# 09_15 - ) = 0.25 Var ( X | Y = 1 ) = 0.21 Var ( X | Y = 2 )...

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STAT 410 Examples for 09/15/2008 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

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Unformatted text preview: ) = 0.25 Var ( X | Y = 1 ) = 0.21 Var ( X | Y = 2 ) = 0.00 Def E ( X | Y = y ) = & x x P ( X = x | Y = y ) = & x x p X | Y ( x | y ) Denote by E ( X | Y ) that function of the random variable Y whose value at Y = y is E ( X | Y = y ). Note that E ( X | Y ) is itself a random variable, it depends on the ( random ) value of Y that occurs. E ( a 1 X 1 + a 2 X 2 | Y ) = a 1 E ( X 1 | Y ) + a 2 E ( X 2 | Y ) E ( E ( X | Y ) ) = E ( X ) E [ g ( Y ) X | Y ] = g ( Y ) E ( X | Y ) E [ g ( Y ) E ( X | Y ) ] = E [ g ( Y ) X ] E [ E ( X | Y ) | Y ] = E ( X | Y )...
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## This note was uploaded on 10/22/2009 for the course STAT 410 taught by Professor Alexeistepanov during the Fall '08 term at University of Illinois at Urbana–Champaign.

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09_15 - ) = 0.25 Var ( X | Y = 1 ) = 0.21 Var ( X | Y = 2 )...

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