409Hw01ans - STAT 409 Homework #1 Fall 2009 (due Friday,...

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Unformatted text preview: STAT 409 Homework #1 Fall 2009 (due Friday, September 4, by 4:00 p.m.) 1. Let > 0 and let X be a random variable with the probability density function f ( x ) = 1 + x , x > 1, zero otherwise. Let W = ln ( X ). What is the probability distribution of W ? w = ln ( x ) x = g 1 ( w ) = e w dw dx = e w x > 1 w > 0 f W ( w ) = f X ( g 1 ( w ) ) dw dx = 1 w w e e + = w e - , w > 0. OR F X ( x ) = + x dy y 1 1 = 1 1 x- ., x > 1. F Y ( y ) = P ( W w ) = P ( X e w ) = F X ( e w ) = w e 1-- , w > 0. f W ( w ) = w e - , w > 0. W has Exponential distribution with mean 1 / . 2. 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. ( 29 < < = otherwise 1 1 X x x f ( 29 < < = otherwise 3 3 1 Y y y f 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 3 1 1 = - 1 3 1 1 dx dy x w = -- - 1 3 1 3 1 1 1 w x w dx dy = 2 6 1 w . = ( 29 1 2 6 1- w . = ( 29 2 4 6 1 1 w-- . f W ( w ) = F W ' ( w ) = = w 3 1 , = 3 1 , = ( 29 4 3 1 w- , 0 < w < 1. 1 < w < 3. 3 < w < 4. OR ( 29 ( 29 ( 29 -- + = dx x w f x f w f Y X Y X ( 29 < < = otherwise 3 3 1 Y y y f ( 29 < <- <- <- = = otherwise 3 3 1 otherwise 3 3 1 Y w x w x w x w f Case 1 : 0 < w < 1. f W ( w ) = w dx 3 1 1 = w 3 1 . Case 2 : 1 < w < 3. f W ( w ) = 1 3 1 1 dx = 3 1 . Case 3 : 3 < w < 4. f W ( w ) = - 1 3 3 1 1 w dx = ( 29 4 3 1 w- . OR ( 29 ( 29 ( 29 -- + = dy y f y w f w f Y X Y X ( 29 < < = otherwise 1 1 X x x f ( 29 < <- = <- < =- otherwise 1 1 otherwise 1 1 X w y w y w y w f Case 1 : 0 < w < 1. f W ( w ) = w dy 3 1 1 = w 3 1 . Case 2 : 1 < w < 3. f W ( w ) = - w w dy 1 3 1 1 = 3 1 ....
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409Hw01ans - STAT 409 Homework #1 Fall 2009 (due Friday,...

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