hw8_sol - ECE 3100 Fall 2009 Homework #8: Solutions 1. (a)...

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Unformatted text preview: ECE 3100 Fall 2009 Homework #8: Solutions 1. (a) The area of the half circle is r 2 2 , since we consider a uniform probability over the semicircle, the joint PDF of X and Y will be f X,Y ( x,y ) = 2 r 2 for ( x,y ) SC = { ( x,y ) | y ,x 2 + y 2 r 2 } (i.e., in the semicircle) and f X,Y ( x,y ) = 0 otherwise. 1. (b) The density of Y can be found using the following formula f Y ( y ) = Z x A y f X,Y ( x,y ) dx, where A y represents the range of possible values of X when Y = y . For any fixed y the range of possible values of X is A y = [- p r 2- y 2 , p r 2- y 2 ] , consequently we deduce that f Y ( y ) = Z r 2- y 2- r 2- y 2 2 r 2 dx = 4 p r 2- y 2 r 2 , for y r, and f Y ( y ) = 0 for y > r . The expectation of Y is computed as follows E [ Y ] = 4 r 2 Z r y p r 2- y 2 dy. Using the change of variable t = r 2- y 2 , we have dt =- 2 ydy , the last integral becomes E [ Y ] = 2 r 2 Z r 2 t 1 2 dt = 2 r 2 [ 2 3 t 3 2 ] r 2 = 4 r 3 . 1. (c) The expectation of Y can be computed using the joint pdf f X,Y ( x,y ) as follows E [ Y ] = Z Z ( x,y ) SC yf X,Y ( x,y ) dxdy = 2 r 2 Z Z ( x,y ) SC ydxdy. In order to compute the last double integral we need to make a change of variables from the cartesian coordinates to the polar coordinates, let x = cos( ) and y = sin( ) , with [0 .r ] and [0 , ] , then since dxdy = dd , we obtain E [ Y ] = 2 r 2 Z Z r sin( ) dd = 2 r 2 Z sin( ) d Z r 2 d = 4 r 3 ....
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hw8_sol - ECE 3100 Fall 2009 Homework #8: Solutions 1. (a)...

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