ACTSC 433 A4 2011-W

ACTSC 433 A4 2011-W - Assignment 4 ACTSC 433/833, Winter...

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Unformatted text preview: Assignment 4 ACTSC 433/833, Winter 2012 (Due at the beginning of the class on March 15) 1. Four losses are observed with values 96, 135, 278, and 396. Six other losses are known to be less than 90. Losses have an inverse exponential distribution with distribution function F ( x ) = e- x , x > , > . (a) Determine the maximum likelihood estimate of . [4 marks] (b) Determine the maximum likelihood estimate of the median of a loss. [2 marks] 2. The following payment data were recorded in a policy with an ordinary deductible of 15, a coinsurance 80%, and a maximum payment of 100: 60 , 20 , 52 , 40 . Also, there were 2 payments for the maximum of 100. The distribution of the ground-up loss has the Pareto distribution with the following pdf f ( x ) = 90 ( x + 90) +1 , x , > 1 . (a) Determine the maximum likelihood estimate for . [4 marks] (b) Determine the maximum likelihood estimate for the probability that a payment paid by the policy will exceed 50. [3 marks] (c) Determine the maximum likelihood estimate for the expected cost per payment. [4 marks] 3. Let the ground-up loss random variable be X with distribution function F ( x ) = 1- 90 90 + x , x , > 1 . In an insurance policy, the amount paid by the policy is the loss truncated from below at a franchise deductible of 10 and censored from above at the policy limit of 60. Leta franchise deductible of 10 and censored from above at the policy limit of 60....
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This note was uploaded on 03/07/2012 for the course ACTSC 433 taught by Professor Johnny during the Winter '09 term at Waterloo.

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ACTSC 433 A4 2011-W - Assignment 4 ACTSC 433/833, Winter...

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