BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: An insurer offers a three-year contract with a guaranteed minimum of the policyholder's original investment.

You are given:
i) The policyholder invests a premium of 1000.
ii) The insurer keeps \(2.5 \%\) of the premium to cover expenses, then invests the remainder in a mutual fund.
iii) The mutual fund investment value follows the lognormal distribution with parameters \(\mu=0.075\) and \(\sigma=0.15\) per year.
iv) The continuously compounded risk-free rate of interest is \(6.5 \%\).
v) The mutual fund does not pay dividends.

Calculate the probability that the insurer's guarantee applies at the maturity date.

Options:
A
0.165
B
0.181
C
0.193
D
0.221
E
0.257

Correct answer: D

Solution: \(S_{0}=1000(1-0.025)=975\)
\(S_{3} / S_{0} \sim L N\left(3 \mu, 3^{0.5} \sigma\right), \ln \left(S_{3} / S_{0}\right) \sim N\left(3 \times 0.075=0.225,3 \times 0.15^{2}\right)\)
Therefore, \(\ln \left(S_{3}\right) \sim N[0.225+\ln (975)=7.10744,0.0675]\)
\(\operatorname{Pr}\left(S_{3}<1000\right)=\operatorname{Pr}\left[\ln \left(S_{3}\right)<\ln (1000)=6.90776\right]\)
\(=\operatorname{Pr}\left[Z<\frac{6.90776-7.10744}{\sqrt{0.0675}}=-0.76857\right]=0.221\)
Question: Losses follow a lognormal distribution with \(\mu=7\) and \(\sigma=2\).

A policy covering these losses makes no payment for losses below 5,000. For losses between 5,000 and 20,000, the policy pays the full amount of the loss. For losses exceeding 20,000, the policy pays 20,000.

Calculate the expected cost per loss.

Options:
A
\(\quad 2,961\)
B
3,109
C
3,273
D
3,307
E
\(\quad 3,583\)

Correct answer: A

Solution: This policy has three components, an ordinary deductible of 5,000 , a maximum covered loss of 20,000 , and a bonus payment of 5,000 whenever there is a payment. The expected cost per loss is
\(E(X \wedge 20,000)-E(X \wedge 5,000)+5,000[1-F(5,000)]\).
\(E(X \wedge x)=\exp \left(\mu+\sigma^{2} / 2\right) \Phi\left(\frac{\ln (x)-\mu-\sigma^{2}}{\sigma}\right)+x[1-F(x)]\)
\(E(X \wedge 20,000)=\exp \left(7+2^{2} / 2\right) \Phi\left(\frac{\ln (20,000)-7-2^{2}}{2}\right)+20,000\left[1-\Phi\left(\frac{\ln (20,000)-7}{2}\right)\right]=3,830\)
\(E(X \wedge 5,000)=\exp \left(7+2^{2} / 2\right) \Phi\left(\frac{\ln (5,000)-7-2^{2}}{2}\right)+5,000\left[1-\Phi\left(\frac{\ln (5,000)-7}{2}\right)\right]=1,989\)
\(5,000[1-F(5,000)]=5,000\left[1-\Phi\left(\frac{\ln (5,000)-7}{2}\right)\right]=1,120\)
\(E(X \wedge 20,000)-E(X \wedge 5,000)+5,000[1-F(5,000)]=3,830-1,989+1,120=2,961\)
Question: You are given the following information about AY2:
i) The age-to-age paid loss development factors are:

\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline \multicolumn{11}{|c|}{ Development Year Factors } \\
\hline \(1 / 0\) & \(2 / 1\) & \(3 / 2\) & \(4 / 3\) & \(5 / 4\) & \(6 / 5\) & \(7 / 6\) & \(8 / 7\) & \(9 / 8\) & \(10 / 9\) & \(\infty / 10\) \\
\hline 1.23 & 1.18 & 1.14 & 1.11 & 1.08 & 1.06 & 1.05 & 1.04 & 1.04 & 1.03 & \(r\) \\
\hline
\end{tabular}
ii) As of December 31, CY8, the estimated loss reserve is 12,200 based on the chainladder method.
iii) During CY9, actual paid loss development equals the expected development.
iv) As of December 31, CY9, the estimated loss reserve is 10,000 based on the chainladder method.

