theory2006f[1] - QUALIFYING EXAM SOLUTIONS STATISTICAL...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
QUALIFYING EXAM SOLUTIONS STATISTICAL THEORY: Saturday, Aug 12, 2006, 8:00 am -12:00 pm 1. Assume h is strictly increasing. (a) This is not invariant. Suppose X U [0 ] for θ > 0. Assume n > 5. Let prior be uniform. Then, the posterior density of θ given X ( n ) is q ( θ | x ( n ) = x ) = - n x n - 1 R x - n x n - 1 = ( n - 1) x n - 1 θ n when θ x ( n ) . The posterior mean is ˆ θ = E ( θ | x ( n ) ) = ( n - 1) x n - 1 Z x 1 θ n - 1 = n - 1 n - 2 x. However, if we choose η = θ 2 . Then, the posterior density of η is q 1 ( η | x ( n ) = x ) = ( n - 1) x n - 1 η n/ 2 = ( n - 1) x n - 1 2 η ( n +1) / 2 and the posterior mean is ˆ η = ( n - 1) x n - 1 2 Z x 2 1 η ( n - 1) / 2 == n - 1 n - 3 x 2 6 = ( n - 1 n - 2 x ) 2 . (b) It is invariant. Proof is easy. Suppose the posterior distribution of θ is Q ( θ | x 1 , ··· ,x n ). Then, the posterior density of η is Q ( h - 1 ( η ) | x 1 , ··· ,x n ) and the posterior median sat- isfies Q ( h - 1 ( η ) h - 1 η M ) | x 1 , ··· ,x n ) = 1 / 2 Q ( θ h - 1 ( η M ) | x 1 , ··· ,x n ) = 1 / 2 . Thus, ˆ η = h ( ˆ θ M ). (c) It is invariant. Look at the likelihood function ( θ | x 1 , ··· ,x n ) = ( h - 1 ( η ) | x 1 , ··· ,x n ) . If
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/22/2011 for the course STAT 598 taught by Professor Tlzhang during the Spring '04 term at Indiana University-Purdue University Fort Wayne.

Page1 / 4

theory2006f[1] - QUALIFYING EXAM SOLUTIONS STATISTICAL...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online