Fall 2015 Q1(c)

How did they come up with R = 1/(1-e^(-x))?

Comments

  • A driver with 0 accident free years has had at least one claim in the last year. Suppose there are N such drivers in this group. A key assumption of Bailey & Simon is the observed claim frequency, lambda, for the class is the same for all sub-classes. So for the sub-class of drivers with 0 accident free years we expect them to have N*lambda claims.

    However, just because they had an accident in the last year doesn't mean they'll have another accident in the next year, some of them will have 0 accidents in the future period. Since all sub-classes have the same frequency assumption and we're told they follow a Poisson distribution, we have the probability of 0 claims is e^-lambda, so the probability of at least one claim is 1- e^-lambda.

    So for our group of N insureds with 0 accident free years, we expect N*(1-e^-lambda) of them to have accidents in the future period. This means their expected claim frequency is N*lambda / [N*(1-e^-lambda)] = lambda/(1-e^-lambda).

    However, the claim frequency for the entire class of all drivers, not just those with 0 accident free years is lambda under the assumption of a Poisson frequency distribution. So the relativity, R, of the 0 accident free years sub-class to the class overall is R = [lambda/(1-e^-lambda)] / lambda = 1/(1-e^-lambda).

  • Thanks for the thorough response, I see it now.

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