracketleftbig (1/3)30 bracketrightbig i i! = 0.130141. 3. 4 · 8 30 = 1.067. 4. 2 summationdisplay i=0 parenleftbigg 12 i parenrightbigg (0.30) i (0.70) 12−i = 0.253. Chapter 5 Review Problems 107 5. parenleftbigg 5 2 parenrightbigg (0.18) 2 (0.82) 3 = 0.179. 6. 1999 summationdisplay i=2 parenleftbigg i − 1 2 − 1 parenrightbigg parenleftBig 1 1000 parenrightBig 2 parenleftBig 999 1000 parenrightBig i−2 = 0.59386. 7. 12 summationdisplay i=7 parenleftbigg 160 i parenrightbiggparenleftbigg 200 12 − i parenrightbigg parenleftbigg 360 12 parenrightbigg = 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 60 = 13 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 λ = 6. The desired probability is P(X≤ 4) = 4 summationdisplay i=0 e −6 6 i i! = 0.285. 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 parenleftbigg 25 2 parenrightbiggparenleftbigg 475 28 parenrightbigg parenleftbigg 500 30 parenrightbigg = 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 = bracketleftbig 1 − (1/2) bracketrightbig /(1/4) = 2. 12. parenleftBig 5 6 parenrightBig 5 parenleftBig 1 6 parenrightBig = 0.067. 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 1 − p − 1 = p 1 − p . 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 parenleftbig F | E parenrightbig = (0.30)(0.85) = 0.255. The random variable X is binomial with parameters 6 and 0.255. Hence P parenleftbig X = i parenrightbig = parenleftbigg 6 i parenrightbigg parenleftbig 0.255 parenrightbig i parenleftbig 1 − 0.255 parenrightbig 6−i ,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. 15. 5 summationdisplay i=3 parenleftbigg 18 i parenrightbiggparenleftbigg 10 5 − i parenrightbigg parenleftbigg 28 5 parenrightbigg = 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 summationdisplay i=0 parenleftbigg 40 i parenrightbigg (0.06) i (0.94) 40−i = 0.91. 18. 1 − 2 summationdisplay i=0 parenleftbigg 20 i parenrightbigg (0.06) i (0.94) 20−i = 0.115. 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 P(X≥ 24) = 1 − 23 summationdisplay i=0 P(X= i) = 1 − 23 summationdisplay i=0 e −18 (18) i i! = 1 − 0.89889 = 0.10111. Chapter 5 Review Problems 109 19. Suppose that the company’s claim is correct. Then the probability of 12 or less drivers using seat belts regularly is 12 summationdisplay i=0 parenleftbigg 20 i parenrightbigg (0.70) i (0.30) 20−i ≈ 0.228. 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. 20. (a) (0.999) 999 (0.001) 1 = 0.000368. (b) parenleftbigg 2999 2 parenrightbigg (0.001) 3 (0.999) 2997 = 0.000224. 21. Let X be the number of children having the disease. We have that the desired probability is P(X= 3 | X ≥ 1) = P(X= 3) P(X≥ 1) = parenleftbigg 5 3 parenrightbigg (0.23) 3 (0.77) 2 1 − (0.77) 5 = 0.0989. 22. (a) parenleftBig w w + b parenrightBig n−1 parenleftBig b w + b parenrightBig . (b) parenleftBig w w + b parenrightBig n−1 . 23. 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 summationdisplay i=0 parenleftbigg n i parenrightbigg (0.75) i (.25) n−i ≤ 0.10. 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 summationtext 4 i=0 parenleftbig n i parenrightbig (0.75) i (.25) n−i 5 0.7627 6 0.4661 7 0.2436 8 0.1139 9 0.0489 110 Chapter 5 Special Discrete Distributions 24. 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(B | A) = P(AB) P(A) = p · parenleftbigg n − 1 k − 1 parenrightbigg p k−1 (1 − p) n−k parenleftbigg n k parenrightbigg p k (1 − p) n−k = parenleftbigg n − 1 k − 1 parenrightbigg parenleftbigg n k parenrightbigg = k n . 25. 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 λ is the average number of seeds per unit area, the expected number of seeds in the area, A,isλA. 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 = λA and p = λA/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 λA and P(X= i) = e −λA (λA) i i! . 26. Let D/N → p, then by the Remark 5.2, for all n, parenleftbigg D x parenrightbiggparenleftbigg N − D n − x parenrightbigg parenleftbigg N n parenrightbigg ≈ parenleftbigg n x parenrightbigg p x (1 − p) n−x . Now since n →∞and nD/N → λ, n is large and np is appreciable, thus parenleftbigg n x parenrightbigg p x (1 − p) n−x ≈ e −λ λ x x! . Chapter 6 Continuous Random Variables 6.1 PROBABILITY DENSITY FUNCTIONS 1. (a) integraldisplay ∞ 0 ce −3x dx = 1 equal1⇒ c = 3. (b) P(0

**0, or, equivalently, B 2 > 4C. Let E = braceleftbig (b, c): 1**

**4c bracerightbig ; 172 Chapter 8 Bivariate Distributions the desired probability is integraldisplayintegraldisplay E 9b 2 c 2 676 dbdc = integraldisplay 3 2 parenleftBig integraldisplay b 2 /4 1 9b 2 c 2 676 dc parenrightBig db ≈ 0.12. (Draw a figure to verify the region of integration.) 23. Note that f X (x) = integraldisplay ∞ −∞ g(x)h(y) dy = g(x) integraldisplay ∞ −∞ h(y) dy, f Y (y) = integraldisplay ∞ −∞ g(x)h(y) dx = h(y) integraldisplay ∞ −∞ g(x)dx. Now f X (x)f Y (y) = g(x)h(y) integraldisplay ∞ −∞ h(y) dy integraldisplay ∞ −∞ g(x)dx = f(x,y) integraldisplay ∞ −∞ integraldisplay ∞ −∞ h(y)g(x) dy dx = f(x,y) integraldisplay ∞ −∞ integraldisplay ∞ −∞ 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) slashbig min(X, Y). Then G(t) = 0ift<1. For t ≥ 1, G(t) = P parenleftBig max(X, Y) min(X, Y) ≤ t parenrightBig = P parenleftBig max(X, Y) ≤ t min(X, Y) parenrightBig = P parenleftBig X ≤ t min(X, Y), Y ≤ t min(X, Y) parenrightBig = P parenleftBig min(X, Y) ≥ X t , min(X, Y) ≥ Y t parenrightBig = P parenleftBig X ≥ X t ,Y≥ X t ,X≥ Y t ,Y≥ Y t parenrightBig = P parenleftBig Y ≥ X t ,X≥ Y t parenrightBig = P parenleftBig X t ≤ Y ≤ tX parenrightBig . This quantity is the area of the region braceleftbig (x, y): 0**|y|, f Y|X (y|x) = (1/2)e −x integraldisplay x −x (1/2)e −x dy = 1 2x , −x5. Using these, we have that E(X | Y = 5) = ∞ summationdisplay x=1 xp X|Y (x|5) = ∞ summationdisplay x=1 x p(x,5) p Y (5) = 4 summationdisplay x=1 1 11 x parenleftBig 11 12 parenrightBig x + ∞ summationdisplay x=6 x parenleftBig 11 12 parenrightBig 4 parenleftBig 1 13 parenrightBigparenleftBig 12 13 parenrightBig x−6 = 0.72932 + parenleftBig 11 12 parenrightBig 4 parenleftBig 1 13 parenrightBig ∞ summationdisplay y=0 (y + 6) parenleftBig 12 13 parenrightBig y = 0.72932 + parenleftBig 11 12 parenrightBig 4 parenleftBig 1 13 parenrightBigbracketleftBig ∞ summationdisplay y=0 y parenleftBig 12 13 parenrightBig y + 6 ∞ summationdisplay y=0 parenleftBig 12 13 parenrightBig y bracketrightBig = 0.702932 + parenleftBig 11 12 parenrightBig 4 parenleftBig 1 13 parenrightBigbracketleftBig 12/13 (1/13) 2 + 6 1 1 − (12/13) bracketrightBig = 13.412. 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 E(X | Y = 10) = 9 summationdisplay x=0 x p(x,10) p Y (10) = 9 summationdisplay x=1 x parenleftbigg 9 x parenrightbigg parenleftBig 12 22 parenrightBig x parenleftBig 10 22 parenrightBig 9−x · parenleftbigg 9 10 − x parenrightbigg parenleftBig 12 22 parenrightBig 10−x parenleftBig 10 22 parenrightBig x−1 parenleftbigg 18 10 parenrightbigg parenleftBig 12 22 parenrightBig 10 parenleftBig 10 22 parenrightBig 8 = 9 summationdisplay x=1 x parenleftbigg 9 x parenrightbiggparenleftbigg 9 10 − x parenrightbigg parenleftbigg 18 10 parenrightbigg = 9 × 10 18 = 5, 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, f X (x) = integraldisplay 1 x n(n − 1)(y − x) n−2 dy = n(1 − x) n−1 . Thus f Y|X (y|x) = f(x,y) f X (x) = n(n − 1)(y − x) n−2 n(1 − x) n−1 = (n − 1)(y − x) n−2 (1 − x) n−1 . Therefore, E(Y | X = x) = integraldisplay 1 x y (n − 1)(y − x) n−2 (1 − x) n−1 dy = n − 1 (1 − x) n−1 integraldisplay 1 x y(y − x) n−2 dy. But integraldisplay 1 x y(y − x) n−2 dy = integraldisplay 1 x (y − x + x)(y − x) n−2 dy = integraldisplay 1 x (y − x) n−1 dy + integraldisplay 1 x x(y − x) n−2 dy = (1 − x) n n + x(1 − x) n−1 n − 1 . Thus E(Y | X = x) = n − 1 n (1 − x)+ x = n − 1 n + 1 n x. 182 Chapter 8 Bivariate Distributions 19. (a) The area of the triangle is 1/2. So f(x,y) = braceleftBigg 2ifx ≥ 0, y ≥ 0, x + y ≤ 1 0 elsewhere. (b) f Y (y) = integraldisplay 1−y 0 2 dx = 2(1 − y), 0 0,v>0 bracerightbig . It has the unique solution x = e −u/2 , y = e −v/2 . Hence J = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle − 1 2 e −u/2 0 0 − 1 2 e −v/2 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 1 4 e −(u+v)/2 negationslash= 0. By Theorem 8.8, g(u, v), the joint probability density function of U and V is g(u, v) = f parenleftbig e −u/2 ,e −v/2 parenrightbig vextendsingle vextendsingle vextendsingle 1 4 e −(u+v)/2 vextendsingle vextendsingle vextendsingle = 1 4 e −(u+v)/2 ,u>0,v>0. 184 Chapter 8 Bivariate Distributions 2. Let f(x,y) be the joint probability density function of X and Y. Clearly, f(x,y) = f 1 (x)f 2 (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 g U (u), its marginal density function. The system of two equations in two unknowns braceleftBigg x/y = u x = v defines a one-to-one transformation of R = braceleftbig (x, y): x>0,y>0 bracerightbig onto the region Q = braceleftbig (u, v): u>0,v>0 bracerightbig . It has the unique solution x = v, y = v/u. Hence J = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 01 − v u 2 1 u vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = v u 2 negationslash= 0. By Theorem 8.8, g(u, v) = f parenleftBig v, v u parenrightBigvextendsingle vextendsingle vextendsingle v u 2 vextendsingle vextendsingle vextendsingle = v u 2 f parenleftBig v, v u parenrightBig = v u 2 f 1 (v)f 2 parenleftBig v u parenrightBig u>0,v>0. Therefore, g U (u) = integraldisplay ∞ 0 v u 2 f 1 (v)f 2 parenleftBig v u parenrightBig dv, u > 0. 3. Let g(r,θ) be the joint probability density function of R and Theta1. We will show that g(r,θ) = g R (r)g Theta1 (θ). This proves the surprising result that R and Theta1 are independent. Let f(x,y) be the joint probability density function of X and Y. Clearly, f(x,y) = 1 2π e −(x 2 +y 2 )/2 , −∞ 0, 0 <θ<2π bracerightbig . It has the unique solution braceleftBigg x = r cos θ y = r sin θ. Hence J = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle cos θ −r sin θ sin θrcos θ vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = r negationslash= 0. By Therorem 8.8, g(r,θ) is given by g(r,θ) = f(rcos θ,r sin θ)|r|= 1 2π re −r 2 /2 0 <θ<2π, r > 0. Now g R (r) = integraldisplay 2π 0 1 2π re −r 2 /2 dθ = re −r 2 /2 ,r>0, and g Theta1 (θ) = integraldisplay ∞ 0 1 2π re −r 2 /2 dr = 1 2π , 0 <θ<2π. Therefore, g(r,θ) = g R (r)g Theta1 (θ), showing that R and Theta1 are independent random variables. The formula for g Theta1 (θ) indicates that Theta1 is a uniform random variable over the interval (0, 2π). The probability density function obtained for R is called Rayleigh. 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) = integraldisplay ∞ −∞ f 1 (x)f 2 (t − x)dx, where f 1 and f 2 are, respectively, the probability density functions of X and Y. Since f 1 (x) = f 2 (x) = braceleftBigg 1 x ∈ (0, 1) 0 elsewhere, the integrand, f 1 (x)f 2 (t − x) is nonzero if 0 t bracerightbig divided by the area of S:1− (2 − t) 2 2 . (Draw figures to verify these regions.) Let G be the probability distribution function of X + Y. We have shown that G(t) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 0 t<0 t 2 2 0 ≤ t<1 1 − (2 − t) 2 2 1 ≤ t<2 1 t ≥ 2. Therefore, g(t) = G prime (t) = ⎧ ⎪ ⎨ ⎪ ⎩ t 0 ≤ t<1 2 − t 1 ≤ t<2 0 otherwise. 5. (a) Clearly, p X (x) = 1/3 for x =−1, 0, 1 and p Y (y) = 1/3 for y =−1, 0, 1. Since P(X+ Y = z) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 1/9 z =−2, +2 2/9 z =−1, +1 3/9 z = 0, the relation P(X+ Y = z) = summationdisplay x p X (x)p Y (z − x) is easily seen to be true. Section 8.4 Transformations of Two Random Variables 187 (b) p(x,y) = p X (x)p Y (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 h(x, y) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 1 x 2 y 2 x ≥ 1,y≥ 1 0 elsewhere. Consider the system of two equations in two unknowns braceleftBigg x/y = u xy = v. (29) This system has the unique solution braceleftBigg x = √ uv y = √ v/u. (30) We have that x ≥ 1 ⇐⇒ √ uv ≥ 1 ⇐⇒ u ≥ 1 v , y ≥ 1 ⇐⇒ √ v/u ≥ 1 ⇐⇒ v ≥ u. Clearly, x ≥ 1, y ≥ 1 imply that v = xy ≥ 1, so 1 v > 0. Therefore, the system of equations (29) defines a one-to-one transformation of R = braceleftbig (x, y): x ≥ 1,y≥ 1 bracerightbig onto the region Q = braceleftBig (u, v): 0 < 1 v ≤ u ≤ v bracerightBig . By (30), J = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 2 radicalbigg v u 1 2 radicalbigg u v − √ v 2u √ u 1 2 √ uv vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 1 2u negationslash= 0. Hence, by Theorem 8.8, g(u, v), the joint probability density function of U and V is given by g(u, v) = h parenleftbigg √ uv, radicalbigg v u parenrightbigg |J|= 1 2uv 2 , 0 < 1 v ≤ u ≤ v. 188 Chapter 8 Bivariate Distributions 7. Let h be the joint probability density function of X and Y. Clearly, h(x, y) = braceleftBigg e −(x+y) x>0,y>0 0 elsewhere. Consider the system of two equations in two unknowns braceleftBigg x + y = u e x = v. (31) This system has the unique solution braceleftBigg x = ln v y = u − ln v. (32) We have that x>0 ⇐⇒ ln v>0 ⇐⇒ v>1, y>0 ⇐⇒ u − ln v>0 ⇐⇒ e u >v. Therefore, the system of equations (31) defines a one-to-one transformation of R = braceleftbig (x, y): x>0,y>0 bracerightbig onto the region Q = braceleftbig (u, v): u>0, 1 0, 1 0,y>0. Consider the system of two equations in two unknowns ⎧ ⎨ ⎩ x + y = u x x + y = v. (33) 190 Chapter 8 Bivariate Distributions Clearly, (33) implies that u>0 and v>0. This system has the unique solution braceleftBigg x = uv y = u − uv. (34) We have that x>0 ⇐⇒ uv > 0 ⇐⇒ u>0 and v>0, y>0 ⇐⇒ u − uv > 0 ⇐⇒ v<1. Therefore, the system of equations (33) defines a one-to-one transformation of R = braceleftbig (x, y): x>0,y>0 bracerightbig onto the region Q = braceleftbig (u, v): u>0, 0 0, 0 0, 0 0,y>0. The system of two equations in two unknowns braceleftBigg x + y = u x/y = v Chapter 8 Review Problems 191 defines a one-to-one transformation of R = braceleftbig (x, y): x>0,y>0 bracerightbig onto the region Q = braceleftbig (u, v): u>0,v>0 bracerightbig . It has he unique solution x = uv/(1 + v), y = u/(1 + v). Hence J = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle v 1 + v u (1 + v) 2 1 1 + v − u (1 + v) 2 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle =− u (1 + v) 2 negationslash= 0. By Theorem 8.8, g(u, v), the joint probability density function of U and V is g(u, v) = f parenleftBig uv 1 + v , u 1 + v parenrightBig |J|= λ 2 u (1 + v) 2 e −λu ,u>0,v>0. This shows that g(u, v) = g U (u)g V (v), where g U (u) = λ 2 ue −λu ,u>0, and g V (v) = 1 (1 + v) 2 ,v>0. 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 p X (x) and p Y (y), the marginal probability mass functions of X and Y. They are given by the following table. x y 123p Y (y) 2 0.05 0.25 0.15 0.45 4 0.14 0.10 0.17 0.41 6 0.10 0.02 0.02 0.14 p X (x) 0.29 0.37 0.34 192 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 p X (x) and p Y (y), the marginal probability mass functions of X and Y are given by the following table. y x 12345 6p X (x) 2 1/36 0000 01/36 3 0 2/36 0 0 0 0 2/36 4 0 1/36 2/36 0 0 0 3/36 5 0 0 2/36 2/36 0 0 4/36 6 0 0 1/36 2/36 2/36 0 5/36 7 0 0 0 2/36 2/36 2/36 6/36 8 0 0 0 1/36 2/36 2/36 5/36 9 00002/36 2/36 4/36 10 00001/36 2/36 3/36 11 000002/36 2/36 12 000001/36 1/36 p Y (y) 1/36 3/36 5/36 7/36 9/36 11/36 (c) E(X) = summationtext 15 x=2 xp X (x) = 7; E(Y) = summationtext 6 y=1 yp Y (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 p X|Y (x|4) = p(x,4) p Y (4) = parenleftbigg 13 x parenrightbiggparenleftbigg 13 4 parenrightbiggparenleftbigg 26 9 − x parenrightbigg parenleftbigg 52 13 parenrightbigg parenleftbigg 13 4 parenrightbiggparenleftbigg 39 9 parenrightbigg parenleftbigg 52 13 parenrightbigg = parenleftbigg 13 x parenrightbiggparenleftbigg 26 9 − x parenrightbigg parenleftbigg 39 9 parenrightbigg , 0 ≤ x ≤ 9. 4. The set of possible values of X and Y, both, is braceleftbig 0, 1, 2, 3 bracerightbig . Let p(x,y)be their joint probability mass function; then p(x,y) = parenleftbigg 13 x parenrightbiggparenleftbigg 13 y parenrightbiggparenleftbigg 26 3 − x − y parenrightbigg parenleftbigg 52 3 parenrightbigg , 0 ≤ x, y, x + y ≤ 3. Chapter 8 Review Problems 193 5. Reducing the sample space, the answer is parenleftbigg 13 x parenrightbiggparenleftbigg 13 6 − x parenrightbigg parenleftbigg 26 6 parenrightbigg , 0 ≤ x ≤ 6. 6. (a) integraldisplay 2 0 parenleftBig integraldisplay x 0 c x dy parenrightBig dx = 1 equal1⇒ c = 1/2. (b) f X (x) = integraldisplay x 0 1 2x dy = 1 2 , 0 0,y>0. Therefore, by symmetry, P(X>2Y)+ P(Y >2X) = 2P(X>2Y)= 2 integraldisplay ∞ 0 parenleftbiggintegraldisplay ∞ 2y 4xye −x 2 e −y 2 dx parenrightbigg dy = 2 5 . 12. We have that f X (x) = integraldisplay 1−x 0 3(x + y)dy =− 3 2 x 2 + 3 2 , 0 1/2) = integraldisplay 1/2 0 bracketleftbiggintegraldisplay 1−x (1/2)−x 3(x + y)dy bracketrightbigg dx + integraldisplay 1 1/2 bracketleftbiggintegraldisplay 1−x 0 3(x + y)dy bracketrightbigg dx = 9 64 + 5 16 = 29 64 . 