Calculate \(r\).

Options:
A
1.092
B
1.141
C
1.162
D
1.187
E
1.220

Correct answer: A

Solution: Let \(X\) represent the claims-paid to-date for AY2, as of December 31, CY8.
Let \(f_{n, m}\) represent the age-to-age (or age-to-ultimate) paid loss development factor from the end of development year \(m\) to the end of development year \(n\) (where \(n\) could be finite or infinite).

In the chain-ladder method, the ultimate loss equals the claims paid-to-date on a given date multiplied by the product of all the loss development factors after this date.

The estimated ultimate loss for AY2 is the claims paid-to-date as of December 31, CY8 (the end of development year 6) multiplied by the product of the loss development factors after the end of development year 6. Therefore, the estimated ultimate loss for AY2 is \(f_{7,6} f_{8,7} f_{9,8} f_{10,9} f_{\infty, 10} X\).

The estimated loss reserve as of any given date equals the estimated ultimate loss minus claims paid to date as of that date, so the estimated loss reserve as of December 31, CY8 is \(\left(f_{7,6} f_{8,7} f_{9,8} f_{10,9} f_{\infty, 10}-1\right) X\).

Since actual paid loss development equals the expected development during CY9, the AY2 estimated claims paid-to-date as of December 31, CY9 (the end of development year 7) is \(f_{7,6} X\). Therefore, the estimated loss reserve as of December 31, CY9 is
\(\left(f_{7,6} f_{8,7} f_{9,8} f_{10,9} f_{\infty, 10}-f_{7,6}\right) X\).
Since the estimated loss reserve as of December 31, CY8 is 12,200 and the estimated loss reserve as of December 31, CY9 is \(10,000, \frac{f_{7,6} f_{8,7} f_{9,8} f_{10,9} f_{\infty, 10}-1}{f_{7,6} f_{8,7} f_{9,8} f_{10,9} f_{\infty, 10}-f_{7,6}}=\frac{12,200}{10,000}\)
which gives \(\frac{1.05(1.04)(1.04)(1.03) r-1}{1.05(1.04)(1.04)(1.03) r-1.05}=1.22\)
which solves for \(r=\frac{1.05(1.22)-1}{1.05(1.04)(1.04)(1.03)(1.22-1)}=1.092\).
Question: You are given the following information to calculate a new average rate, which will be in effect for one year:
i) The current average rate is 300 , which includes an allowance for the following variable expenses:

\begin{tabular}{|c|c|}
\hline Commissions & \(10 \%\) \\
\hline General Expenses & \(12 \%\) \\
\hline Taxes & \(2 \%\) \\
\hline Profit and Contingencies & \(5 \%\) \\
\hline TOTAL & \(29 \%\) \\
\hline
\end{tabular}
ii) The indicated rate change uses two years of data, weighted as shown:

\begin{tabular}{|c|c|c|c|}
\hline \begin{tabular}{c} 
Accident \\
Year
\end{tabular} & \begin{tabular}{c} 
Earned Premium \\
at Current Rates
\end{tabular} & \begin{tabular}{c} 
Ultimate Loss \\
and ALAE
\end{tabular} & Weight \\
\hline AY1 & 300,000 & 190,000 & 0.4 \\
\hline AY2 & 330,000 & 230,000 & 0.6 \\
\hline
\end{tabular}
iii) Annual effective loss trend: \(5 \%\)
iv) Effective date of the new rate (one-year policies): October 1, CY3
v) New legislation, also effective October 1, CY3, mandates that commissions be a fixed rate of 30 per exposure, which gets loaded for general expenses, taxes, and profit and contingencies.

