BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: An insurer writes a large book of home warranty policies. You are given the following information regarding claims filed by insureds against these policies:
(i) A maximum of one claim may be filed per year.
(ii) The probability of a claim varies by insured, and the claims experience for each insured is independent of every other insured.
(iii) The probability of a claim for each insured remains constant over time.
(iv) The overall probability of a claim being filed by a randomly selected insured in a year is 0.10 .
(v) The variance of the individual insured claim probabilities is 0.01 .

An insured selected at random is found to have filed 0 claims over the past 10 years.
Calculate the Bühlmann credibility estimate for the expected number of claims the selected insured will file over the next 5 years.

Options:
A
\(\quad 0.04\)
B
0.08
C
0.17
D
0.22
E
0.25

Correct answer: D

Solution: The number of claims for each insured has a binomial distribution with \(n=1\) and \(q\) unknown. We have
\(\mu(q)=q, v(\theta)=q(1-q)\)
\(\mu=E(q)=0.1\) (see iv)
\(a=\operatorname{Var}(q)=E\left(q^{2}\right)-0.1^{2}=0.01\) (see v) \(\Rightarrow E\left(q^{2}\right)=0.02\)
\(v=E[q(1-q)]=E(q)-E\left(q^{2}\right)=0.1-0.02=0.08\)
\(k=v / a=8, Z=10 /(10+8)=5 / 9\)

Then the expected number of claims in the next one year is \((5 / 9)(0)+(4 / 9)(0.1)=2 / 45\) and the expected number of claims in the next five years is \(5(2 / 45)=2 / 9=0.22\).
Question: For a group of insureds, you are given:
(i) The amount of a claim is uniformly distributed but will not exceed a certain unknown limit \(\theta\).
(ii) The prior distribution of \(\theta\) is \(\pi(\theta)=\frac{500}{\theta^{2}}, \quad \theta>500\).
(iii) Two independent claims of 400 and 600 are observed.

Calculate the probability that the next claim will exceed 550 .

Options:
A
0.19
B
0.22
C
0.25
D
0.28
E
\(\quad 0.31\)

Correct answer: E

Solution: The model distribution is \(f(x \mid \theta)=1 / \theta, 0<x<\theta\). Then the posterior distribution is proportional to
\(\pi(\theta \mid 400,600) \propto \frac{1}{\theta} \frac{1}{\theta} \frac{500}{\theta^{2}} \propto \theta^{-4}, \theta>600\).
It is important to note the range. Being a product, the posterior density function is non-zero only when all three terms are non-zero. Because one of the observations was equal to 600 , the value of the parameter must be greater than 600 for the density function at 600 to be positive. Or, by general reasoning, posterior probability can only be assigned to possible parameter values. Having observed the value 600 we know that parameter values less than or equal to 600 are not possible.

The constant is obtained from \(\int_{600}^{\infty} \theta^{-4} d \theta=\frac{1}{3(600)^{3}}\) and thus the exact posterior density is \(\pi(\theta \mid 400,600)=3(600)^{3} \theta^{-4}, \theta>600\). The posterior probability of an observation exceeding 550 is
\(\operatorname{Pr}\left(X_{3}>550 \mid 400,600\right)=\int_{600}^{\infty} \operatorname{Pr}\left(X_{3}>550 \mid \theta\right) \pi(\theta \mid 400,600) d \theta=\int_{600}^{\infty} \frac{\theta-550}{\theta} 3(600)^{3} \theta^{-4} d \theta=0.3125\).
where the first term in the integrand is the probability of exceeding 550 from the uniform distribution.
Question: You are given the following information about a book of business comprised of 100 insureds:
(i) \(\quad X_{i}=\sum_{j=1}^{N_{i}} Y_{i j}\) is a random variable representing the annual loss of the \(i\) th insured.
(ii) \(\quad N_{1}, N_{2}, \ldots, N_{100}\) are independent random variables distributed according to a negative binomial distribution with parameters \(r\) (unknown) and \(\beta=0.2\).
(iii) The unknown parameter \(r\) has an exponential distribution with mean 2.
(iv) \(Y_{i 1}, Y_{i 2}, \ldots, Y_{i N_{i}}\) are independent random variables distributed according to a Pareto distribution with \(\alpha=3.0\) and \(\theta=1000\).

