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Unformatted text preview: Stat 5101 (Geyer) 2002 Midterm 2 Problem 1 This is like Problem 3.6.4 in DeGroot and Schervish. Since g 1 ( x | y ) = f ( x,y ) f 2 ( y ) (1) First we need to find the marginal of y . f 2 ( y ) = Z 1 f ( x,y ) dx = 2 2- log 2 1 1 + y Z 1 ( x + y ) dx = 2 2- log 2 1 1 + y x 2 2 + xy 1 = 2 2- log 2 1 2 + y 1 + y Plugging back into (1) we get g 1 ( x | y ) = 2 2- log 2 x + y 1+ y 2 2- log 2 1 2 + y 1+ y = x + y 1 2 + y , < x < 1 Alternate Solution (Conditional Probability is Renormalization) The conditional density g 1 ( x | y ) is a probability density though of as a function of the variable(s) in front of the bar (here x ) and is proportional to the joint density, thus g 1 ( x | y ) = c ( x + y ) , < x < 1 . (2) where c is some constant that does not depend on what we are thinking of as the variable (the stuff in front of the bar x ), though it may (and generally does) depend on the stuff behind the bar, which is being treated as constant. All we need to do is choose the constant c so that (2) integrates to one, integrating with respect to the variable x (the stuff behind the bar y...
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This note was uploaded on 10/28/2010 for the course STAT 5101 taught by Professor Staff during the Fall '02 term at Minnesota.
- Fall '02