Calculate the new average rate that combines the indicated rate change and new mandated commission level.

Options:
A
313
B
316
C
320
D
323
E
327

Correct answer: C

Solution: Midpoint of experience period is July 1 of each AY, and midpoint of effective period is October 1, CY4 (e.g., 2.25 years for AY2).

\begin{tabular}{|c|c|c|c|c|c|c|}
\hline \begin{tabular}{c} 
Accident \\
Year
\end{tabular} & \begin{tabular}{c} 
EP at \\
Current \\
Rates
\end{tabular} & \begin{tabular}{c} 
Ult. \\
Loss \& \\
ALAE
\end{tabular} & Trend Factor & \begin{tabular}{c} 
Trended Ult \\
Loss \& ALAE
\end{tabular} & \begin{tabular}{c} 
Loss \& \\
ALAE \\
Ratio
\end{tabular} & Weight \\
\hline AY1 & 300,000 & 190,000 & \(1.05^{3.25}=1.17183\) & 222,648 & 0.74216 & 0.4 \\
\hline AY2 & 330,000 & 230,000 & \(1.05^{2.25}=1.11603\) & 256,687 & 0.77784 & 0.6 \\
\hline
\end{tabular}

Weighted Ratio: \(0.74216 \times 0.4+0.77784 \times 0.6=0.76357\)
The new variable expenses + profit (excludes commission) \(=0.12+0.02+0.05=0.19\)
New commission provision (as a fixed expense ratio to premium) \(=30 / 300\)

Indicated Rate Change Factor \(=\frac{0.76357+\frac{30}{300}}{1-0.19}=1.066136\)
New Rate \(=300(1.066136)=319.84\)
Question: You are asked to consider whether the risk measure \(\rho(X)=E(X)\) is coherent.

Determine which of the following statements is correct.

Options:
A
\(\quad \rho(X)\) does NOT possess subadditivity.
B
\(\quad \rho(X)\) does NOT possess monotonicity.
C
\(\quad \rho(X)\) does NOT possess positive homogeneity.
D
\(\quad \rho(X)\) does NOT possess translation invariance.
E
\(\quad \rho(X)\) is a coherent risk measure.

Correct answer: E

Solution: Subadditivity holds: \(\rho(X+Y)=E(X+Y)=E(X)+E(Y)=\rho(X)+\rho(Y)\)
Monotonicity holds: If \(X \leq Y\), then \(\rho(X)=E(X) \leq \rho(Y)=E(Y)\)
Positive homogeneity holds: \(\rho(c X)=E(c X)=c E(X)=c \rho(X)\)
Translation invariance holds: \(\rho(X+c)=E(X+c)=E(X)+c=\rho(X)+c\)
Since \(\rho(X)=E(X)\) satisfies all four properties, it is coherent.
Question: The number of claims follows a negative binomial distribution with parameters \(\beta\) and \(r\), where \(\beta\) is unknown and \(r\) is known. You wish to estimate \(\beta\) based on \(n\) observations, where \(\bar{x}\) is the mean of these observations.

Determine the maximum likelihood estimate of \(\beta\).

Options:
A
\(\bar{x} / r^{2}\)
B
\(\bar{x} / r\)
C
\(\bar{x}\)
D
\(r \bar{x}\)
E
\(\quad r^{2} \bar{x}\)