13. Since f X|Y (x|y) = f(x,y) f Y (y) = e −y integraltext 1 0 e −y dx = 1, 0 0, we have that E(X n | Y = y) = integraldisplay 1 0 x n · 1 dx = 1 n + 1 ,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) = parenleftbigg 10 x parenrightbigg parenleftBig 1 4 parenrightBig x parenleftBig 3 4 parenrightBig 10−x · parenleftbigg 15 y parenrightbigg parenleftBig 1 4 parenrightBig y parenleftBig 3 4 parenrightBig 15−y = parenleftbigg 10 x parenrightbiggparenleftbigg 15 y parenrightbigg parenleftBig 1 4 parenrightBig x+y parenleftBig 3 4 parenrightBig 25−x−y , 0 ≤ x ≤ 10, 0 ≤ y ≤ 15. 15. integraldisplay 1 0 bracketleftbiggintegraldisplay 1 x cx(1 − x)dy bracketrightbigg dx = 1 equal1⇒ c = 12. Clearly, f X (x) = integraldisplay 1 x 12x(1 − x)dy = 12x(1 − x) 2 , 0 Y parenrightBig P(X>Y) = 2P parenleftBig min(X, Y − X, lscript − Y)< lscript 20 vextendsingle vextendsingle vextendsingle Xx bracerightbig ; that is, 17lscript 20 × 17lscript 20 2 ÷ lscript 2 2 = 0.7225. 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) x−1 (0.10)(0.90) y−1 (0.10) = (0.90) x+y−2 (0.10) 2 . 21. We have that f X (x) = integraldisplay x −x dy = 2x, 0 2 RANDOM VARIABLES 1. Let p(h, d, c, s) be the joint probability mass function of the number of hearts, diamonds, clubs, and spades selected. We have p(h, d, c, s) = parenleftbigg 13 h parenrightbiggparenleftbigg 13 d parenrightbiggparenleftbigg 13 c parenrightbiggparenleftbigg 13 s parenrightbigg parenleftbigg 52 13 parenrightbigg ,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, p(a, h, n, w) = parenleftbigg 8 a parenrightbiggparenleftbigg 7 h parenrightbiggparenleftbigg 3 n parenrightbiggparenleftbigg 20 w parenrightbigg parenleftbigg 38 12 parenrightbigg , 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 p A (a) = parenleftbigg 8 a parenrightbiggparenleftbigg 30 12 − a parenrightbigg parenleftbigg 38 12 parenrightbigg , 0 ≤ a ≤ 8. 3. (a) The desired joint marginal probability mass functions are given by p X,Y (x, y) = 2 summationdisplay z=1 xyz 162 = xy 54 ,x= 4, 5,y= 1, 2, 3. p Y,Z (y, z) = 5 summationdisplay x=4 xyz 162 = yz 18 ,y= 1, 2, 3,z= 1, 2. p X,Z (x, z) = 3 summationdisplay y=1 xyz 162 = xz 27 ,x= 4, 5,z= 1, 2. Section 9.1 Joint Distributions of n > 2 Random Variables 201 (b) E(YZ) = 3 summationdisplay y=1 2 summationdisplay z=1 yzp Y,Z (y, z) = 3 summationdisplay y=1 2 summationdisplay z=1 (yz) 2 18 = 35 9 . 4. (a) The desired marginal joint probability mass functions are given by f X,Y (x, y) = integraldisplay ∞ y 6e −x−y−z dz = 6e −x−2y , 0 0, f Y (y) = integraldisplay ∞ 0 parenleftbiggintegraldisplay ∞ 0 x 2 e −x(1+y+z) dz parenrightbigg dx = 1 (1 + y) 2 ,y>0, and similarly, f Z (z) = 1 (1 + z) 2 ,z>0. Also f X,Y (x, y) = integraldisplay ∞ 0 x 2 e −x(1+y+z) dz = xe −x(1+y) ,y>0. Since f(x,y,z) negationslash= f X (x)f Y (y)f Z (z), X, Y, and Z are not independent. Since f X,Y (x, y) negationslash= f X (x)f Y (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 F X (x) = F(x,∞, ∞) = 1 − e −λ 1 x ,x>0, F Y (y) = F(∞,y,∞) = 1 − e −λ 2 y ,y>0, F Z (z) = F(∞, ∞,z)= 1 − e −λ 3 z ,z>0. Since F(x,y,z) = F X (x)F Y (y)F Z (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 λ 1 , λ 2 , and λ 3 , respectively. Hence their joint probability density functions is given by f(x,y,z) = λ 1 λ 2 λ 3 e −λ 1 x−λ 2 y−λ 3 z . (c) The desired probability is calculated as follows: P(X 2 Random Variables 203 11. Yes, it is because f ≥ 0 and integraldisplay ∞ 0 integraldisplay ∞ x 1 integraldisplay ∞ x 2 ··· integraldisplay ∞ x n−1 e −x n dx n dx n−1 ···dx 1 = integraldisplay ∞ 0 integraldisplay ∞ x 1 integraldisplay ∞ x 2 ··· integraldisplay ∞ x n−2 e −x n−1 dx n−1 ···dx 1 =···= integraldisplay ∞ 0 integraldisplay ∞ x 1 e −x 2 dx 2 dx 1 = integraldisplay ∞ 0 e −x 1 dx 1 = 1. 12. Let f(x 1 ,x 2 ,x 3 ) be the joint probability density function of X 1 , X 2 , and X 3 , the lifetimes of the original, the second, and the third transistors, respectively. We have that f(x 1 ,x 2 ,x 3 ) = 1 5 e −x 1 /5 · 1 5 e −x 2 /5 · 1 5 e −x 3 /5 = 1 125 e −(x 1 +x 2 +x 3 )/5 . Now P(X 1 + X 2 + X 3 < 15) = integraldisplay 15 0 integraldisplay 15−x 1 0 integraldisplay 15−x 1 −x 2 0 1 125 e −(x 1 +x 2 +x 3 )/5 dx 3 dx 2 dx 1 = integraldisplay 15 0 integraldisplay 15−x 1 0 bracketleftbigg 1 25 e −(x 1 +x 2 )/5 − 1 25 e −3 bracketrightbigg dx 2 dx 1 = integraldisplay 15 0 parenleftbigg 1 5 e −x 1 /5 − 4 5 e −3 + 1 25 e −3 x 1 parenrightbigg dx 1 = 1 − 17 2 e −3 = 0.5768. Therefore, the desired probability is P(X 1 + X 2 + X 3 ≥ 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(X 1 >t,X 2 >t,... ,X n >t) = 1 − P(X 1 > t)P(X 2 >t)···P(X n >t)= 1 − e −λ 1 t e −λ 2 t ···e −λ n t = 1 − e −(λ 1 +λ 2 +···+λ n )t ,t>0. Thus X is exponential with parameter λ 1 + λ 2 +···+λ 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 r = P(X= 1) = P(Y ≥ k) = n summationdisplay i=k parenleftbigg n i parenrightbigg p i (1 − p) n−i . 204 Chapter 9 Multivariate Distributions 15. Let X i be the lifetime of the ith part. The time until the item fails is the random variable min(X 1 ,X 2 ,... ,X n ) which by the solution to Exercise 13 is exponentially distributed with parameter nλ. Thus the average life of the item is 1/(nλ). 16. Let X 1 , X 2 , ... be the lifetimes of the transistors selected at random. Clearly, N = min braceleftbig n: X n >s bracerightbig . Note that P parenleftbig X N ≤ t | N = n parenrightbig = P parenleftbig X n ≤ t | X 1 ≤ s, X 2 ≤ s,...,X n−1 ≤ s, X n >s). This shows that for s ≥ t, P parenleftbig X N ≤ t | N = n parenrightbig = 0. For ss) = P(s s) = P(s s) = F(t)− F(s) 1 − F(s) . This relation shows that the probability distribution function of X N given N = n does not depend on n. Therefore, X N and N are independent. 17. Clearly, X = X 1 bracketleftBig 1 − (1 − X 2 )(1 − X 3 ) bracketrightBigbracketleftBig 1 − (1 − X 4 )(1 − X 5 X 6 ) bracketrightBig X 7 = X 1 X 7 parenleftbig X 2 X 4 + X 3 X 4 − X 2 X 3 X 4 + X 2 X 5 X 6 + X 3 X 5 X 6 − X 2 X 3 X 5 X 6 − X 2 X 4 X 5 X 6 − X 3 X 4 X 5 X 6 + X 2 X 3 X 4 X 5 X 6 parenrightbig . The reliability of this system is r = p 1 p 7 parenleftbig p 2 p 4 + p 3 p 4 − p 2 p 3 p 4 + p 2 p 5 p 6 + p 3 p 5 p 6 − p 2 p 3 p 5 p 6 − p 2 p 4 p 5 p 6 − p 3 p 4 p 5 p 6 + p 2 p 3 p 4 p 5 p 6 parenrightbig . 18. Let G and F be the distribution functions of max 1≤i≤n X i and min 1≤i≤n X i , respectively. Let g and f be their probability density functions, respectively. For 0 ≤ t<1, G(t) = P(X 1 ≤ t,X 2 ≤ t,... ,X n ≤ t) = P(X 1 ≤ t)P(X 2 ≤ t)···P(X n ≤ t) = t n . Section 9.1 Joint Distributions of n > 2 Random Variables 205 So G(t) = ⎧ ⎪ ⎨ ⎪ ⎩ 0 t<0 t n 0 ≤ t<1 1 t ≥ 1. Therefore, g(t) = G prime (t) = braceleftBigg nt n−1 0 t parenrightBig = 1 − P(X 1 > t)P(X 2 >t)···P(X n >t) = 1 − (1 − t) n , 0 ≤ t<1. Hence F(t)= ⎧ ⎪ ⎨ ⎪ ⎩ 0 t<0 1 − (1 − t) n 0 ≤ t<1 1 t ≥ 1, and f(t)= braceleftBigg n(1 − t) n−1 0 t parenrightbig = 1 − P(X 1 >t,X 2 >t,... ,X n >t) = 1 − P(X 1 > t)P(X 2 >t)···P(X n >t) = 1 − bracketleftbig 1 − F(t) bracketrightbig n . 206 Chapter 9 Multivariate Distributions 20. We have that P(Y n >x)= P parenleftBig min(X 1 ,X 2 ,... ,X n )> x n parenrightBig = P parenleftBig X 1 > x n ,X 2 > x n ,...,X n > x n parenrightBig = P parenleftBig X 1 > x n parenrightBig P parenleftBig X 2 > x n parenrightBig ···P parenleftBig X n > x n parenrightBig = parenleftBig 1 − x n parenrightBig n . Thus lim n→∞ P(Y n >x)= lim n→∞ parenleftBig 1 − x n parenrightBig n = e −x ,x>0. 21. We have that P(X 2 Random Variables 207 Thus f X (x) = integraldisplay 1−x 0 parenleftbiggintegraldisplay 1−x−y 0 6 dz parenrightbigg dy = 3(1 − x) 2 , 0 0. Applying convolution theorem, we obtain P parenleftbig B 2 − 4AC ≥ 0 parenrightbig = 1 − P parenleftbig B 2 − 4AC < 0 parenrightbig = 1 − integraldisplay ∞ −∞ F −4AC (0 − x)f B 2 (x)dx = 1 − integraldisplay 1 0 parenleftBig 1 − x 4 + x 4 ln x 4 parenrightBig 1 2 √ x dx. Letting y = √ x/2, we get dy = 1 4 √ x dx.So P parenleftbig B 2 − 4AC ≥ 0 parenrightbig = 1 − integraldisplay 1/2 0 (1 − y 2 + y 2 ln y 2 )2dy = 1 − integraldisplay 1/2 0 2dy + 2 integraldisplay 1/2 0 (y 2 − y 2 ln y 2 )dy = 2 integraldisplay 1/2 0 (y 2 − y 2 ln y 2 )dy. Now by integration by parts (u = ln y 2 , dv = y 2 dy), integraldisplay y 2 ln y 2 dy = 1 3 y 3 ln y 2 − 2 9 y 3 . Thus P parenleftbig B 2 − 4AC ≥ 0 parenrightbig = bracketleftBig 10 9 y 3 − 2 3 y 3 ln y 2 bracketrightBig 1/2 0 = 5 36 + 1 6 ln 2 ≈ 0.25. 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, lim x→−∞ f(x) =−∞and lim x→+∞ 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 prime (x) = 3x 2 + 2Ax + B = 0, with roots x =− 1 3 A ± 1 3 radicalbig A 2 − 3B. Let D = √ A 2 − 3B, x 1 =− 1 3 (A + D), and x 2 =− 1 3 (A − D). If A 2 < 3B then the 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 A 2 ≥ 3B. In order for there to be three real roots, counting multiplicities, the local maximum parenleftbig x 1 ,f(x 1 ) parenrightbig and local minimum parenleftbig x 2 ,f(x 2 ) parenrightbig must satisfy f(x 1 ) ≥ 0 and f(x 2 ) ≤ 0; that is, f(x 1 ) =− 1 27 (A 3 + 3A 2 D + 3AD 2 + D 3 ) + 1 9 A(A 2 + 2AD + D 2 ) − 1 3 B(A + D) + C ≥ 0, f(x 2 ) =− 1 27 (A 3 − 3A 2 D + 3AD 2 − D 3 ) + 1 9 A(A 2 − 2AD + D 2 ) − 1 3 B(A − D) + C ≤ 0. Simplifying produces two half-spaces: C ≥ 1 27 parenleftBig − 2A 3 + 9AB − 2(A 2 − 3B) 3/2 parenrightBig , (constraint surface 1); C ≤ 1 27 parenleftBig − 2A 3 + 9AB + 2(A 2 − 3B) 3/2 parenrightBig , (constraint surface 2). These two surfaces intersect at the curve given parametrically by A = t, B = 1 3 t 2 and C = 1 27 t 3 . Note that all points in the intersection of these two half-spaces satisfy B ≤ 1 3 A 2 . Surface 2 intersects the plane C = 0 at the A-axis, but surface 1 intersects the plane C = 0 at the curve B = 1 4 A 2 , which is a quadratic curve in the plane C = 0 located between the A-axis and the upper limit B = 1 3 A 2 . Therefore, V is the region above the plane C = 0 and constraint surface 1, and below constraint surface 2. The volume of V is the volume V 2 under surface 2 minus the volume V 1 under surface 1. Now V 1 = integraldisplay 1 a=0 integraldisplay (1/3)a 2 b=(1/4)a 2 1 27 parenleftBig − 2a 3 + 9ab − 2(a 2 − 3b) 3/2 parenrightBig dbda 210 Chapter 9 Multivariate Distributions = integraldisplay 1 0 1 27 bracketleftbigg − 2a 3 b + 9 2 ab 2 + 4 15 (a 2 − 3b) 5/2 bracketrightbigg (1/3)a 2 b=(1/4)a 2 da = integraldisplay 1 0 1 27 · 7 160 a 5 da = 7 25, 920 , and V 2 = integraldisplay 1 a=0 integraldisplay (1/3)a 2 b=0 1 27 parenleftBig − 2a 3 + 9ab + 2(a 2 − 3b) 3/2 parenrightBig dbda = integraldisplay 1 0 1 27 bracketleftbigg − 2a 3 b + 9 2 ab 2 − 4 15 (a 2 − 3b) 5/2 bracketrightbigg (1/3)a 2 b=0 da = integraldisplay 1 0 1 270 a 5 da = 1 1620 . Thus V = V 2 − V 1 = 1 1, 620 − 7 25, 920 = 1 2, 880 . 9.2 ORDER STATISTICS 1. By Theorem 9.5, we have that f 3 (x) = 4! 2! 1! f(x) bracketleftbig F(x) bracketrightbig 2 bracketleftbig 1 − F(x) bracketrightbig , where f(x)= ⎧ ⎨ ⎩ 10x)dx. Now P parenleftbig X (n) >x parenrightbig = 1 − P parenleftbig X (n) ≤ x parenrightbig = 1 − P(X 1 ≤ x,X 2 ≤ x,... ,X n ≤ x) = 1 − bracketleftbig F(x) bracketrightbig n . So E bracketleftbig X (n) bracketrightbig = integraldisplay ∞ 0 parenleftBig 1 − bracketleftbig F(x) bracketrightbig n parenrightBig dx. 5. To find P parenleftbig X (i) = k parenrightbig ,0≤ k ≤ n, note that P parenleftbig X (i) = k parenrightbig = 1 − P parenleftbig X (i) k parenrightbig . Let N be the number of X j ’s that are less than k. Then N is a binomial random variable with parameters m and p 1 = k−1 summationdisplay l=0 parenleftbigg n l parenrightbigg p l (1 − p) n−l . (35) Let L be the number of X j ’s that are greater than k. Then L is a binomial random variable with parameters m and p 2 = n summationdisplay l=k+1 parenleftbigg n l parenrightbigg p l (1 − p) n−l . (36) 212 Chapter 9 Multivariate Distributions Clearly, P parenleftbig X (i) k parenrightbig = P(L≥ m − i + 1) = m summationdisplay j=m−i+1 parenleftbigg m j parenrightbigg p j 2 (1 − p 2 ) m−j . Thus, for 0 ≤ k ≤ n, P parenleftbig X (i) = k parenrightbig = 1 − m summationdisplay j=i parenleftbigg m j parenrightbigg p j 1 (1 − p 1 ) m−j − m summationdisplay j=m−i+1 parenleftbigg m j parenrightbigg p j 2 (1 − p 2 ) m−j , where p 1 and p 2 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 f 1n (x, y) = n(n − 1)f (x)f (y) bracketleftbig F(y)− F(x) bracketrightbig n−2 ,x0 bracerightbig . It has the unique solution x = u, y = u + v. Hence J = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 10 11 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 1 negationslash= 0. By Thereom 8.8, g(u, v) = f 12 (u, u + v)|J|=2λ 2 e −λ(u+2v) ,u≥ 0,v>0. Since g(u, v) = g U (u)g V (v), where g U (u) = 2λe −2λu ,u≥ 0, and g V (v) = λe −λv ,v>0, we have that U and V are independent. Furthermore, U is exponential with parameter 2λ and V is exponential with parameter λ. 8. Let f 12 (x, y) be the joint probability density function of X (1) and X(2). By Theorem 9.6, f 12 (x, y) = 2! f(x)f(y)= 2 · 1 σ √ 2π e −x 2 /2σ 2 · 1 σ √ 2π e −y 2 /2σ 2 = 1 σ 2 π e −x 2 /2σ 2 · e −y 2 /2σ 2 , −∞ 0 bracerightbig . It has the unique solution x = v − r, y = v. Hence J = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle −11 01 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle =−1 negationslash= 0. By Theorem 8.8, g(u, v) is given by g(r,v) = f 1n (v − r, v)|J| = n(n − 1)f (v − r)f(v) bracketleftbig F(v)− F(v− r) bracketrightbig n−2 , −∞ 0. This implies g R (r) = integraldisplay ∞ −∞ n(n − 1)f (v − r)f(v) bracketleftbig F(v)− F(v− r) bracketrightbig n−2 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 f(v)= ⎧ ⎨ ⎩ 10t)= 1 − P(X 1 >t,X 2 >t,...,X n >t) = 1 − bracketleftbig P(X 1 >t) bracketrightbig n = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 0 t<1 1 − parenleftbig 5 6 parenrightbig n 1 ≤ t<2 1 − parenleftbig 4 6 parenrightbig n 2 ≤ t<3 1 − parenleftbig 3 6 parenrightbig n 3 ≤ t<4 1 − parenleftbig 2 6 parenrightbig n 4 ≤ t<5 1 − parenleftbig 1 6 parenrightbig n 5 ≤ t<6 1 t ≥ 6. The probability mass function of X is p(x) = P(X= x) = parenleftbigg 7 − x 6 parenrightbigg n − parenleftbigg 6 − x 6 parenrightbigg n ,x= 1, 2, 3, 4, 5, 6. 3. Let D 1 , D 2 , ..., D n be the distances of the points selected from the origin. Let D = min(D 1 ,D 2 ,... ,D n ). The desired probability is P(D ≥ r) = P(D 1 ≥ r, D 2 ≥ r,...,D n ≥ r) = bracketleftbig P(D 1 ≥ r) bracketrightbig n = bracketleftbig 1 − P(D 1 0, ¯ F(t)= P parenleftbig min(X 1 ,X 2 ,... ,X n )>t parenrightbig = P(X 1 >t,X 2 >t,... ,X n >t) = P(X 1 > t)P(X 2 >t)···P(X n >t)= ¯ F 1 (t) ¯ F 2 (t) ··· ¯ F n (t). 8. For 1 ≤ i ≤ n, let X i be the lifetime of the ith component. Then max(X 1 ,X 2 ,... ,X n ) 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 parenleftbig max(X 1 ,X 2 ,... ,X n )>t parenrightbig = 1 − P parenleftbig max(X 1 ,X 2 ,... ,X n ) ≤ t parenrightbig = 1 − P(X 1 ≤ t,X 2 ≤ t,... ,X n ≤ t) = 1 − P(X 1 ≤ t)P(X 2 ≤ t)···P(X n ≤ t) = 1 − F 1 (t)F 2 (t) ···F n (t). 9. The problem is equivalent to the following: Two points X and Y are selected independently and at random from the interval (0,lscript). What is the probability that the length of at least one 220 Chapter 9 Multivariate Distributions interval is less than lscript/20? The solution to this problem is as follows: P parenleftBig min(X, Y − X, lscript − Y)< lscript 20 vextendsingle vextendsingle vextendsingle XY parenrightBig P(X>Y) = 2P parenleftBig min(X, Y − X, lscript − Y)< lscript 20 vextendsingle vextendsingle vextendsingle Xx bracerightbig ; that is, 17lscript 20 × 17lscript 20 2 ÷ lscript 2 2 = 0.7225. Therefore, the desired probability is 1 − 0.7225 = 0.2775. 10. Let f 13 (x, y) be the joint probability density function of X (1) and X (3) . By Theorem 9.6, f 13 (x, y) = 6(y − x), 0 x,Y > y)dx dy. 18. Clearly N>iif and only if X 1 ≥ X 2 ≥ X 3 ≥···≥X i . Hence for i ≥ 2, P(N >i)= P parenleftbig X 1 ≥ X 2 ≥ X 3 ≥···≥X i−1 ≥ X i parenrightbig = 1 i! because X i ’s are independent and identically distributed. So, by Theorem 10.2, E(N) = ∞ summationdisplay i=1 P(N ≥ i) = ∞ summationdisplay i=0 P(N >i)= P(N >0) + P(N >1) + ∞ summationdisplay i=2 1 i! = 1 + 1 + ∞ summationdisplay i=2 1 i! = ∞ summationdisplay i=0 1 i! = e. 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) = parenleftBig 1 2 parenrightBig i parenleftBig 1 2 parenrightBig = parenleftBig 1 2 parenrightBig i+1 , 0 ≤ i ≤ 9; P(N = 10) = parenleftBig 1 2 parenrightBig 10 . The desired quantity is E(10 − N) = 9 summationdisplay i=0 (10 − i) parenleftBig 1 2 parenrightBig i+1 + (10 − 10) · parenleftBig 1 2 parenrightBig 10 ≈ 9.001. 20. Clearly, if for some λ ∈ R,X= λY, Cauchy-Schwarz’s inequality becomes equality. We show that the converse of this is also true. Suppose that for random variables X and Y, E(XY) = radicalbig E(X 2 )E(Y 2 ). Then 4 bracketleftbig E(XY) bracketrightbig 2 − 4E(X 2 )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 )λ 2 − 2 bracketleftbig E(XY) bracketrightbig λ + E(X 2 ) = 0. Hence this quadratic equation has exactly one root. On the other hand, E(Y 2 )λ 2 − 2 bracketleftbig E(XY) bracketrightbig λ + E(X 2 ) = E bracketleftbig (X − λY) 2 bracketrightbig . So the equation E bracketleftbig (X − λY) 2 bracketrightbig = 0 has a unique solution. That is, there exists a unique number λ 1 ∈ R such that E bracketleftbig (X − λ 1 Y) 2 bracketrightbig = 0. Since the expected value of a positive random variable is positive, this implies that with probability 1, X − λ 1 Y = 0orX = λ 1 Y. 10.2 COVARIANCE 1. Since X and Y are independent random variables, Cov(X, Y) = 0. 