solutions01 - x y z = f X x f Y y f Z z = 1 2 1 3 e-y...

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Stat 351 Fall 2007 Assignment #1 Solutions 2. (a) If X Unif[0 , 2], then F X ( x ) = x 2 for 0 x 2, and if Y Exp(3), then F Y ( y ) = 1 - e - y/ 3 for y > 0. Since X and Y are independent, we conclude that F X,Y ( x, y ) = F X ( x ) · F Y ( y ) = x 2 ± 1 - e - y/ 3 ² for 0 x 2 and y > 0. We should also note that if x < 0, then F X ( x ) = 0 and if x 2, then F X ( x ) = 1. Furthermore, if y 0, then F Y ( y ) = 0. Combining everything we conclude F X,Y ( x, y ) = x 2 ( 1 - e - y/ 3 ) , if 0 x 2 and y > 0 , 1 - e - y/ 3 , if x > 2 and y > 0 , 0 , if x < 0 or y 0 . (b) We find 2 ∂x∂y F X,Y ( x, y ) = 2 ∂x∂y h x 2 ± 1 - e - y/ 3 ²i = 1 2 · 1 3 e - y/ 3 . Since f X ( x ) = 1 2 and f Y ( y ) = 1 3 e - y/ 3 , we see that 2 ∂x∂y F X,Y ( x, y ) = f X ( x ) · f Y ( y ) as required. (c) If Z ∼ N (0 , 1) is independent of X and Y , then the joint density of ( X, Y, Z ) is given by f X,Y,Z
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Unformatted text preview: ( x, y, z ) = f X ( x ) · f Y ( y ) · f Z ( z ) = 1 2 · 1 3 e-y/ 3 · 1 √ 2 π e-z 2 / 2 = 1 √ 72 π e-1 6 (2 y +3 z 2 ) for 0 ≤ x ≤ 2, y > 0, and-∞ < z < ∞ . 3. If X and Y are both discrete random variables, and their joint mass function is p X,Y ( x, y ), then F X,Y ( x, y ) = X x ≤ x X y ≤ y p X,Y ( x , y ) . If X and Y are both continuous random variables, and their joint density function is f X,Y ( x, y ), then F X,Y ( x, y ) = Z x-∞ Z y-∞ f X,Y ( u, v ) dv du....
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This note was uploaded on 03/26/2012 for the course STAT 351 taught by Professor Michaelkozdron during the Fall '08 term at University of Regina.

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