Calculate the Bühlmann credibility factor, \(Z\), for the book of business.

Options:
A
0.000
B
0.045
C
0.500
D
0.826
E
0.905

Correct answer: E

Solution: \(\mu(r)=E(X \mid r)=E(N) E(Y)=r \beta \theta /(\alpha-1)=100 r\)
\(v(r)=\operatorname{Var}(X \mid r)=\operatorname{Var}(N) E(Y)^{2}+E(N) \operatorname{Var}(Y)\)
\(=\frac{r \beta(1+\beta) \theta^{2}}{(\alpha-1)^{2}}+\frac{r \beta \alpha \theta^{2}}{(\alpha-1)^{2}(\alpha-2)}=210,000 r\)
\(v=E(210,000 r)=210,000(2)=420,000\)
\(a=\operatorname{Var}(100 r)=(100)^{2}(4)=40,000\)
\(k=v / a=10.5, Z=100 /(100+10.5)=0.905\)
Question: You are given the following information on claim frequency of automobile accidents for individual drivers:

\begin{tabular}{|c||c|c||c|c|}
\hline \multirow{2}{*}{} & \multicolumn{2}{c|}{ Business Use } & \multicolumn{2}{c|}{ Pleasure Use } \\
\cline { 2 - 5 } & \begin{tabular}{c} 
Expected \\
Claims
\end{tabular} & \begin{tabular}{c} 
Claim \\
Variance
\end{tabular} & \begin{tabular}{c} 
Expected \\
Claims
\end{tabular} & \begin{tabular}{c} 
Claim \\
Variance
\end{tabular} \\
\hline Rural & 1.0 & 0.5 & 1.5 & 0.8 \\
\hline Urban & 2.0 & 1.0 & 2.5 & 1.0 \\
\hline Total & 1.8 & 1.06 & 2.3 & 1.12 \\
\hline
\end{tabular}

You are also given:
(i) Each driver's claims experience is independent of every other driver's.
(ii) There are an equal number of business and pleasure use drivers.

Calculate the Bühlmann credibility factor for a single driver.

Options:
A
0.05
B
0.09
C
0.17
D
0.19
E
0.27

Correct answer: D

Solution: Because the total expected claims for business use is 1.8 , it must be that \(20 \%\) of business users are rural and \(80 \%\) are urban. Thus the unconditional probabilities of being businessrural and business-urban are 0.1 and 0.4 respectively. Similarly the probabilities of being pleasure-rural and pleasure-urban are also 0.1 and 0.4 respectively. Then,
\(\mu=0.1(1.0)+0.4(2.0)+0.1(1.5)+0.4(2.5)=2.05\)
\(v=0.1(0.5)+0.4(1.0)+0.1(0.8)+0.4(1.0)=0.93\)
\(a=0.1(1.0)^{2}+0.4(2.0)^{2}+0.1(1.5)^{2}+0.4(2.5)^{2}-2.05^{2}=0.2225\)
\(k=v / a=4.18, Z=1 /(1+4.18)=0.193\).
Question: You are given:
(i) The annual number of claims for each policyholder follows a Poisson distribution with mean \(\theta\).
(ii) The distribution of \(\theta\) across all policyholders has probability density function:
\[
f(\theta)=\theta e^{-\theta}, \quad \theta>0
\]
(iii) \(\quad \int_{0}^{\infty} \theta e^{-n \theta} d \theta=\frac{1}{n^{2}}\)

A randomly selected policyholder is known to have had at least one claim last year.

Calculate the posterior probability that this same policyholder will have at least one claim this year.

Options:
A
0.70
B
0.75
C
0.78
D
0.81
E
\(\quad 0.86\)

Correct answer: D

Solution: The posterior density is proportional to the product of the probability of the observed value and the prior density. Thus, \(\pi(\theta \mid N>0) \propto \operatorname{Pr}(N>0 \mid \theta) \pi(\theta)=\left(1-e^{-\theta}\right) \theta e^{-\theta}\).

