Solutions

This page has the worked solutions to the practice problems.

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Solutions to Practice Problems


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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.

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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}

Answer is C.

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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.

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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.

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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.


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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.

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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.

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Solution to Problem 6

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Solution to Problem 7

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Solution to Problem 8

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Solution to Problem 9

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Solution to Problem 10

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Solution to Problem 11

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Solution to Problem 12

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Solution to Problem 13

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Solution to Problem 14

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Solution to Problem 15

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Solution to Problem 16

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Solution to Problem 17

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Solution to Problem 18

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Solution to Problem 19

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Solution to Problem 20

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Solution to Problem 21

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Solution to Problem 22

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Solution to Problem 23

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Solution to Problem 24

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Solution to Problem 25

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Solution to Problem 26

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Solution to Problem 27

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Solution to Problem 28

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Solution to Problem 29

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Solution to Problem 30

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\copyright 2013 to 2018 – Dan Ma

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