May 13'23

Exercise

You are given the following observations on 185 small business policies:

Number of Claims Number of Policies
0 80
1 or more 105


The number of claims per policy follows a Poisson distribution with parameter [math]\lambda [/math].

Using the maximum likelihood estimate of [math]\lambda [/math] , determine the estimated probability of a policy having fewer than two claims.

  • 0.79
  • 0.84
  • 0.89
  • 0.95
  • 0.98

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

May 13'23

Key: A

The likelihood function is

[[math]] L(\alpha ) = (e^{-\lambda})^{80}(1-e^{-\lambda})^{105} [[/math]]

The loglikelihood function is

[[math]] l (\alpha ) = −80\lambda + 105 \ln(1 − e^{-\lambda} ) [[/math]]

Setting [math] l^{'}(\alpha ) = -80 + \frac{105e^{-\lambda}}{1-e^{-\lambda}} [/math] equal to 0, we find:

[[math]]\hat{\lambda} = -\ln \frac{80}{185} = 0.838329[[/math]]

The probability the number of claims, [math]N[/math], is less than 2 is

[[math]] P(N \lt 2) = e^{-\hat{\lambda}} + \hat{\lambda}e^{-\hat{\lambda}} = 0.79495 [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

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