2. E(X) = 3 summationdisplay x=1 4 summationdisplay y=3 1 70 x 2 (x + y) = 17 7 ; E(Y) = summationtext 3 x=1 summationtext 4 y=3 1 70 xy(x + y) = 124 35 ; E(XY) = 3 summationdisplay x=1 4 summationdisplay y=3 1 70 x 2 y(x + y) = 43 5 . Therefore, Cov(X, Y) = E(XY) − E(X)E(Y) = 43 5 − 17 7 · 124 35 =− 1 245 . 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 E(X) = 6 summationdisplay x=1 xp X (x) = 6 summationdisplay 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 · 63 384 + 1 · 120 384 + 2 · 99 384 + 3 · 64 384 + 4 · 29 384 + 5 · 8 384 + 6 · 1 384 = 7 4 . By the same table, E(XY) = 6 summationdisplay x=1 6 summationdisplay y=0 xyp(x,y) = 91/12. Therefore, Cov(X, Y) = E(XY) − E(X)E(Y) = 91 12 − 7 2 · 7 4 = 35 24 > 0. This shows that X and Y are positively correlated. The higher the outcome from rolling the die, the higher the number of tails obtained—a 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 joint probability mass function of X and Y. Then, for 0 ≤ x ≤ 4, 0 ≤ y ≤ 4, 0 ≤ x + y ≤ 4, p(x,y) = parenleftbigg 7 x parenrightbiggparenleftbigg 8 y parenrightbiggparenleftbigg 5 4 − x − y parenrightbigg parenleftbigg 20 4 parenrightbigg ; 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 N = 28 20 = 7 5 . Y is a hypergeometric random variable with parametersn = 4, D = 8, andN = 20. Therefore, E(Y) = nD N = 32 20 = 8 5 . Since E(XY) = 4 summationdisplay x=0 4−x summationdisplay y=0 xyp(x,y) = 168 95 , we have Cov(X, Y) = E(XY) − E(X)E(Y) = 168 95 − 7 5 · 8 5 =− 224 475 < 0. Therefore, X and Y are negatively correlated as expected. Section 10.2 Covariance 229 5. Since Y = n − X, E(XY) = E(nX − X 2 ) = nE(X) − E(X 2 ) = nE(X) − bracketleftbig Var(X) + E(X) 2 bracketrightbig = n · np − bracketleftbig np(1 − p) + n 2 p 2 bracketrightbig = 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(X 2 − Y 2 ) − E(X + Y)E(X− Y) = E(X 2 ) − E(Y 2 ) − bracketleftbig E(X) bracketrightbig 2 + bracketleftbig E(Y) bracketrightbig 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(X 2 Y)− E(X)E(XY) = E(X 2 )E(Y) − bracketleftbig E(X) bracketrightbig 2 E(Y) = E(Y)Var(X). 11. The probability density function of Theta1 is given by f(θ)= ⎧ ⎪ ⎨ ⎪ ⎩ 1 2π if θ ∈[0, 2π] 0 otherwise. Therefore, E(XY) = integraldisplay 2π 0 sin θ cos θ 1 2π dθ = 0, E(X) = integraldisplay 2π 0 sin θ 1 2π dθ = 0, E(Y) = integraldisplay 2π 0 cos θ 1 2π dθ = 0. 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 f(x,y) = ⎧ ⎪ ⎨ ⎪ ⎩ 1 π x 2 + y 2 ≤ 1 0 elsewhere. X and Y are dependent because, for example, P parenleftBig 0 0 ⇐⇒ P(AB) > P(A)P(B) ⇐⇒ P(AB) P(B) > P(A), ⇐⇒ P(A| B)>P(A). The proof that I A and I B are positively correlated if and only if P(B|A) > 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, Cov parenleftBig n summationdisplay i=1 a i X i , m summationdisplay j=1 b j Y j parenrightBig = n summationdisplay i=1 a i Cov parenleftBig X i , m summationdisplay j=1 b j Y j parenrightBig . By Exercise 6 and an induction on m, Cov parenleftBig X i , m summationdisplay j=1 b j Y j parenrightBig = m summationdisplay j=1 b j Cov(X i ,Y j ). The desired identity follows from these two identities. 25. For 1 ≤ i ≤ n, let X i = 1 if the outcome of the ith throw is 1; let X i = 0, otherwise. For 1 ≤ j ≤ n, let Y j = 1 if the outcome of the jth throw is 6; let Y j = 0, otherwise. Clearly, Cov(X i ,Y j ) = 0ifi negationslash= j. By Exercise 24, Cov parenleftBig n summationdisplay i X i , n summationdisplay j=1 Y j parenrightBig = n summationdisplay j=1 n summationdisplay i=1 Cov(X i ,Y j ) = n summationdisplay i=1 Cov(X i ,Y i ) = n summationdisplay i=1 bracketleftbig E(X i Y i ) − E(X i )E(Y i ) bracketrightbig = n summationdisplay i=1 parenleftBig 0 − 1 6 · 1 6 parenrightBig =− n 36 . As expected, in n throws of a fair die, the number of ones and the number of sixes are negatively correlated. 26. Let S n = summationtext n i=1 a i X i , µ i = E(X i ); then E(S n ) = n summationdisplay i=1 a i µ i ,S n − E(S n ) = n summationdisplay i=1 a i (X i − µ i ). Thus Var(S n ) = E parenleftBigbracketleftBig n summationdisplay i=1 a i (X i − µ i ) bracketrightBig 2 parenrightBig = n summationdisplay i=1 a 2 i E bracketleftbig (X i − µ i ) 2 bracketrightbig + 2 summationdisplaysummationdisplay i 2. The time that the admissions office has to wait before doubling its student recruitment efforts is S N+1 = X 1 + X 2 +···+X N+1 . Therefore, E(S N+1 ) = E bracketleftbig E(S N+1 | N) bracketrightbig = ∞ summationdisplay i=0 E(S N+1 | N = i)P(N = i). Now, for i ≥ 0, E(S N+1 | N = i) = E(X 1 + X 2 +···+X i+1 | N = i) = i+1 summationdisplay j=1 E(X j | N = i) = bracketleftBig i summationdisplay j=1 E(X j | X j ≤ 2) bracketrightBig + E(X i+1 | X i+1 > 2), where by Remark 8.1, E(X j | X j ≤ 2) = 1 F(2) integraldisplay 2 0 tf (t) dt, E(X i+1 | X i+1 > 2) = 1 1 − F(2) integraldisplay ∞ 2 tf (t) dt, Section 10.4 Conditioning on Random Variables 245 F and f being the probability distribution and density functions of X i ’s, respectively. That is, for t ≥ 0, F(t)= 1 − e −5t , f(t)= 5e −5t . Thus, for 1 ≤ j ≤ i, E(X j | X j ≤ 2) = 1 1 − e −10 integraldisplay 2 0 5te −5t dt = (1.0000454) bracketleftBigparenleftBig − t − 1 5 parenrightBig e −5t bracketrightBig 2 0 = (1.0000454)(0.19999) = 0.1999092 and, for j = i + 1, E(X i+1 | X i+1 > 2) = 1 e −10 integraldisplay ∞ 2 5te −5t dt = e 10 bracketleftBigparenleftBig − t − 1 5 parenrightBig e −5t bracketrightBig ∞ 2 = 2.2. Thus, for i ≥ 0, E(S N+1 | N = i) = (0.1999092)i + 2.2. To find P(N = i), note that for i ≥ 0, P(N = i) = P(X 1 ≤ 2,X 2 ≤ 2, ... , X i ≤ 2,X i+1 > 2) = bracketleftbig F(2) bracketrightbig i bracketleftbig 1 − F(2) bracketrightbig = (0.9999546) i (0.0000454). Putting all these together, we obtain E(S N+1 ) = ∞ summationdisplay i=0 E(S N+1 | N = i)P(N = i) = ∞ summationdisplay i=0 bracketleftbig (0.1999092)i + 2.2 bracketrightbig (0.9999546) i (0.0000454) = (0.00000908) ∞ summationdisplay i=0 i(0.9999546) i + (0.00009988) ∞ summationdisplay i=0 (0.9999546) i = (0.00000908) · 0.9999546 (1 − 0.9999546) 2 + (0.00009988) · 1 1 − 0.9999546 = 4407.286, where the next to last equality follows from summationtext ∞ i=1 ir i = r/(1 − r) 2 , and summationtext ∞ i=0 r i = 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 X i be the number of calls until Steven has not missed Adam in exactly i consecutive calls. We have that E parenleftbig X i | X i−1 parenrightbig = braceleftBigg X i−1 + 1 with probability p X i−1 + 1 + E(X i ) with probability 1 − p. 246 Chapter 10 More Expectations and Variances Therefore, E(X i ) = E bracketleftbig E(X i | X i−1 ) bracketrightbig = bracketleftbig E(X i−1 ) + 1 bracketrightbig p + bracketleftbig E(X i−1 ) + 1 + E(X i ) bracketrightbig (1 − p). Solving this equation for E(X i ), we obtain E(X i ) = 1 p bracketleftbig 1 + E(X i−1 ) bracketrightbig . Now X 1 is a geometric random variable with parameter p.SoE(X 1 ) = 1/p. Thus E(X 2 ) = 1 p bracketleftbig 1 + E(X 1 ) bracketrightbig = 1 p parenleftBig 1 + 1 p parenrightBig , E(X 3 ) = 1 p bracketleftbig 1 + E(X 2 ) bracketrightbig = 1 p parenleftBig 1 + 1 p + 1 p 2 parenrightBig , . . . E(X k ) = 1 p parenleftBig 1 + 1 p + 1 p 2 +···+ 1 p k−1 parenrightBig = 1 p · (1/p k ) − 1 (1/p) − 1 = 1 − p k p k (1 − 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) + bracketleftbig (x + 2) + E(N) bracketrightbig P(LL). So E(N | X = x) = (x + 1)(0.35) + (x + 2)(0.65)(0.35) + bracketleftbig (x + 2) + E(N) bracketrightbig (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 bracketleftbig E(N | X) bracketrightbig = E(X) + (0.4225)E(N) + 1.65 = 1 0.35 + (0.4225)E(N) + 1.65. Solving this for E(N) gives E(N) = 7.805. Section 10.4 Conditioning on Random Variables 247 16. Since hemophilia is a sex-linked 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 Z = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 0 if Kim is hh 1 if Kim is Hh 2 if Kim is HH. Then E(X) = E bracketleftbig E(X | Z) bracketrightbig = 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) bracketleftbig 2(0.98)(0.02) bracketrightbig + 0 bracketleftbig 0.98)(0.98) bracketrightbig = 0.08. Therefore, on average, 0.08 of the boys and hence 0.08 of the children are expected to have hemophilia. 17. LetX be the number of bags inspected until an unacceptable bag is found. LetK n 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 + K m . We are interested in E(X + K m ) = E(X) + E(K m ). Clearly, X is a geometric random variable with parameter α(1 − p).So E(X) = 1/ bracketleftbig α(1 − p) bracketrightbig . To find E(K m ), note that ∀n, E(K n ) = E bracketleftbig E(K n | K n−1 ) bracketrightbig . Now E(K n | K n−1 = i) = (i + 1)p + bracketleftbig i + 1 + E(K n ) bracketrightbig (1 − p) = (i + 1) + (1 − p)E(K n ). (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(K n ) inspections bracketleftbig a total of i + 1 + E(K n ) inspections bracketrightbig until we get n consecutive acceptable bags of cinnamon. This happens with probability 1 − p. From (41), we have E(K n | K n−1 ) = (K n−1 + 1) + (1 − p)E(K n ). Finding the expected values of both sides of this relation gives E(K n ) = E(K n−1 ) + 1 + (1 − p)E(K n ). 248 Chapter 10 More Expectations and Variances Solving for E(K n ), we obtain E(K n ) = 1 p + E(K n−1 ) p . Noting that E(K 1 ) = 1/p and solving recursively, we find that E(K n ) = 1 p + 1 p 2 +···+ 1 p n . Therefore, the desired quantity is E(X + K m ) = E(X) + E(K m ) = 1 α(1 − p) + 1 p parenleftBig 1 + 1 p +···+ 1 p m−1 parenrightBig = 1 α(1 − p) + 1 p · parenleftBig 1 p parenrightBig m − 1 1 p − 1 = (1 − α)p m + α αp m (1 − p) . 18. For 0 0 and ρ 2 = ρ σ Y σ X · ρ σ X σ Y = 1 4 . Therefore ρ = 1/2. 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 2π exp bracketleftBig − 1 2 parenleftbig z 2 + w 2 parenrightbig bracketrightBig . Let h 1 (z, w) = σ 1 z+µ 1 and h 2 (z, w) = σ 2 parenleftbig ρz+ radicalbig 1 − ρ 2 w parenrightbig +µ 2 . The system of equations ⎧ ⎨ ⎩ σ 1 z + µ 1 = x σ 2 parenleftbig ρz + radicalbig 1 − ρ 2 w parenrightbig + µ 2 = y defines a one-to-one transformation of R 2 in the zw-plane onto R 2 in the xy-plane. It has a unique solution z = x − µ 1 σ 1 , w = 1 radicalbig 1 − ρ 2 bracketleftBig y − µ 2 σ 2 − ρ(x − µ 1 ) σ 1 bracketrightBig for z and w in terms of x and y. Moreover, J = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 σ 1 0 − ρ σ 1 radicalbig 1 − ρ 2 1 σ 2 radicalbig 1 − ρ 2 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 1 σ 1 σ 2 radicalbig 1 − ρ 2 negationslash= 0. Hence, by Theorem 8.8, the joint probability density function of X and Y is given by 1 σ 1 σ 2 radicalbig 1 − ρ 2 f parenleftBig x − µ 1 σ 1 , 1 radicalbig 1 − ρ 2 bracketleftBig y − µ 2 σ 2 − ρ x − µ 1 σ 1 bracketrightBigparenrightBig . Noting that f(z,w) = 1 2π exp bracketleftBig − 1 2 parenleftbig z 2 + w 2 parenrightbig bracketrightBig . Straightforward calculations will result in (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 ρ(X + Y,X − Y)= Cov(X + Y,X − Y) σ X+Y · σ X−Y = Var(X) − Var(Y) σ X+Y · σ X−Y = 0, 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 X i = 1iftheith graduate receives his or her own diploma; 0, otherwise. The number of graduates who will receive their own diploma is X = X 1 + X 2 +···+X n . Since E(X i ) = 1 · 1 n + 0 · parenleftBig 1 − 1 n parenrightBig = 1 n , we have E(X) = E(X 1 ) + E(X 2 ) +···+E(X n ) = n · 1 n = 1. 2. Since E(X) = integraldisplay 2 1 (2x 2 − 2x)dx = 5 3 , and E(X 3 ) = integraldisplay 2 1 (2x 4 − 2x 3 )dx = 49 10 , we have that E(X 3 + 2X − 7) = 49 10 + 10 3 − 7 = 37 30 . 3. Since E(X 2 ) = 1 3 integraldisplay 1 0 integraldisplay 2 0 (3x 5 + x 3 y)dy dx = 1 2 , and E(XY) = 1 3 integraldisplay 1 0 integraldisplay 2 0 (3x 4 y + x 2 y 2 )dydx = 94 135 , we have that E(X 2 + 2XY) = 1 2 + 188 135 = 511 270 . Chapter 10 Review Problems 255 4. Let X 1 , X 2 , ..., X n be geometric random variables with parameters 1, (n − 1)/n, (n − 2)/n, ... ,1/n, respectively. The desired quantity is E(X 1 + X 2 +···+X n ) = 1 + n n − 1 + n n − 2 +···+n = 1 + n parenleftBig 1 n − 1 + 1 n − 2 +···+ 1 2 + 1 parenrightBig = 1 + na n−1 . 5. Let X be the number of tosses until 4 consecutive sixes. Let Y be the number of tosses until the first non-six outcome is obtained. We have E(X) = E bracketleftbig E(X|Y) bracketrightbig = ∞ summationdisplay i=1 E(X | Y = i)P(Y = i) = 4 summationdisplay i=1 E(X | Y = i)P(Y = i)+ ∞ summationdisplay i=5 E(X | Y = i)P(Y = i) = 4 summationdisplay i=1 bracketleftbig i + E(X) bracketrightbig parenleftBig 1 6 parenrightBig i−1 parenleftBig 5 6 parenrightBig + ∞ summationdisplay i=5 4 parenleftBig 1 6 parenrightBig i−1 parenleftBig 5 6 parenrightBig . This equation reduces to E(X) = bracketleftbig 1 + E(X) bracketrightbig 5 6 + bracketleftbig 2 + E(X) bracketrightbig 1 6 · 5 6 + bracketleftbig 3 + E(X) bracketrightbig parenleftBig 1 6 parenrightBig 2 parenleftBig 5 6 parenrightBig + bracketleftbig 4 + E(X) bracketrightbig parenleftBig 1 6 parenrightBig 3 parenleftBig 5 6 parenrightBig + 4 parenleftBig 5 6 parenrightBig (1/6) 4 1 − (1/6) . Solving this equation for E(X), we obtain E(X) = 1554. 6. f(x,y,z) = (2x)(2y)(2z),00,E(X| Y = y) = 1/y, E(X|Y)= 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 bracketleftbig E(Y|X) bracketrightbig = integraldisplay ∞ −∞ E(Y | X = x)f X (x) dx = integraldisplay 30 0 x 2 · 1 30 dx = 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 summationtext N i=1 X i , note that P parenleftBig N summationdisplay i=1 X i ≤ t parenrightBig = ∞ summationdisplay n=1 P parenleftBig N summationdisplay i=1 X i ≤ t vextendsingle vextendsingle N = n parenrightBig P(N = n) = ∞ summationdisplay n=1 P parenleftBig n summationdisplay i=1 X i ≤ t vextendsingle vextendsingle N = n parenrightBig P(N = n) = ∞ summationdisplay n=1 P parenleftBig n summationdisplay i=1 X i ≤ t parenrightBig P(N = n), 260 Chapter 10 More Expectations and Variances where the last inequality follows since N is independent of braceleftbig X 1 ,X 2 ,X 3 ,... bracerightbig .Now summationtext n i=1 X i is a gamma random variable with parameters n and λ. Thus P parenleftBig N summationdisplay i=1 X i ≤ t parenrightBig = ∞ summationdisplay n=1 bracketleftbiggintegraldisplay t 0 λe −λx (λx) n−1 (n − 1)! dx bracketrightbigg (1 − p) n−1 p = ∞ summationdisplay n=1 integraldisplay t 0 λpe −λx bracketleftbig λ(1 − p)x bracketrightbig n−1 (n − 1)! dx = integraldisplay t 0 λpe −λx ∞ summationdisplay n=1 bracketleftbig λ(1 − p)x bracketrightbig n−1 (n − 1)! dx = integraldisplay t 0 λpe −λx e λ(1−p)x dx = integraldisplay t 0 λpe −λpx dx = 1 − e −λpt . This shows that summationtext N i=1 X i is exponential with parameter λp. 17. Let X 1 , X 2 , ..., X i , ..., X 20 be geometric random variables with parameters 1, 19/20, ..., bracketleftbig 20 − (i − 1) bracketrightbig /20, ..., 1/20. The desired quantity is E parenleftBig 20 summationdisplay i=1 X i parenrightBig = 20 summationdisplay i=1 E(X i ) = 20 summationdisplay i=1 20 20 − (i − 1) = 71.9548. Chapter 11 Sums of Independent Random Variables and Limit Theorems 11.1 MOMENT-GENERATING FUNCTIONS 1. M X (t) = E parenleftbig e tX parenrightbig = 5 summationdisplay x=1 e tx p(x) = 1 5 parenleftbig e t + e 2t + e 3t + e 4t + e 5t parenrightbig . 2. (a) For t negationslash= 0, M X (t) = E parenleftbig e tX parenrightbig = integraldisplay 3 −1 1 4 e tx dx = 1 4 parenleftBig e 3t − e −t t parenrightBig , whereas for t = 0, M X (0) = 1. Thus M X (t) = ⎧ ⎪ ⎨ ⎪ ⎩ 1 4 parenleftBig e 3t − e −t t parenrightBig if t negationslash= 0 1ift = 0. Since X is uniform over (−1, 3), E(X) = −1 + 3 2 = 1 and Var(X) = bracketleftbig 3 − (−1) bracketrightbig 2 12 = 4 3 . (b) By the definition of derivative, E(X) = M prime X (0) = lim h→0 M X (h) − M X (0) h = lim h→0 1 h parenleftBig e 3h − e −h 4h − 1 parenrightBig = lim h→0 e 3h − e −h − 4h 4h 2 = lim h→0 3e 3h + e −h − 4 8h = lim h→0 9e 3h − e −h 8 = 1, where the fifth and sixth equalities follow from L’Hôpital’s rule. 262 Chapter 11 Sums of Independent Random Variables and Limit Theorems 3. Note that M X (t) = E parenleftbig e tX parenrightbig = ∞ summationdisplay x=1 e tx · 2 parenleftBig 1 3 parenrightBig x = 2 ∞ summationdisplay x=1 e tx · e −x ln 3 = 2 ∞ summationdisplay x=1 e x(t−ln 3) . Restricting the domain of M X (t) to the set braceleftbig t : t 1 for t ∈ (0, ∞). Therefore, summationtext ∞ x=1 e tx x 2 diverges on (0, ∞) and thus on no interval of the form (−δ,δ), δ > 0, M X (t) exists. 24. For t<1/2, (11.2) implies that M X (t) = ∞ summationdisplay n=0 E(X n ) n! t n = ∞ summationdisplay n=0 (n + 1)(2t) n = 1 2 ∞ summationdisplay n=0 d dt (2t) n+1 = 1 2 d dt bracketleftBig ∞ summationdisplay n=0 (2t) n+1 bracketrightBig = 1 2 · d dt bracketleftBig ∞ summationdisplay n=0 (2t) n − 1 bracketrightBig = 1 2 · d dt bracketleftBig 1 1 − 2t − 1 bracketrightBig = 1 (1 − 2t) 2 = bracketleftBig 1/2 (1/2) − t bracketrightBig 2 . We see that for t<1/2, M X (t) exists; furthermore, it is the moment-generating function of a gamma random variable with parameters r = 2 and λ = 1/2. 