The constant of proportionality is obtained from \(\int_{0}^{\infty} \theta e^{-\theta}-\theta e^{-2 \theta} d \theta=\frac{1}{1^{2}}-\frac{1}{2^{2}}=0.75\).
The posterior density is \(\pi(\theta \mid N>0)=(1 / 0.75)\left(\theta e^{-\theta}-\theta e^{-2 \theta}\right)\).
Then,
\[
\begin{aligned}
& \operatorname{Pr}\left(N_{2}>0 \mid N_{1}>0\right)=\int_{0}^{\infty} \operatorname{Pr}\left(N_{2}>0 \mid \theta\right) \pi\left(\theta \mid N_{1}>0\right) d \theta=\int_{0}^{\infty}\left(1-e^{-\theta}\right)(4 / 3)\left(\theta e^{-\theta}-\theta e^{-2 \theta}\right) d \theta \\
& =\frac{4}{3} \int_{0}^{\infty} \theta e^{-\theta}-2 \theta e^{-2 \theta}+\theta e^{-3 \theta} d \theta=\frac{4}{3}\left(\frac{1}{1^{2}}+\frac{2}{2^{2}}+\frac{1}{3^{2}}\right)=0.8148
\end{aligned}
\]
Question: You are given:
(i) Claim size, \(X\), has mean \(\mu\) and variance 500 .
(ii) The random variable \(\mu\) has a mean of 1000 and variance of 50 .
(iii) The following three claims were observed: 750, 1075, 2000

Calculate the expected size of the next claim using Bühlmann credibility.

Options:
A
1025
B
1063
C
1115
D
1181
E
1266

Correct answer: B

Solution: From item (ii), \(\mu=1000\) and \(a=50\). From item (i), \(v=500\). Therefore, \(k=v / a=10\) and \(Z=3 /(3+10)=3 / 13\). Also, \(\bar{X}=(750+1075+2000) / 3=1275\). Then \(P_{c}=(3 / 13)(1275)+(10 / 13)(1000)=1063.46\).
Question: The number of auto vandalism claims reported per month at Sunny Daze Insurance Company (SDIC) has mean 110 and variance 750. Individual losses have mean 1101 and standard deviation 70. The number of claims and the amounts of individual losses are independent.

Using the normal approximation, calculate the probability that SDIC's aggregate auto vandalism losses reported for a month will be less than 100,000 .

Options:
A
0.24
B
0.31
C
0.36
D
0.39
E
\(\quad 0.49\)

Correct answer: A

Solution: \(E(S)=E(N) E(X)=110(1,101)=121,110\)
\(\operatorname{Var}(S)=E(N) \operatorname{Var}(X)+E(X)^{2} \operatorname{Var}(N)=110\left(70^{2}\right)+1101^{2}(750)=909,689,750\)
\(\operatorname{StdDev}(S)=30,161\)
\(\operatorname{Pr}(S<100,000)=\operatorname{Pr}\left(Z<\frac{100,000-121,110}{30,161}=-0.70\right)=0.242\)
where \(Z\) has standard normal distribution.
Question: WidgetsRUs owns two factories. It buys insurance to protect itself against major repair costs. Profit equals revenues, less the sum of insurance premiums, retained major repair costs, and all other expenses. WidgetsRUs will pay a dividend equal to the profit, if it is positive.

You are given:
(i) Combined revenue for the two factories is 3 .
(ii) Major repair costs at the factories are independent.
(iii) The distribution of major repair costs for each factory is

\begin{tabular}{ccc}
\(k\) & & \(\operatorname{Prob}(k)\) \\
\cline { 1 - 1 } \cline { 3 - 3 } 0 & & 0.4 \\
1 & & 0.3 \\
2 & & 0.2 \\
3 & & 0.1
\end{tabular}
(iv) At each factory, the insurance policy pays the major repair costs in excess of that factory's ordinary deductible of 1 . The insurance premium is \(110 \%\) of the expected claims.
(v) All other expenses are \(15 \%\) of revenues.

Calculate the expected dividend.

