stat210a_HW02

stat210a_HW02 - UC Berkeley Department of Statistics STAT...

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Unformatted text preview: UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 2 - Solutions Fall 2009 Issued: Tuesday, September 8, 2009 Due: Tuesday, September 15, 2009 Problem 2.1 From the distribution of X , we have: P θ ( X = x ) = exp " n X i =1 x i ! log( θ )- nθ # 1 x 1 ! x 2 ! ··· x n ! We have: P θ ( X = x | T ( X ) = S ) = P θ ( X = x and T ( X ) = S ) P θ ( T ( X ) = S ) From the properties of the Poisson distribution: T ( X ) = ∑ n i =1 X i ∼ Poisson( nθ ): P θ ( T ( X ) = S ) = exp[ S log( nθ )- nθ ] 1 S ! Furthermore: P θ ( X = x and T ( X ) = S ) = I ( n X i =1 x i = S )exp " n X i =1 x i ! log( θ )- nθ # 1 x 1 ! x 2 ! ··· x n ! = I ( n X i =1 x i = S )exp[ S log( θ )- nθ ] 1 x 1 ! x 2 ! ··· x n ! which yields: P θ ( X = x | T ( X ) = S ) = exp[- S log( n )] S ! x 1 ! x 2 ! ··· x n ! = 1 n S S ! x 1 ! x 2 ! ··· x n ! Thus, ( X = x | T ( X ) = S ) ∼ M ( S ; 1 n ,..., 1 n ), which does not involve θ proving sufficiency of T . Problem 2.2 From the question, we know, p i = exp( α + βt i ) 1 + exp( α + βt i ) 1 Hence, P ( X 1 = x 1 , ··· ,X n = x n ) = n Y i =1 P ( X i = x i ) = n Y i =1 p x i i (1- p i ) 1- x i = n Y i =1 (exp( α + βt i )) x i (1 + exp( α + βt i )) x i 1 1- x i (1 + exp( α + βt i )) 1- x i = Q n i =1 exp( αx i + βt i x i ) Q n i =1 (1 + exp( α + βt i )) = exp( α ∑ n i =1 x i + β ∑ n i =1 t i x i ) Q n i =1 (1 + exp( α + βt...
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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_HW02 - UC Berkeley Department of Statistics STAT...

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