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Hw4_02_booksol

# Hw4_02_booksol - Probability and Stochastic Processes...

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Probability and Stochastic Processes Homework 4 Book Solutions Fall 2002 Problem Solutions : Yates and Goodman,7.1.5 7.2.2 7.6.3 7.7.1 8.3.1 8.3.2 8.4.1 8.4.2 9.2.1 and 9.3.1 Problem 7.1.5 Since each X i has zero mean, the mean of Y n is E [ Y n ] = E [ X n + X n - 1 + X n - 2 ] / 3 = 0 Since Y n has zero mean, the variance of Y n is Var [ Y n ] = E Y 2 n = E ( X n + X n - 1 + X n - 2 ) 2 / 9 = E X 2 n + X 2 n - 1 + X 2 n - 2 + 2 X n X n - 1 + 2 X n X n - 2 + 2 X n - 1 X n - 2 / 9 = ( 1 + 1 + 1 + 2 / 4 + 0 + 2 / 4 ) / 9 = 4 / 9 Problem 7.2.2 f X , Y ( x , y ) = 1 0 x , y 1 0 otherwise Proceeding as in Problem 7.2.1, we must first find F W ( w ) by integrating over the square defined by 0 x , y 1. Again we are forced to find F W ( w ) in parts as we did in Problem 7.2.1 resulting in the following integrals for their appropriate regions. For 0 w 1, F W ( w ) = w 0 w - x 0 dxdy = w 2 / 2 For 1 w 2, F W ( w ) = w - 1 0 1 0 dxdy + 1 w - 1 w - y 0 dxdy = 2 w - 1 - w 2 / 2 The complete expression for the CDF of W is F W ( w ) = 0 w < 0 w 2 / 2 0 w 1 2 w - 1 - w 2 / 2 1 w 2 1 otherwise With the CDF, we can find f W ( w ) by differentiating with respect to w . f W ( w ) = w 0 w 1 2 - w 1 w 2 0 otherwise 1

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Problem 7.6.3 straightforward random sum of random variables because N and the X i ’s are dependent. In par- ticular, given N = n , then we know that there were exactly 100 heads in N flips. Hence, given N , X 1 + ··· + X N = 100 no matter what is the actual value of N. Hence Y = 100 every time and the PMF of Y is P Y ( y ) = 1 y = 100 0 otherwise Problem 7.7.1 We know that the waiting time, W is uniformly distributed on [0,10] and therefore has the fol- lowing PDF. f W ( w ) = 1 / 10 0 w 10 0 otherwise We also know that the total time is 3 milliseconds plus the waiting time, that is X = W + 3.
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