## Exam C Practice Problem 14 – Examples of Limited Fluctuation Credibility

Problem 14-A

You are given the following:

• The annual number of claims generated from a portfolio of insurance policies follows a Poisson distribution.
• The claim size follows a uniform distribution on $(0,t)$ where $t$ is unknown.
• The number of claims and the claim sizes are independent.

Using limited fluctuation credibility, how many expected claims are required to be 95% certain that actual claim costs will be within 5% of the expected claim costs?

$\text{ }$

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1443$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1579$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1936$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1945$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2050$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

___________________________________________________________________________________

Problem 14-B

You are given the following:

• The annual number of claims generated from a portfolio of insurance policies follows a Poisson distribution.
• The claim size follows a distribution with the following moment generating function.
• $\displaystyle M(t)=\frac{1}{(1-10t)^4}$
• The number of claims and the claim sizes are independent.

What is the least number of expected claims that are required to be 90% certain that actual claim costs will be within 5% of the expected claim costs?

$\text{ }$

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 820$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1230$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1353$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1376$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1396$

___________________________________________________________________________________

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

___________________________________________________________________________________

___________________________________________________________________________________

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

## 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 $\theta$.

$\text{ }$

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{64}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{32}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{16}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{4}$

___________________________________________________________________________________

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

___________________________________________________________________________________

___________________________________________________________________________________

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