Tag Archives: Coefficient of Variation

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?

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      \displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.500

      \displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.600

      \displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.625

      \displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.650

      \displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.667

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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?

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      \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?

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      \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

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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.

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      \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}