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# 09_12_1 - STAT 410 Examples for(part 1 Fall 2011 2 Let the...

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STAT 410 Examples for 09/12/2011 (part 1) Fall 2011 2. Let the joint probability density function for ( X , Y ) be ( ) + = otherwise 0 1 , 1 0 , 1 0 60 , 2 y x y x y x y x f Recall: f X ( x ) = ( ) 2 2 1 30 x x - , 0 < x < 1, E ( X ) = 2 1 , Var ( X ) = 252 9 , E ( X 2 ) = 252 72 = 7 2 ; f Y ( y ) = ( ) 3 1 20 y y - , 0 < y < 1, E ( Y ) = 3 1 , Var ( Y ) = 252 8 , E ( Y 2 ) = 252 36 = 7 1 ; f Y | X ( y | x ) = ( ) ( ) x f y x f , X = ( ) 2 1 2 x y - , 0 < y < 1 – x , 0 < x < 1. f X | Y ( x | y ) = ( ) ( ) y f y x f , Y = ( ) 3 2 1 3 y x - , 0 < x < 1 – y , 0 < y < 1. E ( Y | X = x ) = ( ) - - x dy x y y 1 0 2 1 2 = ( ) 1 3 2 x - , 0 < x < 1. E ( Y | X ) = ( ) X 1 3 2 - . E ( E ( Y | X ) ) = 3 1 = E ( Y ) . E [ ( E ( Y | X ) ) 2 ] = ( ) ( ) ( ) 2 X E X E 2 1 9 4 + - = + - 7 2 2 1 2 1 9 4 = 63 8 . Var ( E ( Y | X ) ) = 2 3 1 63 8 - = 63 1 . OR Var ( E ( Y | X ) ) = ( ) - X 1 3 2 Var = ( ) X Var 9 4 = 252 9 9 4 = 252 4 = 63 1 .

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Var ( E ( Y | X ) ) = 63 1 = 252 4 < 252 8 = Var ( Y ) . Def Var ( Y | X ) ) = E [ ( Y – E ( Y | X ) ) 2 | X ] = E ( Y 2 | X ) [ E ( Y | X ) ] 2 Theorem E ( E ( Y | X ) ) = E ( Y ) Var ( E ( Y | X ) ) Var ( Y ) Furthermore, Var ( Y ) = Var ( E ( Y | X ) ) + E ( Var ( Y | X ) ) E ( Y 2 | X = x ) = ( ) - - x dy x y y 1 0 2 2 1 2 = ( ) 2 1 2 1 x - , 0 < x < 1. Var ( Y | X = x ) = E ( Y 2 | X = x ) [ E ( Y | X = x ) ] 2 = ( ) 2 1 2 1 x - ( ) 2 1 9 4 x - = ( ) 2 1 18 1 x - , 0 < x < 1. Var ( Y | X ) = ( ) 2 X 1 18 1 - . E [ Var ( Y | X ) ] = ( ) ( ) ( ) 2 X E X E 2 1 18 1 + - = + - 7 2 2 1 2 1 18 1 = 63 1 . Var ( E ( Y | X ) ) + E [ Var ( Y | X ) ] = 63 1 63 1 + = 252 8 = Var ( Y ) .
1. Consider the following joint probability distribution p ( x , y ) of two random variables X and Y: y x 0 1 2 p X ( x ) 1 0.15 0.10 0 0.25 2 0.25 0.30 0.20 0.75 p Y ( y ) 0.40 0.40 0.20 x p X | Y ( x | 0 ) x p X | Y ( x | 1 ) x p X | Y ( x | 2 ) 1 0.15 / 0.40 = 0.375 1 0.10 / 0.40 = 0.25 1 0.00 / 0.20 = 0.00 2 0.25 / 0.40 = 0.625 2 0.30 / 0.40 = 0.75 2 0.20 / 0.20 = 1.00 E ( X | Y = 0 ) = 1.625 E ( X | Y = 1 ) = 1.75 E ( X | Y = 2 ) = 2 Var ( X | Y = 0 ) = 0.234375 Var ( X | Y = 1 ) = 0.1875 Var ( X | Y = 2 ) = 0 Def Var ( X | Y ) ) = E [ ( X – E ( X | Y ) ) 2 | Y ] = E ( X 2 | Y ) [ E ( X | Y ) ] 2 y E ( X | Y =

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09_12_1 - STAT 410 Examples for(part 1 Fall 2011 2 Let the...

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