Dr. Hackney STA Solutions pg 68

Dr. Hackney STA Solutions pg 68 - 5-2 Solutions Manual for...

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Unformatted text preview: 5-2 Solutions Manual for Statistical Inference 5.8 a. 1 2n(n - 1) = n n (X i -Xj ) i=1 j=1 n n 2 1 2n(n - 1) 1 2n(n - 1) (Xi - X + X - Xj )2 i=1 j=1 n n = 2 2 (X i -X) -2(X i -X)(X j -X) + (X j -X) i=1 j=1 = n n n n 1 (Xj - X)2 n(Xi - X)2 - 2 (Xi - X) (Xj -X) +n 2n(n - 1) i=1 i=1 j=1 j=1 =0 = n 2n(n - 1) 1 n-1 n n (Xi - X)2 + i=1 n 2n(n - 1) n (Xj - X)2 j=1 = (Xi - X)2 = S 2 . i=1 b. Although all of the calculations here are straightforward, there is a tedious amount of bookkeeping needed. It seems that induction is the easiest route. (Note: Without loss of generality we can assume 1 = 0, so EXi = 0.) 1 (i) Prove the equation for n = 4. We have S 2 = 24 i=1 j=1 (Xi - Xj )2 , and to calculate Var(S 2 ) we need to calculate E(S 2 )2 and E(S 2 ). The latter expectation is straightforward and we get E(S 2 ) = 242 . The expected value E(S 2 )2 = E(S 4 ) contains 256(= 44 ) terms of which 112(= 4 16 + 4 16 - 42 ) are zero, whenever i = j. Of the remaining terms, 2 24 are of the form E(Xi - Xj )4 = 2(4 + 32 ) 2 2 2 96 are of the form E(Xi - Xj ) (Xi - Xk ) = 4 + 32 2 2 2 24 are of the form E(Xi - Xj ) (Xk - X ) = 42 Thus, 4 4 Var(S 2 ) = 1 1 1 2 2 2 2 24 2(4 + 32 ) + 96(4 + 32 ) + 24 44 - (242 ) = 4 - 2 . 2 24 4 3 (ii) Assume that the formula holds for n, and establish it for n+1. (Let Sn denote the variance based on n observations.) Straightforward algebra will establish n n n 1 2 2 2 Sn+1 = (X -Xj ) + 2 (X k -Xn+1 ) 2n(n + 1) i=1 j=1 i k=1 def'n = 1 [A + 2B] 2n(n + 1) where Var(A) = 4n(n - 1)2 4 - n-3 2 n-1 2 (induction hypothesis) (Xk and Xn+1 are independent) (some minor bookkeeping needed) 2 Var(B) = n(n + 1)4 - n(n - 3)2 2 Cov(A, B) = 2n(n - 1) 4 - 2 ...
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