## Exam c Practice Problem 7 – Working with Buhlmann Credibility

Problem 7-A

You are given the following information:

• The number of claims in a calendar year for a given risk follows a Poisson distribution with mean $\theta$.
• The risk parameter $\theta$ follows a Gamma distribution whose coefficient of variation is 0.5.
• After observing the given risk for 5 calendar years, twelve claims are observed.
• Based on the observed data, the posterior distribution of $\theta$ is a continuous distribution whose mean is 2.0.

What is the Buhlmann credibility used in estimating the expected claim frequency for the given risk in the next period? $\text{ }$ $\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.500$ $\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.600$ $\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.625$ $\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.650$ $\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.667$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$

Problem 7-B

You are given the following information:

• The number of claims in a calendar year for a given risk follows a Poisson distribution with mean $\theta$.
• The risk parameter $\theta$ follows a Gamma distribution with mean 1.5.
• The value of Buhlmann’s k is 8.
• After observing the given risk for 4 calendar years, the posterior distribution of $\theta$ is a continuous distribution whose mean is 1.75

What is the number of claims observed in the observation period? $\text{ }$ $\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 7$ $\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8$ $\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9$ $\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 10$ $\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 11$

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

## Exam C Practice Problem 6 – Working with Posterior Distributions

Problem 6-A

You are given the following:

• The number of claims in a calendar year for a given risk follows a Poisson distribution with mean $\theta$.
• The prior distribution of $\theta$ has the Gamma distribution with mean 2 and variance 1.

After observing this risk for five calendar years, a total of 12 claims are observed.

Which of the following is the moment generating function of the posterior distribution of $\theta$? $\text{ }$ $\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M(t)=\biggl(\frac{2}{2-t}\biggr)^{16} \ \ \ \ \ \ \ \ t<2$ $\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M(t)=\biggl(\frac{7}{7-t}\biggr)^{15} \ \ \ \ \ \ \ \ t<7$ $\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M(t)=\biggl(\frac{14}{14-t}\biggr)^{9} \ \ \ \ \ \ \ \ t<14$ $\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M(t)=\biggl(\frac{7}{7-t}\biggr)^{15} \ \ \ \ \ \ \ \ t<2$ $\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M(t)=\biggl(\frac{7}{7-t}\biggr)^{16} \ \ \ \ \ \ \ \ t<7$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$

Problem 6-B

You are given the following:

• The number of claims in a calendar year for a given risk follows a Poisson distribution with mean $\theta$.
• The prior distribution of $\theta$ has the Gamma distribution with mean 4 and variance $\frac{1}{2}$.

After observing this risk for eight calendar years, a total of 32 claims are observed.

Determine the coefficient of variation of the posterior distribution of $\theta$. $\text{ }$ $\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{64}$ $\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{32}$ $\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{16}$ $\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}$ $\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{4}$

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