stat210a_2007_hw4_solutions

stat210a_2007_hw4_solutions - UC Berkeley Department of...

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UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 4- Solutions Fall 2007 Issued: Thursday, September 20 Due: Thursday, September 27 Problem 4.1 (a) P ( X x ) = F ( x ) = ± 1 - ± 1 x c · 1 ( x > 1). f ( x ) = cx - c - 1 · 1 ( x > 1) = exp {- ( c + 1)log x + log c } · 1 ( x > 1) : 1-dimensional exponential family with sufficient statistics T = log X . (b) f T ( t ) = ck c e - ( c +1) t e t · 1 ( t > log k ) = exp {- ct + log( 1 ( t > log k )) + c log k + log c } , where log 0 = 0. Thus, T has an exponential distribution with parameters log k and 1 /c . Problem 4.2 (a) Letting g ( x ; θ ) = 1 θ f ( x θ ) , we know that: log( g ( x ; θ )) = - log( θ ) + log f x θ · Differentiating with respect to θ : ∂θ log( g ( x ; θ )) = - 1 θ " 1 + x θ f 0 ( x θ ) f ( x θ ) # So: I ( θ ) = Z • ∂θ log( g ( x ; θ )) 2 g ( x,θ ) dx = 1 θ 2 Z " 1 + x θ f 0 ( x θ ) f ( x θ ) # 2 1 θ f x θ · dx θdy = dx = 1 θ 2 Z • 1 + yf 0 ( y ) f ( y ) 2 f ( y ) dy 1
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η ( θ ) = log( θ ), we have from Keener 8.1.8 that: I ( η ) = ∂η - 1 ( θ ) ∂θ 2 I ( θ ) = ∂η ( θ ) ∂θ - 2 I ( θ ) = Z • 1 + y f 0 ( y ) f ( y ) 2 f ( y ) dy (c) For the Cauchy distribution C (0 ): p ( x ; θ ) = 1 π θ x 2 + θ 2 = 1 θ f x θ · where f ( y ) = 1 π 1 1 + y 2 so, we can apply the result from (a): f 0 ( y ) = - 1 π 2 y (1 + y 2 ) 2 and as a result: 1 - yf 0 ( y ) f ( y ) = 1 - 2 y 2 1 + y 2 = 1 - y 2 1 + y 2 Finally, from (a): I ( θ ) = 1 πθ 2 Z 1 - y 2 (1 + y 2 ) 2 dy and transforming y = tan( t ): I ( θ ) = 1 πθ 2 Z π 2 - π 2 cos 2 (2 t ) dt = 1 2 θ 2 Problem 4.3 We compute the Fisher information matrix as:
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This note was uploaded on 10/17/2009 for the course STAT 210a taught by Professor Staff during the Fall '08 term at Berkeley.

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stat210a_2007_hw4_solutions - UC Berkeley Department of...

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