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# slides20 - Lecture Stat 302 Introduction to Probability...

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Lecture Stat 302 Introduction to Probability - Slides 20 AD April 2010 AD () April 2010 1 / 14

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Conditional Distributions: Discrete Case Given a joint p.m.f. for two r.v. X , Y it is possible to compute the conditional p.m.f. X given Y = y . Assume X , Y are discrete-valued r.v. with a joint p.m.f. p ( x , y ) then the conditional p.m.f. of X given Y = y is p X j Y ( x j y ) = p ( x , y ) p Y ( y ) = p Y j X ( y j x ) p X ( x ) p Y ( y ) The conditional expectation of g ( X ) is given by E ( g ( X ) j Y = y ) = x g ( x ) ° p X j Y ( x j y ) . E ( g ( X ) j Y = y ) is a function of y and E ( g ( X ) j Y ) is a r.v.. AD () April 2010 2 / 14
Example: Warm-up Let X , Y be two discrete r.v. of joint p.m.f. p ( x , y ) = ( x + y ) 21 for x = 1 , 2 , 3 and y = 1 , 2. (a) Show that X and Y are not independent. (b) Compute p X j Y ( x j y ) . (c) Compute E ( X j Y = y ) . AD () April 2010 3 / 14

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Example: Warm-up (a) X and Y are independent if and only if p ( x , y ) = p X ( x ) p Y ( y ) . We have p X ( x ) = y p ( x , y ) = p ( x , 1 ) + p ( x , 2 ) = ( x + 1 ) 21 + ( x + 2 ) 21 = 2 x 21 + 1 7 and p Y ( y ) = x p ( x , y ) = p ( 1 , y ) + p ( 2 , y ) + p ( 3 , y ) = ( 1 + y ) 21 + ( 2 + y ) 21 + ( 3 + y ) 21 = 2 7 + 3 y 21 . Clearly p ( x , y ) 6 = p X ( x ) p Y ( y ) so X , Y are not independent. Remark: A safety check consists of checking that x p X ( x ) = y p Y ( y ) = 1. AD () April 2010 4 / 14
Example: Warm-up (b) The conditional p.m.f. is given by p X j Y ( x j y ) = p ( x , y ) p Y ( y ) = ( x + y ) / 21 ( 6 + 3 y ) / 21 = x + y 6 + 3 y (c) The conditional mean is given by E ( X j Y = y ) = x x ° p X j Y ( x j y ) = 1 + y 6 + 3 y + 2 ± 2 + y 6 + 3 y + 3 ± 3 + y 6 + 3 y = 14 + 6 y 6 + 3 y AD () April 2010 5 / 14

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slides20 - Lecture Stat 302 Introduction to Probability...

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