ACTSC 433 A5 SOLN W12

# ACTSC 433 A5 SOLN W12 - Solutions to Assignment 5 ACTSC...

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Solutions to Assignment 5 – ACTSC 433/833, Winter 2012 1. (a) For Line 1: S 1 ( x ) = S 0 ( x ) and f 1 ( x ) = α 100 α x α +1 , x > 100. For Line 2: S 2 ( x ) = ( S 0 ( x )) b and f 2 ( x ) = αb 100 αb x αb +1 , x > 100, where b = e β . Thus, L ( α,b ) = f 1 (135) f 1 (160) f 1 (240) f 1 (320) f 1 (380) f 2 (115) f 2 (140) f 2 (205) f 2 (245) f 2 (285) . Then, it follows from ln L ( α,b ) ∂α = 0 and ln L ( α,b ) ∂b = 0 that ˆ α = 1 . 20664 and ˆ b = 1 . 32073. Hence, ˆ β = 0 . 27818. (b) By S 1 ( x ) = 0 . 5, we know that the median loss in line 1 is x = 100 / 0 . 5 1 . Hence, the maximum likelihood estimate for the median loss in line 1 is 100 / 0 . 5 1 / ˆ α = 177 . 64 . (c) The mean loss in line 2 is μ 2 = R 0 S 2 ( x ) dx = 100 + 100 αb - 1 . Thus, ˆ μ 2 = 100 + 100 ˆ α ˆ b - 1 = 268 . 45. (d) The maximum likelihood estimate for the probability that a loss in line 2 will exceed 200 is ˆ S 2 (200) = 0 . 5 ˆ α ˆ b = 0 . 3313 . 2. (a) For Line 1:

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ACTSC 433 A5 SOLN W12 - Solutions to Assignment 5 ACTSC...

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