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410Hw04ansJeff-1

# 410Hw04ansJeff-1 - STAT 410 Fall 2011 Homework#4(due Friday...

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STAT 410 Fall 2011 Homework #4 (due Friday, September 23, by 3:00 p.m.) 1. Let X be a Uniform ( 0, 1 ) and Y be a Uniform ( 0, 3 ) independent random variables. Let W = X + Y. Find and sketch the p.d.f. of W. ( ) < < = otherwise 0 1 0 1 X x x f ( ) < < = otherwise 0 3 0 3 1 Y y y f ( ) ( ) ( ) - - + = dx x w f x f w f Y X Y X ( ) < < = otherwise 0 3 0 3 1 Y y y f ( ) < < - < - < - = = otherwise 0 3 3 1 otherwise 0 3 0 3 1 Y w x w x w x w f Case 1: 0 < w < 1. f W ( w ) = w dx 0 3 1 1 = w 3 1 . Case 2: 1 < w < 3. f W ( w ) = 1 0 3 1 1 dx = 3 1 . Case 3: 3 < w < 4. f W ( w ) = - 1 3 3 1 1 w dx = ( ) 4 3 1 w - .

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OR ( ) ( ) ( ) - - + = dy y f y w f w f Y X Y X ( ) < < = otherwise 0 1 0 1 X x x f ( ) < < - = < - < = - otherwise 0 1 1 otherwise 0 1 0 1 X w y w y w y w f Case 1: 0 < w < 1. f W ( w ) = w dy 0 3 1 1 = w 3 1 . Case 2: 1 < w < 3. f W ( w ) = - w w dy 1 3 1 1 = 3 1 . Case 3: 3 < w < 4. f W ( w ) = - 3 1 3 1 1 w dy = ( ) 4 3 1 w - .
OR F W ( w ) = P ( W w ) = P ( X + Y w ) = Case 1: 0 < w < 1. Case 2: 1 < w < 3. Case 3: 3 < w < 4. = - w x w dx dy 0 0 3 1 1 = - 1 0 0 3 1 1 dx dy x w = - - - 1 3 3 3 1 1 1 w x w dx dy = 2 6 1 w . = ( ) 1 2 6 1 - w . = ( ) 2 4 6 1 1 w - - . f W ( w ) = F W ' ( w ) = = w 3 1 , = 3 1 , = ( ) 4 3 1 w - , 0 < w < 1. 1 < w < 3. 3 < w < 4.

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2. Let random variables X and Y have an Exponential distribution with mean 1 and a Uniform distribution on ( 0 , 1 ) , respectively. Suppose X and Y are independent. Find the probability density function of W = X + Y. ( ) > = - otherwise 0 0 X x x f x e ( ) < < = otherwise 0 1 0 1 Y y y f ( ) < < - = < - < = - otherwise 0 1 1 otherwise 0 1 0 1 Y w x w x w x w f Case 1: w < 0. ( ) w f Y X + = 0. Case 2: 0 < w < 1. Then w – 1 < 0. ( ) ( ) ( ) ( ) - - + = - = w x dx dx x w f x f w f e 0 Y X Y X 1 = 1 – e w .
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