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Dr. Hackney STA Solutions pg 154

Dr. Hackney STA Solutions pg 154 - 9-12 a Solutions Manual...

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9-12 Solutions Manual for Statistical Inference 9.39 Let a be such that a -∞ f ( x ) dx = α/ 2. This value is unique for a unimodal pdf if α > 0. Let μ be the point of symmetry and let b = 2 μ - a . Then f ( b ) = f ( a ) and b f ( x ) dx = α/ 2. a μ since a -∞ f ( x ) dx = α/ 2 1 / 2 = μ -∞ f ( x ) dx . Similarly, b μ . And, f ( b ) = f ( a ) > 0 since f ( a ) f ( x ) for all x a and a -∞ f ( x ) dx = α/ 2 > 0 f ( x ) > 0 for some x < a f ( a ) > 0. So the conditions of Theorem 9.3.2 are satisfied. 9.41 a. We show that for any interval [ a, b ] and > 0, the probability content of [ a - , b - ] is greater (as long as b - > a ). Write a b f ( x ) dx - b - a - f ( x ) dx = b b - f ( x ) dx - a a - f ( x ) dx f ( b - )[ b - ( b - )] - f ( a )[ a - ( a - )] [ f ( b - ) - f ( a )] 0 , where all of the inequalities follow because f ( x ) is decreasing. So moving the interval toward zero increases the probability, and it is therefore maximized by moving a all the way to zero. b. T = Y - μ is a pivot with decreasing pdf f T ( t ) = ne - nt I [0 , ] ( t ). The shortest 1 - α interval on T is [0 , - 1 n log α ], since b 0 ne - nt dt = 1 - α b = - 1 n log α.
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