Options:
A
0.43
B
0.47
C
0.51
D
0.55
E
0.59

Correct answer: E

Solution: Expected insurance benefits per factory \(=E\left[(X-1)_{+}\right]=0.2(1)+0.1(2)=0.4\)
Insurance premium \(=(1.1)(2\) factories \((0.4\) per factory \()=0.88\).
Let \(R=\) retained major repair costs, then
\(f_{R}(0)=0.4^{2}=0.16, f_{R}(1)=2(0.4)(0.6)=0.48, f_{S}(2)=0.6^{2}=0.36\)
Dividend \(=3-0.88-R-0.15(3)=1.67-R\), if positive.
Expected Dividend \(=0.16(1.67-0)+(0.48)(1.67-1)+0.36(0)=0.5888\)
Question: The number of accidents follows a Poisson distribution with mean 12 . Each accident generates 1,2 , or 3 claimants with probabilities \(1 / 2,1 / 3\), and \(1 / 6\), respectively.

Calculate the variance of the total number of claimants.

Options:
A
20
B
25
C
30
D
35
E
40

Correct answer: E

Solution: For a compound Poisson distribution, \(\operatorname{Var}(S)=E(N) E\left(X^{2}\right)\).
\(E\left(X^{2}\right)=(1 / 2)(1)+(1 / 3)(4)+(1 / 6)(9)=10 / 3\)
\(\operatorname{Var}(S)=12(10 / 3)=40\)

Alternatively, the total is the sum \(N_{1}+2 N_{2}+3 N_{3}\) where the \(N\) s are independent Poisson variables with means 6,4 , and 2 . The variance is \(6+4(4)+9(2)=40\).
Question: A claim count distribution can be expressed as a mixed Poisson distribution. The mean of the Poisson distribution is uniformly distributed over the interval \([0,5]\).

Calculate the probability that there are 2 or more claims.

Options:
A
\(\quad 0.61\)
B
\(\quad 0.66\)
C
0.71
D
0.76
E
\(\quad 0.81\)

Correct answer: A

Solution: \(P(0)=\frac{1}{5} \int_{0}^{5} e^{-\lambda} d \lambda=\left.\frac{1}{5}\left(-e^{-\lambda}\right)\right|_{0} ^{5}=\frac{1}{5}\left(1-e^{-5}\right)=0.1987\)
\(P(1)=\frac{1}{5} \int_{0}^{5} \lambda e^{-\lambda} d \lambda=\left.\frac{1}{5}\left(-\lambda e^{-\lambda}-\mathrm{e}^{-\lambda}\right)\right|_{0} ^{5}=\frac{1}{5}\left(1-6 e^{-5}\right)=0.1919\)
\(P(N \geq 2)=1-0.1987-0.1919=0.6094\)
Question: Total hospital claims for a health plan were previously modeled by a two-parameter Pareto distribution with \(\alpha=2\) and \(\theta=500\).

The health plan begins to provide financial incentives to physicians by paying a bonus of \(50 \%\) of the amount by which total hospital claims are less than 500 . No bonus is paid if total claims exceed 500.

Total hospital claims for the health plan are now modeled by a new Pareto distribution with \(\alpha=2\) and \(\theta=K\). The expected claims plus the expected bonus under the revised model equals expected claims under the previous model.

Calculate \(K\).

Options:
A
250
B
300
C
350
D
400
E
450

Correct answer: C

Solution: Expected claims under current distribution \(=500 /(2-1)=500\)
\(K=\) parameter of new distribution
\(X=\) claims
\(E(X)=K /(2-1)=K\)
\(E(\) claims + bonus \()=E\{X+0.5[500-(X \wedge 500)]\}=K+0.5\left[500-K\left(1-\frac{K}{500+K}\right)\right]=500\)
\((K+250-0.5 K)(500+K)+0.5 K^{2}=500(500+K)\)
\(0.5 K^{2}+500 K+125,000+0.5 K^{2}=250,000+500 K\)
\(K^{2}=125,000\)
\(K=\sqrt{125,000}=354\)
Question: Bob is a carnival operator of a game in which a player receives a prize worth \(W=2^{N}\) if the player has \(N\) successes, \(N=0,1,2,3, \ldots\) Bob models the probability of success for a player as follows:
(i) \(\quad N\) has a Poisson distribution with mean \(\Lambda\).
(ii) \(\quad \Lambda\) has a uniform distribution on the interval \((0,4)\).

Calculate \(E[W]\).

