# sfi - Stat 5102(Geyer Final Exam Problem 1 Looking at the...

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Stat 5102 (Geyer) Final Exam Problem 1 Looking at the simplest moment first, E ( X i ) = α λ = θ θ = 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 ) = α λ 2 = θ θ 2 = 1 θ is a simple function of θ and gives the method of moments estimator ˆ θ 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 location-scale family with base density g (Sections 4.1 and 9.2 of the course notes). The variables Y i = X i - μ σ 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 μ 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 = π 2 3 n Thus the ARE is either π 2 / 12 or 12 2 depending on which way you form the ratio. The important point is that X n is the better estimator since π 2 3 = 3 . 2899 is less than 4. 1

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Alternate Solution Things are only a little different if we don’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 = μ + σY i , E ( X i ) = μ and var( X i ) = σ 2 var( Y i ) = π 2 σ 2 3 so X n ≈ N μ, π 2 σ 2 3 n Note that g is symmetric about zero but f is symmetric about μ , so μ is both the population mean and the population median. And the asymptotic variance of e X n is 1 4 nf ( μ ) 2 = 4 σ 2 n and e X n ≈ N μ, 4 σ 2 n The ratio of asymptotic variances is the same as before.
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