Assignment 5- solution

Assignment 5- solution - THE UNIVERSITY OF HONG KONG...

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

View Full Document Right Arrow Icon

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

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

Unformatted text preview: THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT 1302 PROBABILITY AND STATISTICS II (2009-10) Assignment 5 (Sketch Solution) 1. Let ˆ θ be the mle and s . e . ( ˆ θ ) be the s.e. of ˆ θ . Large-sample theory gives that ( ˆ θ- θ ) / s . e . ( ˆ θ ) ∼ N (0 , 1) approximately. To test H : θ = θ vs H 1 : θ 6 = θ at 5%, reject H if fl fl fl ( ˆ θ- θ ) / s . e . ( ˆ θ ) fl fl fl > 1 . 96 . A 95% confidence interval is n θ : fl fl fl ( ˆ θ- θ ) / s . e . ( ˆ θ ) fl fl fl ≤ 1 . 96 o = ˆ θ ± 1 . 96s . e . ( ˆ θ ) . (a) The likelihood is ‘ ( θ ) = θ- n e- ∑ i X i /θ , which is max’d at the mle ˆ θ = ¯ X . Note that i ( θ ) =-{ E [ ∂ 2 ‘ ( θ ) /∂θ 2 ] } = n/θ 2 and we can take s . e . ( ˆ θ ) = ¯ X/ √ n . (b) The likelihood is ‘ ( θ ) = θ 2 n ( Q i X i ) e- θ ∑ i X i , which is max’d at θ = ˆ θ = 2 / ¯ X . As in (a), we find i ( θ ) = 2 n/θ 2 and so s . e . ( ˆ θ ) = ˆ θ/ √ 2 n . (c) The likelihood is ‘ ( θ ) = e- nθ θ ∑ i X i / Q i X i !, which is max’d at θ = ˆ θ = ¯ X . As in (a), we find i ( θ ) = n/θ and so s . e . ( ˆ θ ) = ( ¯ X/n ) 1 / 2 . 2. (a) P ( X 1 /λ ≤ u ) = R λu f ( x | λ ) dx = 1- e- u . (b) (a) implies P (- X 1 / ln α ≤ λ ) = 1- α . Thus we may take λ α ( x 1 ) =- x 1 / ln α , which is a (1- α ) lower confidence bound for λ . 3. Note that the mle of θ is ˆ θ = X/n . If n is large, ( X- nθ ) / p X (1- X/n ) ∼ N (0 , 1) ap- proximately. Find k α such that P ( N (0 , 1) ≥ k α ) = 1- α . Then we may take u α ( X ) = X/n- ( k α /n ) p X (1- X/n ). Illustration: we have k . 1 =- 1 . 28, u α ( X ) = 0 . 1384. answer. 4. Note that P ( | X- θ | ≤ c ) = Z c- c e-| y | / 2 dy = 1 2 Z- c e y dy + Z c e- y dy ¶ = 1- e- c , and consider P ( | X- θ | ≤ c ) = P (- c ≤ X- θ ≤ c ) = 1- e- c ....
View Full Document

This note was uploaded on 05/04/2011 for the course STAT 1302 taught by Professor Smslee during the Spring '10 term at HKU.

Page1 / 5

Assignment 5- solution - THE UNIVERSITY OF HONG KONG...

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