25. (a) At the end of the first period, with probability 1, the investment will grow to A + A X k = A parenleftBig 1 + X k parenrightBig ; at the end of the second period, with probability 1, it will grow to A parenleftBig 1 + X k parenrightBig + A parenleftBig 1 + X k parenrightBig · X k = A parenleftBig 1 + X k parenrightBig 2 ; and, in general, at the end of thenth period, with probability 1, it will grow toA parenleftBig 1+ X k parenrightBig n . (b) 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 ε>0, suppose that the interest is compounded at the end of each period of length ε.Ifε → 0, then the interest is compounded continuously. Since a year is 1/ε periods, each of length ε, the interest rate per period of length ε is the random variable X/(1/ε) = εX. Suppose that at time t, the investment has grown to A(t). Then at t + ε, with probability 1, the investment will be A(t + ε) = A(t) + A(t) · εX. 268 Chapter 11 Sums of Independent Random Variables and Limit Theorems This implies that P parenleftBig A(t + ε) − A(t) ε = XA(t) parenrightBig = 1. Letting ε → 0, yields P parenleftBig lim ε→0 A(t + ε) − A(t) ε = XA(t) parenrightBig = 1 or, equivalently, with probability 1, A prime (t) = XA(t). (c) Part (b) implies that, with probability 1, A prime (t) A(t) = X. Integrating both sides of this equation, we obtain that, with probability 1, ln[A(t)]=tX+ C, or A(t) = e tX+c . Considering the fact that A(0) = A, this equation yields A = e c . Therefore, with probability 1, A(t) = e tX · e c = Ae tX . 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 Ae tX , whose expected value is E(Ae tX ) = AE(e tX ) = AM X (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 av- erage, the money will grow by a factor of M X (t), the moment-generating function of the interest rate. 26. Since X i and X j are binomial with parameters (n, p i ) and (n, p j ), E(X i ) = np i ,E(X j ) = np j , σ X i = radicalbig np i (1 − p i ), σ X j = radicalbig np j (1 − p j ). Section 11.2 Sums of Independent Random Variables 269 To find E(X i X j ), note that M(t 1 ,t 2 ) = E parenleftbig e t 1 X i +t 2 X j parenrightbig = n summationdisplay x i =0 n−x i summationdisplay x j =0 e t 1 x i +t 2 x j P(X i = x i ,X j = x j ) = n summationdisplay x i =0 n−x i summationdisplay x j =0 e t 1 x i +t 2 x j · n! x i ! x j ! (n − x i − x j )! p x i i p x j j (1 − p i − p j ) n−x i −x j = n summationdisplay x i =0 n−x i summationdisplay x j =0 n! x i ! x j ! (n − x i − x j )! parenleftbig e t 1 p i parenrightbig x i parenleftbig e t 2 p j parenrightbig x j (1 − p i − p j ) n−x i −x j = parenleftbig p i e t 1 + p j e t 2 + 1 − p i − p j parenrightbig n , where the last equality follows from multinomial expansion (Theorem 2.6). Therefore, ∂ 2 M ∂t 1 ∂t 2 (t 1 ,t 2 ) = n(n − 1)p i p j e t 1 e t 2 parenleftbig p i e t 1 + p j e t 2 + 1 − p i − p j parenrightbig n−2 , and so E(X i X j ) = ∂ 2 M ∂t 1 ∂t 2 (0, 0) = n(n − 1)p i p j . Thus ρ(X i ,X j ) = n(n − 1)p i p j − (np i )(np j ) √ np i (1 − p i ) · radicalbig np j (1 − p j ) =− radicalbigg p i p j (1 − p i )(1 − p j ) . 11.2 SUMS OF INDEPENDENT RANDOM VARIABLES 1. M αX (t) = E parenleftbig e tαX parenrightbig = M X (tα) = exp bracketleftbig αµt + (1/2)α 2 σ 2 t 2 bracketrightbig . 2. Since M X 1 +X 2 +···+X n (t) = M X 1 (t)M X 2 (t) ···M X n (t) = bracketleftBig pe t 1 − (1 − p)e t bracketrightBig n , X 1 + X 2 +···+X n is negative binomial with parameters (n, p). 3. Since M X 1 +X 2 +···+X n (t) = M X 1 (t)M X 2 (t) ···M X n (t) = parenleftBig λ λ − t parenrightBig n , X 1 + X 2 +···+X n is gamma with parameters n and λ. 270 Chapter 11 Sums of Independent Random Variables and Limit Theorems 4. For 1 ≤ i ≤ n, let X i be negative binomial with parameters r i and p. We have that M X 1 +X 2 +···+X n (t) = M X 1 (t)M X 2 (t) ···M X n (t) = bracketleftBig pe t 1 − (1 − p)e t bracketrightBig r 1 bracketleftBig pe t 1 − (1 − p)e t bracketrightBig r 2 ··· bracketleftBig pe t 1 − (1 − p)e t bracketrightBig r n = bracketleftBig pe t 1 − (1 − p)e t bracketrightBig r 1 +r 2 +···+r n . Thus X 1 + X 2 +···+X r is negative binomial with parameters r 1 + r 2 +···+r n and p. 5. Since M X 1 +X 2 +···+X n (t) = M X 1 (t)M X 2 (t) ···M X n (t) = parenleftBig λ λ − t parenrightBig r 1 parenleftBig λ λ − t parenrightBig r 2 ··· parenleftBig λ λ − t parenrightBig r n = parenleftBig λ λ − t parenrightBig r 1 +r 2 +···+r n , X 1 + X 2 +···+X n is gamma with parameters r 1 + r 2 +···+r n and λ. 6. By Theorem 11.4, the total number of underfilled bottles is binomial with parameters 180 and 0.15. Therefore, the desired probability is parenleftbigg 180 27 parenrightbigg (0.15) 27 (0.85) 153 = 0.083. 7. For j *0) = P parenleftBig X + Y − 5 3 > 0 − 5 3 parenrightBig = 1 − Phi1(−1.67) = Phi1(1.67) = 0.9525, P(X− Y<2) = P parenleftBig X − Y + 3 3 < 2 + 3 3 parenrightBig = Phi1(1.67) = 0.9525, and P(3X + 4Y>20) = P parenleftBig 3X + 4Y − 19 √ 130 > 20 − 19 √ 130 parenrightBig = 1 − Phi1(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 parenleftbigg ¯ X − 110 √ 1.6 ≥ 112 − 110 √ 1.6 parenrightbigg = 1 − Phi1(1.58) = 0.0571. 12. Let ¯ X 1 be the average of the accounts selected at store 1 and ¯ X 2 be the average of the accounts selected at store 2. We have that ¯ X 1 ∼ N parenleftBig 90, 900 10 parenrightBig = N(90, 90) and ¯ X 2 ∼ N parenleftBig 100, 2500 15 parenrightBig = N parenleftBig 100, 500 3 parenrightBig . 272 Chapter 11 Sums of Independent Random Variables and Limit Theorems Therefore, ¯ X 1 − ¯ X 2 ∼ N parenleftBig − 10, 770 3 parenrightBig and so P( ¯ X 1 > ¯ X 2 ) = P( ¯ X 1 − ¯ X 2 > 0) = P parenleftbigg ¯ X 1 − ¯ X 2 + 10 √ 770/3 > 0 + 10 √ 770/3 parenrightbigg = 1 − Phi1(0.62) = 0.2676. 13. By Exercise 6, Section 10.5, X and Y are sums of independent standard normal random variables. Hence αX + βY is a linear combination of independent standard normal random variables. Thus, by Theorem 11.7, αX + βY 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) − 2ρ(X,Y)σ X σ Y = 9 + (2.7) 2 − 2(0.45)(3)(2.7) = 9. Therefore, P(X− Y ≥ 8) = P parenleftBig X − Y − 11 3 ≥ 8 − 11 3 parenrightBig = 1 − Phi1(−1) = Phi1(1) = 0.8413. 15. Let ¯ X be the average of the weights of the 12 randomly selected athletes. Let X 1 , X 2 , ..., X 12 be the weights of these athletes. Since ¯ X ∼ N parenleftBig 225, 25 2 12 parenrightBig = N parenleftBig 225, 625 12 parenrightBig , we have that P(X 1 + X 2 +···+X 12 ≤ 2700) = P parenleftBig ¯ X ≤ 2700 12 parenrightBig = P( ¯ X ≤ 225) = P parenleftbigg ¯ X − 225 √ 625/12 ≤ 225 − 225 √ 625/12 parenrightbigg = Phi1(0) = 1 2 . 16. Let ¯ X 1 and ¯ X 2 be the averages of the final grades of the probability and calculus courses Dr. Olwell teaches, respectively. We have that ¯ X 1 ∼ N parenleftBig 65, 418 22 parenrightBig = N(65, 19) and ¯ X 2 ∼ N parenleftBig 72, 448 28 parenrightBig = N(72, 16). Therefore, ¯ X 1 − ¯ X 2 ∼ N(−7, 35) and hence the desired probability is P parenleftbig | ¯ X 1 − ¯ X 2 |≥2 parenrightbig = P( ¯ X 1 − ¯ X 2 ≥ 2) + P( ¯ X 1 − ¯ X 2 ≤−2) = P parenleftbigg ¯ X 1 − ¯ X 2 + 7 √ 35 ≥ 2 + 7 √ 35 parenrightbigg + P parenleftbigg ¯ X 1 − ¯ X 2 + 7 √ 35 ≤ −2 + 7 √ 35 parenrightbigg = 1 − Phi1(1.52) + Phi1(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 parenleftbig |X − Y|≥1.5 parenrightbig = 2P(X− Y ≥ 1.5). Now X − Y ∼ N(0, 2), hence P parenleftbig |X − Y|≥1.5 parenrightbig = 2P parenleftbigg X − Y − 0 √ 2 ≥ 1.5 − 0 √ 2 parenrightbigg = 2 bracketleftbig 1 − Phi1(1.06) bracketrightbig = 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 parenleftbigg Y − X − Z + 3 √ 3 > 0 + 3 √ 3 parenrightbigg = 2 bracketleftbig 1 − Phi1(1.73) bracketrightbig = 0.0836. 18. Let n be the maximum number of passengers who can use the elevator and X 1 , X 2 , ..., X n be the weights of n random passengers. We must have P(X 1 + X 2 +···X n > 3000)<0.0003 or, equivalently, P(X 1 + X 2 +···+X n ≤ 3000)>0.9997. Let ¯ X be the mean of the weights of the n random passengers. We must have P parenleftbigg ¯ X ≤ 3000 n parenrightbigg > 0.9997. Since ¯ X ∼ N parenleftBig 155, 625 n parenrightBig , we must have P parenleftBig ¯ X − 155 25/ √ n ≤ (3000/n) − 155 25/ √ n parenrightBig > 0.9997, or Phi1 parenleftBig 3000 25 √ n − 155 √ n 25 parenrightBig > 0.9997. Using Table 2 of the Appendix, this gives 3000 25 √ n − 155 √ n 25 ≥ 3.49 or, equivalently, 155n + 87.25 √ n − 3000 ≤ 0. 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 parenleftbig√ n + 4.69 parenrightbigparenleftbig√ n − 4.127 parenrightbig ≤ 0. But √ n + 4.69 > 0, so the inequality is valid if and only if √ n − 4.127 ≤ 0orn ≤ 17.032. Therefore the answer is n = 17. 19. By Remark 9.3, the marginal joint probability mass function of X 1 , X 2 , ..., X k is multinomial with parameters n and (p 1 , p 2 , ... ,p k ,1−p 1 −p 2 −···−p k ). Thus, letting p = p 1 +p 2 + ···+p k and x = x 1 + x 2 +···+x k , we have that p(x 1 ,x 2 ,... ,x k ) = n! x 1 ! x 2 ! ··· x k ! (n − x)! p x 1 1 p x 2 2 ···p x k k (1 − p) n−x . This gives P(X 1 + X 2 +···+X k = i) = summationdisplay x 1 +x 2 +···+x k =i n! x 1 ! x 2 ! ··· x k ! (n − i)! p x 1 1 p x 2 2 ···p x k k (1 − p) n−i = n! i! (n − i)! (1 − p) n−i summationdisplay x 1 +x 2 +···+x k =i i! x 1 ! x 2 ! ··· x k ! p x 1 1 p x 2 2 ···p x k k = parenleftbigg n i parenrightbigg (1 − p) n−i (p 1 + p 2 +···+p k ) i = parenleftbigg n i parenrightbigg p i (1 − p) n−i . This shows that X 1 +X 2 +···+X k is binomial with parameters n and p = p 1 +p 2 +···+p k . 20. First note that if Y 1 and Y 2 are two exponential random variables each with rate λ, min(Y 1 ,Y 2 ) is exponential with rate 2λ. Now let A 1 , A 2 , ..., A 11 be the customers in the line ahead of Kim. Due to the memoryless property of exponential random variables, X 1 , the time until A 1 ’s turn to make a call is exponential with rate 2(1/3) = 2/3. The time until A 2 ’s turn to call is X 1 + X 2 , where X 2 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 X 1 + X 2 +···+X 12 , where {X 1 ,X 2 ,... ,X 12 } 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 1. Let X be the lifetime (in months) of a randomly selected dollar bill. We are given that E(X) = 22. By Markov inequality, Section 11.3 Markov and Chebyshev Inequalities 275 P(X≥ 60) ≤ 22 60 = 0.37. This shows that at most 37% of the one-dollar bills last 60 or more months; that is, at least five years. 2. We have that P(X≥ 2) = 2/5. Hence, by Markov’s inequality, 2 5 = P(X≥ 2) ≤ E(X) 2 . This gives E(X) ≥ 4/5. 3. (a) P(X≥ 11) ≤ E(X) 11 = 5 11 = 0.4545. (b) P(X≥ 11) = P(X− 5 ≥ 6) ≤ P parenleftbig |X − 5|≥6 parenrightbig ≤ σ 2 36 = 42 − 25 36 = 0.472. 4. Let X be the lifetime of the randomly selected light bulb; we have P(X≤ 700) ≤ P parenleftbig |X − 800|≥100 parenrightbig ≤ 2500 10, 000 = 0.25. 5. Let X be the number of accidents that will occur tomorrow. Then (a) P(X≥ 5) ≤ 2 5 = 0.4. (b) P(X≥ 5) = 1 − 4 summationdisplay i=0 e −2 2 i i! = 0.053. (c) P(X≥ 5) = P(X− 2 ≥ 3) ≤ P parenleftbig |X − 2|≥3 parenrightbig ≤ 2 9 = 0.222 6. Let X be the IQ of a randomly selected student from this campus; we have P(X>140) ≤ P parenleftbig |X − 110| > 30 parenrightbig ≤ 15 900 = 0.017. Therefore, less than 1.7% of these students have an IQ above 140. 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(X2µ) = P(X− µ>µ)≤ P parenleftbig |X − µ|≥µ parenrightbig ≤ µ µ 2 = 1 µ . 10. We have that P(38 < ¯ X<46) = P(−4 < ¯ X − 42 < 4) = P parenleftbig | ¯ X − 42| < 4 parenrightbig = 1 − P parenleftbig | ¯ X − 42|≥4 parenrightbig . By (11.3), P parenleftbig | ¯ X − 42|≥4 parenrightbig ≤ 60 16(25) = 3 20 . Hence P(38 < ¯ X<46) ≥ 1 − 3 20 = 17 20 = 0.85. 11. For i = 1, 2,... ,n, let X i be the IQ of the ith student selected at random. We want to find n, so that P parenleftBig − 3 < X 1 + X 2 +···+X n n − µ<3 parenrightBig ≥ 0.92 or, equivalently, P(| ¯ X − µ|≥3) ≤ 0.08. Since E(X i ) = µ and Var(X i ) = 150, by (11.3), P(| ¯ X − µ|≥3) ≤ 150 3 2 · 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 X 1 , X 2 , ..., X n be the random sample, µ be the expected value of the distribution, and σ 2 be the variance of the distribution. We want to find n so that P(| ¯ X − µ| < 2σ)≥ 0.98 or, equivalently, P(| ¯ X − µ|≥2σ)<0.02. By (11.3), P(| ¯ X − µ|≥2σ)≤ σ 2 (2σ) 2 · n = 1 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 ε = 0.05 and α = 0.04;wehave n ≥ 1 4(0.05) 2 (0.04) = 2500. Therefore, at least 2500 independent observations should be made to ensure that (1/n) summationtext 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), n ≥ 1 4(0.05) 2 (0.06) = 1666.67. Therefore, it suffices to flip the coin n = 1667 times independently. 15. P parenleftbig |X − µ|≥α parenrightbig = P parenleftbig |X − µ| 2n ≥ α 2n parenrightbig ≤ E bracketleftbig (X − µ) 2n bracketrightbig α 2n . 16. By Markov’s inequality, P(X>t)= P parenleftbig e kX >e kt parenrightbig ≤ E parenleftbig e kX parenrightbig e kt . 17. By the Corollary of Cauchy-Schwarz Inequality (Theorem 10.3), bracketleftbig E(X − Y) bracketrightbig 2 ≤ E bracketleftbig (X − Y) 2 bracketrightbig = 0. This gives that E(X − Y)= 0. Therefore, Var(X − Y)= E bracketleftbig (X − Y) 2 bracketrightbig − bracketleftbig E(X − Y) bracketrightbig 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 ρ(X,Y) = 1. Suppose that ρ(X,Y) = 1; we show that X=Y with probability 1. Note that E(X) = E(Y) = (n + 1)/2, Var(X) = Var(Y) = (n 2 − 1)/12, and σ X = σ Y = radicalbig (n 2 − 1)/12. These and 1 = ρ(X,Y) = E(XY) − E(X)E(Y) σ X σ Y imply that E(XY) = (2n 2 + 3n + 1)/6. Therefore, E bracketleftbig (X − Y) 2 bracketrightbig = E(X 2 − 2XY + Y 2 ) = E(X 2 ) + E(Y 2 ) − 2E(XY) = Var(X) + bracketleftbig E(X) bracketrightbig 2 + Var(Y) + bracketleftbig E(Y) bracketrightbig 2 − 2E(XY) = n 2 − 1 12 + parenleftBig n + 1 2 parenrightBig 2 + n 2 − 1 12 + parenleftBig n + 1 2 parenrightBig 2 − 2n 2 + 3n + 1 3 = 0. E bracketleftbig (X − Y) 2 bracketrightbig = 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 parenleftBig X ≥ 1 t ln α parenrightBig = P(tX≥ ln α) = P parenleftbig e tX ≥ α parenrightbig ≤ E parenleftbig e tX parenrightbig α = 1 α M X (t). 20. Using gamma function introduced in Section 7.4, E(X) = 1 n! integraldisplay ∞ 0 x n+1 e −x dx = Gamma1(n + 2) n! = (n + 1)! n! = n + 1, E(X 2 ) = 1 n! integraldisplay ∞ 0 x n+2 e −x dx = Gamma1(n + 3) n! = (n + 2)! n! = (n + 1)(n + 2). Hence σ 2 X = (n + 1)(n + 2) − (n + 1) 2 = n + 1. Now P(0*Mwith probability 1, then X 2 >Mwith probability 1 since X 1 and X 2 are identically distributed. Therefore, X 1 + X 2 > 2M>Mwith probability 1. This argument shows that {X 1 >M}⊆{X 1 + X 2 >M}⊆{X 1 + X 2 + X 3 >M}⊆··· . Therefore, by the continuity of probability function (Theorem 1.8), lim n→∞ P(X 1 + X 2 +···+X n >M)= P parenleftbig lim n→∞ X 1 + X 2 +···+X n >M parenrightbig . Section 11.4 Laws of Large Numbers 279 By this relation, it suffices to show that ∀M>0, lim n→∞ X 1 + X 2 +···+X n >M (45) with probability 1. Let S be the sample space over which X i ’s are defined. Let µ = E(X i ); we are given that µ>0. By the central limit theorem, P parenleftBig lim n→∞ X 1 + X 2 +···X n n = µ parenrightBig = 1. Therefore, letting V = braceleftBig ω ∈ S : lim n→∞ X 1 (ω) + X 2 (ω) +···X n (ω) n = µ bracerightBig , we have that P(V)= 1. To establish (45), it is sufficient to show that ∀ω ∈ V , lim n→∞ X 1 (ω) + X 2 (ω) +···X n (ω) =∞. (46) To do so, applying the definition of limit to lim n→∞ X 1 (ω) + X 2 (ω) +···X n (ω) n = µ, we have that for ε = µ/2, there exists a positive integer N (depending on ω) such that ∀n>N, vextendsingle vextendsingle vextendsingle X 1 (ω) + X 2 (ω) +···X n (ω) n − µ vextendsingle vextendsingle vextendsingle <ε= µ 2 or, equivalently, − µ 2 < X 1 (ω) + X 2 (ω) +···X n (ω) n − µ< µ 2 . This yields X 1 (ω) + X 2 (ω) +···X n (ω) n > µ 2 . Thus, for all n>N, X 1 (ω) + X 2 (ω) +···X n (ω) > nµ 2 , which establishes (46). 3. For 0 <ε<1, P parenleftbig |Y n − 0| >ε parenrightbig = 1 − P parenleftbig |Y n − 0|≤ε parenrightbig = 1 − P(X≤ n) = 1 − integraldisplay n 0 f(x)dx. Therefore, lim n→∞ P parenleftbig |Y n − 0| >ε parenrightbig = 1 − integraldisplay ∞ 0 f(x)dx = 1 − 1 = 0, showing that Y n converges to 0 in probability. 280 Chapter 11 Sums of Independent Random Variables and Limit Theorems 4. By the strong law of large numbers, S n /n converges to µ almost surely. Therefore, S n /n converges to µ in probability and hence lim n→∞ P parenleftbig n(µ − ε) ≤ S n ≤ n(µ + ε) parenrightbig = lim n→∞ P parenleftBig µ − ε ≤ S n n ≤ µ + ε parenrightBig = lim n→∞ P parenleftBigvextendsingle vextendsingle vextendsingle S n n − µ vextendsingle vextendsingle vextendsingle ≤ ε parenrightBig = 1 − lim n→∞ P parenleftBigvextendsingle vextendsingle vextendsingle S n n − µ vextendsingle vextendsingle vextendsingle >ε parenrightBig = 1 − 0 = 1. 