You observe the following five ground-up claims from a data set that is truncated from below at 100:

125   150   165   175   250

You fit a ground-up exponential distribution using maximum likelihood estimation.

Calculate the mean of the fitted distribution.

• 73
• 100
• 125
• 156
• 173

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

Key: A

Because the exponential distribution is memoryless, the excess over the deductible is also exponential with the same parameter. So subtracting 100 from each observation yields data from an exponential distribution and noting that the maximum likelihood estimate is the sample mean gives the answer of 73 .

Working from first principles,

[[math]]\begin{aligned} L(\theta)=\frac{f\left(x_{1}\right) f\left(x_{2}\right) f\left(x_{3}\right) f\left(x_{4}\right) f\left(x_{5}\right)}{[1-F(100)]^{5}}&=\frac{\theta^{-1} e^{-125 / \theta} \theta^{-1} e^{-150 / \theta} \theta^{-1} e^{-165 / \theta} \theta^{-1} e^{-175 / \theta} \theta^{-1} e^{-250 / \theta}}{\left(e^{-100 / \theta}\right)^{5}} \\ & =\theta^{-5} e^{-365 / \theta} . \end{aligned}[[/math]]

Taking logarithms and then a derivative gives

[[math]]l(\theta)=-5 \ln (\theta)-365 / \theta, l^{\prime}(\theta)=-5 / \theta+365 / \theta^{2}=0[[/math]]

[[math]]\hat{\theta}=365 / 5=73[[/math]]

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