# Solutions

This page has the worked solutions to the practice problems.

_____________________________________________________________________________________

Solutions to Practice Problems

_____________________________________________________________________________________

Solution to Problem 1

Statement of Problem 1

Problem 1-A
The claim size $X$ is a mixture with equal weights of $X_1$, a Burr distribution with $\alpha=1$ and $\theta=\sqrt{8000}$ and $X_2$, a Pareto distribution with $\alpha=1$ and $\theta=8000$. The unconditional probability $P(X>x)$ is then the weighted average of the corresponding quantities from the two distributions:

\displaystyle \begin{aligned} P(X>x)&=0.5 \times P(X_1>x)+0.5 \times P(X_2>x) \\&=0.5 \biggl( \frac{8000}{8000+x^2} \biggr) + 0.5 \biggl(\frac{8000}{8000+x }\biggr) \end{aligned}

Setting $P(X>x)$ equal to 0.5 and solving for x would give the median.

$\displaystyle 0.5 \biggl( \frac{8000}{8000+x^2} \biggr) + 0.5 \biggl(\frac{8000}{8000+x }\biggr)=0.5$

$\displaystyle \biggl( \frac{8000}{8000+x^2} \biggr) + \biggl(\frac{8000}{8000+x }\biggr)=1$

The median is $x = 400$. Answer is A.

Note the median of a mixture is not the weighted average of the individual medians.

Problem 1-B
The claim size $X$ is a mixture of a Burr distribution with $\alpha=2$ and $\theta=\sqrt{1000}$ and a Pareto distribution with $\alpha=2$ and $\theta=1000$, with 90% weight for the Burr distribution and 10% for the Pareto. Let $Y=1.2X$. Then $P(Y>y)=P(X>\frac{y}{1.2})$.

\displaystyle \begin{aligned} P(Y>50)&=P(X>\frac{50}{1.2}) \\&=0.9 \times P(X_1>\frac{50}{1.2})+0.1 \times P(X_2>\frac{50}{1.2}) \\&=0.9 \biggl( \frac{1000}{1000+(\frac{50}{1.2})^2} \biggr)^2 + 0.1 \biggl(\frac{1000}{1000+\frac{50}{1.2} }\biggr)^2\\&=0.212379537 \end{aligned}

_____________________________________________________________________________________

Solution to Problem 2

Statement of Problem 2

Problem 2-A

For a given risk, the following is the claim count distribution.

$\displaystyle \begin{array}{ccc} n &\text{ } & P(N=n) \\ \text{ } & \text{ } & \text{ } \\ 0 &\text{ } & 0.4 \\ \text{ } & \text{ } & \text{ } \\ 1 &\text{ } & 0.5 \\ \text{ } & \text{ } & \text{ } \\ 2 &\text{ } & 0.1 \end{array}$

The following shows the claim size distributions conditional on the claim count.

$\displaystyle \begin{array}{ccccc} x &\text{ } & P(X=x |N=1) & \text{ } & P(X=x |N=2) \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ 100 &\text{ } & 0.75 & \text{ } & 0.50 \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ 200 &\text{ } & 0.25 & \text{ } & 0.50 \end{array}$

Clearly, the claim count and claim size are not independent. As a result, we cannot use the total variance formula. One possibility is to write out the aggregate distribution. Let $S=X_1+ \cdots + X_N$. The following gives the probability function of the aggregate claims $S$.

$\displaystyle \begin{array}{ccc} s &\text{ } & P(S=s) \\ \text{ } & \text{ } & \text{ } \\ 0 &\text{ } & 0.4 \\ \text{ } & \text{ } & \text{ } \\ 100 &\text{ } & 0.375 \\ \text{ } & \text{ } & \text{ } \\ 200 &\text{ } & 0.15 \\ \text{ } & \text{ } & \text{ } \\ 300 &\text{ } & 0.05 \\ \text{ } & \text{ } & \text{ } \\ 400 &\text{ } & 0.025 \\ \end{array}$

A special note is on $P(S=200)$, which is made up of two cases.

\displaystyle \begin{aligned} P(S=200)&=P(S=200|N=1) \times P(N=1)+ P(S=200|N=2) \times P(N=2)\\&=0.25 \times 0.5+(0.5 \times 0.5) \times 0.1 \\&=0.15 \end{aligned}

Answer is C, where

$E(S)=92.5$
$E(S^2)=18250$
$Var(S)=18250-92.5^2=9693.75$

Problem 2-B
The claim count and claim size in 2-B are independent. As a result, we can use the total variance formula (also called compound variance). Using it means that we only need to calculate the mean and variance of the claim count and the mean and variance of the claim size separately.

