IEOR 6711: Stochastic Models I, Fall 2003, Professor Whitt. Second Midterm Exam, November 13, Based on Chapters 1-3 in Ross

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1 IEOR 6711: Stochastic Models I, Fall 2003, Professor Whitt Second Midterm Exam, November 13, 2003 Based on Chapters 1-3 in Ross There are eight questions. Do as much as you can. Show your work. 1. Alpine Villages (10 points) In a high mountain region there are three villages: Frostbite, Iceburg and Snowbound. There are two direct roads between Frostbite and Iceburg; there are two direct roads between Iceburg and Snowbound; but there are no direct roads between Frostbite and Snowbound. For some strange reason, on a given winter day each of the four roads is blocked by snow with probability p, independently of the others. (a) Find the conditional probability that there is an open road from Frostbite to Iceburg, given that there is no open route from Frostbite to Snowbound. (A route from Frostbite to Snowbound would use one of the two roads between Frostbite and Iceburg plus one of the two roads between Iceburg and Snowbound.) Let A and B be the events that the two roads between Frostbite and Iceburg are closed; let C and D be the events that the two roads between Iceburg and Snowbound are closed. Then the desired conditional probability is P ((AB) c (AB) (CD)) = P ((AB)c and (AB) (CD)) P ((AB) (CD)) = P ((AB)c and (CD)) P ((AB) (CD)) P ((AB) c )P ((CD)) = P ((AB)) + P ((CD)) P ((AB)(CD)) = (1 p2 )p 2 p 2 + p 2 p 4 = (1 p2 )p 2 p 2 (2 p 2 ) = (1 p2 ) (2 p 2 ) = (1 p2 )p 2 1 (1 p 2 ) 2. (b) Suppose that, instead, there are also three direct roads between Frostbite and Snowbound, with each of these roads being blocked by snow with probability q, independent of everything else. Find the conditional probability in part (a) under this additional assumption. (A route between Frostbite and Snowbound could be either one of the new direct one-road routes or one of the previous two-road routes.) The answer is the same as in part (a). The conditioning event implies that none of the direct roads can be open, so it is the same as if they were not there at all.

2 2. Functions of a Random Variable (10 points) Let X be a real-valued random variable with cumulative distribution function (cdf) F and probability density function (pdf) f, i.e., F (x) = x f(t) dt for all x. Suppose that Y = X 2. Find the cdf and pdf of Y. Let G be the cdf of Y. Then G(y) = P (Y y) = P (X 2 y) = P ( y X y) = F ( y) F ( y) for y 0. Hence the cdf G has a pdf g, i.e., G(x) = x g(t) dt. We obtain the pdf g by differentiating the cdf G. To do so, we use the chain rule: g(y) = f( y)((1/2)y 1/2 ) f( y)(( 1/2)y 1/2 ) = f( y)((1/2)y 1/2 ) + f( y)((+1/2)y 1/2 ) = [f( y) + f( y)]((1/2)y 1/2 ) = f( y) + f( y) 2. y 3. Multiple Choice (8 points) Choose the best answer and justify it: For a strictly positive real-valued random variable X, the equation E[1/X] = 1/E[X] (a) is always true. (b) is never true. (c) is sometimes true. (d) is true on Wednesdays, but not on Tuesdays. The correct answer is (c). It is easy to see that the statement is true whenever the random variable X is deterministic; then E[1/X] = 1/X = 1/E[X]. (It is possible to construct nondeterministic counterexamples as well.) However, in general the statement is not true. For example, suppose that P (X = 2) = P (X = 1/2) = 1/2. Then E[1/X] = E[X] = 5/4 4/5 = 1/E[X]. 2

