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Unformatted text preview: 718 Solutions Manual for Statistical Inference To calculate the CramrRao lower bound, we have e E 2 logf (X) 2 = = E E
2 2 1 log e(X) /2 2 2 2 1 log(2)1/2  (X)2 2 2 = E (X) = 1, and () = 2 , [ ()]2 = (2)2 = 42 so the CramrRao Lower Bound for estimating 2 is e [ ()] 42 = . 2 n 2 logf (X)
2 nE Thus, the UMVUE of 2 does not attain the CramrRao bound. (However, the ratio of the e variance and the lower bound 1 as n .) 7.45 a. Because E S 2 = 2 , bias(aS 2 ) = E(aS 2 )  2 = (a  1) 2 . Hence, MSE(aS 2 ) = Var(aS 2 ) + bias(aS 2 )2 = a2 Var(S 2 ) + (a  1)2 4 . b. There were two typos in early printings; = E[X  ]4 / 4 and Var(S 2 ) = See Exercise 5.8b for the proof. c. There was a typo in early printings; under normality = 3. Under normality we have = E[X  ]4 X  =E 4 4 1 n  n3 n1 4 . = E Z 4, where Z n(0, 1). Now, using Lemma 3.6.5 with g(z) = z 3 we have = E Z 4 = E g(Z)Z = 1E(3Z 2 ) = 3E Z 2 = 3. To minimize MSE(S 2 ) in general, write Var(S 2 ) = B 4 . Then minimizing MSE(S 2 ) is equivalent to minimizing a2 B + (a  1)2 . Set the derivative of this equal to 0 (B is not a function of a) to obtain the minimizing value of a is 1/(B + 1). Using the expression in part (b), under normality the minimizing value of a is 1 = B+1 1
1 n 3 n3 n1 = +1 n1 . n+1 d. There was a typo in early printings; the minimizing a is a= n1 (n + 1) +
(3)(n1) n . To obtain this simply calculate 1/(B + 1) with (from part (b)) B= 1 n  n3 n1 . ...
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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.
 Spring '12
 Dr.Hackney
 Statistics

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