hw6_stat210a_solutions

# hw6_stat210a_solutions - UC Berkeley Department of...

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Unformatted text preview: UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 6 Fall 2006 Issued: Thursday, October 5, 2006 Due: Thursday, October 12 , 2006 Problem 6.1 a) Using the Rao-Blackwell theorem and considering the quadratic error loss function L ( θ, δ ) = ( θ- δ ) 2 , it is enough to prove that δ * is unbiased and that it only depends on the data by means of a sufficient complete statistic. To show that N is a complete statistics, Let Z = N + x . From HW5, we rewrite the negative binomial density in its exponential form in terms of N as: p ( n ; θ ) = p ( N = n | θ ) = h ( n ) exp [ η ( θ ) T ( n )- B ( θ )] with h ( n ) = n + x- 1 x- 1 I ( n ∈ { , 1 , . . . } ) η ( θ ) = log(1- θ ) T ( n ) = n B ( θ ) =- x log θ 1- θ Noticing that this corresponds to a full-rank exponential family with complete sufficient statistic T ( N ) = n , it remains to show that δ * is unbiased. Unbiasedness follows from: E θ [ δ * ( N )] = ∞ X N =0 x- 1 N + x- 1 ( N + x- 1)! ( N + x )!( x- 1)! θ x (1- θ ) N = θ ∞ X N =0 ( N + x- 2)! ( N + x )!( x- 2)! θ x- 1 (1- θ ) N = θ b) From item a, we have: log p ( n ; θ ) = log( h ( n )) + x log θ + N log(1- θ ) so: ∂ ∂θ log p ( n ; θ ) = x θ + N 1- θ and ∂ 2 ∂θ 2 log p ( n ; θ ) =- x θ 2 + N (1- θ ) 2 1 Now, we have: I ( θ ) =- E θ ∂ 2 ∂θ 2 log p ( n ; θ ) = x θ 2- E θ N · 1 (1- θ ) 2 ( * ) = x θ 2- x (1- θ ) θ · 1 (1- θ ) 2 = x (1- θ )- xθ θ 2 (1- θ ) The equality in ( * ) follows from the fact that EN = x (1- θ ) θ . To see that, let Z be the total number of trials until x successes occur. We have that Z = ∑ x j =1 X j with X j being the number of trials until one success is observed (and has mean 1 θ from the properties of a geometric distribution). Henceproperties of a geometric distribution)....
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hw6_stat210a_solutions - UC Berkeley Department of...

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