# hw6_soln - Homework 6 Solutions Problem Solutions Yates and...

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Homework 6 Solutions Problem Solutions : Yates and Goodman, 3.7.6 3.8.1 3.8.4 3.9.1 3.9.7 4.1.3 4.2.3 and 4.3.2 Problem 3.7.6 Solution We wish to find a transformation that takes a uniformly distributed random variable on [0,1] to the following PDF for Y . f Y ( y ) = 3 y 2 0 y 1 0 otherwise (1) We begin by realizing that in this case the CDF of Y must be F Y ( y ) = 0 y < 0 y 3 0 y 1 1 otherwise (2) Therefore, for 0 y 1, P [ Y y ] = P [ g ( X ) y ] = y 3 (3) Thus, using g ( X ) = X 1 / 3 , we see that for 0 y 1, P [ g ( X ) y ] = P h X 1 / 3 y i = P X y 3 = y 3 (4) which is the desired answer. Problem 3.8.1 Solution The PDF of X is f X ( x ) = 1 / 10 - 5 x 5 0 otherwise (1) (a) The event B has probability P [ B ] = P [ - 3 X 3] = Z 3 - 3 1 10 dx = 3 5 (2) From Definition 3.15, the conditional PDF of X given B is f X | B ( x ) = f X ( x ) /P [ B ] x B 0 otherwise = 1 / 6 | x | ≤ 3 0 otherwise (3) (b) Given B , we see that X has a uniform PDF over [ a, b ] with a = - 3 and b = 3. From Theorem 3.6, the conditional expected value of X is E [ X | B ] = ( a + b ) / 2 = 0. (c) From Theorem 3.6, the conditional variance of X is Var[ X | B ] = ( b - a ) 2 / 12 = 3. 1

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Problem 3.8.4 Solution From Definition 3.8, the PDF of W is f W ( w ) = 1 32 π e - w 2 / 32 (1) (a) Since W has expected value μ = 0, f W ( w ) is symmetric about w = 0. Hence P [ C ] = P [ W > 0] = 1 / 2. From Definition 3.15, the conditional PDF of W given C is f W | C ( w ) = f W ( w ) /P [ C ] w C 0 otherwise = 2 e - w 2 / 32 / 32 π w > 0 0 otherwise (2) (b) The conditional expected value of W given C is E [ W |
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