Dr. Hackney STA Solutions pg 111

Dr. Hackney STA Solutions pg 111 - 7-14 Solutions Manual...

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Unformatted text preview: 7-14 Solutions Manual for Statistical Inference b. f (xi - t) = i 2 2 2 1 1 1 1 1 x e- 2 i (xi -t) = e- 2 n(-t) e- 2 (n-1)s , (2)n/2 (2)n/2 so ( n/ 2) p (x) = ( n/ 2) 2 1 x te- 2 n(-t) dt - 2 1 x e- 2 n(-t) dt - = x = x. 1 c. f (xi - t) = i i I t- 1 1 xi t + 2 2 x(1) +1/2 x(n) +1/2 = I x(n) - 1 1 t x(1) + 2 2 , so t dt = 1 dt p (x) = x(1) +1/2 x(n) +1/2 x(1) + x(n) . 2 7.37 To find a best unbiased estimator of , first find a complete sufficient statistic. The joint pdf is f (x|) = 1 2 n I(-,) (xi ) = i 1 2 n I[0,) (max|xi |). i By the Factorization Theorem, maxi |Xi | is a sufficient statistic. To check that it is a complete sufficient statistic, let Y = maxi |Xi |. Note that the pdf of Y is fY (y) = ny n-1 /n , 0 < y < . Suppose g(y) is a function such that E g(Y ) = 0 ny n-1 g(y) dy = 0, for all . n Taking derivatives shows that n-1 g() = 0, for all . So g() = 0, for all , and Y = maxi |Xi | is a complete sufficient statistic. Now EY = 0 y ny n-1 n dy = n n+1 E n+1 Y n = . Therefore n+1 maxi |Xi | is a best unbiased estimator for because it is a function of a complete n sufficient statistic. (Note that X(1) , X(n) is not a minimal sufficient statistic (recall Exercise 5.36). It is for < Xi < 2, -2 < Xi < , 4 < Xi < 6, etc., but not when the range is symmetric about zero. Then maxi |Xi | is minimal sufficient.) 7.38 Use Corollary 7.3.15. a. logL(|x) = log x-1 = i i [log + (-1) logxi ] i = i 1 + logxi = -n - i logxi 1 - . n Thus, - i log Xi /n is the UMVUE of 1/ and attains the Cramr-Rao bound. e ...
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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