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Unformatted text preview: 716 Solutions Manual for Statistical Inference Then using () = p and () = 1, () nE
2 2 logf (X) = 1 p(1  p) = = VarX. n/p(1  p) n We know that EX = p. Thus, X attains the CramrRao bound. e 7.41 a. E ( i ai Xi ) = i ai E Xi = i ai = i ai = . Hence the estimator is unbiased. b. Var ( i ai Xi ) = i a2 Var Xi = i a2 2 = 2 i a2 . Therefore, we need to minimize i i i subject to the constraint i ai = 1. Add and subtract the mean of the ai , 1/n, to get a2 = i
i i i a2 , i ai  1 n + 1 n 2 =
i ai  1 n 2 + 1 , n because the crossterm is zero. Hence, i a2 is minimized by choosing ai = 1/n for all i. i Thus, i (1/n)Xi = X has the minimum variance among all linear unbiased estimators. 7.43 a. This one is real hard  it was taken from an American Statistician article, but the proof is not there. A cryptic version of the proof is in Tukey (Approximate Weights, Ann. Math. Statist. 1948, 9192); here is a more detailed version. 2 2 Let qi = qi (1 + ti ) with 0 1 and ti  1. Recall that qi = (1/i )/ j (1/j ) and 2 VarW = 1/ j (1/j ). Then Var qi Wi j qj = = = 1 2 j qj )
2 qi i i 2 2 qi (1 + ti )2 i i 2 j (1/j ) qi (1 + ti )2 , i ( [ [ 1 (1 + t )]2 j j qj 1 qj (1 + tj )]2 j using the definition of qi . Now write qi (1 + ti )2 = 1 + 2 i j q j tj + 2
j qj t2 = [1 + j
j qj tj ]2 + 2 [
j q j t2  ( j
j qj tj )2 ], where we used the fact that [
j j qj = 1. Now since qj (1 + tj )]2 = [1 + j qj tj ]2 , Var qi Wi j qj = 2 [ j qj t2  ( j qj tj )2 ] 1 j 1+ 2) [1 + j qj tj ]2 j (1/j 2 [1  ( 1 1+ 2 [1 + j (1/j )
j j j qj tj )2 ] qj tj ]2 , since j qj t2 1. Now let T = j Var qi Wi j qj qj tj , and 1 2 [1  T 2 ] , 2 ) 1 + [1 + T ]2 j (1/j ...
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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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