BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: The time to an accident follows an exponential distribution. A random sample of size two has a mean time of 6 .

Let \(Y\) denote the mean of a new sample of size two.
Calculate the maximum likelihood estimate of \(\operatorname{Pr}(Y>10)\).

Options:
A
\(\quad 0.04\)
B
0.07
C
0.11
D
0.15
E
0.19

Correct answer: D

Solution: For an exponential distribution, the maximum likelihood estimate of \(\theta\) is the sample mean, 6 . Let \(Y=X_{1}+X_{2}\) where each \(X\) has an exponential distribution with mean 6 . The sample mean is \(Y / 2\) and \(Y\) has a gamma distribution with parameters 2 and 6 . Then
\(\operatorname{Pr}(Y / 2>10)=\operatorname{Pr}(Y>20)=\int_{20}^{\infty} \frac{x e^{-x / 6}}{36} d x\)
\(=-\frac{x e^{-x / 6}}{6}-\left.e^{-x / 6}\right|_{20} ^{\infty}=\frac{20 e^{-20 / 6}}{6}+e^{-20 / 6}=0.1546\).
Question: The time to an accident follows an exponential distribution. A random sample of size two has a sample mean time of 6 .

Let \(Y\) denote the mean of a new sample of size two.
Calculate the delta method approximation of the variance of the maximum likelihood estimator of \(F_{Y}(10)\).

Options:
A
0.08
B
0.12
C
0.16
D
0.19
E
0.22

Correct answer: A

Solution: Let \(W=X_{1}+X_{2}\) where each \(X\) has an exponential distribution with mean \(\theta\). The sample mean is \(Y=W / 2\) and \(W\) has a gamma distribution with parameters 2 and \(\theta\). Then
\(g(\theta)=F_{Y}(10)=\operatorname{Pr}(Y \leq 10)=\operatorname{Pr}(W \leq 20)=\int_{0}^{20} \frac{w e^{-w / \theta}}{\theta^{2}} d w\)
\(=-\frac{w e^{-w / \theta}}{\theta}-\left.e^{-w / \theta}\right|_{0} ^{20}=1-\frac{20 e^{-20 / \theta}}{\theta}-e^{-20 / \theta}=1-\mathrm{e}^{-20 / \theta}\left(1+20 \theta^{-1}\right)\).
\(g^{\prime}(\theta)=-\frac{20}{\theta^{2}} e^{-20 / \theta}\left(1+20 \theta^{-1}\right)+e^{-20 / \theta} \frac{20}{\theta^{2}}=-\frac{400 e^{-20 / \theta}}{\theta^{3}}\).
At the maximum likelihood estimate of \(6, g^{\prime}(6)=-0.066063\).
The maximum likelihood estimator is the sample mean. Its variance is the variance of one observation divided by the sample size. For the exponential distribution the variance is the square of the mean, so the estimated variance of the sample mean is \(36 / 2=18\). The answer is \((-0.066063)^{2}(18)=0.079\).
Question: You are given:
(i) The number of claims in a year for a selected risk follows a Poisson distribution with mean \(\lambda\).
(ii) The severity of claims for the selected risk follows an exponential distribution with mean \(\theta\).
(iii) The number of claims is independent of the severity of claims.
(iv) The prior distribution of \(\lambda\) is exponential with mean 1 .
(v) The prior distribution of \(\theta\) is Poisson with mean 1.
(vi) A priori, \(\lambda\) and \(\theta\) are independent.

Using Bühlmann's credibility for aggregate losses, calculate \(k\).

Options:
A
1
B
\(4 / 3\)
C
2
D
3
E
4

Correct answer: B

Solution: \(\mu(\lambda, \theta)=E(S \mid \lambda, \theta)=\lambda \theta\),
\(v(\lambda, \theta)=\operatorname{Var}(S \mid \lambda, \theta)=\lambda 2 \theta^{2}\),
\(v=E P V=E\left(\lambda 2 \theta^{2}\right)=1(2)(1+1)=4\),
\(a=\operatorname{VHM}=\operatorname{Var}(\lambda \theta)=E\left(\lambda^{2}\right) E\left(\theta^{2}\right)-[E(\lambda) E(\theta)]^{2}=2(2)-1=3\),
\(k=v / a=4 / 3\).
Question: A company insures 100 people age 65 . The annual probability of death for each person is 0.03 . The deaths are independent.

