exercise6512

# exercise6512 - Exercise 6.5.12 To show is a location...

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

Exercise 6.5.12 To show θ is a location parameter see θ Example 2 . 7 . 1 which was done in class. To show that ˆ θ = X (1) see the solution to Problem 55 which is posted with the hints for Problem 60 . Note that if X v EXP (1 , θ ) then P ( X>x )=1 Z x θ e ( u θ ) du =1 h e ( u θ ) | x θ i = e ( x θ ) ,x θ . Therefore P ³ ˆ θ θ q ´ = P ¡ X (1) q + θ ¢ P ¡ X (1) >q + θ ¢ n Q i =1 P ( X i + θ ) n Q i =1 [1 P ( X i q + θ )] n Q i =1 e ( q + θ θ ) e nq ,q 0 . Since P μ ˆ θ + 1 n log (1 p ) θ ˆ θ = P μ 1 n log (1 p ) ˆ θ θ 0 = P μ 0 ˆ θ θ ≤− 1 n log (1 p ) e log(1 p ) (1 p ) = p therefore ³ ˆ θ + 1 n log (1 p ) , ˆ θ ´ is a 100 p % C.I. for θ . 1

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

View Full Document
Since P μ ˆ θ + 1 n log μ 1 p 2 θ ˆ θ + 1 n log μ 1+ p 2 ¶¶ = P μ 1 n log μ 1 p 2 θ ˆ θ 1 n log μ p 2 ¶¶ = P μ 1 n log μ 1 p 2 ˆ θ θ ≥− 1 n log μ p 2 ¶¶ =
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 04/27/2010 for the course STAT 330 taught by Professor Paulasmith during the Fall '08 term at Waterloo.

### Page1 / 2

exercise6512 - Exercise 6.5.12 To show is a location...

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

View Full Document
Ask a homework question - tutors are online