5. Suppose that the bank will never be empty of customers again. We will show a contradiction. Let U n = T 1 +T 2 +···+T n . Then U n is the time the nth new customer arrives. Let W i be the service time of the ith new customer served. Clearly, W 1 , W 2 , W 3 , ... are independent and identicallydistributedrandomvariableswithE(W i ) = 1/µ. LetZ n = T 1 +W 1 +W 2 +···+W n . Since the bank will never be empty of customers, Z n is the departure time of the nth new customer served. By the strong law of large numbers, lim n→∞ U n n = 1 λ and lim n→∞ Z n n = lim n→∞ parenleftBig T 1 n + W 1 + W 2 +···+W n n parenrightBig = lim n→∞ T 1 n + lim n→∞ W 1 + W 2 +···+W n n = 0 + 1 µ = 1 µ . Clearly, the bank will never remain empty of customers again if and only if ∀n, U n+1 0, P parenleftbig |X n − 0|≥ε parenrightbig is the probability that the random point selected from [0, 1] is in bracketleftBig i 2 k , i + 1 2 k bracketrightBig . Now n →∞implies that 2 k →∞ and the length of the interval bracketleftBig i 2 k , i + 1 2 k bracketrightBig → 0, Therefore, lim n→∞ P parenleftbig |X n − 0|≥ε parenrightbig = 0. However, X n does not converge at any point because for all positive natural number N, there are always m>Nand n>N, such that X m = 0 and X n = 1 making it impossible for |X n − X m | to be less than a given 0 <ε<1. 282 Chapter 11 Sums of Independent Random Variables and Limit Theorems 11.5 CENTRAL LIMIT THEOREM 1. Let X 1 , X 2 , ..., X 150 be the random points selected from the interval (0, 1). For 1 ≤ i ≤ 150, X i is uniform over (0, 1). Therefore, E(X i ) = µ = 0.5 and σ X i = 1/ √ 12. We have P parenleftBig 0.48 < X 1 + X 2 +···+X 150 150 < 0.52 parenrightBig = P(72 0) = P parenleftBig X 1 + X 2 +···+X n n > 0 parenrightBig = P parenleftBig X 1 + X 2 +···+X n − n(0) √ 2 √ n > 0 parenrightBig = 1 − Phi1(0) = 0.5. 5. Let µ = E(X i ) and σ = σ X i . Clearly, E(S n ) = nµ and σ S n = σ √ n; thus, by the central limit theorem, P parenleftbig E(S n ) − σ Sn ≤ S n ≤ E(S n ) + σ S n parenrightbig = P parenleftbig nµ − σ √ n ≤ S n ≤ nµ + σ √ n parenrightbig = P parenleftBig − 1 ≤ S n − nµ σ √ n ≤ 1 parenrightBig ≈ Phi1(1) − Phi1(−1) = 2Phi1(1) − 1 = 0.6826. 6. For 1 ≤ i ≤ 300, let X i be the amount of the ith expenditure minus Jim’s ith record; X i is ap- proximatelyuniformover(−1/2, 1/2).HenceE(X i ) = 0 andσ X i = radicalBig bracketleftbig (1/2) − (−1/2) bracketrightbig 2 /12 = 1/(2 √ 3). The desired probability is P(−10 0) = 1 − P(X= 0) = 1 − e −1/2 = 0.393. 7. Note that M (n) X (t) = (−1) n+1 (n + 1)! (1 − t) n+2 . Therefore, E(X n ) = M (n) X (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 heights of 6 randomly selected women. Theorem 10.7 implies that ¯ X ∼ N parenleftBig 173, 40 10 parenrightBig and ¯ Y ∼ N parenleftBig 160, 20 6 parenrightBig ; thus ¯ X − ¯ Y ∼ N parenleftBig 13, 22 3 parenrightBig . Therefore, P( ¯ X − ¯ Y ≥ 5) = P parenleftbigg ¯ X − ¯ Y − 13 √ 22/3 ≥ 5 − 13 √ 22/3 parenrightbigg = Phi1(2.95) = 0.9984. 288 Chapter 11 Sums of Independent Random Variables and Limit Theorems 9. By definition, E parenleftbig e tX parenrightbig = integraldisplay ∞ −∞ 1 2 e −|x| e tx dx = integraldisplay 0 −∞ 1 2 e x · e tx dx + integraldisplay ∞ 0 1 2 e −x · e tx dx = 1 2 integraldisplay 0 −∞ e (1+t)x dx + 1 2 integraldisplay ∞ 0 e x(t−1) dx. Now for these integrals to exist, we must restrict the domain of the moment-generating function of X to {t ∈ R:−1 15) = P(X− 10 > 5) ≤ P parenleftbig |X − 10| > 5 parenrightbig = P parenleftbig |X − E(X)| > 5 parenrightbig ≤ 5/3 25 = 0.0667. 16. P(X≥ 45) ≤ P parenleftbig |X − 0|≥45 parenrightbig ≤ 15 2 /45 2 = 1/9. 290 Chapter 11 Sums of Independent Random Variables and Limit Theorems 17. Suppose that the ith randomly selected book is X i centimeters thick. The desired probability is P(X 1 + X 2 +···+X 31 ≤ 87) = P parenleftbigg X 1 + X 2 +···+X 31 − 3(31) 1 √ 31 ≤ 87 − 3(31) 1 √ 31 parenrightbigg ≈ Phi1 parenleftBig 87 − 93 √ 31 parenrightBig = Phi1(−1.08) = 1 − 0.8599 = 0.1401. 18. For 1 ≤ i ≤ 20, let X i denote the outcome of the ith roll. We have E(X i ) = 6 summationdisplay i=1 i · 1 6 = 7 2 ,E(X 2 i ) = 6 summationdisplay i=1 i 2 · 1 6 = 91 6 . Thus Var(X i ) = 91 6 − 49 4 = 35 12 , and hence P parenleftBig 65 ≤ 20 summationdisplay i=1 X i ≤ 75 parenrightBig = P parenleftbigg 65 − 70 √ 35/12 · √ 20 ≤ summationtext 20 i=1 X i − 70 √ 35/12 · √ 20 ≤ 75 − 70 √ 35/12 · √ 20 parenrightbigg ≈ Phi1(0.65) − Phi1(−0.65) = 2Phi1(0.65) − 1 = 0.4844. 19. By Markov’s inequality, P(X≥ nµ) ≤ µ nµ = 1 n . So nP(X ≥ nµ) ≤ 1. 20. Let X = summationtext 26 i=1 Xi. We have that E(X i ) = 26/51 = 0.5098,E(X 2 i ) = E(X i ) = 0.5098, Var(X i ) = 0.5098 − (0.5098) 2 = 0.2499, E(X i X j ) = P(X i = 1,X j = 1) = P(X i = 1)P(X j = 1 | X i = 1) = 26 51 · 25 49 = 0.2601, and Cov(X i ,X j ) = E(X i X j ) − E(X i )E(X j ) = 0.2601 − (0.5098) 2 = 0.0002. Thus E(X) = 26(0.5098) = 13.2548 and Var(X) = 26 summationdisplay i=1 Var(X i ) + 2 summationdisplaysummationdisplay i 15. The time Linda has to wait before being able to cross the street is 0 if N = 0 (i.e., X 1 > 15), and is S N = X 1 + X 2 +···+X N , otherwise. Therefore, E(S N ) = E bracketleftbig E(S N | N) bracketrightbig = ∞ summationdisplay i=0 E(S N | N = i)P(N = i) = ∞ summationdisplay i=1 E(S N | N = i)P(N = i), 292 Chapter 12 Stochastic Processes where the last equality follows since for N = 0, we have that S N = 0. Now E(S N | N = i) = E(X 1 + X 2 +···+X i | N = i) = i summationdisplay j=1 E(X j | N = i) = i summationdisplay j=1 E(X j | X j ≤ 15), where by Remark 8.1, E(X j | X j ≤ 15) = 1 F(15) integraldisplay 15 0 tf (t) dt; F and f being the probability distribution and density functions of X i ’s, respectively. That is, for t ≥ 0, F(t)= 1 − e −t/7 , f(t)= (1/7)e −t/7 . Thus E(X j | X j ≤ 15) = 1 1 − e −15/7 integraldisplay 15 0 t 7 e −t/7 dt = (1.1329) bracketleftbigg − (t + 7)e −t/7 bracketrightbigg 15 0 = (1.1329)(4.41898) = 5.00631. This gives E(S N | N = i) = 5.00631i. To find P(N = i), note that for i ≥ 1, P(N = i) = P(X 1 ≤ 15,X 2 ≤ 15, ... , X i ≤ 15,X i+1 > 15) = bracketleftbig F(15) bracketrightbig i bracketleftbig 1 − F(15) bracketrightbig = (0.8827) i (0.1173). Putting all these together, we obtain E(S N ) = ∞ summationdisplay i=1 E(S N | N = i)P(N = i) = ∞ summationdisplay i=1 (5.00631i)(0.8827) i (0.1173) = (0.5872) ∞ summationdisplay i=1 i(0.8827) i = (0.5872) · 0.8827 (1 − 0.8827) 2 = 37.6707, where the next to last equality follows from summationtext ∞ i=1 ir i = r/(1 − r) 2 , |r| < 1. Therefore, on 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; braceleftbig N(t): t ≥ 0 bracerightbig is a Poisson process with rate 2. Let X 1 , X 2 , ..., X 6 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(X 1 ,X 2 ,... ,X 6 ). Therefore, the desired probability is P(Y <1) = 1 − P(Y ≥ 1) = 1 − P parenleftbig min(X 1 ,X 2 ,... ,X 6 ) ≥ 1 parenrightbig = 1 − P(X 1 ≥ 1,X 2 ≥ 1,... ,X 6 ≥ 1) = 1 − P(X 1 ≥ 1)P(X 2 ≥ 1) ···P(X 6 ≥ 1) = 1 − parenleftBig 3 4 parenrightBig 6 = 0.822. Section 12.2 More on Poisson Processes 293 5. Let S 1 , S 2 , and S 3 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 S 1 , S 2 , and S 3 is f S 1 ,S 2 ,S 3 |N(200) (t 1 ,t 2 ,t 3 | 3) = 3! 8, 000, 000 , 0 t and S>T)= P(T >t)P(S >t)= e −λt · e −µt = e −(λ+µ)t , 294 Chapter 12 Stochastic Processes P(B)= P(S >T)= integraldisplay ∞ 0 P(S >T | T = u)λe −λu du = integraldisplay ∞ 0 P(S >u| T = u)λe −λu du = integraldisplay ∞ 0 P(S > u)λe −λu du = λ integraldisplay ∞ 0 e −µu · e λu du = λ λ + µ . A similar calculation shows that P(AB) = P(S >T >t)= integraldisplay ∞ t P(S >T | T = u)λe −λu du = integraldisplay ∞ t e −µu · λe −λu du = λ λ + µ e −(λ+µ)t = P(A)P(B). 8. (a) Let X be the number of customers arriving to the queue during a service period S. Then P(X= n) = integraldisplay ∞ 0 P(X= n | S = t)µe −µt dt = integraldisplay ∞ 0 e −λt (λt) n n! µe −µt dt = λ n µ n! integraldisplay ∞ 0 t n e −(λ+µ)t dt = λ n µ n! (λ + µ) integraldisplay ∞ 0 t n (λ + µ)e −(λ+µ)t dt. Note that (λ + µ)e −(λ+µ)t is the probability density function of an exponential random variable Z with parameter λ + µ. Hence P(X= n) = λ n µ n! (λ + µ) E(Z n ). By Example 11.4, E(Z n ) = n! (λ + µ) n . Therefore, P(X= n) = λ n µ (λ + µ) n+1 = parenleftBig 1 − λ λ + µ parenrightBig n parenleftBig µ λ + µ parenrightBig ,n≥ 0. This is the probability mass function of a geometric random variable with parameter µ/(λ + µ). (b) Due to the memoryless property of exponential random variables, the remaining service time of the customer being served is also exponential with parameter µ. 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 µ/(λ+µ), 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 µ/(λ+µ), which is negative binomial with parameters n + 1 and µ/(λ + µ). 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 braceleftbig M(t): t ≥ 0 bracerightbig is a Poisson process. To find its rate, note that, for 0 ≤ k<∞, P parenleftbig M(t) = k parenrightbig = ∞ summationdisplay n=k P parenleftbig M(t) = k | N(t) = n parenrightbig P parenleftbig N(t) = n parenrightbig = ∞ summationdisplay n=k parenleftbigg n k parenrightbigg p k (1 − p) n−k · e −λt (λt) n n! = e −λt p k k! (1 − p) k ∞ summationdisplay n=k bracketleftbig λt(1 − p) bracketrightbig n (n − k)! = e −λt p k k! (1 − p) k · bracketleftbig λt(1 − p) bracketrightbig k ∞ summationdisplay n=k bracketleftbig λt(1 − p) bracketrightbig n−k (n − k)! = e −λt p k k! (λt) k e λt(1−p) = (λpt) k k! e −λpt . This shows that the parameter of braceleftbig M(t): t ≥ 0 bracerightbig is λp. 10. Note that P parenleftbig V i = min(V 1 ,V 2 ,... ,V k ) parenrightbig 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.Ifwe label the time point u as t = 0, then from that point on, by stationarity and the independent- increments 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 parenleftbig V i = min(V 1 ,V 2 ,... ,V k ) parenrightbig . To find this probability, note that, for 1 ≤ j ≤ k, V j ’s, are independent exponential random variables, and the probability density function of V j is λ j e −λ j t . Thus P(V j >u)= e −λ j u . By conditioning on V i ,wehave P parenleftbig V i = min(V 1 ,... ,V k ) parenrightbig = integraldisplay ∞ 0 P parenleftbig min(V 1 ,... ,V k ) = V i | V i = u parenrightbig λ i e −λ i u du = λ i integraldisplay ∞ 0 P parenleftbig min(V 1 ,... ,V k ) = u | V i = u parenrightbig e −λ i u du = λ i integraldisplay ∞ 0 P(V 1 ≥ u,... ,V i−1 ≥ u, V i+1 ≥ u,... ,V k ≥ u | V i = u)e −λ i u du = λ i integraldisplay ∞ 0 P(V 1 ≥ u,... ,V i−1 ≥ u, V i+1 ≥ u,... ,V k ≥ u)e −λ i u du = λ i integraldisplay ∞ 0 P(V 1 ≥ u) ···P(V i−1 ≥ u)P(V i+1 ≥ u) ···P(V k ≥ u)e −λ i u du 296 Chapter 12 Stochastic Processes = λ i integraldisplay ∞ 0 e −λ 1 u ···e −λ i−1 u · e −λ i+1 u ···e −λ k u · e −λ i u du = λ i integraldisplay ∞ 0 e −(λ 1 +···+λ k )u du = λ i integraldisplay ∞ 0 e −λu du = λ i λ . 12.3 MARKOV CHAINS 1. {X n : n = 1, 2,...} is not a Markov chain. For example, P(X 4 = 1) depends on all the values of X 1 , X 2 , and X 3 , and not just X 3 . 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(X n = j) = ∞ summationdisplay i=0 P(X n = j | X 0 = i)P(X 0 = i) = ∞ summationdisplay i=0 p n ij p(i), where p n ij is the ijth entry of the matrix P n . 3. The transition probability matrix of this Markov chain is P = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 01/20001/2 1/201/2000 01/201/20 0 001/201/20 0001/201/2 1/20001/20 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . By calculating P 4 and P 5 , we will find that, (a) the probability that in 4 transitions the Markov chain returns to 1 is P 4 11 = 3/8; (b) the probability that, in 5 transitions, the Markov chain enters 2 or 6 is p 5 12 + p 5 16 = 11 32 + 11 32 = 11 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) = parenleftBig 1 2 parenrightBig n−1 parenleftBig 1 4 parenrightBig ,n= 1, 2,... . The probability that the process ever enters 1 is P(Z <∞) = ∞ summationdisplay n=1 parenleftBig 1 2 parenrightBig n−1 parenleftBig 1 4 parenrightBig = 1/4 1 − (1/2) = 1 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, p 21 + p 23 = 1/2 + 1/6 = 2/3. (b) The desired probability is p 21 p 11 + p 21 p 13 + p 23 p 31 + p 23 p 33 = 1/4. 6. Let X n denote the number of balls in urn I after n transfers. The stochastic process {X n : n = 0, 1,...} is a Markov chain with state space {0, 1,... ,5} and transition probability matrix P = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 010000 1/504/50 0 0 02/503/50 0 003/502/50 0004/501/5 000010 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . Direct calculations show that P (6) = P 6 = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 241 3125 0 2044 3125 0 168 625 0 0 5293 15625 0 9492 15625 0 168 3125 1022 15625 0 9857 15625 0 4746 15625 0 0 4746 15625 0 9857 15625 0 1022 15625 168 3125 0 9492 15625 0 5293 15625 0 0 168 625 0 2044 3125 0 241 3125 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . 298 Chapter 12 Stochastic Processes Hence, by Theorem 12.5, P(X 6 = 4) = 0 · 168 625 + 1 15 · 0 + 2 15 · 4746 15625 + 3 15 · 0 + 4 15 · 5293 15625 + 5 15 · 0 = 0.1308. 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 Z n be the outcome of the nth toss. Then X n+1 = max(X n ,Z n+1 ) shows that {X n : n = 1, 2,...} is a Markov chain. Its state space is {1, 2,... ,6}, and its transition probability matrix is given by P = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 1/61/61/61/61/61/6 02/61/61/61/61/6 003/61/61/61/6 0004/61/61/6 00005/61/6 000001 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . 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: ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 001/201/2000 10000000 01000000 001/32/30000 000002/503/5 00 0 0 1/201/20 000003/502/5 00 0 0 1/302/30 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . 10. For 1 ≤ i ≤ 7, starting from state i, let x i be the probability that the Markov chain will eventually be absorbed into state 4. We are interested in x 6 . Applying the law of total Section 12.3 Markov Chains 299 probability repeatedly, we obtain the following system of linear equations: ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ x 1 = (0.3)x 1 + (0.7)x 2 x 2 = (0.3)x 1 + (0.2)x 2 + (0.5)x 3 x 3 = (0.6)x 4 + (0.4)x 5 x 4 = 1 x 5 = x 3 x 6 = (0.1)x 1 + (0.3)x 2 + (0.1)x 3 + (0.2)x 5 + (0.2)x 6 + (0.1)x 7 x 7 = 0. Solving this system of equations, we obtain ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ x 1 = x 2 = x 3 = x 4 = x 5 = 1 x 6 = 0.875 x 7 = 0. Therefore, the probability is 0.875 that, starting from state 6, the Markov chain will eventually be absorbed into state 4. 11. Let π 1 , π 2 , and π 3 be the long-run probabilities that the sportsman devotes to horseback riding, sailing, and scuba diving, respectively. Then, by Theorem 12.7, π 1 , π 2 , and π 3 are obtained from solving the system of equations. ⎛ ⎝ π 1 π 2 π 3 ⎞ ⎠ = ⎛ ⎝ 0.20 0.32 0.60 0.30 0.15 0.13 0.50 0.53 0.27 ⎞ ⎠ ⎛ ⎝ π 1 π 2 π 3 ⎞ ⎠ along with π 1 + π 2 + π 3 = 1. The matrix equation above gives us the following system of equations ⎧ ⎪ ⎨ ⎪ ⎩ π 1 = 0.20π 1 + 0.32π 2 + 0.60π 3 π 2 = 0.30π 1 + 0.15π 2 + 0.13π 3 π 3 = 0.50π 1 + 0.53π 2 + 0.27π 3 . By choosing any two of these equations along with π 1 + π 2 + π 3 = 1, we obtain a system of three equations in three unknowns. Solving that system yields π 1 = 0.38856, π 2 = 0.200056, and π 3 = 0.411383. Hence the long-run probability that on a randomly selected vacation day the sportsman sails is approximately 0.20. 12. For n ≥ 1, let X n = braceleftBigg 1 if the nth fish caught is trout 0 if the nth fish caught is not trout. 300 Chapter 12 Stochastic Processes Then {X n : n = 1, 2,...} is a Markov chain with state space {0, 1} and transition probability matrix parenleftbigg 10/11 1/11 8/91/9 parenrightbigg Let π 0 be the fraction of fish in the lake that are not trout, and π 1 be the fraction of fish in the lake that are trout. Then, by Theorem 12.7, π 0 and π 1 satisfy parenleftbigg π 0 π 1 parenrightbigg = parenleftbigg 10/11 8/9 1/11 1/9 parenrightbiggparenleftbigg π 0 π 1 parenrightbigg , which gives us the following system of equations ⎧ ⎨ ⎩ π 0 = (10/11)π 0 + (8/9)π 1 π 1 = (1/11)π 0 + (1/9)π 1 . By choosing any one of these equations along with the relation π 0 + π 1 = 1, we obtain a system of two equations in two unknown. Solving that system yields π 0 = 88/97 ≈ 0.907 and π 1 = 9/97 ≈ 0.093. Therefore, approximately 9.3% of the fish in the lake are trout. 13. Let X n = ⎧ ⎪ ⎨ ⎪ ⎩ 1 if the nth card is drawn by player I 2 if the nth card is drawn by player II 3 if the nth card is drawn by player III. {X n : n = 1, 2,...} is a Markov chain with probability transition matrix P = ⎛ ⎝ 48/52 4/52 0 039/52 13/52 12/52 0 40/52 ⎞ ⎠ . Let π 1 , π 2 , and π 3 be the proportion of cards drawn by players I, II, and III, respectively. π 1 , π 2 , and π 3 are obtained from ⎛ ⎝ π 1 π 2 π 3 ⎞ ⎠ = ⎛ ⎝ 12/13 0 3/13 1/13 3/40 01/410/13 ⎞ ⎠ ⎛ ⎝ π 1 π 2 π 3 ⎞ ⎠ and π 1 + π 2 + π 3 = 1, which gives π 1 = 39/64 ≈ 0.61, π 2 = 12/64 ≈ 0.19, and π 3 = 13/64 ≈ 0.20. 