Use the inversion method to simulate the number of deaths in a year for three years using the uniform \((0,1)\) random numbers \(0.20,0.03\), and 0.09 .

Calculate the average of the simulated values.

Options:
A
\(1 / 3\)
B
1
C
\(5 / 3\)
D
\(7 / 3\)
E
3

Correct answer: B

Solution: The distribution is binomial with \(m=100\) and \(q=0.03\). The first three probabilities are:
\(p_{0}=0.97^{100}=0.04755, p_{1}=100(0.97)^{99}(0.03)=0.14707\),
\(p_{2}=\frac{100(99)}{2}(0.97)^{98}(0.03)^{2}=0.22515\).
Values between 0 and 0.04755 simulate a 0 , between 0.04755 and 0.19462 simulate a 1 , and between 0.19462 and 0.41977 simulate a 2 . The three simulated values are 2,0 , and 1 . The mean is 1 .
Question: For the random variable \(X\), you are given:
(i) \(E[X]=\theta, \quad \theta>0\)
(ii) \(\quad \operatorname{Var}(X)=\frac{\theta^{2}}{25}\)
(iii) \(\quad \hat{\theta}=\frac{k}{k+1} X, \quad k>0\)
(iv) \(\operatorname{MSE}_{\hat{\theta}}(\theta)=2\left[\operatorname{bias}_{\hat{\theta}}(\theta)\right]^{2}\)

Calculate \(k\).

Options:
A
0.2
B
0.5
C
2
D
5
E
25

Correct answer: D

Solution: \[
\begin{aligned}
& \operatorname{bias}=E(\hat{\theta})-\theta=\frac{k}{k+1} \theta-\theta=-\frac{\theta}{k+1}, \quad \operatorname{Var}(\hat{\theta})=\operatorname{Var}\left(\frac{k \theta}{k+1}\right)=\frac{k^{2} \theta^{2}}{25(k+1)^{2}} \\
& \operatorname{MSE}=\operatorname{Var}(\hat{\theta})+\operatorname{bias}^{2}=\frac{k^{2} \theta^{2}}{25(k+1)^{2}}+\frac{\theta^{2}}{(k+1)^{2}} \\
& \operatorname{MSE}=2 \operatorname{bias}^{2}=\frac{2 \theta^{2}}{(k+1)^{2}}
\end{aligned}
\]

Setting the last two equal and canceling the common terms gives \(\frac{k^{2}}{25}+1=2\) for \(k=5\).
Question: You are given:
(i) The annual number of claims for a policyholder follows a Poisson distribution with mean \(\Lambda\).
(ii) The prior distribution of \(\Lambda\) is gamma with probability density function:
\[
f(\lambda)=\frac{(2 \lambda)^{5} e^{-2 \lambda}}{24 \lambda}, \quad \lambda>0
\]

An insured is selected at random and observed to have \(X_{1}=5\) claims during Year 1 and \(X_{2}=3\) claims during Year 2.

Calculate \(E\left[\Lambda \mid x_{1}=5, x_{2}=3\right]\).

Options:
A
3.00
B
3.25
C
3.50
D
3.75
E
4.00

Correct answer: B

Solution: \(f(\lambda \mid 5,3) \propto \frac{e^{-\lambda} \lambda}{5!} \frac{e^{-\lambda} \lambda^{3}}{3!} \frac{2^{5} \lambda^{5} e^{-2 \lambda}}{24 \lambda} \propto \lambda^{12} e^{-4 \lambda}\). This is a gamma distribution with parameters 13 and 0.25 . The expected value is \(13(0.25)=3.25\).

