NAME: ECE 302 Division MWF 0:30-:20 (Prof. Pollak) Final Exam Solutions, 5/3/2004. Please read the instructions carefully before proceeding. If you are not in Prof. Pollak s section, you may not take this exam. No questions will be answered during the exam. If you think you found a mistake in a problem statement, write so in your solution and explain your reasoning. This is a closed-book exam. A formula sheet is provided. No calculators are allowed. You have two hours to complete FIVE problems. Be sure to fully and clearly explain all your answers. There will not be any discussion of grades. All re-grade requests must be submitted in writing. Problem Points Score 60 2 35 3 20 4 20 5 5 TOTAL 50
Some random variables, their distributions, and associated transforms: Random variable PMF or PDF Mean Variance Transform Bernoulli p for k =; p for k =0. p p( p) p + pe s Discrete uniform n, k = k 0 +,k 0 +2,...,k 0 + n k 0 + n+ 2 Geometric ( p) k p, k =, 2, 3,... p Binomial Pascal of order k Poisson n k t k n 2 2 p 2 p e s (e (k 0 +n)s e k 0 s ) n(e s ) pe s ( p)e s ( p) n k p k, k =0,,...,n pn np( p) ( p + pe s ) n Continuous uniform b a, a x b p k ( p) t k, t = k, k +,... k p ( ) ( k p 2 p ) pe s k ( p)e s e λ λk k!, k =0,, 2,... λ λ eλ(es ) b+a 2 (b a) 2 2 e sb e sa (b a)s Exponential λe λx, x 0 λ λ 2 λ λ s Normal (Gaussian) e (x µ)2 2σ 2 µ σ 2 e σ2 s 2 +µs 2 2πσ Erlang of order k λ k y k e λy (k )!, y 0 k λ k λ 2 ( ) k λ λ s where ( n k ) = n! (n k)!k!. 2
Problem (60 points). Two independent continuous random variables X and Y have the following probability density functions: { 0 x f X (x) = 0 otherwise { 0 y f Y (y) = 0 otherwise Let W = X + Y. a (8 points). Find E[W ], the expected value of W. b (8 points). Find var(w ), the variance of W. c (8 points). Find M W (s), the transform associated with W. d (8 points). Find ρ XY, the correlation coefficient of X and Y. e (7 points). Given X = 0.5, find the conditional probability density function of W, i.e. find f W X (w 0.5). f (7 points). Find the least-squares estimate of W based on observing X = 0.5. g (7 points). Find the probability of the event W<0.5. h (7 points). Find f W (w), the probability density function of W. Solution. a. E[W ]=E[X]+E[Y ]=. b. Because of independence, var(w ) = var(x)+var(y )=/6. c. Because of independence, M W (s) =M X (s)m Y (s) = ( ) e s 2. s d. Since X and Y are independent, they are uncorrelated, and so the correlation coefficient is zero. e. Conditioned on X =0.5, W = Y +0.5 which is uniform between 0.5 and.5. f. The least-squares estimate is the conditional mean E[W X = 0.5] which, according to Part e, is equal to. g. Integrating the joint pdf over the set of points for which x + y<0.5, we get /8. h. Since X and Y are independent, the PDF of their sum is the convolution of f X and f Y, i.e., the following hat function: w, 0 w f W (w) = 2 w, w 2 0, otherwise 3
Problem 2 (35 points). Consider the Markov chain below. /2 /4 /2 2 3 /2 /4 /2 4 /2 5 /3 0 /3 /3 Given that this process is in state 0 just before the first transition, determine the probability that: a (7 points) The process enters state 4 for the first time as the result of the K-th transition. Solution. By inspecting the transition graph, we can see that it takes at least 2 steps to reach state 4 from state 0. If K 2, the only way from state 0 to state 2 with exactly K transitions is: The probability of this is: 0 3 } 3 3{{ 3} 4 K 2 transitions P K = p 03 p K 2 33 p 34 =(/3)(/4) K 2 (/2) = (/6)(/4) K 2, K =2, 3,... b (7 points.) The process never enters state 3. Solution. From zero, the process can only go to 3,, or 5. In the two latter cases, it will never reach 3. Therefore, the probability to never enter 3 is 2/3. c (7 points) The process enters state 3, and leaves state 3 on the transition immediately after it entered state 3. Solution. This is the probability to enter state 3, which is /3, times the probability to immediately leave state 3, which is /4+/2 = 3/4. Answer: /3 3/4 =/4. d (7 points) The process enters state 5 for the first time on the third transition. Solution. The only path that takes the process from state 0 to state 5 in exactly 3 steps is: 0 3 4 5 The probability of this is: p 03 p 34 p 45 = 3 2 2 = 2 4