Options:
A
5
B
7
C
9
D
11
E
13

Correct answer: E

Solution: Begin with \(E(W)=E\left(2^{N}\right)=E\left[E\left(2^{N} \mid \Lambda\right)\right]\). The inner expectation is the probability generating function of Poisson distribution evaluated at \(2, P_{N}(2)=e^{\Lambda(2-1)}=e^{\Lambda}\). Then,
\(E\left(e^{\Lambda}\right)=\int_{0}^{4} e^{\lambda}(0.25) d \lambda=\left.0.25 e^{\lambda}\right|_{0} ^{4}=0.25\left(e^{4}-1\right)=13.4\).
If the pgf is not recognized, the inner expectation can be derived as
\(E\left(2^{N} \mid \Lambda\right)=\sum_{n=0}^{\infty} 2^{n} \frac{e^{-\Lambda} \Lambda^{n}}{n!}=\sum_{n=0}^{\infty} \frac{e^{-\Lambda}(2 \Lambda)^{n}}{n!}=e^{\Lambda} \sum_{n=0}^{\infty} \frac{e^{-2 \Lambda}(2 \Lambda)^{n}}{n!}=e^{\Lambda}\) noting that the sum is over the probabilities for a Poisson distribution with mean \(2 \Lambda\) and so must be 1 .
Question: You are given:
(i) The annual number of claims for an insured has probability function:
\[
p(x)=\binom{3}{x} q^{x}(1-q)^{3-x}, \quad x=0,1,2,3
\]
(ii) The prior density is \(\pi(q)=2 q, \quad 0<q<1\).

A randomly chosen insured has zero claims in Year 1.
Using Bühlmann credibility, calculate the estimate of the number of claims in Year 2 for the selected insured.

Options:
A
0.33
B
0.50
C
1.00
D
1.33
E
1.50

Correct answer: C

Solution: \(E(X \mid q)=3 q, \operatorname{Var}(X \mid q)=3 q(1-q)\)
\(\mu=E(3 q)=\int_{0}^{1} 3 q 2 q d q=\left.2 q^{3}\right|_{0} ^{1}=2\)
\(v=E[3 q(1-q)]=\int_{0}^{1} 3 q(1-q) 2 q d q=2 q^{3}-\left.1.5 q^{4}\right|_{0} ^{1}=0.5\)
\(a=\operatorname{Var}(3 q)=E\left(9 q^{2}\right)-\mu^{2}=\int_{0}^{1} 9 q^{2} 2 q d q-2^{2}=\left.4.5 q^{4}\right|_{0} ^{1}-4=4.5-4=0.5\)
\(k=v / a=0.5 / 0.5=1, Z=\frac{1}{1+1}=0.5\)
The estimate is \(0.5(0)+0.5(2)=1\).
Question: Members of three classes of insureds can have 0,1 or 2 claims, with the following probabilities:

\begin{tabular}{|c|c|c|c|}
\hline \multirow{2}{*}{ Class } & \multicolumn{3}{|c|}{ Number of Claims } \\
\cline { 2 - 4 } & 0 & 1 & 2 \\
\hline I & 0.9 & 0.0 & 0.1 \\
\hline II & 0.8 & 0.1 & 0.1 \\
\hline III & 0.7 & 0.2 & 0.1 \\
\hline
\end{tabular}

A class is chosen at random, and varying numbers of insureds from that class are observed over 2 years, as shown below:

\begin{tabular}{|c|c|c|}
\hline Year & Number of Insureds & Number of Claims \\
\hline 1 & 20 & 7 \\
\hline 2 & 30 & 10 \\
\hline
\end{tabular}

Calculate the Bühlmann-Straub credibility estimate of the number of claims in Year 3 for 35 insureds from the same class.

