# 461s11t2s - Math 461 Test 2, Spring 2011, Solutions...

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Math 461 Test 2, Spring 2011, Solutions Calculators, books, notes and extra papers are not allowed on this test! Show all work to qualify for full credits 1. (15 points) The joint density of X and Y is given by f ( x,y ) = 6 7 ( x 2 + xy 2 ) , 0 < x < 1 , 0 < y < 2 , 0 , otherwise . Find (a) E [ X ]; (b) E [ Y ]; (c) Cov( X,Y ). Solution . (a) E [ X ] = 6 7 Z 1 0 ±Z 2 0 ( x 3 + x 2 y 2 ) dy dx = 6 7 Z 1 0 (2 x 3 + x 2 ) dx = 5 7 . (b) E [ Y ] = 6 7 Z 1 0 ±Z 2 0 ( x 2 y + xy 2 2 ) dy dx = 6 7 Z 1 0 (2 x 2 + 4 3 x ) dx = 8 7 . (c) E [ XY ] = 6 7 Z 1 0 ±Z 2 0 ( x 3 y + x 2 y 2 2 ) dy dx = 6 7 Z 1 0 (2 x 3 + 4 3 x 2 ) dx = 17 21 . Thus Cov( X,Y ) = 17 21 - 40 49 . 2. (12 points) Suppose that X is an absolutely continuous with density given by f ( x ) = 1 π (1 + x 2 ) , x ( -∞ , ) . Find the density of Y = X 4 . Solution . For y > 0, P ( Y y ) = P ( X 4 y ) = P ( - y 1 / 4 X y 1 / 4 ) = Z y 1 / 4 - y 1 / 4 1 π (1 + x 2 ) dx. Thus by the chain rule we get

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## This note was uploaded on 04/03/2012 for the course STAT 461 taught by Professor Renmingsong during the Fall '11 term at University of Illinois, Urbana Champaign.

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461s11t2s - Math 461 Test 2, Spring 2011, Solutions...

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