Alternatively, if the Poisson-gamma relationships are known, begin with the prior parameters \(\alpha=5\) and \(\beta=2\) where \(\beta=1 / \theta\) if the parameterization from Loss Models is considered. Then the posterior parameters are \(\alpha^{\prime}=5+5+3=13\) where the second 5 and the 3 are the observations and \(\beta^{\prime}=2+2=4\) where the second 2 is the number of observations. The posterior mean is then \(13 / 4=3.25\).
Question: You are given:

\begin{tabular}{|l|l|l|l|l|l|}
\hline & Group & Year 1 & Year 2 & Year 3 & Total \\
\hline Total Claims & 1 & & 10,000 & 15,000 & 25,000 \\
\hline Number in Group & & & 50 & 60 & 110 \\
\hline Average & & & 200 & 250 & 227.27 \\
\hline Total Claims & 2 & 16,000 & 18,000 & & 34,000 \\
\hline Number in Group & & 100 & 90 & & 190 \\
\hline Average & & 160 & 200 & & 178.95 \\
\hline Total Claims & & & & & 59,000 \\
\hline Number in Group & & & & & 300 \\
\hline Average & & & & & 196.67 \\
\hline
\end{tabular}

You are also given \(\hat{a}=651.03\).
Calculate the nonparametric empirical Bayes credibility factor for Group 1.

Options:
A
\(\quad 0.48\)
B
0.50
C
0.52
D
0.54
E
0.56

Correct answer: B

Solution: \[
\begin{aligned}
\hat{v} & =\frac{50(200-227.27)^{2}+60(250-227.27)^{2}+100(160-178.95)^{2}+90(200-178.95)^{2}}{1+1} \\
& =71,985.647 \\
\hat{k} & =71,985.647 / 651.03=110.57 \\
\hat{Z} & =\frac{110}{110+110.57}=0.499
\end{aligned}
\]
Question: You are given the following 20 bodily injury losses (before the deductible is applied):

\begin{tabular}{|r|c|c|c|}
\hline Loss & \begin{tabular}{c} 
Number of \\
Losses
\end{tabular} & Deductible & Policy Limit \\
\hline 750 & 3 & 200 & \(\infty\) \\
200 & 3 & 0 & 10,000 \\
300 & 4 & 0 & 20,000 \\
\(>10,000\) & 6 & 0 & 10,000 \\
400 & 4 & 300 & \(\infty\) \\
\hline
\end{tabular}

Past experience indicates that these losses follow a Pareto distribution with parameters \(\alpha\) and \(\theta=10,000\).

Calculate the maximum likelihood estimate of \(\alpha\).

Options:
A
Less than 2.0
B
At least 2.0 , but less than 3.0
C
At least 3.0 , but less than 4.0
D
At least 4.0 , but less than 5.0
E
At least 5.0

Correct answer: C

Solution: \[
\begin{aligned}
& L=\left[\frac{f(750)}{1-F(200)}\right]^{3} f(200)^{3} f(300)^{4}[1-F(10,000)]^{6}\left[\frac{f(400)}{1-F(300)}\right]^{4} \\
& =\left[\frac{\alpha 10,200^{\alpha}}{10,750^{\alpha+1}}\right]^{3}\left[\frac{\alpha 10,000^{\alpha}}{10,200^{\alpha+1}}\right]^{3}\left[\frac{\alpha 10,000^{\alpha}}{10,300^{\alpha+1}}\right]^{4}\left[\frac{10,000^{\alpha}}{20,000^{\alpha}}\right]^{6}\left[\frac{\alpha 10,300^{\alpha}}{10,400^{\alpha+1}}\right]^{4} \\
& =\alpha^{14} 10,200^{-3} 10,000^{13 \alpha} 10,300^{-4} 10,750^{-3 \alpha-3} 20,000^{-6 \alpha} 10,400^{-4 \alpha-4} \\
& \propto \alpha^{14} 10,000^{13 \alpha} 10,750^{-3 \alpha} 20,000^{-6 \alpha} 10,400^{-4 \alpha} \\
& \ln L=14 \ln \alpha+13 \alpha \ln (10,000)-3 \alpha \ln (10,750)-6 \alpha \ln (20,000)-4 \alpha \ln (10,400) \\
& =14 \ln \alpha-4.5327 \alpha
\end{aligned}
\]

The derivative is \(14 / \alpha-4.5327\) and setting it equal to zero gives \(\hat{\alpha}=3.089\).
Question: You are given:
(i) The annual number of claims on a given policy has the geometric distribution with parameter \(\beta\).
(ii) One-third of the policies have \(\beta=2\), and the remaining two-thirds have \(\beta=5\).

