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Probset9Sol

# Probset9Sol - Probability and Stochastic Processes Problem...

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Probability and Stochastic Processes Problem Set 9 Solutions Spring’2010 Problem Solutions : Yates and Goodman, 4.9.1 4.9.4 4.9.5 4.9.9 4.9.13 4.10.1 4.10.3 4.10.7 4.10.9 4.10.15 4.11.1 4.11.3 4.11.4 and 4.11.7 Problem 4.9.1 Solution The main part of this problem is just interpreting the problem statement. No calculations are necessary. Since a trip is equally likely to last 2, 3 or 4 days, P D ( d ) = braceleftbigg 1 / 3 d = 2 , 3 , 4 0 otherwise (1) Given a trip lasts d days, the weight change is equally likely to be any value between d and d pounds. Thus, P W | D ( w | d ) = braceleftbigg 1 / (2 d + 1) w = d, d + 1 , . . . , d 0 otherwise (2) The joint PMF is simply P D,W ( d, w ) = P W | D ( w | d ) P D ( d ) (3) = braceleftbigg 1 / (6 d + 3) d = 2 , 3 , 4; w = d, . . . , d 0 otherwise (4) Problem 4.9.4 Solution Random variables X and Y have joint PDF Y X 1 1 f X,Y ( x, y ) = braceleftbigg 2 0 y x 1 0 otherwise (1) For 0 y 1, f Y ( y ) = integraldisplay −∞ f X,Y ( x, y ) dx = integraldisplay 1 y 2 dx = 2(1 y ) (2) Also, for y < 0 or y > 1, f Y ( y ) = 0. The complete expression for the marginal PDF is f Y ( y ) = braceleftbigg 2(1 y ) 0 y 1 0 otherwise (3) 1

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By Theorem 4.24, the conditional PDF of X given Y is f X | Y ( x | y ) = f X,Y ( x, y ) f Y ( y ) = braceleftbigg 1 1 y y x 1 0 otherwise (4) That is, since Y X 1, X is uniform over [ y, 1] when Y = y . The conditional expectation of X given Y = y can be calculated as E [ X | Y = y ] = integraldisplay −∞ xf X | Y ( x | y ) dx (5) = integraldisplay 1 y x 1 y dx = x 2 2(1 y ) vextendsingle vextendsingle vextendsingle vextendsingle 1 y = 1 + y 2 (6) In fact, since we know that the conditional PDF of X is uniform over [ y, 1] when Y = y , it wasn’t really necessary to perform the calculation. Problem 4.9.5 Solution Random variables X and Y have joint PDF Y X 1 1 f X,Y ( x, y ) = braceleftbigg 2 0 y x 1 0 otherwise (1) For 0 x 1, the marginal PDF for X satisfies f X ( x ) = integraldisplay −∞ f X,Y ( x, y ) dy = integraldisplay x 0 2 dy = 2 x (2) Note that f X ( x ) = 0 for x < 0 or x > 1. Hence the complete expression for the marginal PDF of X is f X ( x ) = braceleftbigg 2 x 0 x 1 0 otherwise (3) The conditional PDF of Y given X = x is f Y | X ( y | x ) = f X,Y ( x, y ) f X ( x ) = braceleftbigg 1 /x 0 y x 0 otherwise (4) Given X = x , Y has a uniform PDF over [0 , x ] and thus has conditional expected value E [ Y | X = x ] = x/ 2. Another way to obtain this result is to calculate integraltext −∞ yf Y | X ( y | x ) dy . Problem 4.9.9 Solution Random variables N and K have the joint PMF P N,K ( n, k ) = 100 n e - 100 ( n +1)! k = 0 , 1 , . . . , n ; n = 0 , 1 , . . . 0 otherwise (1) 2
We can find the marginal PMF for N by summing over all possible K . For n 0, P N ( n ) = n summationdisplay k =0 100 n e 100 ( n + 1)! = 100 n e 100 n ! (2) We see that N has a Poisson PMF with expected value 100. For n 0, the conditional PMF of K given N = n is P K | N ( k | n ) = P N,K ( n, k ) P N ( n ) = braceleftbigg 1 / ( n + 1) k = 0 , 1 , . . . , n 0 otherwise (3) That is, given N = n , K has a discrete uniform PMF over { 0 , 1 , . . . , n } . Thus, E [ K | N = n ] = n summationdisplay k =0 k/ ( n + 1) = n/ 2 (4) We can conclude that E [ K | N ] = N/ 2. Thus, by Theorem 4.25, E [ K ] = E [ E [ K | N ]] = E [ N/ 2] = 50 . (5) Problem 4.9.13 Solution The key to solving this problem is to find the joint PMF of M and N . Note that N M .

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