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HW6solution

# HW6solution - Econ 3190 Fall 2011 HW#6 Solution 1#3.41(c 0...

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Econ 3190 Fall 2011 HW #6 Solution 1. #3.41(c) From (b), we know the marginal pdf of creams is f X ( x ) = & 0 elsewhere 12(1 x ) 2 0 ± x ± 1 (See solutions to HW5) f Y j X ( y j x ) = f X;Y ( x;y ) f X ( x ) = 24 xy 12 x (1 x ) 2 = 2 y (1 x ) 2 for 0 ± x ± 1 ; 0 ± y ± 1 ; x + y ± 1 f Y j X ( y j x ) = 0 elsewhere Given X = 3 = 4 ; we know the conditional pdf of Y given X is, f Y j X =3 = 4 ( y j 3 = 4) = 32 y 0 ± y ± 1 = 4 0 elsewhere P ( Y < 1 = 8 j X = 3 = 4) = R 1 = 8 0 f Y j X =3 = 4 ( y j 3 = 4) dy = R 1 = 8 0 32 ydy = 16 y 2 j 1 = 8 0 = 1 = 4 : 2. #3.42 First, we need to compute the marginal density function of Y: f Y ( y ) = R 1 0 e ( x + y ) dx = R 1 0 e x ² e y dx = e y R 1 0 e x dx = e y ² ( e x ) j 1 0 = e y for y > 0 f Y ( y ) = 0 ; elsewhere f X j Y ( x j y ) = f X;Y ( x;y ) f Y ( y ) = e ( x + y ) e y = e x for x > 0 ; y > 0 f X j Y ( x j y ) = 0 ; elsewhere. Here we can verify that X and Y are independent since the conditional distribution of X given Y has nothing to do with

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