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Exercise


May 14'23

Answer

Key: A

These observations are truncated at 500. The contribution to the likelihood function is

[[math]] \frac{f(x)}{1-F(500)} = \frac{\theta^{-1}e^{-x/\theta}}{e^{-500/\theta}}. [[/math]]

Then the likelihood function is

[[math]] \begin{aligned} &L(\theta) = \frac{\theta^{-1}e^{-600/\theta}\theta^{-1}e^{-700/\theta}\theta^{-1}e^{-900/\theta}}{\left( e^{-500/\theta}\right)^3} = \theta^{-3}e^{-700/\theta} \\ & l (\theta ) = \ln L(\theta ) = −3\ln \theta − 700\theta^{−1} \\ & l^{'}(\theta) = -3\theta^{-1} + 700\theta^{-2} = 0 \\ & \theta = 700 / 3 = 233.33. \end{aligned} [[/math]]

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

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