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Unformatted text preview: STAT 410 Homework #8 Fall 2011 (due Friday, October 28, by 3:00 p.m.) 1. 4.3.16 Hint: g G + + + = 1 2 1 2 n n t n t e o n t for large n . a) M X 1 ( t ) = ( 1 t ) 1 , t < 1. M Y n ( t ) = ( ) g g G  1 X E n n t e = ( ) g g G + + + n t n t n e e 2 1 X ... X X E = n n t n t e 1 X M g g G g g G  = n n t n t e g g G  1 = n n t n t e n t e g g G  , 1 < n t . b) g G + + + = n n t n t e o n t 1 2 1 2 . M Y n ( t ) = n o n n t n t n t n t g g G g G + + + 1 2 1 2 2 = n o n n t g g G g G + 1 2 1 2 . As n , M Y n ( t ) g G g g g 2 2 exp t = M Z ( t ), where Z has Standard Normal N ( 0, 1 ) distribution. Z Y D n , Z ~ N ( 0, 1 ). 2. 4.3.17 Hint: Use Theorem 4.3.9. From 4.3.16, Y n = ( ) 1 X n n D N ( 0, 1 ). Let g ( x ) = x . Then g ( x ) is differentiable, g ' ( x ) = x 2 1 , and g ' ( 1 ) = 2 1 . Then, by Theorem 4.3.9, ( ) 1 X n n ( ) ( ) 1 1 , 2 ' g N D = 4 1 , N . 3. Let Y n be 2 ( n ). What is the limiting distribution of Z n = n n Y ? Hint: We already know that ( ) n n n 2 Y = g g G  1 Y 2 n n n ( ) 1 , N D . Z n = n n Y = g g G  1 Y n n n . ( ) n n n 2 Y = g g G  1 Y 2 n n n ( ) 1 , N D . g g G  1 Y n n n ( ) 2 , N D . Let g ( x ) = x . Then g ( x ) is differentiable, g ' ( x ) = x 2 1 , and g ' ( 1 ) = 2 1 . Then, by Theorem 4.3.9, g g G  1 Y n n n ( ) ( ) g G 2 1 , 2 ' g N D = g g G g G 2 2 1 , 2 N . Z n = n n Y = g g G  1 Y n n n g G 2 1 , N D . 4. Let X 1 , X 2 , , X n be a random sample from the distribution with probability density function ( ) ( ) ( ) g 1 1 g g ; X x x f + = , 0 < x < 1, > 1. Recall: g ~ = 2 X 1 is the method of moments estimator of . Show that g ~ is asymptotically normally distributed ( as n )....
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This note was uploaded on 01/26/2012 for the course STAT 410 taught by Professor Monrad during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Monrad
 Statistics

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