3 I add an addendum by Wanmo, observing that equality holds for a strictly positive random variable X if and only if the random variable X is deterministic. From Holder inequality, we have a stronger conclusion: [ ] 1 E = 1 iff X is deterministic. X E[X] Basically we have [ ] X 1 1 = E [ ] 1 E[X] E X X where the last inequality is equality iff there exist α and β such that αβ 0 and α( ( ) 1 2 X) 2 = β X a.s. or X 2 = β/α = constant a.s. or X = constant a.s.. Hence the last statement says the conclusion. Note: The following is the exact statement from page 182 of Gerald Folland s Real Analysis : Modern Techniques and Their Applications, Second Edition : Holder sinequality. Suppose 1 < p < and 1 p + 1 q = 1. If f and g are measurable functions on Ω, then fg 1 f p g g. In particular, if f L p and g L q, then fg L 1 and in this case equality holds in the above inequality iff α f p = β g q a.e. for some constants α, β with αβ 0. Note also that if we use Jensen s inequality from the convexity of 1/x, we get the inequality, but we can not sharpen the conclusion into an equality. 4. The Conditional Variance (15 points) (a) How is the conditional variance of X, given Y, defined? It was put on the V ar(x Y ) E[(X E[X Y ]) 2 Y ]. (b) Show that V ar(x) = E[V ar(x Y )] + V ar(e[x Y ]). This is Problem 1.22 in the text, which was a homework problem. Alternative Midterm Exam Emphasizing Proofs. The conditional variance of X given Y is defined as - 3

4 A key relation is EX = E[E[X Y ]] for any random variables X and Y. It is easier to start with the two pieces and put them together. First, Next, Therefore, E[V ar(x Y )] = E[E[(X E[X Y ]) 2 Y ]] = E[E[X 2 2XE[X Y ] + E[X Y ] 2 Y ]] = E[E[X 2 Y ] 2E[X Y ] 2 + E[X Y ] 2 ] = E[E[X 2 Y ] E[X Y ] 2 ] = E[E[X 2 Y ]] E[E[X Y ] 2 ]] = E[X 2 ] E[E[X Y ] 2 ]]. V ar(e[x Y ]) = E[E[X Y ] 2 ] (E[E[X Y ]]) 2 = E[E[X Y ] 2 ] (E[X]) 2. E[V ar(x Y )]+V ar(e[x Y ]) = (E[X 2 ] E[E[X Y ] 2 ])+(E[E[X Y ] 2 ] (E[X]) 2 ) = E[X 2 ] (E[X]) 2 = V ar(x) Patterns in Rolls of a Die (15 points) Consider successive independent rolls of a six-sided die, where each of the sides numbered 1 through 6 is equally likely to appear. ( Die is the singular of dice. ) (a) What is the expected number of rolls until the pattern (1, 1, 2, 6, 5, 6, 1, 1, 2) first appears? This is just like Problem 8 on the first midterm exam, which in turn is just like Problem 3.22 in the text. The general idea is to exploit renewal theory. It is easier to start with part (b). Let T be the time until the pattern first appears. Then ET = = 10, 077, = 10, 077, 912. (b) What is the expected number of rolls after the pattern (1, 1, 2, 6, 5, 6, 1, 1, 2) first appears until it appears again? The probability that this pattern ends at any specified time (roll) n (for n 9) is (1/6) 9. Hence the expected number of rolls is 6 9 = 10, 077, 696. (c) What is the probability that the pattern A = (1, 1, 2, 2, 1, 1) occurs before the pattern B = (1, 1, 2, 6, 5, 6, 1, 1, 2)? Let the first pattern be pattern A and let the second pattern be pattern B. Let T A and T B be the time until these patterns first appear. Let T A B be the time after pattern A appears 4

5 until pattern B appears; T B A be the time after pattern B appears until pattern A appears. We have seen that ET B = 10, 077, 954 by part (a). By the same reasoning, so that ET B = ET (1,1) + ET (1,1) B = ET (1,1) + ET A B, ET A B = ET B ET (1,1) = p 9 + p 3 p 2 p 1 = 10, 077, so that Similarly, ET A = ET (1,1,2) + ET (1,1,2) A = ET (1,1,2) + ET B A, ET B A = ET A ET (1,1,2) = p 6 + p 2 + p 1 p 3. Finally, we have the formula (derived on p. 127 of the text) P (T A < T B ) = ET B ET A + ET B A ET A B + ET B A. Plug in the four answers above to get the final result, getting P (T A < T B ) = Cars and Trucks Vehicles pass a recording checkpoint on the highway according to a Poisson process with arrival rate 40 per minute. Suppose that 20% of the vehicles are trucks and 80% of the vehicles are cars. Suppose that each successive vehicles is a truck with probability 0.20, independently of the types of past vehicles. (a) What is the expected number of trucks that pass the checkpoint in any given 5 minute interval? The number of trucks that pass the checkpoint is again a Poisson process, but with arrival rate = 8 per minute. Hence the expected number of trucks that pass the checkpoint in any given 5 minute interval is 5 8 = 40 trucks. (b) What is the probability that no trucks pass the checkpoint in any given 5-minute interval? As stated in part (a) above, the number of trucks that pass the checkpoint is again a Poisson process with arrival rate = 8 per minute. Hence the probability that no trucks pass the checkpoint in any given 5 minute interval is e (8 5) = e 40. (c) After a truck passes the checkpoint, what is the expected time until two more trucks have passed the checkpoint? 5

