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# 400Hw06ans - STAT 400 Fall 2011 Homework#6(due Friday...

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STAT 400 Fall 2011 Homework #6 (due Friday, October 14, by 3:00 p.m.) 1. Suppose the joint probability density function of ( X , Y ) is ( ) = otherwise 0 1 0 , 2 x y y x C y x f a) Find the value of C that would make ( ) y x f , a valid probability density function. 1 = ( ) ∫ ∫ - - dy dx y x f , = ∫ ∫ 1 0 0 2 dx dy y x C x = 1 0 0 2 dx dy y x C x = 1 0 3 0 3 1 dx x y x C = 1 0 4 3 dx x C = 0 1 5 1 3 5 x C = 15 C . C = 15 . b) Find the marginal densities of X and Y. Are X and Y independent? ( ) ( ) - = dy y x f x f , X = x dy y x 0 2 15 = x dy y x 0 2 15 = 0 3 1 15 3 x y x = 5 x 4 , if 0 x 1. ( ) ( ) - = dx y x f y f , Y = 1 2 15 y dx y x = y x y 1 2 1 15 2 2 = ( ) 2 2 1 2 15 y y - , if 0 y 1. X and Y are NOT independent .

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2. Suppose the joint probability density function of ( X , Y ) is ( ) = otherwise 0 1 0 15 , 2 x y y x y x f a) Find P ( X > 2 Y ) . P ( X > 2 Y ) = 1 0 2 0 2 15 dx dy y x x = 1 0 2 0 2 15 dx dy y x x = 1 0 3 0 2 3 1 15 dx x y x = 1 0 4 8 5 dx x = ( ) 0 1 8 1 5 x = 8 1 = 0.125 . b) Find P ( X + Y < 1 ) . P ( X + Y < 1 ) = - 2 1 0 1 2 15 dy dx y x y y = - 2 1 0 1 2 2 2 15 dy dx x y y y = ( ) ( ) - - 2 1 0 2 2 2 1 2 15 dy y y y = ( ) - 2 1 0 2 2 1 2 15 dy y y = - 2 1 0 3 2 15 2 15 dy y y = 64 5 = - = - 64 15 16 5 0 2 1 4 15 2 5 4 3 y y .
c) Find Cov ( X, Y ) . E ( X ) = ( ) - dx x f x X = 1 0 4 5 dx x x = 0 1 6 5 6 x = 6 5 . E ( Y ) = ( ) - dy y f y Y = - 1 0 2 2 1 15 dy y y y = 0 1 12 15 8 15 6 4 - y y = 8 5 . E ( X Y ) = ∫ ∫ 1 0 0 2 15 dx dy y x y x x = ∫ ∫ 1 0 0 6 4 15 dx dy x x = 28 15 .

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400Hw06ans - STAT 400 Fall 2011 Homework#6(due Friday...

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