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hw9soln - ORIE 3500/5500 Fall'10 HW 9 Solutions Assignment...

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ORIE 3500/5500, Fall ’10 HW 9 Solutions Assignment 9 Solutions Problem 1 Clearly, Y takes values in [0,1]. So F Y ( y ) = 0 for y 0 and F Y ( y ) = 1 for y 1. Now for y (0 , 1) F Y ( y ) = P ( Y y ) = P ( X y or 1 X y ) = P ( X y ) + P ( X 1 y ) = 1 - e - y + (1 - (1 - e - 1 y )) = 1 - e - y + e - 1 y Problem 2 Define a function T : R \{ 0 } 7→ R \{ 0 } as T ( x ) = x, if x > 0 , - x 2 / 4 , if x < 0 . So, T is a one-one function with inverse function given by T - 1 ( y ) = y, if y > 0 , - - 4 y, if y < 0 . Also, note that X is a continuous random variable and hence, P ( X = 0) = 0. So, the density of Y = T ( X ) is given by f Y ( y ) = f X ( T - 1 ( y )) dT - 1 ( y ) dy = (1 - y )1 { 0 <y< 1 } · 1 , if y > 0 , (1 - - 4 y )1 {- 1 < - - 4 y< 0 } · 1 - y , if y < 0 . = (1 - y ) , if 0 < y < 1 , (( - y ) - 1 / 2 - 2) , if - 1 4 < y < 0 . Problem 3 The table below presents values of Z for certain values of X and Y X/Y 0 1 2 3 0 0 1 2 3 1 1 2 3 4 2 2 3 4 5 3 3 4 5 6 4 4 5 6 7 5 5 6 7 8 1
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ORIE 3500/5500, Fall ’10 HW 9 Solutions So P ( Z = 0) = P ( X = 0 , Y = 0) = . 4 P ( Z = 1) = P ( X = 1 , Y = 0) + P ( X = 0 , Y = 1) = . 38 Continue adding up, we get z 0 1 2 3 4 5 6 7 8 P ( Z = z ) .4 .38 .1 .054 .029 .021 .011 .004 .001 Problem 4 (a) Let us define a one-one transformation Tr : ( x, y ) 7→ ( xy, x/y ) defined on (0 , ) × (0 , ) (Note the space where the function Tr is defined). Then, ( Tr ) - 1 ( u, v ) = ( uv, p u/v ) = ( h 1 ( u, v ) , h 2 ( u, v )). Therefore, the jacobian is given by J ( T r ) - 1 = det ∂h 1 ∂u ∂h 1 ∂v ∂h 2 ∂u ∂h 2 ∂v = det 1 2
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