Dr. Hackney STA Solutions pg 121

Dr. Hackney STA Solutions pg 121 - 7-24 Solutions Manual...

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Unformatted text preview: 7-24 Solutions Manual for Statistical Inference c. (|x) = n(, 2 /n). So, substituting into the formula for a normal mgf, we find E e-c = x 2 2 x e-c+ c /2n , and the LINEX posterior loss is x E (L(, a)|x) = ec(a-)+ 2 2 2 2 c /2n - c(a - x) - 1. x Substitute E e-c = e-c+ c /2n into the formula in part (b) to find the Bayes rule is 2 x - c /2n. d. For an estimator X + b, the LINEX posterior loss (from part (c)) is E (L(, x + b)|x) = ecb ec 2 2 2 2 /2n - cb - 1. For X the expected loss is ec /2n - 1, and for the Bayes estimator (b = -c 2 /2n) the expected loss is c2 2 /2n. The marginal distribution of X is m() = 1, so the Bayes risk is x infinite for any estimator of the form X + b. 2 e. For X + b, the squared error risk is E (X + b) - = 2 /n + b2 , so X is better than the Bayes estimator. The Bayes risk is infinite for both estimators. 7.66 Let S = i Xi binomial(n, ). ^ a. E 2 = E S2 = n b. Tn = (i) j=i 2 1 2 n2 E S 2 = 1 n2 (n(1 - ) + (n)2 ) = n + (i) n-1 2 n . Xj (n - 1)2 . For S values of i, Tn = (S - 1)2 /(n - 1)2 because the Xi (i) that is dropped out equals 1. For the other n - S values of i, Tn = S 2 /(n - 1)2 because the Xi that is dropped out equals 0. Thus we can write the estimator as JK(Tn ) = n S2 n-1 - n2 n S (S - 1) (n - 1) 2 2 + (n - S) n2 2 -n 2 n(n-1) S2 (n - 1) 2 = S 2 -S . n(n - 1) c. E JK(Tn ) = 1 n(n-1) (n(1 - ) + (n)2 - n) = = 2 . d. For this binomial model, S is a complete sufficient statistic. Because JK(Tn ) is a function of S that is an unbiased estimator of 2 , it is the best unbiased estimator of 2 . ...
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