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ieor_hw3_sol - IEOR 165 Engineering Statistics Quality...

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IEOR 165: Engineering Statistics, Quality Control and Forecasting, Spring 2008 Homework 3 Solution Extra Question (a) The first sample moment is μ = n i =1 X i n . Looking at the expected value of the first moment for a single uniform random variable, E [ X ] = a 2 a = 2 · E [ X ] . Therefore A MOM = 2 μ = 2 n i =1 X i n . (b) E [ A MOM ] = 2 n n X i =1 E [ X i ] = 2 n n · a 2 = a V ar ( A MOM ) = 4 n 2 n X i =1 V ar ( X i ) = 4 n n 2 · a 2 12 = a 2 3 n . The estimator is unbiased. (c) Noting that the density of X i is f X i ( x ) = 1 | a | I { a < x < 0 } : a MLE = argmax a f ( x 1 , . . . , x n | a ) = argmax a n Y i =1 f X i ( x i | a ) = argmax 1 | a | ! n n Y i =1 I { a < x i < 0 } = argmax I { a < min { x 1 , . . . , x n }} | a | n = min { x 1 , . . . . x n } Thus, A MLE = min { X 1 , . . . . X n } . (d) P (min { X 1 , . . . , X n } ≤ u ) = 1 - P (min { X 1 , . . . , X n } > u ) = 1 - n Y i =1 P ( X i > u ) = 1 - u a n f min { X 1 ,...,X n } ( u ) = - nu n - 1 a n E [ A MLE ] = - n Z 0 a u u n - 1 a n du = - n n n +1 ( n + 1) a n ! 0 u = a = n n + 1 a E [ A 2 MLE ] = - n Z 0 a u 2 u n - 1 a n du = n n + 2 a 2 V ar ( A MLE ) = n n + 2 a 2 - n n + 1 a 2 = n ( n + 2)( n + 1) 2 a 2 This estimator is biased. The unbiased alternative is: n +1 n A MLE (e) We use the mean square error MSE ( ˆ θ ) = E ( ˆ θ - θ ) 2 = V ar
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