A randomly selected policy had two claims in Year 1.
Calculate the Bayesian expected number of claims for the selected policy in Year 2.

Options:
A
\(\quad 3.4\)
B
3.6
C
3.8
D
4.0
E
4.2

Correct answer: C

Solution: For the geometric distribution, \(\operatorname{Pr}\left(X_{1}=2 \mid \beta\right)=\frac{\beta^{2}}{(1+\beta)^{3}}\) and the expected value is \(\beta\).
\[
\begin{aligned}
& \operatorname{Pr}\left(\beta=2 \mid X_{1}=2\right)=\frac{\operatorname{Pr}\left(X_{1}=2 \mid \beta=2\right) \operatorname{Pr}(\beta=2)}{\operatorname{Pr}\left(X_{1}=2 \mid \beta=2\right) \operatorname{Pr}(\beta=2)+\operatorname{Pr}\left(X_{1}=2 \mid \beta=5\right) \operatorname{Pr}(\beta=5)} \\
& =\frac{\frac{4}{27} \frac{1}{3}}{\frac{4}{27} \frac{1}{3}+\frac{25}{216} \frac{2}{3}}=0.39024 .
\end{aligned}
\]

The expected value is then \(0.39024(2)+0.60976(5)=3.83\).
Question: For an insurance:
(i) Losses have density function
\[
f(x)= \begin{cases}0.02 x, & 0<x<10 \\ 0, & \text { elsewhere }\end{cases}
\]
(ii) The insurance has an ordinary deductible of 4 per loss.
(iii) \(\quad Y^{P}\) is the claim payment per payment random variable. Calculate \(E\left[Y^{P}\right]\).

Options:
A
2.9
B
3.0
C
3.2
D
3.3
E
\(\quad 3.4\)

Correct answer: E

Solution: \[
\begin{aligned}
& S_{X}(4)=1-\int_{0}^{4} f_{X}(s) d x=1-\int_{0}^{4} 0.02 x d x=1-\left.0.01 x^{2}\right|_{0} ^{4}=0.84 \\
& f_{Y^{p}}(y)=\frac{f_{X}(y+4)}{S_{X}(4)}=\frac{0.02(y+4)}{0.84}=0.0238(y+4)^{2} \\
& E\left(Y^{P}\right)=\int_{0}^{6} y[0.0238(y+4)] d y=\left.0.0238\left(\frac{y^{3}}{3}+\frac{4 y^{2}}{2}\right)\right|_{0} ^{6}=3.4272
\end{aligned}
\]
Question: In 2005 a risk has a two-parameter Pareto distribution with \(\alpha=2\) and \(\theta=3000\). In 2006 losses inflate by \(20 \%\).

An insurance on the risk has a deductible of 600 in each year. \(P_{i}\), the premium in year \(i\), equals 1.2 times the expected claims.

The risk is reinsured with a deductible that stays the same in each year. \(R_{i}\), the reinsurance premium in year \(i\), equals 1.1 times the expected reinsured claims.
\(\frac{R_{2005}}{P_{2005}}=0.55\)

Calculate \(\frac{R_{2006}}{P_{2006}}\).

