test567-10sol

test567-10sol - A-AE 567 Quiz Spring 2010 Solutions Problem...

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A-AE 567 Quiz Spring 2010 Solutions Problem 1. Assume that x is uniform over { x : 1 ≤ | x | ≤ 2 } , that is, f x ( x ) = γ if 1 ≤ | x | ≤ 2 = 0 otherwise . Let y = x 2 . Compute: E y = E x 2 = 7 3 . (1.1) Notice that γ = 1 / 2. Thus E y = E x 2 = -∞ x 2 f x ( x ) dx = 1 2 - 1 - 2 x 2 dx + 1 2 2 1 x 2 dx = x 3 6 ± ± ± ± - 1 - 2 + x 3 6 ± ± ± ± 2 1 = 7 6 + 7 6 = 14 6 = 7 3 . The density f y ( y ) is given by f y ( y ) = 1 2 y if 1 y 4 = 0 otherwise . 1
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x is a continuous random variable, that is, f x ( x ) contains no delta Dirac functions, we have F y ( y ) = P ( y y ) = P ( x 2 y ) = P ( - y x y ) = P ( - y < x y ) = F x ( y ) - F x ( - y ) . The fourth equality uses the fact that x is a continuous random variable. Hence F y ( y ) = F x ( y ) - F x ( - y ). By taking the derivative, we obtain f y ( y ) = d dy F y ( y ) = d dy ( F x ( y ) - F x ( - y )) = f x ( y ) 2 y + f x ( - y ) 2 y = f x ( y ) y . The last equality follows from the fact that f x ( x ) is symmetric about the origin. Thus f y ( y ) = f x ( y ) / y . Since y = x 2 , we must have y 0. Therefore f y ( y ) = f x ( y ) y = 1 2 y if 1 y 2 , or equivalently, 1 y 4 = 0 otherwise . Finally, it is noted that
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test567-10sol - A-AE 567 Quiz Spring 2010 Solutions Problem...

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