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 ) = γe x if x < = α − γe − x if x ≥ . Here α and γ are constants. Then compute • µ x = E x = 0 • σ 2 x = E ( x − µ x ) 2 = E x 2 = 2 By taking the derivative of F x ( x ) we obtain f x ( x ) = γe −| x | = γe x if x ≤ = γe − x if x ≥ . Since the area under f x ( x ) equals one, we must have 1 = ∫ ∞ −∞ f x ( x ) dx = γ ∫ −∞ e x dx + γ ∫ ∞ e − x dx = 2 γ. Hence γ = 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, µ 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 ) = γ | x | if | x | ≤ 1 = 0 otherwise where γ 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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