Options:
A
\(\quad 0.46\)
B
0.52
C
0.55
D
0.58
E
0.66

Correct answer: D

Solution: For any deductible \(d\) and the given severity distribution
\(E\left[(X-d)_{+}\right]=E(X)-E(X \wedge d)=3000-3000\left(1-\frac{3000}{3000+d}\right)=3000\left(\frac{3000}{3000+d}\right)=\frac{9,000,000}{3000+d}\)
So \(P_{2005}=1.2 \frac{9,000,000}{3000+600}=3000\)
Let \(r\) denote the reinsurer's deductible relative to insured losses. Thus, the reinsurer's deductible is \(600+r\) relative to losses. Thus
\(R_{2005}=1.1\left(\frac{9,000,000}{3000+600+r}\right)=0.55 P_{2005}=0.55(3000)=1650 \Rightarrow r=2400\)
In 2006, after 20\% inflation, losses will have a two-parameter Pareto distribution with \(\alpha=2\) and \(\theta=1.2(3000)=3600\). The general formula for expected claims with a deductible of \(d\) is
\(E\left[(X-d)_{+}\right]=3600\left(\frac{3600}{3600+d}\right)=\frac{12,960,000}{3600+d}\)
\(P_{2006}=1.2 \frac{12,960,000}{3000+600}=3703, R_{2006}=1.1 \frac{12,960,000}{3000+600+2400}=2160, \frac{R_{2006}}{R_{2005}}=\frac{2160}{3703}=0.583\)
Question: You are given:
(i) The conditional distribution of the number of claims per policyholder is Poisson with mean \(\lambda\).
(ii) The variable \(\lambda\) has a gamma distribution with parameters \(\alpha\) and \(\theta\).
(iii) For policyholders with 1 claim in Year 1, the credibility estimate for the number of claims in Year 2 is 0.15 .
(iv) For policyholders with an average of 2 claims per year in Year 1 and Year 2, the credibility estimate for the number of claims in Year 3 is 0.20.

Calculate \(\theta\).

Options:
A
Less than 0.02
B
At least 0.02 , but less than 0.03
C
At least 0.03 , but less than 0.04
D
At least 0.04 , but less than 0.05
E
At least 0.05

Correct answer: A

Solution: \(E(X \mid \lambda)=\operatorname{Var}(X \mid \lambda)=\lambda\)
\(\mu=v=E(\lambda)=\alpha \theta ; a=\operatorname{Var}(\lambda)=\alpha \theta^{2} ; k=v / a=1 / \theta\)
\(Z=\frac{n}{n+1 / \theta}=\frac{n \theta}{n \theta+1}\)
\(0.15=\frac{\theta}{\theta+1}(1)+\frac{1}{\theta+1} \mu=\frac{\theta+\mu}{\theta+1}\)
\(0.20=\frac{2 \theta}{2 \theta+1}(2)+\frac{1}{2 \theta+1} \mu=\frac{4 \theta+\mu}{2 \theta+1}\)
From the first equation,
\(0.15 \theta+0.15=\theta+\mu\) and so \(\mu=0.15-0.85 \theta\)
Then the second equation becomes
\(0.4 \theta+0.2=4 \theta+0.15-0.85 \theta\)
\(0.05=2.75 \theta ; \theta=0.01818\)
Question: For a portfolio of policies, you are given:
(i) There is no deductible and the policy limit varies by policy.
(ii) A sample of ten claims is:
\(350 \quad 350 \quad 500 \quad 500 \quad 500+\quad 1000 \quad 1000+\quad 1000+\quad 1200 \quad 1500\)
where the symbol + indicates that the loss exceeds the policy limit.
(iii) \(\quad \hat{S}_{1}(1250)\) is the Kaplan-Meier product-limit estimate of \(S(1250)\).
(iv) \(\quad \hat{S}_{2}(1250)\) is the maximum likelihood estimate of \(S(1250)\) under the assumption that the losses follow an exponential distribution.

Calculate the absolute difference between \(\hat{S}_{1}(1250)\) and \(\hat{S}_{2}(1250)\).

Options:
A
\(\quad 0.00\)
B
0.03
C
0.05
D
0.07
E
0.09

Correct answer: E

Solution: Begin with

\begin{tabular}{|c|c|c|c|c|c|}
\hline\(y\) & 350 & 500 & 1000 & 1200 & 1500 \\
\hline\(s\) & 2 & 2 & 1 & 1 & 1 \\
\hline\(r\) & 10 & 8 & 5 & 2 & 1 \\
\hline
\end{tabular}

