This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Stat 5102 (Geyer) Final Exam Problem 1 Looking at the simplest moment first, E ( X i ) = = = 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 ) = 2 = 2 = 1 is a simple function of and gives the method of moments estimator 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 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 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 = 2 3 n Thus the ARE is either 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 2 3 = 3 . 2899 is less than 4. 1 Alternate Solution Things are only a little different if we dont realize we can give the answer for Y n and e Y n instead of X n and e X n . Since X i = + Y i , E ( X i ) = and var( X i ) = 2 var( Y i ) = 2 2 3 so X n N , 2 2 3 n Note that g is symmetric about zero but f is symmetric about , so is both the population mean and the population median. And the asymptotic variance of e X n is 1 4 nf ( ) 2 = 4 2 n and e X n N , 4 2 n The ratio of asymptotic variances is the same as before.The ratio of asymptotic variances is the same as before....
View Full
Document
 Spring '02
 Staff

Click to edit the document details