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stat611w09midterm_sol

stat611w09midterm_sol - Stats 611 Midterm Exam...

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Stats 611 Midterm Exam Solutions (1) (i) Prove by induction. For n = 1, the equality obviously holds. Suppose that it holds for n = k , i.e., k n =1 n 2 = k ( k + 1)(2 k + 1) / 6. Then, k +1 X n =1 n 2 = k X n =1 n 2 + ( k + 1) 2 = 1 6 k ( k + 1)(2 k + 1) + ( k + 1) 2 = 1 6 ( k + 1) (2 k 2 + k ) + 6( k + 1) = 1 6 ( k + 1)(2 k 2 + 7 k + 6) = 1 6 ( k + 1)( k + 2) { 2( k + 1) + 1 } . (ii) Note that E[ n i =1 X i ] = 0 and s 2 n Var( n i =1 X i ) = n ( n +1)(2 n +1) 6 . So if Lindeberg condition is satisfied, we have n i =1 X i s n = ¯ X n q 1 6 2 n 2 +3 n +2 n L N (0 , 1) . Since 1 - q 1 6 2 n 2 +3 n +2 n q 3 n n →∞ 0, it follows r 3 n ¯ X n L N (0 , 1) . Now, let’s check the Lindeberg condition: for > 0, 1 2 s 2 n n X i =1 E[ X 2 i 1 {| X i |≥ s n } ] n →∞ 0 . Since X i L = iX 1 , it follows 1 2 s 2 n n X i =1 E h i 2 X 2 1 1 {| X 1 |≥ sn i } i 1 2 s 2 n n X i =1 E h i 2 X 2 1 1 {| X 1 |≥ sn n } i = 1 2 E h X 2 1 1 {| X 1 |≥ sn n } i n →∞ 0 1

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since s n n = O ( n 1 / 2 ). (2) (a) By taking Taylor expansion of n i =1 ˙ ( X i , ˆ θ n ) = 0 about θ 0 , we have n X i =1 ˙ ( X i , θ 0 ) + ( ˆ θ n - θ 0 ) n X i =1 ¨ ( X i , ˜ θ n ) = 0 ( ˆ θ n - θ 0 ) n X
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stat611w09midterm_sol - Stats 611 Midterm Exam...

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