Assignment 5- solution

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

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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 0 : θ = θ 0 vs H 1 : θ 6 = θ 0 at 5%, reject H 0 if fl fl fl ( ˆ θ - θ 0 ) / 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 . 96 s . e . ( ˆ θ ) . (a) The likelihood is ( θ ) = θ - n e - i X i , 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 - 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 u ) = R λu 0 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 - ) / 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 0 . 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 0 - c e y dy + Z c 0 e - y dy = 1 - e - c , and consider P ( | X - θ | ≤ c ) = P ( - c X - θ c ) = 1 - e - c . Equating 1 - e - c to 1 - α gives c = - ln α , therefore P (ln α X - θ ≤ - ln α ) = 1 - α . Thus the required C.I. is [ X + ln α, X - ln α ].

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