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Unformatted text preview: Stat 5102 (Geyer) Final Exam Problem 1 Looking at the simplest moment first, E ( X i ) = Î± Î» = Î¸ Î¸ = 1 , we see that it is not a function of the unknown parameter and hence useless for finding a method of moments estimator. Moving up to the second central moment, var( X i ) = Î± Î» 2 = Î¸ Î¸ 2 = 1 Î¸ is a simple function of Î¸ and gives the method of moments estimator Ë† Î¸ n = 1 V n . (or the same with V n replaced by S 2 n ). Problem 2 The formula for f should be familiar. It defines the locationscale family with base density g (Sections 4.1 and 9.2 of the course notes). The variables Y i = X i Î¼ Ïƒ are i. i. d. with density g . We get the same answer for ARE whether we compare e X n and X n as estimators of Î¼ or whether we compare e Y n and Y n as estimators of zero. e Y n is asymptotically normal with variance 1 4 ng (0) 2 = 4 n (Corollary 7.28 in the notes), and Y n is asymptotically normal with variance var( Y ) n = Ï€ 2 3 n Thus the ARE is either Ï€ 2 / 12 or 12 /Ï€ 2 depending on which way you form the ratio. The important point is that X n is the better estimator since Ï€ 2 3 = 3 . 2899 is less than 4. 1 Alternate Solution Things are only a little different if we donâ€™t realize we can give the answer for Y n and e Y n instead of X n and e X n . Since X i = Î¼ + ÏƒY i , E ( X i ) = Î¼ and var( X i ) = Ïƒ 2 var( Y i ) = Ï€ 2 Ïƒ 2 3 so X n â‰ˆ N Î¼, Ï€ 2 Ïƒ 2 3 n Note that g is symmetric about zero but f is symmetric about Î¼ , so Î¼ is both the population mean and the population median. And the asymptotic variance of e X n is 1 4 nf ( Î¼ ) 2 = 4 Ïƒ 2 n and e X n â‰ˆ N Î¼, 4 Ïƒ 2 n The ratio of asymptotic variances is the same as before.The ratio of asymptotic variances is the same as before....
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This note was uploaded on 10/28/2010 for the course STAT 5101 taught by Professor Staff during the Spring '02 term at Minnesota.
 Spring '02
 Staff

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