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ACTSC 433 A3 SOLN W12

ACTSC 433 A3 SOLN W12 - Solutions to Assignment 3 ACTSC...

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Solutions to Assignment 3 – ACTSC 433/833, Winter 2012 1. (a) ˆ S (60) = 1 10 (0 + 0 + 0 + 1 10 + 4 10 + 1 + 1 + 1 + 1) = 0 . 45. (b) ˆ S (60) = 1 10 (0 + 0 + 2 × 7 30 + 11 30 + 14 30 + 23 30 + 23 30 + 25 30 + 29 30 + 1) = 0 . 486667. 2. (a) ˆ S (220) = 1 5 0 + (190+75 - 220) 2 2 × 75 2 + 1 - (220 - 241+75) 2 2 × 75 2 + 1 + 1 = 0 . 58416. (b) ˆ S (220) = 1 5 0 + (190+100 - 220) 2 2 × 100 2 + 1 - (220 - 241+100) 2 2 × 100 2 + 1 + 1 = 0 . 58659. 3. (a) We have L ( θ ) = θ n e - θ n i =1 x i . Then, it follows from ln L ∂θ = 0 that ˆ θ = n n i =1 X i . (b) Note that S = n i =1 X i has the gamma distribution with pdf f S ( y ) = θ n y n - 1 e - θy Γ( n ) , y > 0. Thus, E ( ˆ θ ) = n R 0 1 y f S ( y ) dy = n n - 1 θ θ as n → ∞ and E ( ˆ θ 2 ) = n 2 Z 0 1 y 2 f S ( y ) dy = n 2 ( n - 1)( n - 2) θ 2 . Hence, V ar ( ˆ θ ) 0 as n → ∞ . Thus, ˆ θ is consistent for θ . 4. (a) i. When μ is known, the log-likelihood function is l ( σ 2 ) = - n 2 ln(2 π ) - n 2 ln σ 2 - 1 2 σ 2 n X j =1 ( x j - μ ) 2 . It follows from d 2 l ( σ 2 ) = 0 that ˆ σ 2 = 1 n n j =1 ( X j - μ ) 2 . ii. Note that n i =1 X i - μ σ 2 has the chi-square distribution χ 2 ( n ). Thus, E ( ˆ σ 2 ) = σ 2 and V ar ( ˆ σ 2 ) = 2 σ 4 n 0 as n → ∞ . Hence, ˆ σ 2
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