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
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]]