6 As stated in part (a) above, the number of trucks that pass the checkpoint is a Poisson process with arrival rate = 8 per minute. Hence the time between successive arrivals (passings of the checkpoint) of trucks is an exponential random variable with mean 1/8 minute. Thus the expected time until two more trucks have passed the checkpoint is 1/4 minute = 15 seconds. (d) What is the probability that exactly 4 trucks and 8 cars pass the checkpoint in any given 15-second interval? By the stated assumptions, the number of trucks that pass the checkpoint and the number of cars that pass the checkpoint are independent Poisson processes. The rate of the truck Poisson process is 8 per minute, as indicated above. The rate of the car Poisson process is = 32 per minute. Let N C (t) and N T (t) be the numbers of cars and trucks, respectively, that pass the checkpoint in an interval of length t minutes. Then, since 15 seconds = 1/4 minute, the desired probability is P (N T (1/4) = 4, N C (1/4) = 8) = e e ! 8! (e) What is the conditional probability that exactly 4 trucks pass the checkpoint in any given 15-second interval, given that 8 cars pass the checkpoint during that same 15-second interval? Since the arrival process for trucks and cars are independent Poisson processes, the conditioning event has no influence. Thus, P (N T (1/4) = 4 N C (1/4) = 8) = P (N T (1/4) = 4) = e ! = (f) What is the approximate probability that at least 256 trucks have passed the checkpoint in a time period of 36 minutes? Apply the central limit theorem to get a normal approximation. By part (c) above, the interval between successive trucks passing the checkpoint has an exponential distribution with mean 1/8 minute. The variance is thus 1/64 minute. The interval between the passing of 256 trucks thus has approximately a normal distribution with mean 256 (1/8) = 32 minutes and variance 256 (1/64) = 4. Let S 256 denote the time interval required for 256 trucks to pass the checkpoint. Then P (S ) P (N(32, 4) 36) = P (N(0, 1) 2) = Mark s Car Ownership (15 points) Mark buys a new car as soon as his old car breaks down or reaches the age of 8 years, whichever occurs first. Suppose that a new car costs $20, 000 and an additional cost of $4000 6

7 is incurred whenever a car breaks down. Suppose that an 8-year old car (that has not failed) has a resale value of $5000. Suppose that the lifetime of each car is uniformly distributed in the interval between 2 years and 12 years. (Ignore the time value of money; i.e., assume that money is worth the same throughout time.) (a) What is the expected interval between the times that Mark gets a new car? Let L be the lifetime of each car. Let X be the interval between successive times that Mark gets a new car. Then X = min{l, 8} and EX = E[min{L, 8}] = 8 2 x 1 dx + 8P (L > 8) = = 6.2 years 10 (b) What is Mark s long-run average cost for car ownership (assuming an endless sequence of car replacements according to the model specified above)? where Thus, The long-run average cost = E[cost per cycle] E[length of cycle], E[cost per cycle] = $20, (0.6 $4, 000) (0.4 $5, 000) = $20, 400 The long-run average cost = - $20, = $3, 290 per year. (c) At what rate does Mark have to replace cars that breakdown (again assuming an endless sequence of car replacements according to the model specified above)? Let B be the time between successive breakdowns. The desired rate is 1/EB. By conditioning on what happens during the first car lifetime, we develop an equation for EB: so that EB = 8 The desired rate is then 3/ x 1 dx + P (L > 8)(8 + EB) = (8 + EB), 10 (0.6)EB = 6.2 and EB = The Elementary Renewal Theorem (12 points) 7

8 (a) State the Elementary Renewal Theorem. (b) Prove the Elementary Renewal Theorem. See p. 107 in the text. This problem was on the Alternative Midterm Exam Emphasizing Proofs, handed out on Thursday, 10/23. 8