# 06_24 - STAT 410 Examples for Summer 2009 1 Let f x y = e x...

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STAT 410 Examples for 06/24/2009 Summer 2009 1. Let f ( x , y ) = e x y , 0 < x < , 0 < y < , zero elsewhere, be the pdf of X and Y . Determine the probability distribution of the following random variables: a) W = X + Y ; b) V = X Y ; c) U = Y X X + . a) 2.1.6 Let f ( x , y ) = e – x – y , 0 < x < , 0 < y < , zero elsewhere, be the p.d.f. of X and Y. Then if Z = X + Y, … What is the p.d.f. of Z? F Z ( z ) = P ( Z z ) = P ( Y z – X ) = - - - z d x z d y x x y e 0 0 = - - - z d x z d y x x y e e 0 0 = ( 29 + - - - z d x z x x e e 0 1 = - - - z d z z d x x e x e 0 0 = z z e z e - - - - 1 , z > 0. f Z ( z ) = F Z ' ( z ) = e z e z + z e z = z e z , z > 0. Another approach: X and Y are two independent Exponential random variables with mean 1. M X ( t ) = t - 1 1 , t < 1. M Y ( t ) = t - 1 1 , t < 1. M X + Y ( t ) = M X ( t ) M Y ( t ) = 2 1 1 - t , t < 1. Z = X + Y has a Gamma distribution with α = 2, θ = 1. f Z ( z ) = ( 29 1 1 2 2 1 2 1 z e z - - Γ = z e z , z > 0.

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b) F V ( v ) = P ( V v ) = P ( Y v X ) = ( 29 - - 0 dy dx v y y x e = ( 29 - - 0 dy dx v y x y e e = - - 0 dy v y y e e = ( 29 + - 0 1 1 dy y v e = v v + 1 = v + - 1 1 1 , v > 0. f V ( v ) = ( 29 2 1 1 v + , v > 0. c) F U ( u ) = P ( U u ) = P ( X u ( X + Y ) ) = P ( Y u u - 1 X ) = 1 1 1 u u - + = u , 0< u < 1. P ( Y v X ) = v + - 1 1 1 P ( Y v X ) = v + 1 1 , then v = u u - 1 . f U ( u ) = 1, 0< u < 1. U has a Uniform ( 0, 1 ) distribution. 2. Let X and Y be two independent Exponential random variables with mean 1. Find the p.d.f. of
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06_24 - STAT 410 Examples for Summer 2009 1 Let f x y = e x...

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