test567-11sol

test567-11sol - A-AE 567 Quiz Spring 2011 Clearly write...

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Unformatted text preview: A-AE 567 Quiz Spring 2011 Clearly write your final answer on the exam. NAME: 1 Problem 1. Consider the random variable x given by the distribution: F x ( x ) = e x if x < = e x if x . Here and are constants. Then compute x = E x = 0 2 x = E ( x x ) 2 = E x 2 = 2 By taking the derivative of F x ( x ) we obtain f x ( x ) = e | x | = e x if x = e x if x . Since the area under f x ( x ) equals one, we must have 1 = f x ( x ) dx = e x dx + e x dx = 2 . Hence = 1 / 2 and f x ( x ) = 1 2 e | x | . Because f x ( x ) = e | x | / 2 is symmetric about the origin, the mean of x equals zero, that is, x = 0. For second part notice that E x 2 = x 2 f x ( x ) dx = 1 2 x 2 e | x | dx = x 2 e x dx = [ x 2 e x + 2 xe x + 2 e x ] = 2 . Therefore E x 2 = 2. 2 Problem 2. Let x be the random variable defined by the density f x ( x ) = | x | if | x | 1 = 0 otherwise where is a constant. Let y be the random variable defined by y = x 2 . Find f y ( y ). Answer: f y ( y ) = 1 if 0 x 1 = 0 otherwise In other words, y is a uniform random variable over [0 , 1] Therefore for any positive integer k 0: E y k = 1 y k dy = 1 k +1 To verify this notice that F y ( y ) = P ( y y ) = P ( x 2 y ) = P ( y x y ) = P ( y = x ) + P ( y < x y ) = P ( y < x y ) =...
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test567-11sol - A-AE 567 Quiz Spring 2011 Clearly write...

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