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610-hw4sol

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

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Homework 4 Solutions (1) Recall that t -pivot is n ( X - μ ) s t n - 1 , where s = n i =1 ( X 2 i - X ) 2 / ( n - 1). Now, β 2 = α - β 1 , and let F be the CDF of t -distribution with df = n - 1. Then, P F - 1 ( β 1 ) n ( X - μ ) s F - 1 (1 + β 1 - α ) , from which we obtain the confidence interval X - s n F - 1 (1 + β 1 - α ) , X - s n F - 1 ( β 1 ) . In particular, the length of the interval is L = s n { F - 1 (1 + β 1 - α ) - F - 1 ( β 1 ) } , To find the minimizer, take the derivative w.r.t. β 2 and set it to zero. By using the inverse function theorem, 0 = ∂L ∂β 1 = s n 1 g ( F - 1 (1 + β 1 - α )) - 1 g ( F - 1 ( β 1 )) , which gives g ( F - 1 (1 + β 1 - α )) = g ( F - 1 ( β 1 )). Since g is even, F - 1 (1 + β 1 - α ) = ± F - 1 ( β 1 ). Since F - 1 ( · + 1 2 ) is odd , 1 2 + β 1 - α = ± ( β 1 - 1 2 ) . Since α < 1, 1 2 + β 1 - α = - β 1 + 1 2 , which gives β 1 = β 2 = α/ 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 - α/ 2) = q * so that F ( - q * ) = α/ 2. Then, the power function is β ( μ ) = P n ( X - μ 0 ) s + n ( μ 0 - μ ) s ≤ - q * + P q * n ( X - μ 0 ) s + n ( μ 0 - μ ) s = Z 0 F - q * - n ( μ 0 - μ ) s G ( ds ) + Z 0 F - q * + n ( μ 0 - μ ) 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 Basu’s theorem.) 1

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(a) β ( μ 0 ) = F ( - q * ) + F ( - q * ) = α (b) By the monotone convergence theorem, lim μ →∞ = R 0 F ( ) G ( ds )+ R 0 F ( -∞ ) G ( ds ) = 1. (c) By interchanging the integration and the differentiation, (Justify this!) β 0 ( μ ) = Z 0 n s f - q * - n ( μ 0 - μ ) s G ( ds ) - Z 0 n s f - q * + n ( μ 0 - μ ) s G ( ds ) . Now, it is clear that β is even and β 0 is odd about μ 0 . (d) Omitted. (2) (i) Note that L ( X 1 , . . . , X n | ( μ, σ )) = 1 (2 πσ 2 ) n/ 2 exp " - 1 2 σ 2 n X i =1 ( X i - μ ) 2 # . To compute λ n , first maximize the numerator with ( μ, σ ) varying.
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