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Unformatted text preview: 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 ≥ ,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 ≤ 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 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 t 1 2 dt = 2 πr 2 [ 2 3 t 3 2 ] r 2 = 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 Z r ρ sin( θ ) ρdρdθ = 2 πr 2 Z π sin( θ ) dθ Z r ρ 2 dρ = 4 r 3 π ....
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This note was uploaded on 01/16/2010 for the course ECE 3100 at Cornell.
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