stat210a_2007_hw3_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 3- Solutions Fall 2007 Issued: Thursday, September 13 Due: Thursday, September 20 Problem 3.1 (a) Let F X be distribution of X . Then, F X ( x ) = 1- e- x for all > 0. Thus, F Y ( y ) = P ( Y y ) = P ( X y ) = F X ( y ) = 1- e- y . i.e. Y has an exponential distribution with failure rate 1. (b) Because X is 1-dimensional full rank exponential family, X is a complete sufficient statistics. Let Y i = X i , i = 1 ,...,n . Then, X 2 1 + + X 2 n X 2 = Y 2 1 + + Y 2 n Y 2 : not depend on . Thus, X 2 1 + + X 2 n X 2 is an ancillary statistics. By Basus theorem, X and X 2 1 + + X 2 n X 2 are independent. (c) X (1) X ( n ) = Y (1) Y ( n ) : not depend on . By Basus theorem, X and X (1) X ( n ) are independent. Problem 3.2 To determine the natural parameter space of each of the families, we must determine the set A T,h = { : R h ( x )exp[ T ( x )] dx < } . (a) In this case A h,T = R . To prove that, notice that: Z exp x- x 2 / dx = exp 2 4 Z exp - 2 + x 2 dx = 2exp 2 4 Z exp - y 2 2 dy = 2 exp 2 4 Z exp 1 2 - y 2 2 dy = 2 exp 2 4 < , R (b) First, notice that: C ( ) = Z exp[ x- | x | ] dx = Z x> exp[( - 1) x ] dx + Z x< exp[( + 1) x ] dx 1 If <- 1, lim x - exp[( + 1) x ] > 0 so the integral over x < 0 diverges. Likewise, if > 1, lim x + exp[( + 1) x ] > 0 so the integral over x > 0 diverges. Now, if- 1 < < 1, both integrals can be rewritten as R exp[- ax ] dx with a > 0 which is finite for any value of a . Now if = 1 ( alt. =- 1), the integral over the positive branch (alt. negative branch) diverges. As a result the natural parameter space is given by (- 1 , 1)....
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stat210a_2007_hw3_solutions - UC Berkeley Department of...

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