14. For 1 ≤ i ≤ 9, let π 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 π i ’s satisfy ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ π 1 π 2 π 3 π 4 π 5 π 6 π 7 π 8 π 9 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 01/301/300000 1/201/201/40000 01/30 0 01/3000 1/20001/401/20 0 01/301/301/301/30 001/201/40001/2 0001/30 0 01/30 00001/401/201/2 000001/301/30 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ π 1 π 2 π 3 π 4 π 5 π 6 π 7 π 8 π 9 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . Solving this system of equations along with summationtext 9 i=1 π 1 , we obtain π 1 = π 3 = π 7 = π 9 = 1/12, π 2 = π 4 = π 6 = π 8 = 1/8, π 5 = 1/6. 15. Let X n denote the number of balls in urn I after n transfers. The stochastic process {X n : n = 0, 1,...} is a Markov chain with state space {0, 1,... ,5} and transition probability matrix P = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 010000 1/504/50 0 0 02/503/50 0 003/502/50 0004/501/5 000010 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . Clearly, {X n : n = 0, 1,...} is an irreducible recurrent Markov chain; since it is finite-state, it is positive recurrent. However, {X n : 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 π i be the fraction of time urn I contains i balls. Then with this interpretation, π i ’s satisfy the following equations ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ π 0 π 1 π 2 π 3 π 4 π 5 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 01/50 0 00 102/50 00 04/503/500 003/504/50 00 02/501 00 0 01/50 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ π 0 π 1 π 2 π 3 π 4 π 5 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ , summationtext 5 i=0 π i = 1. Solving these equations, we obtain π 0 = π 5 = 1/31, π 1 = π 4 = 5/31, π 2 = π 3 = 10/31. 302 Chapter 12 Stochastic Processes Therefore, the fraction of time an urn is empty is π 0 +π 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 X n be the number of balls in urn I immediately before the nth game begins. Then {X n : n = 1, 2,...} is a Markov chain with state space {0, 1,... ,7} and transition probability matrix P = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 3/41/4000000 1/41/21/400000 01/41/21/40000 001/41/21/40 0 0 0001/41/21/40 0 00001/41/21/40 000001/41/21/4 0000001/43/4 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . Since the transition probability matrix is doubly stochastic; that is, the sum of each column is also 1, for i = 0, 1,... ,7, π i , the long-run 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 long-run probability mass function of the number of balls in urn I or II is 1/8 for i = 0, 1,... ,7. Solution 2: Let X n be the number of balls in the urn selected at step 1 of the nth game. Then {X n : n = 1, 2,...} is a Markov chain with state space {0, 1,... ,7} and transition probability matrix P = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 1/20000001/2 1/41/40 0 0 01/41/4 01/41/40 01/41/40 001/41/41/41/40 0 0001/21/20 0 0 001/41/41/41/40 0 01/41/40 01/41/40 1/41/40 0 0 01/41/4 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . Since the transition probability matrix is doubly stochastic; that is, the sum of each column is also 1, for i = 0, 1,... ,7, π i , the long-run 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 long-run 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 p 00 > 0 makes 0 aperiodic), all states are recurrent and aperiodic. Let π k be the long-run 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 ⎛ ⎜ ⎜ ⎜ ⎝ π 0 π 1 π 2 . . . ⎞ ⎟ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎜ ⎝ p 1 100... p 2 010... p 3 001... . . . ⎞ ⎟ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎜ ⎝ π 0 π 1 π 2 . . . ⎞ ⎟ ⎟ ⎟ ⎠ along with summationtext ∞ i=0 π i = 1, we obtain π 0 = 1 1 + summationtext ∞ i=1 (1 − p 1 − p 2 −···−p i ) , π k = 1 − p 1 − p 2 −···−p k 1 + summationtext ∞ i=1 (1 − p 1 − p 2 −···−p i ) ,k≥ 1. 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, ND, and NN similarly, and label the states DD, DN, ND, and NN by 0, 1, 2, and 3, respectively. Let X n = 0 if the nth and (n − 1)st movies Mr. Gorfin watched were both dramas. Define X n = 1, 2, and 3 similarly. Then {X n : n = 1, 2,...} is a Markov chain with state space {0, 1, 2, 3} and transition probability matrix P = ⎛ ⎜ ⎜ ⎝ 7/81/80 0 001/21/2 1/21/20 0 001/87/8 ⎞ ⎟ ⎟ ⎠ . (a) If the first two movies Mr. Gorfin watched last weekend were dramas, the probability that the fourth one is a drama is p 2 00 + p 2 02 . Since P 2 = ⎛ ⎜ ⎜ ⎝ 49/64 7/64 1/16 1/16 1/41/41/16 7/16 7/16 1/16 1/41/4 1/16 1/16 7/64 49/64 ⎞ ⎟ ⎟ ⎠ , the desired probability is (49/64) + (1/16) = 53/64. (b) Let π 0 denote the long-run probability that Mr. Gorfin watches two dramas in a row. Define π 1 , π 2 , and π 3 similarly. We have that, ⎛ ⎜ ⎜ ⎝ π 0 π 1 π 2 π 3 ⎞ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎝ 7/801/20 1/801/20 01/201/8 01/207/8 ⎞ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎝ π 0 π 1 π 2 π 3 ⎞ ⎟ ⎟ ⎠ . Solving this system along with π 0 +π 1 +π 2 +π 3 = 1, we obtain π 0 = 2/5, π 1 = 1/10, π 2 = 1/10, and π 3 = 2/5. Hence the probability that Mr. Gorfin watches two dramas in a row is 2/5. 304 Chapter 12 Stochastic Processes 19. Clearly, X n+1 = braceleftBigg 0 if the (n + 1)st outcome is 6 1 + X n otherwise. This relation shows that {X n : n = 1, 2,...} is a Markov chain. Its transition probability matrix is given by P = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 1/65/6000... 1/605/60 0... 1/60 05/60... 1/60005/6 ... . . . ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . It is readily seen that all states communicate with 0. Therefore, by transitivity of the com- munication property, all states communicate with each other. Therefore, the Markov chain is irreducible. Clearly, 0 is recurrent. Since p 00 > 0, it is aperiodic as well. Hence all states are recurrent and aperiodic. On the other hand, starting at 0, the expected number of transi- tions 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, π i = parenleftBig 5 6 parenrightBig i parenleftBig 1 6 parenrightBig ,i= 0, 1, 2,... . This can also be shown by solving the following system of equations: ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ π 0 π 1 π 2 π 3 . . . ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 1/61/61/61/6 ... 5/6000... 05/60 0... 005/60... . . . ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ π 0 π 1 π 2 π 3 . . . ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ π 0 + π 1 + π 2 +···=1. 20. (a) Let X n = braceleftBigg 1 if Alberto wins the nth game 0 if Alberto loses the nth game. Then {X n : n = 1, 2,...} is a Markov chain with state space {0, 1}. Its transition probability matrix is P = parenleftBig 1 − pp p 1 − p parenrightBig . Using induction, we will now show that P (n) = P n = ⎛ ⎜ ⎜ ⎝ 1 2 + 1 2 (1 − 2p) n 1 2 − 1 2 (1 − 2p) n 1 2 − 1 2 (1 − 2p) n 1 2 + 1 2 (1 − 2p) n ⎞ ⎟ ⎟ ⎠ . Section 12.3 Markov Chains 305 Clearly, for n = 1, P (1) = P. Suppose that P (n) = ⎛ ⎜ ⎜ ⎜ ⎝ 1 2 + 1 2 (1 − 2p) n 1 2 − 1 2 (1 − 2p) n 1 2 − 1 2 (1 − 2p) n 1 2 + 1 2 (1 − 2p) n ⎞ ⎟ ⎟ ⎟ ⎠ . We will show that P n+1 = ⎛ ⎜ ⎜ ⎜ ⎝ 1 2 + 1 2 (1 − 2p) n+1 1 2 − 1 2 (1 − 2p) n+1 1 2 − 1 2 (1 − 2p) n+1 1 2 + 1 2 (1 − 2p) n+1 ⎞ ⎟ ⎟ ⎟ ⎠ . To do so, note that P (n+1) = parenleftbigg p 00 p 01 p 10 p 11 parenrightbiggparenleftbigg p n 00 p n 01 p n 10 p n 11 parenrightbigg = parenleftbigg p 00 p n 00 + p 01 p n 10 p 00 p n 01 + p 01 p n 11 p 10 p n 00 + p 11 p n 10 p 10 p n 01 + p 11 p n 11 parenrightbigg . Thus p n+1 11 = p 10 p n 01 + p 11 p n 11 = p bracketleftBig 1 2 − 1 2 (1 − 2p) n bracketrightBig + (1 − p) bracketleftBig 1 2 + 1 2 (1 − 2p) n bracketrightBig = 1 2 bracketleftbig p + (1 − p) bracketrightbig + 1 2 (1 − 2p) n bracketleftbig − p + (1 − p) bracketrightbig = 1 2 + 1 2 (1 − 2p) n+1 . This establishes what we wanted to show. The proof that p n+1 00 = 1 2 + 1 2 (1 − 2p) n+1 is identical to what we just showed. We have P n+1 01 = 1 − P n+1 00 = 1 − bracketleftBig 1 2 + 1 2 (1 − 2p) n bracketrightBig = 1 2 − 1 2 (1 − 2p) n . Similarly, p n+1 10 = 1 − p n+1 11 = 1 2 − 1 2 (1 − 2p) n . (b) Let π 0 and π 1 be the long-run probabilities that Alberto loses and wins a game, respec- tively. Then parenleftbigg π 0 π 1 parenrightbigg = parenleftbigg 1 − pp p 1 − p parenrightbiggparenleftbigg π 0 π 1 parenrightbigg , and π 0 + π 1 = 1 imply that π 0 = π 1 = 1/2. Therefore, the expected number of games Alberto will play between two consecutive wins is 1/π 1 = 2. 306 Chapter 12 Stochastic Processes 21. For each j ≥ 0, lim n→∞ p n ij exists and is independent of i if the following system of equations, in π 0 , π 1 , ..., have a unique solution. ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ π 0 π 1 π 2 π 3 . . . ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 1 − p 1 − p 0000... p 01− p 000... 0 p 01− p 00... 00p 01− p 0 ... . . . ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ π 0 π 1 π 2 π 3 . . . ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ π 0 + π 1 + π 2 +···=1. From the matrix equation, we obtain π i = parenleftBig p 1 − p parenrightBig i π 0 ,i= 0, 1,... . For these quantities to satisfy summationtext ∞ i=0 π i = 1, we need the geometric series ∞ summationdisplay i=0 parenleftBig p 1 − p parenrightBig i to converge. Hence we must have p<1 − p,orp<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, π 0 ∞ summationdisplay i=0 parenleftBig p 1 − p parenrightBig i = 1 yields π 0 = 1 − p 1 − p . Thus the limiting probabilities are π i = parenleftBig p 1 − p parenrightBig i parenleftBig 1 − p 1 − p parenrightBig ,i= 0, 1, 2,... . 22. Let Y n be Carl’s fortune after the nth game. Let X n be Stan’s fortune after the nth game. Let Z n = Y n − X n . The {Z n : n = 0, 1,...} is a random walk with state space {0, ±2, ±4,...}. We have that Z 0 = 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| Z 1 = 2)P(Z 1 = 2) + P(A| Z 1 =−2)P(Z 1 =−2) = 1 · (0.46) + bracketleftbig P(A) bracketrightbig 2 · (0.54). To show that P(A| Z 1 =−2) = bracketleftbig P(A) bracketrightbig 2 , 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,wehave P(A| Z =−2) = P(EA)= P(E)P(A)= bracketleftbig P(A) bracketrightbig 2 . Section 12.3 Markov Chains 307 We have shown that P(A), the quantity we are interested in, satisfies (0.54) bracketleftbig P(A) bracketrightbig 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.Form = 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(X n+m = j | X 0 = i 0 ,X 1 = i 1 ,... ,X n = i n ) = summationdisplay i∈S P(X n+m = j | X 0 = i 0 ,... ,X n = i n ,X n+m−1 = i) P(X n+m−1 = i | X 0 = i 0 ,... ,X n = i n ) = summationdisplay i∈S P(X n+m = j | X n+m−1 = i)P(X n+m−1 = i | X n = i n ) = summationdisplay i∈S P(X n+m = j | X n+m−1 = i, X n = i n )P(X n+m−1 = i | X n = i n ) = P(X n+m = j | X n = i n ), where the following relations are valid from the definition of Markov chain: given the present state, the process is independent of the past. P(X n+m = j | X 0 = i 0 ,... ,X n = i n ,X n+m−1 = i) = P(X n+m = j | X n+m−1 = i), P(X n+m = j | X n+m−1 = i) = P(X n+m = j | X n+m−1 = i, X n = i n ). 24. Let (0, 0), the origin, be denoted by O. It should be clear that, for all n ≥ 0, P 2n+1 OO = 0. Now, for n ≥ 1, let Z 1 , Z 2 , Z 3 , and Z 4 be the number of transitions to the right, left, up, and down, respectively. The joint probability mass function of Z 1 , Z 2 , Z 3 , and Z 4 is multinomial. We have P 2n OO = n summationdisplay i=0 P(Z 1 = i,Z 2 = i,Z 3 = n − i,Z 4 = n − i) = n summationdisplay i=0 (2n)! i! i! (n − i)! (n − i)! parenleftBig 1 4 parenrightBig i parenleftBig 1 4 parenrightBig i parenleftBig 1 4 parenrightBig n−i parenleftBig 1 4 parenrightBig n−i = n summationdisplay i=0 (2n)! n! n! · n! i! (n − i)! · n! i! (n − i)! parenleftBig 1 4 parenrightBig 2n = parenleftBig 1 4 parenrightBig 2n parenleftbigg 2n n parenrightbigg n summationdisplay i=0 parenleftbigg n i parenrightbigg 2 . 308 Chapter 12 Stochastic Processes By Example 2.28, n summationdisplay i=0 parenleftbigg n i parenrightbigg 2 = parenleftbigg 2n n parenrightbigg . Thus P n OO = parenleftBig 1 4 parenrightBig 2n parenleftbigg 2n n parenrightbigg 2 . Now, by Theorem 2.7 (Stirling’s formula), parenleftBig 1 4 parenrightBig 2n parenleftbigg 2n n parenrightbigg 2 = parenleftBig 1 4 parenrightBig 2n · bracketleftBig (2n)! n! n! bracketrightBig 2 ∼ 1 4 2n · bracketleftBig √ 4πn(2n) 2n e −2n ( √ 2πn· n n · e −n ) 2 bracketrightBig 2 = 1 πn . Therefore, ∞ summationdisplay n=1 P n OO = ∞ summationdisplay n=1 parenleftBig 1 4 parenrightBig 2n parenleftbigg 2n n parenrightbigg 2 is convergent if and only if ∞ summationdisplay n=1 1 πn is convergent. Since 1 π ∞ summationdisplay n=1 1 n is divergent, summationtext ∞ n=1 P n OO is divergent, showing that the state (0, 0) is recurrent. 25. Clearly, P(X n+1 = 1 | X n = 0) = 1. For i ≥ 1, given X n = i, either X n+1 = i + 1 in which case we say that a transition to the right has occurred, or X n+1 = i − 1 in which case we say that a transition to the left has occurred. For i ≥ 1, given X n = 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 µ and λ, respectively. For i ≥ 1, P(X n+1 = i + 1 | X n = i) = P(T ~~T | T = t)λe −λt dt = integraldisplay ∞ 0 P(S > t)λe −λt dt = integraldisplay ∞ 0 e −µt · λe −λt dt = λ λ + µ . Therefore, P(X n+1 = i − 1 | X n = i) = P(T >S)= 1 − λ λ + µ = µ λ + µ . These calculations show that knowing X n , the next transition does not depend on the values of X j for j~~ 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 ijth element of PQis the product of the ith row of P with the jth column of Q. Thus it is summationdisplay lscript p ilscript q lscriptj . To show that the sum of each row of PQis 1, we will now calculate the sum of the elements of the ith row of PQ, which is summationdisplay j summationdisplay lscript p ilscript q lscriptj = summationdisplay lscript summationdisplay j p ilscript q lscriptj = summationdisplay lscript parenleftBig p ilscript summationdisplay j q lscriptj parenrightBig = summationdisplay lscript p ilscript = 1. Note that summationdisplay j q lscriptj = 1 and summationdisplay lscript p ilscript = 1 since the sum of the elements of the lscriptth 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 = i 1 ,i 2 ,i 3 , ... , i n = j from i to j.Ifn ≤ K, we are done. If n>K, by the pigeonhole principle, there must exist k and lscript (k 0} and J ={n ≥ 1: p n jj > 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.Ifd(i) negationslash= 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) 0 and p m ji > 0. Let k ∈ J; then p k jj > 0. We have p n+m ii ≥ p n ij p m ji > 0, 310 Chapter 12 Stochastic Processes and p n+k+m ii ≥ p n ij p k jj p m ji > 0. 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)ioccurs 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 P(Z >i)= 1 − i k ,i= 1, 2,... ,k− 1. Using this fact, for 0 ≤ i ≤ k − 2, we obtain p i(i+1) = P(X n+1 = i + 1 | X n = i) = P(X n = i, X n+1 = i + 1) P(X n = i) = P(Z >i+ 1) P(Z >i) = 1 − i + 1 k 1 − i k = k − i − 1 k − i , p i0 = P(X n+1 = 0 | X n = i) = 1 − k − i − 1 k − i = 1 k − i , p (k−1)0 = P(X n+1 = 0 | X n = k − 1) = 1. Hence the transition probability matrix of {X n : n = 1, 2,...} is given by Section 12.3 Markov Chains 311 P = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 1 k 1 − 1 k 000... 00 1 k − 1 01− 1 k − 1 00... 00 1 k − 2 001− 1 k − 2 0 ... 00 1 k − 3 00 01− 1 k − 3 ... 00 . . . 1 2 0000... 0 1 2 10000... 00 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . It should be clear that the Markov chain under consideration is irreducible, aperiodic, and positively recurrent. For 0 ≤ i ≤ k − 1, let π i be the long-run probability that Liz says no to Bob for i consecutive weeks. π 0 , π 1 , ..., π k−1 are obtained from solving the following matrix equation along with summationtext k−1 i=0 π i = 1. ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ π 0 π 1 π 2 π 3 . . . π k−2 π k−1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 1 k 1 k − 1 1 k − 2 1 k − 3 ... 1 2 1 1 − 1 k 000... 00 01− 1 k − 1 00... 00 001− 1 k − 2 0 ... 00 00 01− 1 k − 3 ... 00 . . . 0000... 1 2 0 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ π 0 π 1 π 2 π 3 . . . π k−2 π k−1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . The matrix equation gives π i = k − i k π 0 ,i= 1, 2,... ,k− 1. 312 Chapter 12 Stochastic Processes Using summationtext k−1 i=0 π i = 1, we obtain π 0 k−1 summationdisplay i=0 k − i k = 1 or, equivalently, π 0 k bracketleftBig k−1 summationdisplay i=0 k − k−1 summationdisplay i=0 i bracketrightBig = 1. This implies that π 0 k bracketleftBig k 2 − (k − 1)k 2 bracketrightBig = 1, which gives π 0 = 2/(k + 1). Hence π i = 2(k − i) k(k + 1) ,i= 0, 1, 2,... ,k− 1. 30. Let X i be the amount of money player A has after i games. Clearly, X 0 = a and {X n : n = 0, 1,...