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Unformatted text preview: UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Solutions  Problem Set 9 Fall 2006 Issued: Thursday, November 2, 2006 Due: Thursday, November 9, 2006 Graded exercises Problem 9.1 (a) First, notice that g a ( ) = P ( X 1 a ) = P ( X 1 a ) = ( a ). Additionally, by the CLT, we have that: n (  X n ) = n ( ( a X n ) ( a ) ) d N (0 , 1) Letting h ( . ) = ( . ) and using the delta method yields: n ( ( a X n ) ( a ) ) d N , ( a ) 2 So: n ( ( a X n ) g a ( ) ) d N , [ ( a )] 2 To get to the result, it is enough to prove that n ( X n ) ( a X n ) p 0. This follows from continuity of and the fact that: r n n 1 ( a X n ) ( a X n ) = r n n 1 1 ( a X n ) p (b) We know that I ( X i a ) is a Bernoulli variable with mean P ( X i a ) = F X ( a ) = g a ( ) and variance g a ( )(1 g a ( )). Using the central limit theorem: n ( ( X ) g a ( )) = n h P n i =1 I ( X i a ) g a n i d N (0 ,g a ( )(1 g a ( ))) Under normality, g a ( ) = ( a ) and the result follows. (c) The asymptotic relative efficiency between n and n is: ARE ( , ) = ( a )[1 ( a )] 2 ( a ) = ( a )(  a ) 2 ( a ) From the plots below, we can see that the nonparametric estimator is less efficient that the parametric one. Furthermore, the efficiency of the nonparametric estimate degrades exponentially fast as we move towards the tails of the distribution.degrades exponentially fast as we move towards the tails of the distribution....
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This note was uploaded on 10/17/2009 for the course STAT 210a taught by Professor Staff during the Fall '08 term at University of California, Berkeley.
 Fall '08
 Staff
 Statistics

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