09_09ans - STAT 410 Examples for 09/09/2011 Fall 2011 2.3...

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STAT 410 Examples for 09/09/2011 Fall 2011 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 0.10 0 0.25 2 0.25 0.30 0.20 0.75 p Y ( y ) 0.40 0.40 0.20 a) 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 , and E ( E ( X | Y ) ). x p X | Y ( x | 0 ) x p X | Y ( x | 1 ) x p X | Y ( x | 2 ) 1 0.15 / 0.40 = 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 ) = 2.0 Def E ( X | Y = y ) = x x P ( X = x | Y = y ) = x x p X | Y ( x | y ) discrete E ( X | Y = y ) = ( ) - dx y x f x | Y | X continuous 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.
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y E ( X | Y = y ) p Y ( y ) 0 1.625 0.40 0.65 E ( E ( X | Y ) ) = 1.75. 1 1.75 0.40 0.70 2 2.0 0.20 0.40 Recall: E ( X ) = 1.75. E ( a 1 X 1 + a 2 X 2 | Y ) = a 1 E ( X 1 | Y ) + a 2 E ( X 2 | Y ) E [ g ( Y ) | Y ] = g ( Y ) E ( E ( X | Y ) ) = E ( X ) E [ E ( X | Y ) | Y ] = E ( X | Y ) E [ g ( Y ) X | Y ] = g ( Y ) E ( X | Y ) b) 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 , and E ( E ( Y | X ) ). y
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This note was uploaded on 10/31/2011 for the course MATH 464 taught by Professor Monrad during the Fall '08 term at University of Illinois, Urbana Champaign.

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09_09ans - STAT 410 Examples for 09/09/2011 Fall 2011 2.3...

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