} is a Markov chain with state space {0, 1,... ,a,a+ 1,... ,a+ b}. For 0 ≤ i ≤ a + b, let m i = E(T | X 0 = i). Let F be the event that A wins the first game. Then, for 1 ≤ i ≤ a + b − 1, E(T | X 0 = i) = E(T | X 0 = i,F)P(F | X 0 = i)+ E(T | X 0 = i,F c )P(F c | X 0 = i). This gives m i = (1 + m i+1 ) 1 2 + (1 + m i−1 ) 1 2 , 1 ≤ i ≤ a + b − 1, or, equivalently, 2m i = 2 + m i+1 + m i−1 , 1 ≤ i ≤ a + b − 1. Now rewrite this relation as m i+1 − m i =−2 + m i − m i−1 , 1 ≤ i ≤ a + b − 1, and, for 1 ≤ i ≤ a + b, let y i = m i − m i−1 . Then y i+1 =−2 + y i , 1 ≤ i ≤ a + b − 1, and, for 1 ≤ i ≤ a + b, m i = y 1 + y 2 +···+y i . Clearly, m 0 = 0, m a+b = 0,y 1 = m 1 , and y 2 =−2 + y 1 =−2 + m 1 , y 3 =−2 + y 2 =−2 + (−2 + m 1 ) =−4 + m 1 . . . y i =−2(i − 1) + m 1 , 1 ≤ i ≤ a + b. Section 12.3 Markov Chains 313 Hence, for 1 ≤ i ≤ a + b, m i = y 1 + y 2 +···+y i = im 1 − 2 bracketleftbig 1 + 2 +···+(i − 1) bracketrightbig = im 1 − i(i − 1) = i(m 1 − i + 1). This and m a+b = 0 imply that (a + b)(m 1 − a − b + 1) = 0, or m 1 = a + b − 1. Therefore, m i = i(a + b − i), and hence the desired quantity is E(T | X 0 = a) = m a = ab. 31. Let q be a positive solution of the equation x = summationtext ∞ i=0 α i x i . Then q = summationtext ∞ i=0 α i q i . We will show that ∀n ≥ 0, P(X n = 0) ≤ q. This implies that p = lim n→∞ P(X n = 0) ≤ q. To establish that P(X n = 0) ≤ q, we use induction. For n = 0, P(X 0 = 0) = 0 ≤ q is trivially true. Suppose that P(X n = 0) ≤ q.We have P(X n+1 = 0) = ∞ summationdisplay i=0 P(X n+1 = 0 | X 1 = i)P(X 1 = i). It should be clear that P(X n+1 = 0 | X 1 = i) = bracketleftbig P(X n = 0 | X 0 = 1) bracketrightbig i . However, since P(X 0 = 1) = 1, P(X n = 0 | X 0 = 1) = P(X n = 0). Therefore, P(X n+1 = 0 | X 1 = i) = bracketleftbig P(X n = 0) bracketrightbig i . Thus P(X n+1 = 0) = ∞ summationdisplay i=0 bracketleftbig P(X n = 0) bracketrightbig i P(X 1 = i) ≤ ∞ summationdisplay i=0 q i α i = q. This establishes the theorem. 314 Chapter 12 Stochastic Processes 32. Multiplying P successively, we obtain p 12 = 1 13 p 2 12 = parenleftBig 9 13 parenrightBigparenleftBig 1 13 parenrightBig + 1 13 , p 3 12 = parenleftBig 9 13 parenrightBig 2 parenleftBig 1 13 parenrightBig + parenleftBig 9 13 parenrightBigparenleftBig 1 13 parenrightBig + 1 13 , and in general, p n 12 = 1 13 bracketleftBigparenleftBig 9 13 parenrightBig n−1 + parenleftBig 9 13 parenrightBig n−2 +···+1 bracketrightBig = 1 13 · 1 − parenleftBig 9 13 parenrightBig n 1 − 9 13 = 1 4 bracketleftBig 1 − parenleftBig 9 13 parenrightBig n bracketrightBig . Hence the desired probability is lim n→∞ p n 12 = 1/4. 33. We will use induction. Let n = 1; then, for 1 + j − i to be nonnegative, we must have i − 1 ≤ j. For the inequality 1 + j − i 2 ≤ 1 to be valid, we must have j ≤ i + 1. Therefore, i − 1 ≤ j ≤ i + 1. But, for j = i,1+ j − i is not even. Therefore, if 1 + j − i is an even nonnegative integer satisfying 1 + j − i 2 ≤ 1, we must have j = i − 1orj = i + 1. For j = i − 1, n + j − i 2 = 1 + i − 1 − i 2 = 0 and n − j + i 2 = 1 − i + 1 + i 2 = 1. Hence P(X 1 = i − 1 | X 0 = i) = 1 − p = parenleftbigg 1 0 parenrightbigg p 0 (1 − p) 1 , showing that the relation is valid. For j = i + 1, n + j − i 2 = 1 + i + 1 − i 2 = 1 and n − j + i 2 = 1 − i − 1 + i 2 = 0. Hence P(X 1 = i + 1 | X 0 = i) = p = parenleftbigg 1 1 parenrightbigg p 1 (1 − p) 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(X 1 = j | X 0 = i) = 0. Section 12.4 Continuous-Time 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 X n : P(X n+1 = j | X 0 = i) = P(X n+1 = j | X 0 = i, X n = j − 1)P(X n = j − 1 | X 0 = i) + P(X n+1 = j | X 0 = i, X n = j + 1)P(X n = j + 1 | X 0 = i) = P(X n+1 = j | X n = j − 1)P(X n = j − 1 | X 0 = i) + P(X n+1 = j | X n = j + 1)P(X n = j + 1 | X 0 = i) = p · parenleftbigg n n + j − 1 − i 2 parenrightbigg p (n+j−1−i)/2 (1 − p) (n−j+1+i)/2 + (1 − p) parenleftbigg n n + j + 1 − i 2 parenrightbigg p (n+j+1−i)/2 (1 − p) (n−j−1+i)/2 = bracketleftbiggparenleftbigg n n − 1 + j − i 2 parenrightbigg + parenleftbigg n n + 1 + j − i 2 parenrightbiggbracketrightbigg p (n+1+j−i)/2 (1 − p) (n+1−j+i)/2 = parenleftbigg n + 1 n + 1 + j − i 2 parenrightbigg p (n+1+j−i)/2 (1 − p) (n+1−j+i)/2 . 12.4 CONTINUOUS-TIME MARKOV CHAINS 1. By Chapman-Kolmogorov equations, p ij (t + h) − p ij (t) = ∞ summationdisplay k=0 p ik (h)p kj (t) − p ij (t) = summationdisplay knegationslash=i p ik (h)p kj (t) + p ii (h)p ij (t) − p ij (t) = summationdisplay knegationslash=i p ik (h)p kj (t) + p ij (t) bracketleftbig p ii (h) − 1 bracketrightbig . Thus p ij (t + h) − p ij (t) h = summationdisplay knegationslash=i p ik (h) h p kj (t) − p ij (t) 1 − p ii (h) h . Letting h → 0, by (12.13) and (12.14), we have p prime ij (t) = summationdisplay knegationslash=i q ik p kj (t) − ν i p ij (t). 316 Chapter 12 Stochastic Processes 2. Clearly, braceleftbig X(t): t ≥ bracerightbig is a continuous-time Markov chain. Its balance equations are as follows: State Input rate to = Output rate from f µπ 0 = λπ f 0 λπ f + µπ 1 + µπ 2 + µπ 3 = µπ 0 + λπ 0 1 λπ 0 = λπ 1 + µπ 1 2 λπ 1 = λπ 2 + µπ 2 3 λπ 2 = µπ 3 . Solving these equations along with π f + π 0 + π 1 + π 2 + π 3 = 1 we obtain π f = µ 2 λ(λ + µ) ,π 0 = µ λ + µ , π 1 = λµ (λ + µ) 2 ,π 2 = λ 2 µ (λ + µ) 3 , π 3 = parenleftBig λ λ + µ parenrightBig 3 . 3. The fact that braceleftbig X(t): t ≥ 0 bracerightbig is a continuous-time Markov chain should be clear. The balance equations are State Input rate to = Output rate from (0, 0) µπ (1,0) + λπ (0,1) = λπ (0,0) + µπ (0,0) (n, 0) µπ (n+1,0) + λπ (n−1,0) = λπ (n,0) + µπ (n,0) ,n≥ 1 (0,m) λπ (0,m+1) + µπ (0,m−1) = λπ (0,m) + µπ (0,m) m ≥ 1. 4. Let X(t) be the number of customers in the system at time t. Then the process braceleftbig X(t): t ≥ 0 bracerightbig is a birth and death process with λ n = λ, n ≥ 0, and µ n = nµ, n ≥ 1. To find π 0 , the probability that the system is empty, we will first calculate the sum in (12.18). We have ∞ summationdisplay n=1 λ 0 λ 1 ···λ n−1 µ 1 µ 2 ···µ n = ∞ summationdisplay n=1 λ n n! µ n = ∞ summationdisplay n=1 1 n! parenleftBig λ µ parenrightBig n =−1 + ∞ summationdisplay n=0 1 n! parenleftBig λ µ parenrightBig n =−1 + e λ/µ . Hence, by (12.18), π 0 = 1 1 − 1 + e λ/µ = e −λ/µ . Section 12.4 Continuous-Time Markov Chains 317 By (12.17), π n = λ n π 0 n!µ n = (λ/µ) n e −λ/µ n! ,n= 0, 1, 2,... . This shows that the long-run number of customers in such an M/M/∞ queueing system is Poisson with parameter λ/µ. The average number of customers in the system is, therefore, λ/µ. 5. Let X(t) be the number of operators busy serving customers at time t. Clearly, braceleftbig X(t): t ≥ 0 bracerightbig is a finite-state birth and death process with state space {0, 1,... ,c}, birth rates λ n = λ, n = 0, 1,... ,c, and death rates µ n = nµ, n = 0, 1,... ,c. Let π 0 be the proportion of time that all operators are free. Let π c be the proportion of time all of them are busy serving customers. (a) π c is the desired quantity. By (12.22), π 0 = 1 1 + c summationdisplay n=1 λ n n! µ n = 1 c summationdisplay n=0 1 n! parenleftBig λ µ parenrightBig n . By (12.21), π c = 1 c! (λ/µ) c summationtext c n=0 1 n! (λ/µ) n . This formula is called Erlang’s loss formula. (b) We want to find the smallest c for which 1/c! summationtext 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 H n be the time, starting from n, until the process enters state n + 1 for the first time. Clearly, E(H 0 ) = 1/λ and, by Lemma 12.2, E(H n ) = 1 λ + E(H n−1 ), n ≥ 1. 318 Chapter 12 Stochastic Processes Hence E(H 0 ) = 1 λ , E(H 1 ) = 1 λ + 1 λ = 2 λ , E(H 2 ) = 1 λ + 2 λ = 3 λ . Continuing this process, we obtain, E(H n ) = n + 1 λ ,n≥ 0. The desired quantity is j−1 summationdisplay n=i E(H n ) = j−1 summationdisplay n=i n + 1 λ = 1 λ bracketleftbig (i + 1) + (i + 2) +···+j bracketrightbig = 1 λ bracketleftbig (1 + 2 +···+j)− (1 + 2 +···+i) bracketrightbig = 1 λ bracketleftBig j(j + 1) 2 − i(i + 1) 2 bracketrightBig = j(j + 1) − i(i + 1) 2λ . 8. Suppose that a birth occurs each time that an out-of-order machine is repaired and begins to operate, and a death occurs each time that a machine breaks down. The fact that braceleftbig X(t): t ≥ 0 bracerightbig is a birth and death process with state space {0, 1,... ,m} should be clear. The birth and death rates are λ n = braceleftBigg kλ n = 0, 1,... ,m− k (m − n)λ n = m − k + 1,m− k + 2,... ,m, µ n = nµ n = 0, 1,... ,m. 9. The Birth rates are braceleftBigg λ 0 = λ λ n = α n λ, n ≥ 1. The death rates are braceleftBigg µ 0 = 0 µ n = µ + (n − 1)γ, n ≥ 1. 10. Let X(t) be the population size at time t. Then braceleftbig X(t): t ≥ 0 bracerightbig is a birth and death process with birth rates λ n = nλ + γ , n ≥ 0, and death rates µ n = nµ, n ≥ 1. For i ≥ 0, let H i Section 12.4 Continuous-Time 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(H 0 ) + E(H 1 ) + E(H 2 ). Note that, by Lemma 12.2, E(H i ) = 1 λ i + µ i λ i E(H i−1 ), i ≥ 1. Since E(H 0 ) = 1/γ , E(H 1 ) = 1 λ + γ + µ λ + γ · 1 γ = µ + γ γ(λ+ γ) , and E(H 2 ) = 1 2λ + γ + 2µ 2λ + γ · µ + γ γ(λ+ γ) = γ(λ+ γ)+ 2µ(µ + γ) γ(λ+ γ)(2λ + γ) . Thus the desired quantity is E(H 0 ) + E(H 1 ) + E(H 2 ) = (λ + γ)(2λ + γ)+ (µ + γ)(2λ + 2µ + γ)+ γ(λ+ γ) γ(λ+ γ)(2λ + γ) . 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 braceleftbig X(t): t ≥ 0 bracerightbig is a Poisson process with rate µ as long as the population is not extinct. Therefore, for 0 0, j = 0, we have p i0 (t) = 1 − i summationdisplay j=1 p ij (t) = 1 − i summationdisplay j=1 e −µt (µt) i−j (i − j)! = 1 − 1 summationdisplay j=i e −µt (µt) i−j (i − j)! . Letting k = i − j yields p i0 (t) = 1 − i−1 summationdisplay k=0 e −µt (µt) k k! = ∞ summationdisplay k=i e −µt (µt) 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 braceleftbig X(t): t ≥ 0 bracerightbig is a birth and death process with state space {0, 1, 2}. Clearly, X(t) = 0ifatt 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) = 2ifatt none of the physicians is available to answer patients’ calls. We have that λ 0 = 2λ, λ 1 = λ, λ 2 = 0, 320 Chapter 12 Stochastic Processes µ 0 = 0,µ 1 = µ, µ 2 = 2µ. Therefore, ν 0 = 2λ, ν 1 = λ + µ, ν 2 = 2µ. Also, p 01 = p 21 = 1,p 02 = p 20 = 0,p 10 = µ λ + µ ,p 12 = λ λ + µ . Therefore, q 01 = ν 0 p 01 = 2λ, q 10 = ν 1 p 10 = µ, q 12 = ν 1 p 12 = λ, q 21 = ν 2 p 21 = 2µ, q 02 = q 20 = 0. Substituting these quantities in the Kolmogorov backward equations p prime ij (t) = summationdisplay knegationslash=i q ik p kj (t) − ν i p ij (t), we obtain p prime 00 (t) = 2λp 10 (t) − 2λp 00 (t) p prime 01 (t) = 2λp 11 (t) − 2λp 01 (t) p prime 02 (t) = 2λp 12 (t) − 2λp 02 (t) p prime 10 (t) = λp 20 (t) + µp 00 (t) − (λ + µ)p 10 (t) p prime 11 (t) = λp 21 (t) + µp 01 (t) − (λ + µ)p 11 (t) p prime 12 (t) = λp 22 (t) + µp 02 (t) − (λ + µ)p 12 (t) p prime 20 (t) = 2µp 10 (t) − 2µp 20 (t) p prime 21 (t) = 2µp 11 (t) − 2µp 21 (t) p prime 22 (t) = 2µp 12 (t) − 2µp 22 (t). 13. Let X(t) be the number of customers in the system at time t. Then braceleftbig X(t): n ≥ 0 bracerightbig is a birth and death process with λ n = λ, for n ≥ 0, and µ n = braceleftBigg nµ n = 0, 1,... ,c cµ n>c. By (12.21), for n = 1, 2,...c, π n = λ n n! µ n π 0 = 1 n! parenleftBig λ µ parenrightBig n π 0 ; for n>c, π n = λ n c! µ c (cµ) n−c π 0 = λ n c! c n−c µ n π 0 = c c c! parenleftBig λ cµ parenrightBig n π 0 = c c c! ρ n π 0 . Section 12.4 Continuous-Time Markov Chains 321 Noting that summationtext c n=0 π n + summationtext ∞ n=c+1 π n = 1, we have π 0 c summationdisplay n=0 1 n! parenleftBig λ µ parenrightBig n + π 0 c c c! ∞ summationdisplay n=c+1 ρ n = 1. Since ρ<1, we have summationtext ∞ n=c+1 ρ n = ρ c+1 1 − ρ . Therefore, π 0 = 1 c summationdisplay n=0 1 n! parenleftBig λ µ parenrightBig n + c c c! ∞ summationdisplay n=c+1 ρ n = c! (1 − ρ) c! (1 − ρ) c summationdisplay n=0 1 n! parenleftBig λ µ parenrightBig n + c c ρ c+1 . 14. Let s,t > 0. If j

*j.Therefore, for j*

*i. Then ∞ summationdisplay k=0 p ik (s)p kj (t) = j summationdisplay k=i p ik (s)p kj (t) = j summationdisplay k=i e −λs (λs) k−i (k − i)! · e −λt (λt) j−k (j − k)! = e −λ(t+s) (j − i)! j summationdisplay k=i (j − i)! k − i)! (j − k)! (λs) k−i (λt) j−k = e −λ(t+s) (j − i)! j−i summationdisplay lscript=0 (j − i)! lscript! (j − i − lscript)! (λs) lscript (λt) (j−i)−lscript = e −λ(t+s) (j − i)! j−i summationdisplay lscript=0 parenleftbigg j − i lscript parenrightbigg (λs) lscript (λt) (j−i)−lscript = e −λ(t+s) (j − i)! (λs + λt) j−i where the last equality follows by Theorem 2.5, the binomial expansion. Since e −λ(t+s) (j − i)! bracketleftbig λ(t + s) bracketrightbig j−i = p ij (s + t), we have shown that the Chapman-Kolmogorov 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, braceleftbig X(t): t ≥ 0 bracerightbig is a continuous-time Markov chain with state space {1, 2,...} and ν i = iλ, i ≥ 1. parenleftbig In fact, braceleftbig X(t): t ≥ 0 bracerightbig is a pure birth process, but that fact will not help us solve this exercise. parenrightbig Clearly, for i ≥ 1, j ≥ 1, p ij = braceleftBigg 1ifj = i + 1 0ifj negationslash= i + 1. Hence q ij = braceleftBigg ν i if j = i + 1 0ifj negationslash= i + 1. We are interested in finding p 1n (t). This is the desired probability. For n = 1, p 11 (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 λ,wehavep 11 (t) = e −λt .Forn ≥ 2, by the Kolmogorov’s forward equation, p prime 1n (t) = summationdisplay knegationslash=n q kn p 1k (t) − ν n p 1n (t) = q (n−1)n p 1(n−1) (t) − ν n p 1n (t) = ν n−1 p 1(n−1) (t) − ν n p 1n (t). Therefore, p prime 1n (t) = (n − 1)λp 1(n−1) (t) − nλp 1n (t). (49) For n = 2, this gives p prime 12 (t) = λp 11 (t) − 2λp 12 (t) or, equivalently, p prime 12 (t) = λe −λt − 2λp 12 (t). Solving this first order linear differential equation with boundary condition p 12 (0) = 0, we obtain p 12 (t) = e −λt (1 − e −λt ). For n = 3, by (49), p prime 13 (t) = 2λp 12 (t) − 3λp 13 (t) or, equivalently, p prime 13 (t) = 2λe −λt (1 − e −λt ) − 3λp 13 (t). Solving this first order linear differential equation with boundary condition p 13 (0) = 0 yields p 13 (t) = e −λt (1 − e −λt ) 2 . Continuing this process, and using induction, we obtain that p 1n (t) = e −λt (1 − e −λt ) n−1 n ≥ 1. Section 12.4 Continuous-Time Markov Chains 323 16. It is straightforward to see that π (i,j) = parenleftBig λ µ 1 parenrightBig i parenleftBig 1 − λ µ 1 parenrightBigparenleftBig λ µ 2 parenrightBig j parenleftBig 1 − λ µ 2 parenrightBig ,i,j≥ 0, satisfy the following balance equations for the tandem queueing system under consideration. Hence, by Example 12.43, π (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 Input rate to = Output rate from (0, 0) µ 2 π (0,1) = λπ (0,0) (i, 0), i ≥ 1 µ 2 π (i,1) + λπ (i−1,0) = λπ (i,0) + µ 1 π (i,0) (0,j), j ≥ 1 µ 2 π (0,j+1) + µ 1 π (1,j−1) = λπ (0,j) + µ 2 π (0,j) (i,j), i,j ≥ 1 µ 2 π (i,j+1) + µ 1 π (i+1,j−1) + λπ (i−1,j) = λπ (i,j) + µ 1 π (i,j) + µ 2 π (i,j) . 17. Clearly, braceleftbig X(t): t ≥ 0 bracerightbig is a birth and death process with birth rates λ i = iλ, i ≥ 0, and death rates µ i = iµ + γ , i>0; µ 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 mλh+o(h), it is m − 1 with probability (mµ + γ)h+ o(h), and it is still m with probability 1 − mλh − o(h) − (mµ + γ)h− o(h) = 1 − (mλ + mµ + γ)h+ o(h). Therefore, E bracketleftbig X(t + h) | X(t) = m bracketrightbig = (m + 1) bracketleftbig mλh + o(h) bracketrightbig + (m − 1) bracketleftbig (mµ + γ)h+ o(h) bracketrightbig + m bracketleftbig 1 − (mλ + mµ + γ)h+ o(h) bracketrightbig = m + bracketleftbig m(λ − µ) − γ bracketrightbig h + o(h). This relation implies that E bracketleftbig X(t + h) | X(t) bracketrightbig = X(t)+ bracketleftbig (λ − µ)X(t) − γ bracketrightbig h + o(h). Equating the expected values of both sides, and noting that E bracketleftBig E bracketleftbig X(t + h) | X(t) bracketrightbig bracketrightBig = E bracketleftbig X(t + h) bracketrightbig , we obtain E bracketleftbig X(t + h) bracketrightbig = E bracketleftbig X(t) bracketrightbig + h(λ − µ)E bracketleftbig X(t) bracketrightbig − γh+ o(h). For simplicity, let g(t) = E bracketleftbig X(t) bracketrightbig . We have shown that g(t + h) = g(t) + h(λ − µ)g(t) − γh+ o(h) 324 Chapter 12 Stochastic Processes or, equivalently, g(t + h) − g(t) h = (λ − µ)g(t) − γ + o(h) h . As h → 0, this gives g prime (t) = (λ − µ)g(t) − γ. If λ = µ, then g prime (t) =−γ .Sog(t) =−γt + c. Since g(0) = n, we must have c = n,or g(t) =−γt+ n.Ifλ negationslash= µ, to solve the first order linear differential equation, g prime (t) = (λ − µ)g(t) − γ, let f(t)= (λ − µ)g(t) − γ.Then 1 λ − µ f prime (t) = f(t), or f prime (t) f(t) = λ − µ. This yields ln |f(t)|=(λ − µ)t + c, or f(t)= e (λ−µ)t+c = Ke (λ−µ)t , where K = e c . Thus g(t) = K λ − µ e (λ−µ)t + γ λ − µ . Now g(0) = n implies that K = n(γ − µ) − γ.Thus g(t) = E bracketleftbig X(t) bracketrightbig = ne (λ−µ)t + γ λ − µ bracketleftbig 1 − e (λ−µ)t bracketrightbig . 