Dr. Hackney STA Solutions pg 115

Dr. Hackney STA Solutions pg 115 - 7-18 Solutions Manual...

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Unformatted text preview: 7-18 Solutions Manual for Statistical Inference To calculate the Cramr-Rao 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 Cramr-Rao Lower Bound for estimating 2 is e [ ()] 42 = . 2 n 2 logf (X|) 2 -nE Thus, the UMVUE of 2 does not attain the Cramr-Rao 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 - n-3 n-1 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- n-3 n-1 = +1 n-1 . n+1 d. There was a typo in early printings; the minimizing a is a= n-1 (n + 1) + (-3)(n-1) n . To obtain this simply calculate 1/(B + 1) with (from part (b)) B= 1 n - n-3 n-1 . ...
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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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