Options:
A
10.6
B
10.9
C
11.1
D
11.4
E
11.6

Correct answer: C

Solution: Let \(X\) be the number of claims.
\(E(X \mid I)=0.9(0)+0.1(2)=0.2\)
\(E(X \mid I I)=0.8(0)+0.1(1)+0.1(2)=0.3\)
\(E(X \mid I I I)=0.7(0)+0.2(1)+0.1(2)=0.4\)
\(\operatorname{Var}(X \mid I)=0.9(0)+0.1(4)-0.2^{2}=0.36\)
\(\operatorname{Var}(X \mid I I)=0.8(0)+0.1(1)+0.1(4)-0.3^{2}=0.41\)
\(\operatorname{Var}(X \mid I I I)=0.7(0)+0.2(1)+0.1(4)-0.4^{2}=0.44\).
\(\mu=(1 / 2)(0.2+0.3+0.4)=0.3\)
\(v=(1 / 3)(0.36+0.41+0.44)=0.403333\)
\(a=(1 / 3)\left(0.2^{2}+0.3^{2}+0.4^{2}\right)-0.3^{2}=0.006667\)
\(k=0.403333 / 0.006667=60.5\)
\(Z=\frac{50}{50+60.5}=0.45249\).
For one insured the estimate is \(0.45249(17 / 50)+0.54751(0.3)=0.31810\).
For 35 insureds the estimate is \(35(0.31810)=11.13\).
Question: You are given:
(i) The number of claims observed in a 1 -year period has a Poisson distribution with mean \(\theta\).
(ii) The prior density is:
\[
\pi(\theta)=\frac{e^{-\theta}}{1-e^{-k}}, \quad 0<\theta<k
\]
(iii) The unconditional probability of observing zero claims in 1 year is 0.575 . Calculate \(k\).

Options:
A
1.5
B
1.7
C
1.9
D
2.1
E
2.3

Correct answer: C

Solution: \[
\begin{aligned}
& 0.575=\operatorname{Pr}(N=0)=\int_{0}^{k} \operatorname{Pr}(N=0 \mid \theta) \pi(\theta) d \theta \\
& =\int_{0}^{k} e^{-\theta} \frac{e^{-\theta}}{1-e^{-k}} d \theta=-\left.\frac{e^{-2 \theta}}{2\left(1-e^{-k}\right)}\right|_{0} ^{k}=-\frac{e^{-2 k}}{2\left(1-e^{-k}\right)}+\frac{1}{2\left(1-e^{-k}\right)} \\
& =\frac{1-e^{-2 k}}{2\left(1-e^{-k}\right)}=\frac{1+e^{-k}}{2} \\
& e^{-k}=2(0.575)-1=0.15 \\
& k=1.90
\end{aligned}
\]
Question: You are given:
(i) In a portfolio of risks, each policyholder can have at most one claim per year.
(ii) The probability of a claim for a policyholder during a year is \(q\).
(iii) The prior density is \(\pi(q)=\frac{q^{3}}{0.07}, \quad 0.6<q<0.8\)

A randomly selected policyholder has one claim in Year 1 and zero claims in Year 2.

For this policyholder, calculate the posterior probability that \(0.7<q<0.8\).

Options:
A
Less than 0.3
B
At least 0.3 , but less than 0.4
C
At least 0.4 , but less than 0.5
D
At least 0.5 , but less than 0.6
E
At least 0.6

Correct answer: D

Solution: The posterior density function is proportional to the product of the likelihood function and prior density. That is, \(\pi(q \mid 1,0) \propto f(1 \mid q) f(0 \mid q) \pi(q) \propto q(1-q) q^{3}=q^{4}-q^{5}\). To get the exact posterior density, integrate this function over its range:
\(\int_{0.6}^{0.8} q^{4}-q^{5} d q=\frac{q^{5}}{5}-\left.\frac{q^{6}}{6}\right|_{0.6} ^{0.8}=0.014069\) and so \(\pi(q \mid 1,0)=\frac{q^{4}-q^{5}}{0.014069}\). Then,
\(\operatorname{Pr}(0.7<q<0.8 \mid 1,0)=\int_{0.7}^{0.8} \frac{q^{4}-q^{5}}{0.014069} d q=0.5572\).
Question: You are given a random sample of observations:
\[
\begin{array}{ccccc}
0.1 & 0.2 & 0.5 & 0.7 & 1.3
\end{array}
\]

You test the hypothesis that the probability density function is:
\[
f(x)=\frac{4}{(1+x)^{5}}, \quad x>0
\]

Calculate the Kolmogorov-Smirnov test statistic.