Correct answer: B

Solution: The likelihood is:
\(L=\prod_{j=1}^{n} \frac{r(r+1) \cdots\left(r+x_{j}-1\right) \beta^{x_{j}}}{x_{j}!(1+\beta)^{r+x_{j}}} \propto \prod_{j=1}^{n} \beta^{x_{j}}(1+\beta)^{-r-x_{j}}\).
The loglikelihood is:
\(l=\sum_{j=1}^{n}\left[x_{j} \ln \beta-\left(r+x_{j}\right) \ln (1+\beta)\right]\)
\(l^{\prime}=\sum_{j=1}^{n}\left[\frac{x_{j}}{\beta}-\frac{r+x_{j}}{1+\beta}\right]=0\)
\(0=\sum_{j=1}^{n}\left[x_{j}(1+\beta)-\left(r+x_{j}\right) \beta\right]=\sum_{j=1}^{n} x_{j}-r n \beta=n \bar{x}-r n \beta\)
\(\hat{\beta}=\bar{x} / r\).
Question: Aggregate losses for a portfolio of policies are modeled as follows:
(i) The number of losses before any coverage modifications follows a Poisson distribution with mean \(\lambda\).
(ii) The severity of each loss before any coverage modifications is uniformly distributed between 0 and \(b\).

The insurer would like to model the effect of imposing an ordinary deductible, \(d\) ( \(0<d<b\) ), on each loss and reimbursing only a percentage, \(c(0<c \leq 1)\), of each loss in excess of the deductible.

It is assumed that the coverage modifications will not affect the loss distribution.
The insurer models its claims with modified frequency and severity distributions. The modified claim amount is uniformly distributed on the interval \([0, c(b-d)]\).

Determine the mean of the modified frequency distribution.

Options:
A
\(\lambda\)
B
\(\quad \lambda c\)
C
\(\lambda \frac{d}{b}\)
D
\(\lambda \frac{b-d}{b}\)
E
\(\quad \lambda c \frac{b-d}{b}\)

Correct answer: D

Solution: The modified severity, \(X^{*}\), represents the conditional payment amount given that a payment occurs. Given that a payment is required ( \(X>d\) ), the payment must be uniformly distributed between 0 and \(c(b-d)\).

The modified frequency, \(N^{*}\), represents the number of losses that result in a payment. The deductible eliminates payments for losses below \(d\), so only \(1-F_{X}(d)=\frac{b-d}{b}\) of losses will require payments. Therefore, the Poisson parameter for the modified frequency distribution is \(\lambda \frac{b-d}{b}\). (Reimbursing \(c \%\) after the deductible affects only the payment amount and not the frequency of payments).
Question: Losses come from a mixture of an exponential distribution with mean 100 with probability \(p\) and an exponential distribution with mean 10,000 with probability \(1-p\).

Losses of 100 and 2000 are observed.
Determine the likelihood function of \(p\).

Options:
A
\(\quad\left(\frac{p e^{-1}}{100} \cdot \frac{(1-p) e^{-0.01}}{10,000}\right) \cdot\left(\frac{p e^{-20}}{100} \cdot \frac{(1-p) e^{-0.2}}{10,000}\right)\)
B
\(\quad\left(\frac{p e^{-1}}{100} \cdot \frac{(1-p) e^{-0.01}}{10,000}\right)+\left(\frac{p e^{-20}}{100} \cdot \frac{(1-p) e^{-0.2}}{10,000}\right)\)
C
\(\quad\left(\frac{p e^{-1}}{100}+\frac{(1-p) e^{-0.01}}{10,000}\right) \cdot\left(\frac{p e^{-20}}{100}+\frac{(1-p) e^{-0.2}}{10,000}\right)\)
D
\(\quad\left(\frac{p e^{-1}}{100}+\frac{(1-p) e^{-0.01}}{10,000}\right)+\left(\frac{p e^{-20}}{100}+\frac{(1-p) e^{-0.2}}{10,000}\right)\)
E
\(\quad p\left(\frac{e^{-1}}{100}+\frac{e^{-0.01}}{10,000}\right)+(1-p)\left(\frac{e^{-20}}{100}+\frac{e^{-0.2}}{10,000}\right)\)

