hw7_soln

hw7_soln - Homework 7 Solutions Problem Solutions : Yates...

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Homework 7 Solutions Problem Solutions : Yates and Goodman, 4.4.1 4.4.3 4.5.2 4.6.1 4.6.7 4.7.1 and 4.7.4 Problem 4.4.1 Solution (a) The joint PDF of X and Y is Y X Y + X = 1 1 1 f X,Y ( x, y ) = ± c x + y 1 , x, y 0 0 otherwise (1) To ±nd the constant c we integrate over the region shown. This gives Z 1 0 Z 1 - x 0 c dy dx = cx - cx 2 ² ² ² 1 0 = c 2 = 1 (2) Therefore c = 2. (b) To ±nd the P [ X Y ] we look to integrate over the area indicated by the graph Y X X=Y 1 1 X Y £ P [ X Y ] = Z 1 / 2 0 Z 1 - x x dy dx (3) = Z 1 / 2 0 (2 - 4 x ) dx (4) = 1 / 2 (5) (c) The probability P [ X + Y 1 / 2] can be seen in the ±gure. Here we can set up the following integrals Y X Y + X = 1 Y + X = ½ 1 1 P [ X + Y 1 / 2] = Z 1 / 2 0 Z 1 / 2 - x 0 2 dy dx (6) = Z 1 / 2 0 (1 - 2 x ) dx (7) = 1 / 2 - 1 / 4 = 1 / 4 (8) Problem 4.4.3 Solution The joint PDF of X and Y is f X,Y ( x, y ) = ± 6 e - (2 x +3 y ) x 0 , y 0 , 0 otherwise . (1) 1
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(a) The probability that X Y is: Y X X Y P [ X Y ] = Z 0 Z x 0 6 e - (2 x +3 y ) dy dx (2) = Z 0 2 e - 2 x ± - e - 3 y ² ² y = x y =0 ³ dx (3) = Z 0 [2 e - 2 x - 2 e - 5 x ] dx = 3 / 5 (4) The P [ X + Y 1] is found by integrating over the region where X + Y 1 Y X X+Y 1 1 1 P [ X + Y 1] = Z 1 0 Z 1 - x 0 6 e - (2 x +3 y ) dy dx (5) = Z 1 0 2 e - 2 x h - e - 3 y ² ² y =1 - x y =0 i dx (6) = Z 1 0 2 e - 2 x h 1 - e - 3(1 - x ) i dx (7) = - e - 2 x - 2 e x - 3 ² ² 1 0 (8) = 1 + 2 e - 3 - 3 e - 2 (9) (b) The event min(
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This note was uploaded on 04/03/2008 for the course EEE 350 taught by Professor Duman during the Fall '08 term at ASU.

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hw7_soln - Homework 7 Solutions Problem Solutions : Yates...

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