Probset9Sol

Probset9Sol - Probability and Stochastic Processes Problem...

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Unformatted text preview: Probability and Stochastic Processes Problem Set 9 Solutions Spring2010 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 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 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 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 otherwise (3) 1 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 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 wasnt 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 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 x 1 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 y x 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)!...
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This note was uploaded on 02/13/2011 for the course 332 226 taught by Professor Staff during the Spring '08 term at Rutgers.

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Probset9Sol - Probability and Stochastic Processes Problem...

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