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tutprac2ans - 620-202 Tutorial/Computing Laboratory 2...

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620-202 Tutorial/Computing Labo- ratory 2. Answers. 1. (a) . (b) The likelihood is L ( θ ) = θ n n Y i =1 x θ - 1 i . Hence the log-likelihood is ( θ ) = n ln θ + ( θ - 1) n X i =1 ln( x i ) , and the score function is s ( θ ) = n θ + n X i =1 ln( x i ) , so that setting s ( θ ) = 0 yields ˆ θ = - n n i =1 ln( X i ) = - n ln( Q n i =1 X i ) (c) Hence ˆ θ X = 0 . 549, ˆ θ Y = 2 . 210, ˆ θ Z = 0 . 959. To find the method of moments estimators solve ¯ x = θ/ ( θ + 1) which yields ˜ θ = ¯ x/ (1 - ¯ x ) and hence ˜ θ X = 0 . 598, ˜ θ Y = 2 . 400 and ˜ θ Z = 0 . 865. 2. Recall that for a random sample X 1 , · · · , X n , E ( ¯ X ) = 1 n n X i =1 E ( X i ) = E ( X 1 ) and Var( ¯ X ) = 1 n 2 n X i =1 Var( X i ) = Var( X 1 ) n 1
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(a) Now, the X i are i.i.d. exponential random variables with mean θ . Hence E ( ¯ X ) = E ( X i ) = θ and ¯ X is unbiased. (b) σ 2 = Var( X i ) = θ 2 . Hence Var( ¯ X ) = σ 2 /n = θ 2 /n as required. A good estimator of θ is ˆ θ = ¯ x = 3 . 48. (This is a good estimator as it is unbiased and its variance tends to zero as n → ∞ .) 3. S 2 = n X i =1 ( X i - ¯ X ) 2 n - 1 = 1 n - 1 n X i =1 ( X 2 i - 2 X i ¯ X + ¯ X 2 ) = 1 n - 1 n X i =1 X 2 i - 2 n ¯ X 2 + n ¯ X 2 !
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