Then \(\hat{S}_{1}(1250)=\frac{8}{10} \frac{6}{8} \frac{4}{5} \frac{1}{2}=0.24\)
The likelihood function is
\(L(\theta)=\left[\theta^{-1} e^{-350 / \theta}\right]^{2}\left[\theta^{-1} e^{-500 / \theta}\right]^{2} e^{-500 / \theta} \theta^{-1} e^{-1000 / \theta}\left[e^{-1000 / \theta}\right]^{2} \theta^{-1} e^{-1200 / \theta} \theta^{-1} e^{-1500 / \theta}=\theta^{-7} e^{-7900 / \theta}\)
\(l(\theta)=-7 \ln \theta-\frac{7900}{\theta} ; l^{\prime}(\theta)=-\frac{7}{\theta}+\frac{7900}{\theta^{2}}=0 ; \hat{\theta}=7900 / 7\)
\(\hat{S}_{2}(1250)=e^{-1250(7) / 7900}=0.33\)
The absolute difference is 0.09 .
Question: The random variable \(X\) has survival function:
\[
S_{X}(x)=\frac{\theta^{4}}{\left(\theta^{2}+x^{2}\right)^{2}}
\]

Two values of \(X\) are observed to be 2 and 4 . One other value exceeds 4 .
Calculate the maximum likelihood estimate of \(\theta\).

Options:
A
Less than 4.0
B
At least 4.0 , but less than 4.5
C
At least 4.5 , but less than 5.0
D
At least 5.0 , but less than 5.5
E
At least 5.5

Correct answer: E

Solution: \(f(x)=-S^{\prime}(x)=\frac{4 x \theta^{4}}{\left(\theta^{2}+x^{2}\right)^{3}}\)
\(L(\theta)=f(2) f(4) S(4)=\frac{4(2) \theta^{4}}{\left(\theta^{2}+2^{2}\right)^{3}} \frac{4(4) \theta^{4}}{\left(\theta^{2}+4^{2}\right)^{3}} \frac{\theta^{4}}{\left(\theta^{2}+4^{2}\right)^{2}}=\frac{128 \theta^{12}}{\left(\theta^{2}+4\right)^{3}\left(\theta^{2}+16\right)^{5}}\)
\(l(\theta)=\ln 128+12 \ln \theta-3 \ln \left(\theta^{2}+4\right)-5 \ln \left(\theta^{2}+16\right)\)
\(l^{\prime}(\theta)=\frac{12}{\theta}-\frac{6 \theta}{\theta^{2}+4}-\frac{10 \theta}{\theta^{2}+16}=0 ; 12\left(\theta^{4}+20 \theta^{2}+64\right)-6\left(\theta^{4}+16 \theta^{2}\right)-10\left(\theta^{4}+4 \theta^{2}\right)=0\)
\(0=-4 \theta^{4}+104 \theta^{2}+768=\theta^{4}-26 \theta^{2}-192\)
\(\theta^{2}=\frac{26 \pm \sqrt{26^{2}+4(192)}}{2}=32 ; \theta=5.657\)
Question: You are given:
(i) Fifty claims have been observed from a lognormal distribution with unknown parameters \(\mu\) and \(\sigma\).
(ii) The maximum likelihood estimates are \(\hat{\mu}=6.84\) and \(\hat{\sigma}=1.49\).
(iii) The covariance matrix of \(\hat{\mu}\) and \(\hat{\sigma}\) is:
\[
\left[\begin{array}{cc}
0.0444 & 0 \\
0 & 0.0222
\end{array}\right]
\]
(iv) The partial derivatives of the lognormal cumulative distribution function are:
\[
\frac{\partial F}{\partial \mu}=\frac{-\phi(z)}{\sigma} \text { and } \frac{\partial F}{\partial \sigma}=\frac{-z \phi(z)}{\sigma}
\]
(v) An approximate \(95 \%\) confidence interval for the probability that the next claim will be less than or equal to 5000 is \([L, U]\)

Calculate \(L\).

