stat210a_HW02

# stat210a_HW02 - UC Berkeley Department of Statistics STAT...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This note was uploaded on 10/17/2009 for the course STAT 210a taught by Professor Staff during the Fall '08 term at Berkeley.

### Page1 / 4

stat210a_HW02 - UC Berkeley Department of Statistics STAT...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online