18. For n ≥ 0, let E n be the event that, starting from state n, eventually extinction will occur. Let α n = P(E n ). Clearly, α 0 = 1. We will show that α n = 1, for all n.Forn ≥ 1, starting from n, let Z n be the state to which the process will move. Then Z n is a discrete random variable with set of possible values {n − 1,n+ 1}. Conditioning on Z n yields P(E n ) = P(E n | Z n = n − 1)P(Z n = n − 1) + P(E n | Z n = n + 1)P(Z n = n + 1). Hence α n = α n−1 · µ n λ n + µ n + α n+1 · λ n λ n + µ n ,n≥ 1, or, equivalently, λ n (α n+1 − α n ) = µ n (α n − α n−1 ), n ≥ 1. Section 12.4 Continuous-Time Markov Chains 325 For n ≥ 0, let y n = α n+1 − α n .Wehave λ n y n = µ n y n−1 ,n≥ 1, or y n = µ n λ n y n−1 ,n≥ 1. Therefore, y 1 = µ 1 λ 1 y 0 y 2 = µ 2 λ 2 y 1 = µ 1 µ 2 λ 1 λ 2 y 0 . . . y n = µ 1 µ 2 ···µ n λ 1 λ 2 ···λ n y 0 .n≥ 1. On the other hand, by y n = α n+1 − α n , n ≥ 0, α 1 = α 0 + y 0 = 1 + y 0 α 2 = α 1 + y 1 = 1 + y 0 + y 1 . . . α n+1 = 1 + y 0 + y 1 +···+y n . Hence α n+1 = 1 + y 0 + n summationdisplay k=1 y k = 1 + y 0 + y 0 n summationdisplay k=1 µ 1 µ 2 ···µ k λ 1 λ 2 ···λ k = 1 + y 0 parenleftBig 1 + n summationdisplay k=1 µ 1 µ 2 ···µ k λ 1 λ 2 ···λ k parenrightBig = 1 + (α 1 − 1) parenleftBig 1 + n summationdisplay k=1 µ 1 µ 2 ···µ k λ 1 λ 2 ···λ k parenrightBig . Since ∞ summationdisplay k=1 µ 1 µ 2 ···µ k λ 1 λ 2 ···λ k =∞, the sequence n summationdisplay k=1 µ 1 µ 2 ···µ k λ 1 λ 2 ···λ k increases without bound. For α n ’s to exist, this requires that α 1 = 1, which in turn implies that α n+1 = 1, for n ≥ 1. 326 Chapter 12 Stochastic Processes 12.5 BROWNIAN MOTION 1. (a) By the independent-increments property of Brownian motions, the desired probability is P parenleftbig − 1/2*ε parenrightBig = P parenleftbig |X(t)| >εt parenrightbig = P parenleftbig X(t)>εt parenrightbig + P parenleftbig X(t) < −εt parenrightbig = P parenleftBig Z> εt σ √ t parenrightBig + P parenleftBig Z<− εt σ √ t parenrightBig = P parenleftBig Z> ε √ t σ parenrightBig + P parenleftBig Z<− ε √ t σ parenrightBig = 1 − Phi1 parenleftbig ε √ t/σ parenrightbig + Phi1 parenleftbig − ε √ t/σ parenrightbig = 1 − Phi1 parenleftbig ε √ t/σ parenrightbig + 1 − Phi1 parenleftbig ε √ t/σ parenrightbig = 2 − 2Phi1 parenleftbig ε √ t/σ parenrightbig . This implies that lim t→0 P parenleftBig |X(t)| t >ε parenrightBig = 2 − 1 = 1. whereas lim t→∞ P parenleftBig |X(t)| t >ε parenrightBig = 2 − 2 = 0, 4. Let F be the probability distribution function of 1/Y 2 . Let Z ∼ N(0, 1).Wehave F(t)= P parenleftbig 1/Y 2 ≤ t parenrightbig = P parenleftbig Y 2 ≥ 1/t parenrightbig = P parenleftbig Y ≥ 1/ √ t parenrightbig + P parenleftbig Y ≤−1/ √ t parenrightbig = P parenleftBig Z ≥ α σ √ t parenrightBig + P parenleftBig Z ≤− α σ √ t parenrightBig = 1 − Phi1 parenleftBig α σ √ t parenrightBig + Phi1 parenleftBig − α σ √ t parenrightBig = 2 bracketleftBig 1 − Phi1 parenleftBig α σ √ t parenrightBigbracketrightBig , which, by (12.35), is also the distribution function of T α . 5. Clearly, P(T t, by Theorem 12.10, P(T 0, the probability density function of Z(t) is φ t (x) = 1 σ √ 2πt exp bracketleftBig − x 2 2σ 2 t bracketrightBig . Section 12.5 Brownian Motion 329 Therefore, E bracketleftbig V(t) bracketrightbig = E bracketleftbig |Z(t)| bracketrightbig = integraldisplay ∞ −∞ |x|φ t (x) dx = 2 integraldisplay ∞ 0 xφ t (x) dx = 2 integraldisplay ∞ 0 x σ √ 2πt e −x 2 /(2σ 2 t) dx. Making the change of variable u = x σ √ t yields E bracketleftbig V(t) bracketrightbig = σ radicalbigg 2t π integraldisplay ∞ 0 ue −u 2 /2 du = σ radicalbigg 2t π bracketleftBig − e −u 2 /2 bracketrightBig ∞ 0 = σ radicalbigg 2t π . Var bracketleftbig V(t) bracketrightbig = E bracketleftbig V(t) 2 bracketrightbig − parenleftbig E bracketleftbig V(t) bracketrightbigparenrightbig 2 = E bracketleftbig Z(t) 2 bracketrightbig − 2σ 2 t π = σ 2 t − 2σ 2 t π = σ 2 t parenleftBig 1 − 2 π parenrightBig , since E bracketleftbig Z(t) 2 bracketrightbig = Var bracketleftbig Z(t) bracketrightbig + parenleftbig E bracketleftbig Z(t) bracketrightbigparenrightbig 2 = σ 2 t + 0 = σ 2 t. To find P parenleftbig V(t)≤ z | V(0) = z 0 parenrightbig , note that, by (12.27), P parenleftbig V(t)≤ z | V(0) = z 0 parenrightbig = P parenleftbig |Z(t)|≤z | V(0) = z 0 parenrightbig = P parenleftbig − z ≤ Z(t) ≤ z | V(0) = z 0 parenrightbig = integraldisplay z −z 1 σ √ 2πt e −(u−z 0 ) 2 /(2σ 2 t) du. Letting U ∼ N(z 0 ,σ 2 t) and Z ∼ N(0, 1), this implies that P parenleftbig V(t)≤ z | V(0) = z 0 parenrightbig = P(−z ≤ U ≤ z) = P parenleftBig −z − z 0 σ √ t ≤ z ≤ z − z 0 σ √ t parenrightBig = Phi1 parenleftBig z − z 0 σ √ t parenrightBig − Phi1 parenleftBig −z − z 0 σ √ t parenrightBig = Phi1 parenleftBig z + z 0 σ √ t parenrightBig + Phi1 parenleftBig z − z 0 σ √ t parenrightBig − 1. 10. Clearly, D(t) = radicalbig 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 σ 2 t,wehave E bracketleftbig D(t) bracketrightbig = integraldisplay ∞ −∞ integraldisplay ∞ −∞ integraldisplay ∞ −∞ radicalbig x 2 + y 2 + z 2 · 1 σ √ 2πt e −x 2 /(2σ 2 t) · 1 σ √ 2πt e −y 2 /(2σ 2 t) · 1 σ √ 2πt e −z 2 /(2σ 2 t) dxdydz = 1 2πσ 3 t √ 2πt integraldisplay ∞ −∞ integraldisplay ∞ −∞ integraldisplay ∞ −∞ radicalbig x 2 + y 2 + z 2 · e −(x 2 +y 2 +z 2 )/(2σ 2 t) dxdydz. We now make a change of variables to spherical coordinates: x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ, ρ 2 = x 2 + y 2 + z 2 , dxdydz = ρ 2 sin φdρdφdθ,0 ≤ ρ<∞, 0 ≤ φ ≤ π, and 0 ≤ θ ≤ 2π. We obtain E bracketleftbig D(t) bracketrightbig = 1 2πσ 3 t √ 2πt integraldisplay 2π 0 integraldisplay π 0 integraldisplay ∞ 0 ρe −ρ 2 /(2σ 2 t) · ρ 2 sin φdρdφ,dθ = 1 2πσ 3 t √ 2πt integraldisplay 2π 0 bracketleftBig integraldisplay π 0 parenleftBig integraldisplay ∞ 0 ρ 3 e −ρ 2 /(2σ 2 t) dρ parenrightBig sin φdφ bracketrightBig dθ = 1 2πσ 3 t √ 2πt integraldisplay 2π 0 parenleftBig integraldisplay π 0 bracketleftBig − σ 2 t(ρ 2 + 2σ 2 t)e −ρ 2 /(2σ 2 t) bracketrightBig ∞ 0 sin φdφ parenrightBig dθ = 1 2πσ 3 t √ 2πt · 2σ 4 t 2 integraldisplay 2π 0 parenleftBig integraldisplay π 0 sin φdφ parenrightBig dθ = 2σ radicalbigg 2t π . 11. Noting that √ 5.29 = 2.3, we have V(t)= 95e −2t+2.3W(t) , where braceleftbig W(t): t ≥ 0 bracerightbig is a standard Brownian motion. Hence W(t) ∼ N(0,t). The desired probability is P parenleftbig V(0.75)<80 parenrightbig = P parenleftbig 95e −2(0.75)+2.3W(0.75) < 80 parenrightbig = P parenleftbig e 2.3W(0.75) < 3.774 parenrightbig = P parenleftbig W(0.75)<0.577 parenrightbig = P parenleftBig W(0.75) − 0 √ 0.75 < 0.577 √ 0.75 parenrightBig = P(Z <0.67) = Phi1(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 parenleftbig S 10 ≥ 160 | N(180) = 10 parenrightbig . Let X 1 , X 2 , ..., X 10 be 10 independent random variables uni- formly distributed over the interval [0, 180]. Let Y = max(X 1 ,... ,X 10 ). By Theorem 12.4, P parenleftBig S 10 > 160 | N(180) = 10 parenrightbig = P(Y >160) = 1 − P(Y ≤ 160) = 1 − P parenleftbig max(X 1 ,... ,X 10 ) ≤ 160 parenrightbig = 1 − P(X 1 ≤ 160)P(X 2 ≤ 160) ···P(X 10 ≤ 160) = 1 − parenleftBig 160 180 parenrightBig 10 = 0.692. 2. For all positive integer n, we have that P 2n = parenleftbigg 10 01 parenrightbigg and P 2n+1 = parenleftbigg 01 10 parenrightbigg . Therefore, {X n : 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. 5. By drawing a transition graph, it is readily seen that this Markov chain consists of two recurrent classes {3, 5} and {4}, and two transient classes {1} and {2}. 6. We have that X n+1 = braceleftBigg X n if the (n + 1)st outcome is not 6 1 + X n if the (n + 1)st outcome is 6. This shows that {X n : n = 1, 2,...} is a Markov chain with state space {0, 1, 2,...}. Its transition probability matrix is given by P = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 5/61/6000... 05/61/60 0... 005/61/60... 0005/61/6 ... . . . ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . 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 p 11 p 11 + p 11 p 12 + p 12 p 22 + p 12 p 21 + p 21 p 11 + p 21 p 12 + p 22 p 21 + p 22 p 22 = (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: P = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 00100 000 1 0 00 0 0 00 00010 000 01/2001/2000 00001/32/300 00000 010 00000 001 00000 100 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . 9. For n ≥ 1, let X n = braceleftBigg 1 if the nth golfball produced is defective 0 if the nth golfball produced is good. Then {X n : n = 1, 2,...} is a Markov chain with state space {0, 1} and transition probability matrix parenleftBig 15/18 3/18 11/12 1/12 parenrightBig . Let π 0 be the fraction of golfballs produced that are good, and π 1 be the fraction of the balls produced that are defective. Then, by Theorem 12.7, π 0 and π 1 satisfy parenleftbigg π 0 π 1 parenrightbigg = parenleftbigg 15/18 11/12 3/18 1/12 parenrightbiggparenleftbigg π 0 π 1 parenrightbigg , which gives us the following system of equations ⎧ ⎨ ⎩ π 0 = (15/18)π 0 + (11/12)π 1 π 1 = (3/18)π 0 + (1/12)π 1 . By choosing any one of these equations along with the relation π 0 + π 1 = 1, we obtain a system of two equations in two unknowns. Solving that system yields π 0 = 11 13 ≈ 0.85 and π 1 = 2 13 ≈ 0.15. Therefore, approximately 15% of the golfballs produced have no logos. 10. Let X n = ⎧ ⎪ ⎨ ⎪ ⎩ 1 if the nth ball is drawn by Carmela 2 if the nth ball is drawn by Daniela 3 if the nth ball is drawn by Lucrezia. Chapter 12 Review Problems 333 The process {X n : n = 1, 2,...} is an irreducible, aperiodic, positive recurrent Markov chain with transition probability matrix P = ⎛ ⎝ 7/31 11/31 13/31 7/31 11/31 13/31 7/31 11/31 13/31 ⎞ ⎠ . Let π 1 , π 2 , and π 3 be the long-run 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 π 0 + π 1 + π 3 = 1. ⎛ ⎝ π 1 π 2 π 3 ⎞ ⎠ = ⎛ ⎝ 7/31 7/31 7/31 11/31 11/31 11/31 13/31 13/31 13/31 ⎞ ⎠ ⎛ ⎝ π 1 π 2 π 3 ⎞ ⎠ . 11. Let π 1 and π 2 be the long-run probabilities that Francesco devotes to playing golf and playing tennis, respectively. Then, by Theorem 12.7, π 1 and π 2 are obtained from solving the system of equations parenleftbigg π 1 π 2 parenrightbigg = parenleftbigg 0.30 0.58 0.70 0.42 parenrightbiggparenleftbigg π 1 π 2 parenrightbigg along with π 1 + π 2 = 1. The matrix equation above gives the following system of equations: braceleftBigg π 1 = 0.30π 1 + 0.58π 2 π 2 = 0.70π 1 + 0.42π 2 . By choosing any one of these equations along with the relation π 1 + π 2 = 1, we obtain a system of two equations in two unknowns. Solving that system yields π 1 = 0.453125 and π 2 = 0.546875. Therefore, the long-run 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 X 1 be the time until the first passenger arrives at the station after t = 0. Let X 2 be the additional time it will take until a train arrives at the station, X 3 be the time after that until a passenger arrives, and so on. Clearly, X 1 , X 2 , ... are the times between consecutive change of states. By the memoryless property of exponential random variables, {X 1 ,X 2 ,...} is a sequence of independent and identically distributed exponential random variables with mean 1/λ. Hence, by Remark 7.2, braceleftbig N(t): t ≥ 0 bracerightbig is a Poisson process with rate λ. Therefore, N(t) is a Poisson random variable with parameter λt. 13. Let X(t) be the number of components working at time t. Clearly, braceleftbig X(t): t ≥ 0 bracerightbig is a continuous-time Markov chain with state space {0, 1, 2}. Let π 0 , π 1 , and π 2 be the long-run proportion of time the process is in states 0, 1, and 2, respectively. The balance equations for braceleftbig X(t): t ≥ 0 bracerightbig are as follows: 334 Chapter 12 Stochastic Processes State Input rate to = Output rate from 0 λπ 1 = µπ 0 1 2λπ 2 + µπ 0 = µπ 1 + λπ 1 2 µπ 1 =2λπ 2 From these equations, we obtain π 1 = µ λ π 0 and π 2 = µ 2 2λ 2 π 0 . Using π 0 + π 1 + π 2 = 1 yields π 0 = 2λ 2 2λ 2 + 2λµ + µ 2 . Hence the desired probability is 1 − π 0 = µ(2λ + µ) 2λ 2 + 2λµ + µ 2 . 14. Suppose that every time an out-of-order 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 braceleftbig X(t): t ≥ 0 bracerightbig is a birth and death process should be clear. The birth and death rates are λ n = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ kλ n = 0, 1,... ,m+ s − k (m + s − n)λ n = m + s − k + 1,m+ s − k + 2,... ,m+ s 0 n ≥ m + s; µ n = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ nµ n = 0, 1,... ,m mµ 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 π i be the long-run proportion of time that there are exactly i machines operating. Suppose that a birth occurs each time that an out-of-order machine is repaired and begins to operate, and a death occurs each time that a machine breaks down. Then braceleftbig X(t): t ≥ 0 bracerightbig is a birth and death process with state space {0, 1,... ,m}, and birth and death rates, respectively, given by λ i = (m − i)λ and µ i = iµ for i = 0, 1,... ,m. To find π 0 , first we will calculate the following sum: m summationdisplay i=1 λ 0 λ 1 ···λ i−1 µ 1 µ 2 ···µ i = m summationdisplay i=1 (mλ) bracketleftbig (m − 1)λ bracketrightbigbracketleftbig (m − 2)λ bracketrightbig ··· bracketleftbig (m − i + 1)λ bracketrightbig µ(2µ)(3µ) ···(iµ) = m summationdisplay i=1 m P i λ i i! µ i = m summationdisplay i=1 parenleftbigg m i parenrightbigg parenleftBig λ µ parenrightBig i =−1 + m summationdisplay i=0 parenleftbigg m i parenrightbigg parenleftBig λ µ parenrightBig i 1 m−i =−1 + parenleftBig 1 + λ µ parenrightBig m , Chapter 12 Review Problems 335 where m P i is the number of i-element permutations of a set containing m objects. Hence, by (12.22), π 0 = parenleftBig 1 + λ µ parenrightBig −m = parenleftBig λ + µ µ parenrightBig −m = parenleftBig µ λ + µ parenrightBig m . By (12.21), π i = λ 0 λ 1 ···λ i−1 µ 1 µ 2 ···µ i π 0 = m P i λ i i! µ i π 0 = parenleftbigg m i parenrightbigg parenleftBig λ µ parenrightBig i parenleftBig µ λ + µ parenrightBig m = parenleftbigg m i parenrightbigg parenleftBig λ µ parenrightBig i parenleftBig µ λ + µ parenrightBig i parenleftBig µ λ + µ parenrightBig m−i = parenleftbigg m i parenrightbigg parenleftBig λ λ + µ parenrightBig i parenleftBig 1 − λ λ + µ parenrightBig m−i , 0 ≤ i ≤ m. Therefore, in steady-state, the number of machines that are operating is binomial with param- eters m and λ/(λ + µ). 16. Let X(t) be the number of cars at the center, either being inspected or waiting to be inspected, at time t. Clearly, braceleftbig X(t): t ≥ 0 bracerightbig is a birth and death process with rates λ n = λ/(n + 1), n ≥ 0, and µ n = µ, n ≥ 1. Since ∞ summationdisplay n=1 λ 0 λ 1 ···λ n−1 µ 1 µ 2 ···µ n = ∞ summationdisplay n=1 λ · λ 2 · λ 3 ··· λ n µ n =−1 + ∞ summationdisplay n=0 1 n! parenleftBig λ µ parenrightBig n = e λ/µ − 1. By (12.18), π 0 = e −λ/µ . Hence, by (12.17), π n = λ · λ 2 · λ 3 ··· λ n µ n e −λ/µ = (λ/µ) n e −λ/µ n! ,n≥ 0. Therefore, the long-run probability that there are n cars at the center for inspection is Poisson with rate λ/µ. 17. Let X(t)be the population size at time t. Then braceleftbig X(t): t ≥ 0 bracerightbig is a birth and death process with birth rates λ n = nλ, n ≥ 1, and death rates µ n = nµ, n ≥ 0. For i ≥ 0, let H i be the time, starting from i, until the population size reaches i + 1 for the first time. We are interested in summationtext 4 i=1 E(H i ). Note that, by Lemma 12.2, E(H i ) = 1 λ i + µ i λ i E(H i−1 ), i ≥ 1. Since E(H 0 ) = 1/λ, E(H 1 ) = 1 λ + µ λ · 1 λ = 1 λ + µ λ 2 , 336 Chapter 12 Stochastic Processes E(H 2 ) = 1 2λ + 2µ 2λ · parenleftBig 1 λ + µ λ 2 parenrightBig = 1 2λ + µ λ 2 + µ 2 λ 3 , E(H 3 ) = 1 3λ + 3µ 3λ parenleftBig 1 2λ + µ λ 2 + µ 2 λ 3 parenrightBig = 1 3λ + µ 2λ 2 + µ 2 λ 3 + µ 3 λ 4 , E(H 4 ) = 1 4λ + 4µ 4λ parenleftBig 1 3λ + µ 2λ 2 + µ 2 λ 3 + µ 3 λ 4 parenrightBig = 1 4λ + µ 3λ 2 + µ 2 2λ 3 + µ 3 λ 4 + µ 4 λ 5 . Therefore, the answer is 4 summationdisplay i=1 E(H i ) = 25λ 4 + 34λ 3 µ + 30λ 2 µ 2 + 24λµ 3 + 12µ 4 12λ 5 . 18. Let X(t) be the population size at time t. Then braceleftbig X(t): t ≥ 0 bracerightbig is a birth and death process with rates λ n = γ,n ≥ 0, and µ n = nµ, n ≥ 1. To find π i ’s, we will first calculate the sum in the relation (12.18): ∞ summationdisplay n=1 λ 0 λ 1 ···λ n−1 µ 1 µ 2 ···µ n = ∞ summationdisplay n=1 γ n n! µ n =−1 + ∞ summationdisplay n=0 1 n! parenleftBig γ µ parenrightBig n =−1 + e γ/µ . Thus, by (12.18), π 0 = e −γ/µ and, by (12.17), for i ≥ 1, π i = γ n n! µ n e −γ/µ = (γ/µ) n e −γ/µ n! . Hence the steady-state probability mass function of the population size is Poisson with pa- rameter γ/µ. 19. By applying Theorem 12.9 to braceleftbig Y(t): t ≥ 0 bracerightbig with t 1 = 0, t 2 = t, y 1 = 0, y 2 = y, and t = s, we have E bracketleftbig Y(s)| Y(t)= y bracketrightbig = 0 + y − 0 t − 0 (s − 0) = s t y, and Var bracketleftbig Y(s)| Y(t)= y bracketrightbig = σ 2 · (t − s)(s − 0) t − 0 = σ 2 (t − s) s t . 20. First, suppose that sy)= P parenleftbig no zeros in (x, y) parenrightbig = 1 − 2 π arccos radicalbigg x y . 22. Let the current price of the stock, per share, be v 0 . Noting that √ 27.04 = 5.2, we have V(t)= v 0 e 3t+5.2W(t) , where braceleftbig W(t): t ≥ 0 bracerightbig is a standard Brownian motion. Hence W(t) ∼ N(0,t). The desired probability is calculated as follows: P parenleftbig V(2) ≥ 2v 0 parenrightbig = P parenleftbig v 0 e 6+5.2W(2) ≥ 2v 0 parenrightbig = P parenleftbig 6 + 5.2W(2) ≥ ln 2 parenrightbig = P parenleftbig W(2) ≥−1.02 parenrightbig = P parenleftBig W(2) − 0 √ 2 ≥−0.72 parenrightBig = P(Z ≥−0.72) = 1 − P(Z <−0.72) = 1 − Phi1(−0.72) = 0.7642. artsci aaa.pdf
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