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May 13'23
Exercise
You are given:
- Low-hazard risks have an exponential claim size distribution with mean [math]\theta[/math].
- Medium-hazard risks have an exponential claim size distribution with mean [math]2 \theta [/math].
- High-hazard risks have an exponential claim size distribution with mean [math]3 \theta [/math] .
- No claims from low-hazard risks are observed.
- Three claims from medium-hazard risks are observed, of sizes 1, 2 and 3.
- One claim from a high-hazard risk is observed, of size 15.
Calculate the maximum likelihood estimate of [math]\theta[/math].
- 1
- 2
- 3
- 4
- 5
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]