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09_17 - STAT 410 Examples for Fall 2008 2.3 1 Conditional...

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STAT 410 Examples for 09/17/2008 Fall 2008 2.3 Conditional Distributions and Expectations. 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 0.15 0 0.30 2 0.15 0.35 0.20 0.70 p Y ( y ) 0.30 0.50 0.20 f) Find the conditional probability distributions p X | Y ( x | y ) = ( ) ( ) y p y x p , Y of X given Y = y , conditional expectation E ( X | Y = y ) of X given Y = y , conditional variance Var ( X | Y = y ) of X given Y = y , E ( E ( X | Y ) ) , and Var ( E ( X | Y ) ) . x p X | Y ( x | 0 ) x p X | Y ( x | 1 ) x p X | Y ( x | 2 ) 1 0.15 / 0.30 = 0.50 1 0.15 / 0.50 = 0.30 1 0.00 / 0.20 = 0.00 2 0.15 / 0.30 = 0.50 2 0.35 / 0.50 = 0.70 2 0.20 / 0.20 = 1.00 E ( X | Y = 0 ) = 1.5 E ( X | Y = 1 ) = 1.7 E ( X | Y = 2 ) = 2.0 Var ( X | Y = 0 ) = 0.25 Var ( X | Y = 1 ) = 0.21 Var ( X | Y = 2 ) = 0.00 E ( X | Y = y ) p Y ( y ) Var ( X | Y = y ) p Y ( y ) 1.5 0.30 0.25 0.30 1.7 0.50 0.21 0.50 2.0 0.20 0.00 0.20 E ( E ( X | Y ) ) = 1.7 = E ( X ) E ( Var ( X | Y ) ) = 0.18 Var ( E ( X | Y ) ) = 0.03 < 0.21 = Var ( X ) . 0.21 = 0.03 + 0.18.
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Def Var ( X | Y ) ) = E [ ( X – E ( X | Y ) ) 2 | Y ] = E ( X 2 | Y ) [ E ( X | Y ) ] 2 Theorem E ( E ( X | Y ) ) = E ( X ) Var ( E ( X | Y ) ) Var ( X ) Furthermore, Var ( X ) = Var ( E ( X | Y ) ) + E ( Var ( X | Y ) ) g) Find the conditional probability distributions p Y | X ( y | x ) = ( ) ( ) x p y x p , X of Y given X = x , conditional expectation E ( Y | X = x ) of Y given X = x , conditional variance Var ( Y | X = x ) of Y given X = x , E ( E ( Y | X ) ) , and Var ( E
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