06_26_2 - ( X , Y ) be ( 29 + = otherwise 1 , 1 , 1 24 , y...

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STAT 410 Examples for 06/26/2009 Summer 2009 2.3 Conditional Distributions. 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 ) = ( 29 ( 29 y p y x p , Y of X given Y = y , y = 0, 1, 2. 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 g) Find the conditional probability distributions p Y | X ( y | x ) = ( 29 ( 29 x p y x p , X of Y given X = x , x = 1, 2. y p Y | X ( y | 1 ) y p Y | X ( y | 2 ) 0 0.15 / 0.30 = 0.50 0 0.15 / 0.70 = 3 / 14 1 0.15 / 0.30 = 0.50 1 0.35 / 0.70 = 7 / 14 2 0.00 / 0.30 = 0.00 2 0.20 / 0.70 = 4 / 14
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2. Let the joint probability density function for
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Unformatted text preview: ( X , Y ) be ( 29 + = otherwise 1 , 1 , 1 24 , y x y x y x y x f Recall: f X ( x ) = ( 29 2 1 12 x x-, 0 < x < 1. f Y ( y ) = ( 29 2 1 12 y y-, 0 < y < 1. f Y | X ( y | x ) = ( 29 ( 29 x f y x f , X = ( 29 2 1 2 x y-, 0 < y < 1 x , 0 < x < 1. P ( Y > 1 / 3 | X = 1 / 2 ) = ( 29 2 1 3 1 2 2 1 2 dy y = 2 1 3 1 8 dy y = 9 5 . P ( Y > 1 / 4 | X = 1 / 3 ) = ( 29 3 2 4 1 2 3 2 2 dy y = 64 55 . f X | Y ( x | y ) = ( 29 ( 29 y f y x f , Y = ( 29 2 1 2 y x-, 0 < x < 1 y , 0 < y < 1. P ( X < 1 / 2 | Y = 1 / 3 ) = ( 29 2 1 2 3 2 2 dx x = 16 9 = 0.5625....
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This note was uploaded on 10/29/2009 for the course STAT 410 taught by Professor Alexeistepanov during the Summer '08 term at University of Illinois at Urbana–Champaign.

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06_26_2 - ( X , Y ) be ( 29 + = otherwise 1 , 1 , 1 24 , y...

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