610-hw4sol - Homework 4 Solutions (1) Recall that t-pivot...

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Unformatted text preview: Homework 4 Solutions (1) Recall that t-pivot 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 t-distribution 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 t-pivot 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.

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610-hw4sol - Homework 4 Solutions (1) Recall that t-pivot...

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