# 410Hw03ansJeff - STAT 410 Fall 2011 Homework#3(due Friday...

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STAT 410 Fall 2011 Homework #3 (due Friday, September 16, by 3:00 p.m.) 1. Let X and Y have the joint p.d.f. f X , Y ( x , y ) = 20 x 2 y 3 , 0 < x < 1, 0 < y < x , zero elsewhere. a) Find f X | Y ( x | y ) . Recall: f Y ( y ) = ( ) 9 3 3 20 y y - , 0 < y < 1. f X | Y ( x | y ) = ( ) ( ) y f y x f , Y Y X, = 6 2 1 3 y x - , y 2 < x < 1. b) Find E ( X | Y = y ) . E ( X | Y = y ) = - 1 6 2 2 1 3 y dx y x x = 6 8 1 1 4 3 y y - - , 0 < y < 1. c) Find f Y | X ( y | x ) . Recall: f X ( x ) = 5 x 4 , 0 < x < 1. f Y | X ( y | x ) = ( ) ( ) x f y x f , X Y X, = 2 3 4 x y , 0 < y < x . d) Find E ( Y | X = x ) . E ( Y | X = x ) = x dy x y y 0 2 3 4 = x 5 4 , 0 < x < 1.

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2. Once a car accident is reported to an insurance company, the company makes an initial estimate, X, of the amount it will pay to the claimant. When the claim is finally settled, the company pays an amount, Y, to the claimant. The company has determined that X and Y have the joint p.d.f. f ( x , y ) = ( ) ( ) ( ) 1 1 2 2 1 2 - - - - x x y x x , x > 1, y > 1. a) Given that the initial claim estimated by the company is 1.5, determine the probability that the final settlement amount exceeds 2. Recall: - 1 dy y b a = 1 - a b , a > 1. f X ( x ) = ( ) ( ) ( ) - - - - 1 1 1 2 2 1 2 dy y x x x x = ( ) 1 1 1 2 1 2 2 - - - - x x x x = 3 2 x , x > 1. f Y | X ( y | x ) = ( ) ( ) x f y x f , X = ( ) ( ) 1 1 2 1 - - - - x x y x x , y > 1. f Y | X ( y | x = 1.5 ) = 4 3 - y , y > 1. P ( Y > 2 | X = 1.5 ) = - 2 4 3 dy y = 8 1 = 0.125 . b) Find E ( Y | X = x ) . E ( Y | X = x ) = ( ) ( ) - - - - 1 1 1 2 1 dy y x x y x x = ( ) - - - 1 1 1 dy y x x x x = 1 1 1 - - - x x x x = x , x > 1.
3. When you leave your car at Honest Harry’s Car Repair Shop , first it takes X weeks for needed parts to arrive, and then Y more weeks for the repairs to be finished. Thus the total wait is W = X + Y weeks. Suppose that X and Y are independent, the p.d.f. of X is f X ( x ) = 2 x , 0 < x < 1, zero otherwise, and Y has a Uniform distribution on interval ( 0, 1 ) . Find the p.d.f. of W, f W ( w ) = f X + Y ( w ) . 0 < x < 1, 0 < y < 1 0 < x + y < 2 ( ) ( ) ( ) - = - + dx x w f x f w f Y X Y X . ( ) < < = otherwise 0 1 0 2 X x x x f ( ) < < = 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. 0 < w < 1 w – 1 < 0 Case 2. 1 < w < 2 0 < w – 1 < 1 ( ) ( ) = + w dx x w f 0 Y X 1 2 = w 2 , 0 < w < 1. ( ) ( ) - + = 1 1 Y X 1 2 w dx x w f = 1 – ( w – 1 ) 2 = 2 w w 2 , 1 < w < 2. OR

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( ) ( ) ( ) - = - + dy y f y w f w f Y X Y X . Case 1. 0 < w < 1 w – 1 < 0 Case 2. 1 < w < 2 0 < w – 1 < 1 ( ) ( ) - = + w dy y w w f 0 Y X 2 1 = w 2 , 0 < w < 1.
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