Correct answer: C

Solution: \[
\begin{aligned}
f(x) & =p \frac{1}{100} e^{-x / 100}+(1-p) \frac{1}{10,000} e^{-x / 10,000} \\
L(100,200) & =f(100) f(2000) \\
& =\left(\frac{p e^{-1}}{100}+\frac{(1-p) e^{-0.01}}{10,000}\right)\left(\frac{p e^{-20}}{100}+\frac{(1-p) e^{-0.2}}{10,000}\right)
\end{aligned}
\]
Question: You are given:
(i) The annual number of claims for a policyholder has a binomial distribution with probability function:
\[
p(x \mid q)=\binom{2}{x} q^{x}(1-q)^{2-x}, \quad x=0,1,2
\]
(ii) The prior distribution is:
\[
\pi(q)=4 q^{3}, \quad 0<q<1
\]

This policyholder had one claim in each of Years 1 and 2.
Calculate the Bayesian estimate of the number of claims in Year 3.

Options:
A
Less than 1.1
B
At least 1.1, but less than 1.3
C
At least 1.3, but less than 1.5
D
At least 1.5 , but less than 1.7
E
At least 1.7

Correct answer: C

Solution: \(\pi(q \mid 1,1) \propto p(1 \mid q) p(1 \mid q) \pi(q)=2 q(1-q) 2 q(1-q) 4 q^{3} \propto q^{5}(1-q)^{2}\)
\(\int_{0}^{1} q^{5}(1-q)^{2} d q=1 / 168, \pi(q \mid 1,1)=168 q^{5}(1-q)^{2}\).
The expected number of claims in a year is \(E(X \mid q)=2 q\) and so the Bayesian estimate is \(E(2 q \mid 1,1)=\int_{0}^{1} 2 q(168) q^{5}(1-q)^{2} d q=4 / 3\).
The answer can be obtained without integrals by recognizing that the posterior distribution of \(q\) is beta with \(a=6\) and \(b=3\). The posterior mean is \(E(q \mid 1,1)=a /(a+b)=6 / 9=2 / 3\) The posterior mean of \(2 q\) is then \(4 / 3\).
Question: A particular line of business has three types of claim. The historical probability and the number of claims for each type in the current year are:

\begin{tabular}{|c|c|c|}
\hline Type & \begin{tabular}{c} 
Historical \\
Probability
\end{tabular} & \begin{tabular}{c} 
Number of Claims \\
in Current Year
\end{tabular} \\
\hline X & 0.2744 & 112 \\
Y & 0.3512 & 180 \\
Z & 0.3744 & 138 \\
\hline
\end{tabular}

You test the null hypothesis that the probability of each type of claim in the current year is the same as the historical probability.

Calculate the chi-square goodness-of-fit test statistic.

Options:
A
Less than 9
B
At least 9 , but less than 10
C
At least 10 , but less than 11
D
At least 11 , but less than 12
E
At least 12

Correct answer: B

Solution: There are 430 observations. The expected counts are \(430(0.2744)=117.99,430(0.3512)=\) 151.02 , and \(430(0.3744)=160.99\). The test statistic is
\(\frac{(112-117.99)^{2}}{117.99}+\frac{(180-151.02)^{2}}{151.02}+\frac{(138-160.99)^{2}}{160.99}=9.15\).
Question: You are given:
(i) The probability that an insured will have at least one loss during any year is \(p\).
(ii) The prior distribution for \(p\) is uniform on \([0,0.5]\).
(iii) An insured is observed for 8 years and has at least one loss every year.

Calculate the posterior probability that the insured will have at least one loss during Year 9.

