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hw8_sol - ECE 3100 Homework#8 Solutions 1(a The area of the...

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ECE 3100 Fall 2009 Homework #8: Solutions 1. (a) The area of the half circle is πr 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 πr 2 for ( x, y ) SC = { ( x, y ) | y 0 , 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 πr 2 dx = 4 p r 2 - y 2 πr 2 , for 0 y r, and f Y ( y ) = 0 for y > r . The expectation of Y is computed as follows E [ Y ] = 4 πr 2 Z r 0 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 πr 2 Z r 2 0 t 1 2 dt = 2 πr 2 [ 2 3 t 3 2 ] r 2 0 = 4 r 3 π . 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 πr 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 = ρ cos( θ ) and y = ρ sin( θ ) , with ρ [0 .r ] and θ [0 , π ] , then since dxdy = ρdρdθ , we obtain E [ Y ] = 2 πr 2 Z π 0 Z r 0 ρ sin( θ ) ρdρdθ = 2 πr 2 Z π 0 sin( θ ) Z r 0 ρ 2 = 4 r 3 π . 2. (a) Yes X is a continuous random variable, it has the following PDF f X ( x ) = 1 if 0 x 1 , 0 otherwise .
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