This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: UC Berkeley Department of Statistics STAT 210A: Introduction to Mathematical Statistics Problem Set 2 Fall 2006 Issued: Thursday, September 7, 2006 Due: Thursday, September 14, 2006 Graded exercises 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 ! (a) 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 ! which does not involve θ proving sufficiency of T . (b) Define: g ( S, θ ) = exp [ S log( θ ) nθ ] h ( x ) = 1 x 1 ! x 2 ! ··· x n ! It is then possible to write P θ ( X = x ) as: P θ ( X = x ) = g ( T ( X ) , θ ) h ( x ) so sufficiency of T follows from the factorization theorem. 1 Problem 2.2 The result is a direct consequence of the RaoBlackwell theorem. It can be proven directly using Jensen’s inequality by noticing that L ( δ, θ ) = ( δ θ ) 2 is convex in δ . Using iterated expectations and Jensen’s inequality: E [ L ( δ ( X ) , θ )] = E [ E [ L ( δ ( X ) , θ )  T ( X )]] ≥ E [ L ( E ( δ ( X )  T ( X )) , θ )] = E h L ˜ δ ( X ) , θ i We now prove that δ ( X ) = min i X i +max i X i 2 . To do that, we first consider the expected value of X j given M = min i X i and W = max i X i . There is a 1 n chance that X j = M and a 1 n chance that X j = W . The remaining mass n 2 n is uniformly spread between M and W . Hence: E [ X j  M, W ] = 1 n M + 1 n W + n 2 n Z W M ( x M ) ( W M ) dx = 1 n ( M + W ) + n 2 2 n ( M + W ) = M + W 2 Problem 2.3 We have: p ( X 1 = x 1 ) = exp x 1 log θ 1 1 θ 1 + log(1 θ 1 ) p ( X i = x i  X i 1 = x i 1 ) = exp (1 x i 1 ) x i log θ 10 1 θ 10 + log(1 θ 10 ) + x i 1 x i log θ 11 1 θ 11 + log(1 θ 11 ) = exp x i log θ 10 1 θ 10 + x i 1 x i log θ 11 (1 θ 10 ) (1 θ 11 ) θ 10 + x i 1 log(1 θ 11 ) + log(1 θ 10 ) Hence: P ( X = x ) = p ( X 1 = x 1 ) n Y i =2 p ( X i = x i  X i 1 = x i 1 ) = exp " x 1 log θ 1 1 θ 1 + ( n X i =2 x i ) log θ 10 1 θ 10 + ( n X i =2 x i 1 x i ) log θ 11 (1 θ 10 ) (1 θ 11 ) θ 10 + ( n X i =2 x i 1 ) log(1 θ 10 + log(1 θ 1 ) + ( n 1) log(1 θ 10 ) + ( n 1) log 1 θ 11 1 θ 10...
View
Full
Document
 Fall '08
 Staff
 Statistics

Click to edit the document details