# Exam C Practice Problem 19 – Buhlmann Credibility Estimates

Problem 19-A

The number of claims in a year for an insurance policy in a large pool of insurance policies has a distribution with mean $\theta$ and variance $\lambda$.

The following provides more information about the large pool of insurance policies.

• For half of the insurance policies in the large pool $\theta=1$, while for the other half $\theta=0.5$.
• For three-quarters of the insurance policies in the large pool $\lambda=0.5$, while for the other one-quarter $\lambda=0.375$.

An insurance policy is randomly selected from the large pool. Insurance company records indicate that there are 6 claims in last 5 years.

Determine the Buhlmann credibility estimate of the number of claims for the selected insurance policy in the next year.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.82$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.85$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.88$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.93$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.02$

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Problem 19-B

The number of claims in a year for an insurance policy in a large pool of insurance policies has a distribution with mean $\theta$ and variance $\lambda$.

The following provides more information about the large pool of insurance policies.

• For three-quarters of the insurance policies in the large pool $\theta=1$, while for the other one-quarter $\theta=0.5$.
• For one-quarter of the insurance policies in the large pool $\lambda=0.5$, while for the other three-quarters $\lambda=0.375$.

An insurance policy is randomly selected from the large pool.

Determine the Buhlmann credibility factor assigned to 5 years of claim data from the selected insurance policy.

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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{15}{41}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{49}{133}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{19}{41}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{62}{133}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{130}{133}$

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$\copyright \ 2013 \ \ \text{Dan Ma}$