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
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 ( X = x ) = exp " n X i =1 x i ! log( ) n # 1 x 1 ! x 2 ! x n ! (a) We have: P ( X = x  T ( X ) = S ) = P ( X = x and T ( X ) = S ) P ( T ( X ) = S ) From the properties of the Poisson distribution: T ( X ) = n i =1 X i Poisson( n ): P ( T ( X ) = S ) = exp [ S log( n ) n ] 1 S ! Furthermore: P ( X = x and T ( X ) = S ) = I ( n X i =1 x i = S ) exp " n X i =1 x i ! log( ) n # 1 x 1 ! x 2 ! x n ! = I ( n X i =1 x i = S ) exp [ S log( ) n ] 1 x 1 ! x 2 ! x n ! which yields: P ( 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 proving sufficiency of T . (b) Define: g ( S, ) = exp [ S log( ) n ] h ( x ) = 1 x 1 ! x 2 ! x n ! It is then possible to write P ( X = x ) as: P ( X = x ) = g ( T ( X ) , ) 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 Jensens inequality by noticing that L ( , ) = (  ) 2 is convex in . Using iterated expectations and Jensens inequality: E [ L ( ( X ) , )] = E [ E [ L ( ( X ) , )  T ( X )]] E [ L ( E ( ( X )  T ( X )) , )] = E h L ( X ) , i We now prove that ( 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 1 1 1 + log(1 1 ) p ( X i = x i  X i 1 = x i 1 ) = exp (1 x i 1 ) x i log 10 1 10 + log(1 10 ) + x i 1 x i log 11 1 11 + log(1 11 ) = exp x i log 10 1 10 + x i 1 x i log 11 (1 10 ) (1 11 ) 10 + x i 1 log(1 11 ) + log(1 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 1 1 1 + ( n X i =2 x i ) log 10 1 10 + ( n X i =2 x i 1 x i ) log 11 (1 10 ) (1 11 ) 10 + ( n X i =2 x i 1 ) log(1 10 + log(1 1 ) + ( n 1) log(1 10 ) + ( n 1) log 1 11 1 10...
View Full
Document
 Fall '08
 Staff
 Statistics

Click to edit the document details