Unformatted text preview: Second Edition 95 9.13 a. For Y = (log X ) 1 , the pdf of Y is f Y ( y ) = θ y 2 e θ/y , 0 < y < ∞ , and P ( Y/ 2 ≤ θ ≤ Y ) = Z 2 θ θ θ y 2 e θ/y dy = e θ/y 2 θ θ = e 1 / 2 e 1 = . 239 . b. Since f X ( x ) = θx θ 1 , 0 < x < 1, T = X θ is a good guess at a pivot, and it is since f T ( t ) = 1, < t < 1. Thus a pivotal interval is formed from P ( a < X θ < b ) = b a and is θ : log b log x ≤ θ ≤ log a log x . Since X θ ∼ uniform(0 , 1), the interval will have confidence .239 as long as b a = . 239. c. The interval in part a) is a special case of the one in part b). To find the best interval, we minimize log b log a subject to b a = 1 α , or b = 1 α + a . Thus we want to minimize log(1 α + a ) log a = log ( 1+ 1 α a ) , which is minimized by taking a as big as possible. Thus, take b = 1 and a = α , and the best 1 α pivotal interval is n θ : 0 ≤ θ ≤ log α log x o . Thus the interval in part a) is nonoptimal. A shorter interval with confidence coefficient .239 isthe interval in part a) is nonoptimal....
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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.
 Spring '12
 Dr.Hackney
 Statistics

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