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Unformatted text preview: STAT 410 Homework #9 Spring 2009 (due Monday, July 27, by 4:00 p.m.) 1. 4.3.16 Hint: n n e t n t n t e z n t 6 2 1 3 2 + + + = for some z between 0 and n t . a) M X 1 ( t ) = ( 1 – t ) – 1 , t < 1. M Y n ( t ) = ( 29  1 X E n n t e = ( 29 + + + n t n t n e e 2 1 X ... X X E = n n t n t e 1 X M  = n n t n t e  1 = n n t n t e n t e  , 1 < n t . b) n n e t n t n t e u n t 6 2 1 3 2 + + + = for some u between 0 and n t . Also n e t n t e v n t 2 1 2 + + = for some v between 0 and n t . Thus, for some u between 0 and n t and for some v between 0 and n t , M Y n ( t ) = n v u n n e t n t n t n n e t n t n t  + + + 2 6 2 1 3 2 3 2 = n v u n n e t n n e t n t  + 2 6 2 1 3 3 2 . As n → ∞ , M Y n ( t ) → 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 = ( 29 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, ( 29 1 X n n ( 29 ( 29 → ⋅ 1 1 , 2 ' g N D = 4 1 , N . 3. 4.3.19 F ( x ) = x 5 , 0 < x < 1. F Y 1 ( x ) = 1 – ( 1 – x 5 ) n , 0 < x < 1. F Z n ( z ) = P ( n p Y 1 ≤ z ) = F Y 1 ( z / n p ) = n p n z 5 5 1 1  , 0 < z < n p . If p = 1 / 5 , F Z n ( z ) → 1 – e – z 5 , z > 0, as n → ∞ , and Z n converges in distribution. If p < 1 / 5 , as n → ∞ , n p n z 5 5 1  → 0, and F Z n ( z ) → 1, z > 0, and Z n P → 0. If p > 1 / 5 , as n → ∞ , n p n z 5 5 1  → 1, and F Z n ( z ) → 0, z > 0, and Z n does not have a limiting distribution. 4. 5.1.3 “Hint”: Table II ( p. 673 ) gives quantiles (percentiles) of χ 2 distribution....
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This note was uploaded on 10/29/2009 for the course STAT 420 taught by Professor Stepanov during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 STEPANOV
 Statistics

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