$\displaystyle \begin{array}{ccc} n &\text{ } & P(N=n) \\ \text{ } & \text{ } & \text{ } \\ 0 &\text{ } & 0.4 \\ \text{ } & \text{ } & \text{ } \\ 1 &\text{ } & 0.5 \\ \text{ } & \text{ } & \text{ } \\ 2 &\text{ } & 0.1 \end{array} \displaystyle \begin{array}{cccc} | & x &\text{ } & P(X=x) \\ | & \text{ } & \text{ } & \text{ } \\ | & 100 &\text{ } & 0.75 \\ | & \text{ } & \text{ } & \text{ } \\ | & 200 &\text{ } & 0.25 \\ | & \text{ } & \text{ } & \text{ } \\ | & \text{ } & \text{ } & \text{ } \\ \end{array}$

The following gives the individual calculations.

$E(N)=0.5+0.2=0.7$

$E(N^2)=0.5+0.4=0.9$

$Var(N)=0.9-0.7^2=0.41$

$E(X)=100(0.75)+200(0.25)=125$

$E(X^2)=100^2(0.75)+200^2(0.25)=17500$

$Var(X)=17500-125^2=1875$

The variance of the total claims (aggregate claims) is:

\displaystyle \begin{aligned} Var(S)&=E(N) Var(X)+Var(N) E(X)^2 \\&=0.7(1875)+0.41 (125^2) \\&=7718.75 \end{aligned}

The answer is D.

_____________________________________________________________________________________

Solution to Problem 3

Statement of Problem 3

Problem 3-A
Before observing any experience data, the mean claim frequency would be:

$E(X |\text{Class 1})=0 \cdot 0.5+1 \cdot 0.25 + 2 \cdot 0.12+3 \cdot 0.08+4 \cdot 0.05=0.93$

$E(X |\text{Class 2})=0 \cdot 0.2+1 \cdot 0.25 + 2 \cdot 0.30+3 \cdot 0.15+4 \cdot 0.10=1.70$

\displaystyle \begin{aligned} E(X)&=E(X |\text{Class 1}) \times P(\text{Class 1}) +E(X |\text{Class 2}) \times P(\text{Class 2})\\&=0.93 \cdot 0.6+1.7 \cdot 0.4 \\&=1.238 \end{aligned}

Without knowing any experience data, there are two choices for an estimate of the mean claim count in the next period – using the unconditional mean claim frequency, which is 1.238 or use the observed mean claim frequency, which is 2.5 (=(2 + 3)/2). A more “balanced” answer is probably somewhere in between. We can use observed claim data to interpolate these two estimates. This problem presents two methods – Bayesian estimate (Problem 3-A) and Buhlmann estimate (Problem 3-B).

The following calculates the marginal probability $P(X_1=2, X_2=3)$, which is the probability of the occurrence of the experience data.

\displaystyle \begin{aligned} P(X_1=2, X_2=3)&=P(X_1=2, X_2=3 |\text{Class 1}) \times P(\text{Class 1}) \\& \ \ \ + \ \ \ \ P(X_1=2, X_2=3 |\text{Class 2}) \times P(\text{Class 2}) \\&=(0.12 \cdot 0.08) \cdot 0.6 + (0.30 \cdot 0.15) \cdot 0.4 \\&=0.00576+0.018 \\&=0.02376 \end{aligned}

Given that the experience data (2 claims in year 1 and 3 claims in year 2) has occurred, there is a 24% chance that the risk is from Class 1 and there is a 76% chance that it is from Class 2.

$\displaystyle P(\text{Class 1} | X_1=2, X_2=3)=\frac{0.00576}{0.02376}=0.242424242$

$\displaystyle P(\text{Class 2} | X_1=2, X_2=3)=\frac{0.018}{0.02376}=0.757575758$

The above new weights of Class 1 and Class 2 form the posterior distribution. The Bayesian estimate of the claim frequency of the next year is the weighted average of the conditional mean claim frequency using the posterior weights:

\displaystyle \begin{aligned}E(X |X_1=2, X_2=3)&=E(X |\text{Class 1}) \times P(\text{Class 1}|X_1=2, X_2=3) \\& \ \ \ + \ \ \ E(X |\text{Class 2}) \times P(\text{Class 2}|X_1=2, X_2=3) \\&=0.93 \cdot 0.242424242+1.7 \cdot 0.757575758 \\&=1.513333 \end{aligned}

The answer is D.

Problem 3-B
The Buhlmann method takes a different approach. First calculate the conditional mean and variance in each class.