Options:
A
0.73
B
0.76
C
0.79
D
0.82
E
0.85

Correct answer: C

Solution: The quantity of interest is \(P=\operatorname{Pr}(X \leq 5000)=\Phi\left(\frac{\ln 5000-\mu}{\sigma}\right)\). The point estimate is \(\Phi\left(\frac{\ln 5000-6.84}{1.49}\right)=\Phi(1.125)=0.87\).
For the delta method:
\(\frac{\partial P}{\partial \mu}=\frac{-\phi(1.125)}{1.49}=-0.1422 ; \frac{\partial P}{\partial \sigma}=\frac{-1.125 \phi(1.125)}{1.49}=-0.1600\) where \(\phi(z)=\frac{1}{\sqrt{2 \pi}} e^{-z^{2} / 2}\).
Then the variance of \(\hat{\boldsymbol{P}}\) is estimated as \((-0.1422)^{2} 0.0444+(-0.16)^{2} 0.0222=0.001466\) and the lower limit is \(P_{L}=0.87-1.96 \sqrt{0.001466}=0.79496\).
Question: For a particular policy, the conditional probability of the annual number of claims given \(\Theta=\theta\), and the probability distribution of \(\Theta\) are as follows:

\begin{tabular}{|l|c|c|c|}
\hline Number of Claims & 0 & 1 & 2 \\
\hline Probability & \(2 \theta\) & \(\theta\) & \(1-3 \theta\) \\
\hline
\end{tabular}

\begin{tabular}{|l|l|l|}
\hline\(\theta\) & 0.10 & 0.30 \\
\hline Probability & 0.80 & 0.20 \\
\hline
\end{tabular}

One claim was observed in Year 1.
Calculate the Bayesian estimate of the expected number of claims for Year 2.

Options:
A
Less than 1.1
B
At least 1.1, but less than 1.2
C
At least 1.2, but less than 1.3
D
At least 1.3, but less than 1.4
E
At least 1.4

Correct answer: A

Solution: \(\operatorname{Pr}\left(\theta=0.1 \mid X_{1}=1\right)=\frac{\operatorname{Pr}\left(X_{1}=1 \mid \theta=0.1\right) \operatorname{Pr}(\theta=0.1)}{\operatorname{Pr}\left(X_{1}=1 \mid \theta=0.1\right) \operatorname{Pr}(\theta=0.1)+\operatorname{Pr}\left(X_{1}=1 \mid \theta=0.3\right) \operatorname{Pr}(\theta=0.3)}\)
\(=\frac{0.1(0.8)}{0.1(0.8)+0.3(0.2)}=\frac{4}{7}\)
Then,
\(E\left(X_{2} \mid \theta=0.1\right)=0(0.2)+1(0.1)+2(0.7)=1.5\)
\(E\left(X_{2} \mid \theta=0.3\right)=0(0.6)+1(0.3)+2(0.1)=0.5\)
\(E\left(X_{2} \mid X_{1}=1\right)=(1.5)(4 / 7)+(0.5)(3 / 7)=1.071\)
Question: For a portfolio of independent risks, the number of claims for each risk in a year follows a Poisson distribution with means given in the following table:

\begin{tabular}{|c|c|c|}
\hline Class & \begin{tabular}{c} 
Mean Number of \\
Claims per Risk
\end{tabular} & Number of Risks \\
\hline 1 & 1 & 900 \\
\hline 2 & 10 & 90 \\
\hline 3 & 20 & 10 \\
\hline
\end{tabular}

You observe \(x\) claims in Year 1 for a randomly selected risk.
The Bühlmann credibility estimate of the number of claims for the same risk in Year 2 is 11.983.

Calculate \(x\).

Options:
A
13
B
14
C
15
D
16
E
17

Correct answer: B

Solution: \(\mu=E[E(X \mid \lambda)]=E(\lambda)=1(0.9)+10(0.09)+20(0.01)=2\)
\(E P V=v=E[\operatorname{Var}(X \mid \lambda)]=E(\lambda)=2\)
\(\operatorname{VHM}=a=\operatorname{Var}[E(X \mid \lambda)]=\operatorname{Var}(\lambda)=1(0.9)+100(0.09)+400(0.01)-2^{2}=9.9\)
\(Z=\frac{1}{1+2 / 9.9}=0.83193 ; \quad 11.983=0.83193 x+0.16807(2) ; \quad x=14\)
Question: You are given:
(i) A region is comprised of three territories. Claims experience for Year 1 is as follows:

\begin{tabular}{|c|c|c|}
\hline Territory & Number of Insureds & Number of Claims \\
\hline X & 10 & 4 \\
\hline Y & 20 & 5 \\
\hline Z & 30 & 3 \\
\hline
\end{tabular}
(ii) The number of claims for each insured each year has a Poisson distribution.
(iii) Each insured in a territory has the same expected claim frequency.
(iv) The number of insureds is constant over time for each territory.

