Dr. Hackney STA Solutions pg 166

Dr. Hackney STA Solutions pg 166 - Second Edition 10-9 and...

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Unformatted text preview: Second Edition 10-9 and (1 - )F (a) = (1 - )F (a) + = 1 2 1 2 a = F -1 a = F -1 1 2(1 - ) - 2(1 - ) 1 2 c. The limit is a - a0 = a |=0 0 lim 1 2(1-) , by the definition of derivative. Since F (a ) = d d 1 F (a ) = d d 2(1 - ) or f (a )a = 1 1 a = . 2(1 - )2 2(1 - )2 f (a ) Since a0 = m, the result follows. The other limit can be calculated in a similar manner. 10.29 a. Substituting cl for makes the ARE equal to 1. b. For each distribution is the case that the given function is equal to cl , hence the resulting M-estimator is asymptotically efficient by (10.2.9). 10.31 a. By the CLT, n1 p1 - p1 ^ p1 (1 - p1 ) n(0, 1) and n2 p 2 - p2 ^ p2 (1 - p2 ) n(0, 1), so if p1 and p2 are independent, under H0 : p1 = p2 = p, ^ ^ p1 - p 2 ^ ^ 1 n1 n(0, 1) + 1 n2 p(1 - p) ^ ^ where we use Slutsky's Theorem and the fact that p = (S1 + S2 )/(n1 + n2 ) is the MLE of ^ p under H0 and converges to p in probability. Therefore, T 2 . 1 b. Substitute pi s for Si and Fi s to get ^ T = n2 (^1 - p) ^ n2 (^ - p) p ^ 1 p + 2 2 n1 p ^ n2 p ^ + 2 2 = ^ ^ n2 [(1 - p1 ) - (1 - p)] n2 [(1 - p2 ) - (1 - p)] ^ ^ 1 + 2 n1 (1 - p) ^ n2 p ^ 2 2 n1 (^1 - p) p ^ n2 (^2 - p) p ^ + p(1 - p) ^ ^ p(1 - p) ^ ^ 2 2 Write p = (n1 p1 + n2 p2 )/(n1 + n2 ). Substitute this into the numerator, and some algebra ^ ^ ^ will get (^1 - p2 )2 p ^ n1 (^1 - p)2 + n2 (^2 - p)2 = 1 p ^ p ^ 1 , n1 + n2 so T = T . ...
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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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