Options:
A
Less than 0.05
B
At least 0.05 , but less than 0.15
C
At least 0.15 , but less than 0.25
D
At least 0.25 , but less than 0.35
E
At least 0.35

Correct answer: E

Solution: The cdf is \(F(x)=\int_{0}^{x} 4(1+t)^{-5} d t=-\left.(1+t)^{-4}\right|_{0} ^{x}=1-\frac{1}{(1+x)^{4}}\).

\begin{tabular}{|c|c|c|c|}
\hline Observation \((x)\) & \(F(x)\) & compare to: & Maximum difference \\
\hline 0.1 & 0.317 & \(0,0.2\) & 0.317 \\
\hline 0.2 & 0.518 & \(0.2,0.4\) & 0.318 \\
\hline 0.5 & 0.802 & \(0.4,0.6\) & 0.402 \\
\hline 0.7 & 0.880 & \(0.6,0.8\) & 0.280 \\
\hline 1.3 & 0.964 & \(0.8,1.0\) & 0.164 \\
\hline
\end{tabular}

K-S statistic is 0.402 .
Question: A loss, \(X\), follows a 2 -parameter Pareto distribution with \(\alpha=2\) and unspecified parameter \(\theta\). You are given:
\[
E[X-100 \mid X>100]=\frac{5}{3} E[X-50 \mid X>50]
\]

Calculate \(E[X-150 \mid X>150]\).

Options:
A
150
B
175
C
200
D
225
E
250

Correct answer: B

Solution: \(E[X-d \mid X>d]\) is the expected payment per payment with an ordinary deductible of \(d\) It can be evaluated (for Pareto) as
\(\frac{E(X)-E(X \wedge d)}{1-F(d)}=\frac{\frac{\theta}{\alpha-1}-\frac{\theta}{\alpha-1}\left[1-\left(\frac{\theta}{d+\theta}\right)^{\alpha-1}\right]}{1-\left[1-\left(\frac{\theta}{d+\theta}\right)^{\alpha}\right]}=\frac{\frac{\theta}{\alpha-1}\left(\frac{\theta}{d+\theta}\right)^{\alpha-1}}{\left(\frac{\theta}{d+\theta}\right)^{\alpha}}=\frac{d+\theta}{\alpha-1}=d+\theta\)
\(E[X-100 \mid X>100]=(5 / 3) E[X-50 \mid \mathrm{X}>50]\)
\(100+\theta=(5 / 3)(50+\theta)\)
\(300+3 \theta=250+5 \theta, 50=2 \theta, \theta=25\)
\(E[X-150 \mid X>150]=150+\theta=150+25=175\)
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
\(\quad 0.08\)
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) Annual claim frequencies follow a Poisson distribution with mean \(\lambda\).
(ii) The prior distribution of \(\lambda\) has probability density function:
\[
\pi(\lambda)=(0.4) \frac{1}{6} e^{-\lambda / 6}+(0.6) \frac{1}{12} e^{-\lambda / 12}, \quad \lambda>0
\]

Ten claims are observed for an insured in Year 1.
Calculate the Bayesian expected number of claims for the insured in Year 2.

Options:
A
\(\quad 9.6\)
B
9.7
C
9.8
D
\(\quad 9.9\)
E
10.0

Correct answer: D

Solution: The posterior distribution can be found from
\(\pi(\lambda \mid 10) \propto \frac{e^{-\lambda} \lambda^{10}}{10!}\left(\frac{0.4}{6} e^{-\lambda / 6}+\frac{0.6}{12} e^{-\lambda / 12}\right) \propto \lambda^{10}\left(0.8 e^{-7 \lambda / 6}+0.6 e^{-13 \lambda / 12}\right)\).
The required constant is
\(\int_{0}^{\infty} \lambda^{10}\left(0.8 e^{-7 \lambda / 6}+0.6 e^{-13 \lambda / 12}\right) d \lambda=0.8(10!)(6 / 7)^{11}+0.6(10!)(12 / 13)^{11}=0.395536(10!)\).
The posterior mean is
\[
\begin{gathered}
E(\lambda \mid 10)=\frac{1}{0.395536(10!)} \int_{0}^{\infty} \lambda^{11}\left(0.8 e^{-7 \lambda / 6}+0.6 e^{-13 \lambda / 12}\right) d \lambda \\
=\frac{0.8(11!)(6 / 7)^{12}+0.6(11!)(12 / 13)^{12}}{0.395536(10!)}=9.88
\end{gathered}
\]