Calculate the Bühlmann-Straub empirical Bayes estimate of the credibility factor \(Z\) for Territory A.

Options:
A
Less than 0.4
B
At least 0.4 , but less than 0.5
C
At least 0.5 , but less than 0.6
D
At least 0.6 , but less than 0.7
E
At least 0.7

Correct answer: A

Solution: \(\hat{\mu}=\bar{x}=12 / 60=0.2, E V P V=\hat{v}=\bar{x}=0.2\)
\(V H M=\hat{a}=\frac{10(0.4-0.2)^{2}+20(0.25-0.2)^{2}+30(0.1-0.2)^{2}-(3-1)(0.2)}{60-\frac{10^{2}+20^{2}+30^{2}}{60}}=0.009545\)
\(\hat{k}=20.9524 ; \quad Z=\frac{10}{10+20.9524}=0.323\)
Question: Losses for a warranty product follow the lognormal distribution with underlying normal mean and standard deviation of 5.6 and 0.75 respectively.

The following are four uniform \((0,1)\) random numbers:
\[
\begin{array}{llll}
0.6217 & 0.9941 & 0.8686 & 0.0485
\end{array}
\]

Using these numbers and the inversion method, calculate the average payment per loss for a contract with a deductible of 100 .

Options:
A
Less than 630
B
At least 630, but less than 680
C
At least 680, but less than 730
D
At least 730 , but less than 780
E
At least 780

Correct answer: A

Solution: \begin{tabular}{cclcc}
\(U\) & \(z\) & \multicolumn{1}{c}{\(x\)} & lognormal & with deductible \\
\hline 0.6217 & 0.31 & 5.8325 & 341.21 & 241.21 \\
0.9941 & 2.52 & 7.49 & 1790.05 & 1690.05 \\
0.8686 & 1.12 & 6.44 & 626.41 & 526.41 \\
0.0485 & -1.66 & 4.355 & 77.87 & 0 \\
\hline & & & Average & 614.42
\end{tabular}

The value of \(z\) is obtained by inversion from the standard normal table. That is, \(u=\operatorname{Pr}(Z \leq z)\). The value of \(x\) is obtained from \(x=0.75 z+5.6\). The lognormal value is obtained by exponentiating \(x\) and the final column applies the deductible.
Question: For a group of auto policyholders, you are given:
(i) The number of claims for each policyholder has a conditional Poisson distribution.
(ii) During Year 1, the following data are observed for 8000 policyholders:

\begin{tabular}{|c|c|}
\hline Number of Claims & Number of Policyholders \\
\hline 0 & 5000 \\
\hline 1 & 2100 \\
\hline 2 & 750 \\
\hline 3 & 100 \\
\hline 4 & 50 \\
\hline \(5+\) & 0 \\
\hline
\end{tabular}

A randomly selected policyholder had one claim in Year 1.
Calculate the semiparametric empirical Bayes estimate of the number of claims in Year 2 for the same policyholder.

Options:
A
Less than 0.15
B
At least 0.15 , but less than 0.30
C
At least 0.30 , but less than 0.45
D
At least 0.45 , but less than 0.60
E
At least 0.60

Correct answer: D

Solution: \(\bar{x}=\frac{5000(0)+2100(1)+750(2)+100(3)+50(4)}{8000}=0.5125\) and
\(s^{2}=\frac{5000(0.5125)^{2}+2100(0.4875)^{2}+750(1.4875)^{2}+100(2.4875)^{2}+50(3.4875)^{2}}{7999}=0.5874\).
Then, \(\hat{\mu}=\hat{v}=\bar{x}=0.5125\) and \(\hat{a}=s^{2}-\bar{x}=0.0749\). The credibility factor is
\(Z=\frac{1}{1+0.5125 / 0.0749}=0.1275\) and the estimate is \(0.1275(1)+0.8725(0.5125)=0.5747\).