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Quiz_2_Solutions

# Quiz_2_Solutions - MSE ˆ θ = E ˆ θ-θ 2 = V ar ˆ θ...

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IEOR 165: Engineering Statistics, Quality Control and Forecasting, Spring 2007 Quiz 2 Solution Question 1 (a) The expected value of the ﬁrst moment for a single uniform random variable E [ X ] = Z b -∞ xe x - b dx = Z 0 -∞ ( y + b ) e y dy = Z 0 -∞ yde y + be y | 0 -∞ = b - 1 Therefore b MOM = E [ X ] + 1 = 1 n n X i =1 X i + 1 . (b) E [ b MOM ] = 1 n n X i =1 E [ X i ] + 1 = n n ( b - 1) + 1 = b E [ X 2 ] = Z b -∞ x 2 e x - b dx = e - b Z b -∞ x 2 e x dx = b 2 - 2 b + 2 V ar ( b MOM ) = V ar 1 n n X i =1 X i + 1 ! = 1 n 2 n X i =1 V ar ( X i ) = n n 2 · ( b 2 - 2 b + 2 - ( b - 1) 2 ) = 1 n . The estimator is unbiased. (c) Noting that the density of X i is f X i ( x ) = e x - b I { x b } : b MLE = argmax b f ( x 1 ,...,x n | b ) = argmax b n Y i =1 f X i ( x i | b ) = argmax b n Y i =1 e x i - b I { x i b } = argmax b I { b max { x 1 ,...,x n }} e n i =1 x i - nb = max { x 1 ,.... x n } (d) P (max { X 1 ,...,X n } ≤ u ) = n Y i =1 P ( X i u ) = e n ( u - b ) f max { X 1 ,...,X n } ( u ) = ne n ( u - b ) E [ b MLE ] = n Z b -∞ ue n ( u - b ) du = b - 1 n E [ b 2 MLE ] = n Z b -∞ u 2 e n ( u - b ) = b 2 - 2 b n + 2 n 2 V ar ( b MLE ) = b 2 - 2 b n + 2 n 2 - ± b - 1 n ² 2 = 1 n 2 . This estimator is biased. The unbiased alternative is: b MLE + 1 n (e) We use the mean square error

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Unformatted text preview: MSE ( ˆ θ ) = E ( ˆ θ-θ ) 2 = V ar ( ˆ θ ) + bias ( ˆ θ 2 ) to compare the estimators. MSE ( b MOM ) = 1 n 1 MSE ( b MLE ) = 1 n 2 + ± b-1 n-b ² 2 = 2 n 2 MSE ± b MLE + 1 n ² = 1 n 2 Since MSE ³ b MLE + 1 n ´ ≤ MSE ( b MLE ) ≤ MSE ( b MOM ), b MLE + 1 n is the better of the three using this criterion. Also note that A MLE has fundamental ﬂaws. Depending on your sample points, the estimator may end up being larger than some of the observed values, which is impossible under the uniform distribution. 2...
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Quiz_2_Solutions - MSE ˆ θ = E ˆ θ-θ 2 = V ar ˆ θ...

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