{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

411_hw07_soln

# 411_hw07_soln - Stat 411 Homework 07 Solutions 1 The...

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

Stat 411 – Homework 07 Solutions 1. The distribution N ( θ, 1) is a regular one-parameter exponential family problem with K ( x ) = x . Therefore, T = n i =1 X i is a complete sufficient statistic for θ and, consequently, the MVUE of θ is X = T/n . It is easy to check that ˆ η = X 2 - 1 /n is an unbiased estimator of η = θ 2 . Since it’s a function of the complete sufficient statistic, by the Lehmann–Scheffe theorem, it must be the MVUE. 2. Problem 7.5.3 from the text . (a) The beta distribution with PDF f θ ( x ) = θx θ - 1 , with x (0 , 1) and θ > 0, is a regular exponential family. That is, f θ ( x ) = exp { log θ + ( θ - 1) log x } . Since K ( x ) = log x here, a complete sufficient statistic, based on X 1 , . . . , X n iid f θ ( x ), is T 0 = n i =1 log X i . So too is T = exp { T 0 /n } = ( X 1 X 2 · · · X n ) 1 /n , the geometric mean, since the function t 7→ e t/n is one-to-one. (b) The log-likelihood function ( θ ) = n i =1 log f θ ( X i ) is ( θ ) = n log θ + ( θ - 1) n X i =1 log X i . Differentiating and setting equal to zero gives the likelihood equation: n θ + n X i =1 log X i = 0 . Therefore, the MLE is ˆ θ = - n/ n i =1 log X i = - 1 / log T , a function of the geometric mean T from part (a).

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.

{[ snackBarMessage ]}

### Page1 / 3

411_hw07_soln - Stat 411 Homework 07 Solutions 1 The...

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

View Full Document
Ask a homework question - tutors are online