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Unformatted text preview: Homework 4 Solutions (1) Recall that tpivot is n ( X ) s t n 1 , where s = n i =1 ( X 2 i X ) 2 / ( n 1). Now, 2 =  1 , and let F be the CDF of tdistribution with df = n 1. Then, P F 1 ( 1 ) n ( X ) s F 1 (1 + 1 ) , from which we obtain the confidence interval X s n F 1 (1 + 1 ) , X s n F 1 ( 1 ) . In particular, the length of the interval is L = s n { F 1 (1 + 1 ) F 1 ( 1 ) } , To find the minimizer, take the derivative w.r.t. 2 and set it to zero. By using the inverse function theorem, 0 = L 1 = s n 1 g ( F 1 (1 + 1 )) 1 g ( F 1 ( 1 )) , which gives g ( F 1 (1 + 1 )) = g ( F 1 ( 1 )). Since g is even, F 1 (1 + 1 ) = F 1 ( 1 ). Since F 1 ( + 1 2 ) is odd , 1 2 + 1 = ( 1 1 2 ) . Since < 1, 1 2 + 1 = 1 + 1 2 , which gives 1 = 2 = / 2. (Why is this not the maximizer but the minimizer? Consider what happens when you shift the interval to the right or left.) Now, for simplicity, denote F 1 (1 / 2) = q * so that F ( q * ) = / 2. Then, the power function is ( ) = P n ( X ) s + n (  ) s  q * + P q * n ( X ) s + n (  ) s = Z F q * n (  ) s G ( ds ) + Z F q * + n (  ) s G ( ds ) , where G is a CDF of s . The second inequality is due to the power property of the conditional expectation. (We used the independence of tpivot and s there. Why? One way to prove that is using Basus theorem.) 1 (a) ( ) = F ( q * ) + F ( q * ) = (b) By the monotone convergence theorem, lim = R F ( ) G ( ds )+ R F ( ) G ( ds ) = 1. (c) By interchanging the integration and the differentiation, (Justify this!) ( ) = Z n s f q * n (  ) s G ( ds ) Z n s f q * + n (  ) s G ( ds ) ....
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This note was uploaded on 04/14/2010 for the course STATS 610 taught by Professor Moulib during the Fall '09 term at University of Michigan.
 Fall '09
 moulib
 TDistribution

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