This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Homework 6 Solutions (i) Note that l ( x 1 ,Î¾ ) â‰¡ ln p ( x 1 ,Î¾ ) = Î¾T ( x 1 ) A ( Î¾ ) + ln g ( x 1 ) l ( x 1 ,Î¾ ) = T ( x 1 ) A ( Î¾ ) l 00 ( x 1 ,Î¾ ) = A 00 ( Î¾ ) . Since E Î¾ [ l ( X 1 ,Î¾ )] = 0, E Î¾ T ( X 1 ) = A ( Î¾ ) . Also, I ( Î¾ ) = E Î¾ [ l 00 ( X 1 ,Î¾ )] = A 00 ( Î¾ ) > . Then, Var Î¾ ( T ( X 1 )) = E Î¾ [( T ( X 1 ) A ( Î¾ )) 2 ] = E Î¾ [ l ( X 1 ,Î¾ ) 2 ] = I ( Î¾ ) = A 00 ( Î¾ ) . A moment estimator for Î¾ : \ A ( Î¾ ) = \ E Î¾ ( T ( X 1 )) = 1 n n X i =1 T ( X i ) â‰¡ T ( X ) = â‡’ \ Î¾ ,mom = ( A ) 1 ( TX ) â‰¡ H (( TX )) . (ii) Consider the loglikelihood function: l n ( Î¾  x 1 ,...,x n ) = n X i =1 l ( x i ,Î¾ ) = Î¾ n X i =1 T ( x i ) nA ( Î¾ ) + n X i =1 ln h ( x i ) . This is strictly concave in Î¾ ( A is convex) and therefore a stationary point must give the unique maximum. Solving âˆ‚ âˆ‚Î¾ n X i =1 l ( x i ,Î¾ ) = 0 gives n X i =1 T ( X i ) nA ( Î¾ ) = 0, i.e., A ( b Î¾ MLE ) = T ( X ). Thus, b Î¾ MLE = H ( T ( X ))....
View
Full
Document
This note was uploaded on 04/14/2010 for the course STATS 610 taught by Professor Moulib during the Fall '09 term at University of Michigan.
 Fall '09
 moulib

Click to edit the document details