Assignment 5 Solution

# Assignment 5 Solution - 4/2011 THE UNIVERSITY OF HONG KONG...

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Unformatted text preview: 4/2011 THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT1302 PROBABILITY AND STATISTICS II 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) approximately. 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. Substitute this into above to get the answer. 4. Note that P ( | X- θ | ≤ c ) = R c- c e-| y | / 2 dy = 1- e- c . Equating this to 1- α gives c =- ln α . Thus the required C.I. is [ X + ln α,X- ln α ]....
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Assignment 5 Solution - 4/2011 THE UNIVERSITY OF HONG KONG...

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