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Unformatted text preview: 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 ≤ y ≤ 1 otherwise (1) We begin by realizing that in this case the CDF of Y must be F Y ( y ) = y < y 3 ≤ 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 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 otherwise = 1 / 6  x  ≤ 3 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 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...
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This note was uploaded on 04/03/2008 for the course EEE 350 taught by Professor Duman during the Fall '08 term at ASU.
 Fall '08
 Duman

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