$E(X |\text{Class 1})=0 \cdot 0.5+1 \cdot 0.25 + 2 \cdot 0.12+3 \cdot 0.08+4 \cdot 0.05=0.93$

$E(X^2 |\text{Class 1})=0^2 \cdot 0.5+1^2 \cdot 0.25 + 2^2 \cdot 0.12+3^2 \cdot 0.08+4^2 \cdot 0.05=2.25$

$Var(X |\text{Class 1})=2.25-0.93^2=1.3851$

$E(X |\text{Class 2})=0 \cdot 0.2+1 \cdot 0.25 + 2 \cdot 0.30+3 \cdot 0.15+4 \cdot 0.10=1.70$

$E(X |\text{Class 2})=0^2 \cdot 0.2+1^2 \cdot 0.25 + 2^2 \cdot 0.30+3^2 \cdot 0.15+4^2 \cdot 0.10=4.40$

$Var(X |\text{Class 2})=4.4-1.7^2=1.51$

The following calculates the Buhlmann credibility factor.

$\displaystyle \text{EPV}=1.3851 \cdot 0.6+1.51 \cdot 0.4=1.43506$

$\displaystyle \text{VHM}=0.93^2 \cdot 0.6+1.7^2 \cdot 0.4-1.238^2=0.142296$

$\displaystyle k=\frac{\text{EPV}}{\text{VHM}}=\frac{1.43506}{0.142296}=10.08503401$

$\displaystyle Z=\frac{2}{2+k}=\frac{2}{2+10.08503401}=0.165493949$

The credibility assigned to the observed rate is about 16.55%. Then the Buhlmann credibility estimate of the claim frequency for next year is:

\displaystyle \begin{aligned}E(X |X_1=2, X_2=3)&=Z \cdot \overline{x}+(1-Z) \cdot E(X) \\&=0.165493949 \cdot 2.5 + (1-0.165493949) \cdot 1.238 \\&=1.446853363 \end{aligned}

The answer is C.

_____________________________________________________________________________________

Solution to Problem 4

Statement of Problem 4

Problem 4-A
The hypothetical mean is $E[X| \theta]=\theta$ and the process variance is $Var[X|\theta]=\theta$ since this is the Poisson model. Next, calculate EPV (expected value of process variance) and VHM (variance of hypothetical means).

$\displaystyle EPV = E[Var[X|\theta]]=E[\theta]=\frac{0.5+2.5}{2}=1.5$

$\displaystyle E[X]=E[E[X| \theta]]=E[\theta]=1.5$

$\displaystyle VHM = Var[E[X|\theta]]=Var[\theta]=\frac{(2.5-0.5)^2}{12}=\frac{1}{3}$

the following calculates the Buhlmann credibility factor (based on three years of experience data).

$\displaystyle K=\frac{EPV}{VHM}=\frac{9}{2}$

$\displaystyle Z=\frac{3}{3+K}=\frac{6}{6+9}=\frac{6}{15}=\frac{2}{5}=0.4$

The Buhlmann estimate of claim frequency for year 4 is:

$\displaystyle Z \times \overline{x}+(1-Z) \times E(X)=0.4 \times \frac{1+2+3}{3}+0.6 \times 1.5=1.7$

The answer is C.

Problem 4-B
4-B is similar to 4-A in that the claim frequency model is Poisson, except that the prior distribution of the parameter $\theta$ is no longer a uniform distribution. To calculate EPV and VHM, we need to use the given density function of the $\theta$.

$\displaystyle EPV=E[Var[X|\theta]] =\int_0^2 \theta \frac{1}{2} \ (2-\theta) \ d \theta=\frac{2}{3}$

$\displaystyle E[X]=E[E[X| \theta]]=E[\theta]=\frac{2}{3}$

\displaystyle \begin{aligned} VHM&=Var[E[X|\theta]]=Var[\theta] \\&=\int_0^2 \theta^2 \frac{1}{2} \ (2-\theta) \ d \theta - \biggl[ \frac{2}{3}\biggr]^2 =\frac{2}{9} \end{aligned}

$\displaystyle K=\frac{EPV}{VHM}=3$

$\displaystyle Z=\frac{6}{6+K}=\frac{6}{9}=\frac{2}{3}$

$\displaystyle Z \times \overline{x}+(1-Z) \times E(X)=\frac{2}{3} \times \frac{10}{6}+\frac{1}{3} \times \frac{2}{3}=\frac{4}{3}$

The answer is C.

_____________________________________________________________________________________

Solution to Problem 5

Statement of Problem 5

Problem 5-A
To incorporate the observed data of $X_1=4$ into the estimate of the next period, first find the marginal probability of $X_1=4$.

