May 13'23

Exercise

You are given:

  1. Low-hazard risks have an exponential claim size distribution with mean [math]\theta[/math].
  2. Medium-hazard risks have an exponential claim size distribution with mean [math]2 \theta [/math].
  3. High-hazard risks have an exponential claim size distribution with mean [math]3 \theta [/math] .
  4. No claims from low-hazard risks are observed.
  5. Three claims from medium-hazard risks are observed, of sizes 1, 2 and 3.
  6. One claim from a high-hazard risk is observed, of size 15.

Calculate the maximum likelihood estimate of [math]\theta[/math].

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

May 13'23

Key: B

The likelihood function is

[[math]]\frac{e^{-1 /(2 \theta)}}{2 \theta} \frac{e^{-2 /(2 \theta)}}{2 \theta} \frac{e^{-3 /(2 \theta)}}{2 \theta} \frac{e^{-15 /(3 \theta)}}{3 \theta}=\frac{e^{-8 / \theta}}{24 \theta^{4}}[[/math]]

. The loglikelihood function is [math]-\ln (24)-4 \ln (\theta)-8 / \theta[/math]. Differentiating with respect to [math]\theta[/math] and setting the result equal to 0 yields

[[math]]-\frac{4}{\theta}+\frac{8}{\theta^{2}}=0[[/math]]

which produces [math]\hat{\theta}=2[/math]

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

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