Dr. Hackney STA Solutions pg 68

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

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5-2 Solutions Manual for Statistical Inference 5.8 a. 1 2 n ( n - 1) n i =1 n j =1 ( X i - X j ) 2 = 1 2 n ( n - 1) n i =1 n j =1 ( X i - ¯ X + ¯ X - X j ) 2 = 1 2 n ( n - 1) n i =1 n j =1 ( X i - ¯ X ) 2 - 2( X i - ¯ X )( X j - ¯ X ) + ( X j - ¯ X ) 2 = 1 2 n ( n - 1) n i =1 n ( X i - ¯ X ) 2 - 2 n i =1 ( X i - ¯ X ) n j =1 ( X j - ¯ X ) =0 + n n j =1 ( X j - ¯ X ) 2 = n 2 n ( n - 1) n i =1 ( X i - ¯ X ) 2 + n 2 n ( n - 1) n j =1 ( X j - ¯ X ) 2 = 1 n - 1 n i =1 ( X i - ¯ X ) 2 = S 2 . b. Although all of the calculations here are straightforward, there is a tedious amount of book- keeping needed. It seems that induction is the easiest route. (Note: Without loss of generality we can assume θ 1 = 0, so E X i = 0.) (i) Prove the equation for n = 4. We have S 2 = 1 24 4 i =1 4 j =1 ( X i - X j ) 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 ) = 24 θ 2 . The expected value E( S 2 ) 2 = E( S 4 ) contains 256(= 4 4 ) terms of which 112(= 4 × 16 + 4 × 16 - 4 2 ) are zero, whenever i = j . Of the remaining terms, 24 are of the form E( X i - X j ) 4 = 2( θ 4 + 3 θ 2 2 ) 96 are of the form E( X i - X j ) 2 ( X i - X k ) 2 = θ 4 + 3 θ 2 2
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