$\displaystyle P[X_1=4]=\int_1^\infty \frac{e^{-\theta} \theta^4}{4!} \ \frac{2}{\theta^3} \ d \theta=\int_1^\infty \frac{1}{12} \theta \ e^{-\theta} \ d \theta=\frac{1}{12} \ 2e^{-1}$

The posterior distribution is $\theta$ is:

$\displaystyle \pi(\theta | X_1=4)=\frac{\frac{1}{12} \theta \ e^{-\theta}}{\frac{1}{12} \ 2e^{-1}}=\frac{1}{2e^{-1}} \ \theta \ e^{-\theta} \ \ \ \ \ \ \ \theta>1$

The Bayesian estimate of the claim frequency for the next period is:

\displaystyle \begin{aligned} E[X_2| X_1=4]&=\int_1^\infty E[\theta|X_1=4] \ \pi(\theta | X_1=4) \ d \theta \\&=\int_1^\infty \theta \ \frac{1}{2e^{-1}} \ \theta \ e^{-\theta} \ d \theta \\&=\int_1^\infty \frac{1}{2e^{-1}} \ \theta^2 \ e^{-\theta} \ d \theta \\&=\frac{1}{e^{-1}} \int_1^\infty \frac{1}{2} \ \theta^2 \ e^{-\theta} \ d \theta \\&=\frac{1}{e^{-1}} [e^{-1}+e^{-1}+\frac{1}{2}e^{-1}]=\frac{5}{2} \end{aligned}

The answer is B.

Problem 5-B
5-B is similar to 5-A, except that the experience data consist of $X_1=2,X_2=2$. The marginal probability is:

$\displaystyle P[X_1=2,X_2=2]=\int_1^\infty \frac{e^{-\theta} \theta^2}{2!} \ \frac{e^{-\theta} \theta^2}{2!} \ \frac{2}{\theta^3} \ d \theta=\int_1^\infty \frac{1}{2} \theta \ e^{-2 \theta} \ d \theta=\frac{1}{8} \ e^{-2}$

The posterior distribution is $\theta$ is:

$\displaystyle \pi(\theta | X_1=2,X_2=2)=\frac{\frac{1}{2} \theta \ e^{-2 \theta}}{\frac{3}{8} \ e^{-2}}=\frac{4}{3e^{-2}} \ \theta \ e^{-2 \theta} \ \ \ \ \ \ \ \theta>1$

The Bayesian estimate of the claim frequency for the next period is:

\displaystyle \begin{aligned} E[X_3| X_1=2,X_2=2]&=\int_1^\infty E[\theta|X_1=2,X_2=2] \ \pi(\theta | X_1=2,X_2=2) \ d \theta \\&=\int_1^\infty \theta \ \frac{4}{3e^{-2}} \ \theta \ e^{-2 \theta} \ d \theta \\&=\int_1^\infty \frac{4}{3e^{-2}} \ \theta^2 \ e^{-2 \theta} \ d \theta \\&=\frac{1}{3e^{-2}} \int_1^\infty \frac{8}{2} \ \theta^2 \ e^{-\theta} \ d \theta \\&=\frac{1}{3e^{-2}} [e^{-2}+2e^{-2}+\frac{4}{2}e^{-2}]=\frac{5}{3} \end{aligned}

The answer is C.

_____________________________________________________________________________________

Solution to Problem 6

_____________________________________________________________________________________

Solution to Problem 7

_____________________________________________________________________________________

Solution to Problem 8

_____________________________________________________________________________________

Solution to Problem 9

_____________________________________________________________________________________

Solution to Problem 10

_____________________________________________________________________________________

Solution to Problem 11

_____________________________________________________________________________________

Solution to Problem 12

_____________________________________________________________________________________

Solution to Problem 13

_____________________________________________________________________________________

Solution to Problem 14

_____________________________________________________________________________________

Solution to Problem 15

_____________________________________________________________________________________

Solution to Problem 16

_____________________________________________________________________________________

Solution to Problem 17

_____________________________________________________________________________________

Solution to Problem 18

_____________________________________________________________________________________

Solution to Problem 19

_____________________________________________________________________________________

Solution to Problem 20

_____________________________________________________________________________________

Solution to Problem 21

_____________________________________________________________________________________

Solution to Problem 22

_____________________________________________________________________________________

Solution to Problem 23

_____________________________________________________________________________________

Solution to Problem 24

_____________________________________________________________________________________

Solution to Problem 25

_____________________________________________________________________________________

Solution to Problem 26

_____________________________________________________________________________________

Solution to Problem 27

_____________________________________________________________________________________

Solution to Problem 28

_____________________________________________________________________________________

Solution to Problem 29

$\copyright$ 2013 to 2018 – Dan Ma