e (7 points.) The process is in state 3 as a result of the N-th transition. Solution. The only way for this to happen is: The probability of this sequence is: 0 3 3 3 }{{} N transitions p 03 p N 33 =(/3)(/4) N 5
Problem 3 (20 points). Ships traveling through a wide canal are being observed from a point A in the canal. An indicator at point A is always pointing in the direction of travel of the most recent ship to pass it. Eastbound ships arrive at point A according to a Poisson process with an average arrival rate λ E ships per day. Westbound ships arrive according to a Poisson process with an average arrival rate λ W ships per day. We begin observing at an arbitrary time. a (5 points) What is the probability that the next ship to arrive at A will be westbound? Solution. We are observing the merged Poisson process of two types of arrivals: eastbound and westbound. Therefore, the probability that a particular arrival will be westbound is λ W λ E + λ W b (5 points) Given that the indicator is pointing west, what is the PDF for the remaining time until the indicator changes direction? Solution. The time when the indicator changes direction is the first arrival time for the eastbound Poisson process which is exponential with parameter λ E : { λe e λex, x 0 0, otherwise, c (5 points) Let T be the time we have to continue observing until we see the eighth eastbound ship. Determine the PDF for T. Solution. T is the eighth arrival time for the eastbound Poisson process. Therefore, it is an Erlang random variable of order 8 with parameter λ E : { λ 8 E t 7 e λ E t f T (t) = 7! t 0 0, otherwise, d (5 points) Let Y be the time we have to continue observing until we see the eighth ship (westbound or eastbound). Determine E[Y ]. Solution. Y is the eighth arrival time for the merged Poisson process. Therefore, it is an Erlang random variable of order 8 with parameter λ E + λ W whose expectation is 8/(λ E + λ W ). 6
Problem 4 (20 points). Let X,X 2,...be independent, identically distributed, exponential random variables with parameter λ =4. Forn =, 2,..., let S n = X +...+ X n Y n = S n n W n = 4S n n n a (7 points.) Is Y n convergent in probability? If so, to what value? Explain. b (7 points.) Is S n convergent in probability? If so, to what value? Explain. c (6 points.) Let a n = P(W n < 0), for n =, 2,... Does the sequence of numbers a,a 2,a 3,... converge? If so, to what value? Explain. Solution. a,b. According to the weak law of large numbers, Y n converges in probability to the expectation of X n which is /4. Therefore, since S n = ny n, lim S n = lim ny n = lim n/4 =, n n n which means that S n does not converge in probability. c. Note that E[X n ]=/4, and therefore E[S n ]=n/4, and E[W n ]=(4E[S n ] n)/ n = 0. Also, E[X n ]=/6, and therefore var(s n )=n/6, var(4s n n) =n, and var(w n ) =. We can therefore apply the central limit theorem to yield: lim a n = lim P(W n < 0) = Φ(0) = 0.5. n n 7
Problem 5 (5 points). In each part, obtain the simplest possible answer. Hint. This problem does not require much calculation. a (5 points). Evaluate (x 2) 2 6x3 e 2x dx 0 6 Solution. Note that Erlang distribution of order k = 4 with parameter λ = 2 is: 2 4 x 4 e 2x (4 )! = 6x3 e 2x, 6 and, moreover, the expected value for this distribution is k/λ = 2. Therefore, the given integral is simply the variance, which is equal to k/λ 2 =. b (5 points). Evaluate n=4 0.5 n ( n 3 )( ) 4 9 ( ) 8 n 4 9 You can use 0.5 n = e n log e 0.5. Solution. This expression is the transform M X (s), where X is a Pascal random variable of order k = 4 with p =/9. Since 0.5 n = e log e 0.5n = e n log e 0.5, in this case s = log e 0.5, and so the expression is equal to: c (5 points). Let ( ) (/9)e log e 0.5 4 M X (log e 0.5) = (8/9)e log e 0.5 = f(x) = { 2e 2x, x 0 0, otherwise, ( ) /8 4 = 8/8 0 4 =0.000. Find f(x) f(x) f(x) f(x) where stands for convolution. Solution. Note that f is an exponential PDF with parameter λ = 2. The convolution of four f s is therefore the PDF of the sum of four independent identical exponential random variables with parameter λ = 2, which is Erlang PDF of order k =4: { 6x 3 e 2x 6, x 0 0, otherwise, 8