HW5solution (1)

# HW5solution (1) - Econ 3190 Fall 2010 HW#5 Solution 1(a P(X...

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Econ 3190 Fall 2010 HW#5 Solution 1. (a) P ( X + Y 1 2 ) = Z 1 2 0 Z 1 2 x 0 24 xydxdy = 24 Z 1 2 0 x Z 1 2 x 0 ydydx = 12 Z 1 2 0 x ( 1 2 ± x ) 2 dx = 12( 1 8 x 2 ± 1 3 x 3 + 1 4 x 4 ) j 1 2 0 = 1 16 (b) f X ( x ) = Z 1 x 0 24 xydy = 12 xy 2 j 1 x 0 = 12 x (1 ± x ) 2 for x 2 [0 ; 1] (c) f ( y j x ) = f ( x; y ) f X ( x ) = 24 xy 12 x (1 ± x ) 2 = 2 y (1 ± x ) 2 for 0 y 1 ± x; 0 x 1 x = 3 4 f ( y j 3 4 ) = 2 y 1 = 16 = 32 y for 0 y 1 4 so P ( Y < 1 8 j X = 3 4 ) = Z 1 8 0 f ( y j 3 4 ) dy = Z 1 8 0 32 ydy = 1 4 1

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(a) f X ( x ) = Z 1 x 2 dy = 2(1 x ) for x 2 (0 ; 1) f Y ( y ) = Z y 0 2 dx = 2 y for y 2 (0 ; 1) (b) Since f X ( x ) ± f Y ( y ) 6 = f ( x; y ) 8 x; y; X and Y are NOT independent. (c) f ( x j y ) = f ( x; y ) f Y ( y ) = 2 2 y = 1 y for 0 < x < y < 1 y = 3 4 f ( x j 3 4 ) = 4 3 for 0 < x < 3 4 (1) so P ( 1 4 < X < 1 2 j Y = 3 4 ) = Z 1 2 1 4 f ( x j 3 4 ) dx = Z 1 2 1 4 4 3 dx = 1 3 (a) f X ( x ) = 8 < : f
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## This note was uploaded on 04/06/2011 for the course ECON 3190 taught by Professor Hong during the Fall '07 term at Cornell.

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HW5solution (1) - Econ 3190 Fall 2010 HW#5 Solution 1(a P(X...

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