hw2_sol - IEOR 165: Engineering Statistics, Quality Control...

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Unformatted text preview: IEOR 165: Engineering Statistics, Quality Control and Forecasting, Summer 2009 Homework 2 Solution Extra Question (1) The first sample moment is = 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 = 2 n i =1 X i n . (2) 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. (3) Noting that the density of X i is f X i ( x ) = 1 | a | I { a < x < } : 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 < } = argmax I { a < min { x 1 ,...,x n }} | a | n = min { x 1 ,....x n } Thus, A MLE = min { X 1 ,....X n } . (4) 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 a u u n- 1 a...
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This note was uploaded on 07/22/2009 for the course IEOR 165 taught by Professor Shanthikumar during the Summer '08 term at University of California, Berkeley.

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hw2_sol - IEOR 165: Engineering Statistics, Quality Control...

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