Options:
A
0.450
B
0.475
C
0.500
D
0.550
E
0.625

Correct answer: A

Solution: \(\pi(p \mid 1,1,1,1,1,1,1,1) \propto \operatorname{Pr}(1,1,1,1,1,1,1,1 \mid p) \pi(p) \propto p^{8}\)
\(\pi(p \mid 1,1,1,1,1,1,1,1)=\frac{p^{8}}{\int_{0}^{0.5} p^{8} d p}=\frac{p^{8}}{0.5^{9} / 9}=9\left(0.5^{-9}\right) p^{8}\)
\(\operatorname{Pr}\left(X_{9}=1 \mid 1,1,1,1,1,1,1,1\right)=\int_{0}^{0.5} \operatorname{Pr}\left(X_{9}=1 \mid p\right) \pi(p \mid 1,1,1,1,1,1,1,1) d p\)
\(=\int_{0}^{0.5} p 9\left(0.5^{-9}\right) p^{8} d p=9\left(0.5^{-9}\right)\left(0.5^{10}\right) / 10=0.45\).
Question: You are given:
(i) The probability that an insured will have exactly one claim is \(\theta\).
(ii) The prior distribution of \(\theta\) has probability density function:
\[
\pi(\theta)=\frac{3}{2} \sqrt{\theta}, \quad 0<\theta<1
\]

A randomly chosen insured is observed to have exactly one claim.
Calculate the posterior probability that \(\theta\) is greater than 0.60 .

Options:
A
0.54
B
0.58
C
0.63
D
0.67
E
0.72

Correct answer: E

Solution: \(\pi(\theta \mid 1) \propto \theta\left(1.5 \theta^{0.5}\right) \propto \theta^{1.5}\). The required constant is the reciprocal of \(\int_{0}^{1} \theta^{1.5} d \theta=0.4\) and so \(\pi(\theta \mid 1)=2.5 \theta^{1.5}\). The requested probability is
\(\operatorname{Pr}(\theta>0.6 \mid 1)=\int_{0.6}^{1} 2.5 \theta^{1.5} d \theta=\left.\theta^{2.5}\right|_{0.6} ^{1}=1-0.6^{2.5}=0.721\).
Question: You are given:
(i) Each risk has at most one claim each year.
(ii)

\begin{tabular}{|c|c|c|}
\hline Type of Risk & Prior Probability & \begin{tabular}{c} 
Annual Claim \\
Probability
\end{tabular} \\
\hline I & 0.7 & 0.1 \\
\hline II & 0.2 & 0.2 \\
\hline III & 0.1 & 0.4 \\
\hline
\end{tabular}

One randomly chosen risk has three claims during Years 1-6.
Calculate the posterior probability of a claim for this risk in Year 7.

Options:
A
0.22
B
0.28
C
0.33
D
\(\quad 0.40\)
E
0.46

Correct answer: B

Solution: The probabilities are from a binomial distribution with 6 trials. Three successes were observed.
\[
\begin{aligned}
& \operatorname{Pr}(3 \mid I)=\binom{6}{3}(0.1)^{3}(0.9)^{3}=0.01458 \\
& \operatorname{Pr}(3 \mid I I)=\binom{6}{3}(0.2)^{3}(0.8)^{3}=0.08192 \\
& \operatorname{Pr}(3 \mid I I I)=\binom{6}{3}(0.4)^{3}(0.6)^{3}=0.27648
\end{aligned}
\]

The probability of observing three successes is \(0.7(.01458)+0.2(.08192)+0.1(.27648)=\) 0.054238 . The three posterior probabilities are:
\[
\begin{aligned}
& \operatorname{Pr}(I \mid 3)=\frac{0.7(0.01458)}{0.054238}=0.18817 \\
& \operatorname{Pr}(I I \mid 3)=\frac{0.2(0.08192)}{0.054238}=0.30208 \\
& \operatorname{Pr}(I I I \mid 3)=\frac{0.1(0.27648)}{0.054238}=0.50975
\end{aligned}
\]

The posterior probability of a claim is then \(0.1(0.18817)+0.2(0.30208)+0.4(0.50975)=\) 0.28313 .
Question: You are given:
(i) The number of claims made by an individual insured in a year has a Poisson distribution with mean \(\lambda\).
(ii) The prior distribution for \(\lambda\) is gamma with parameters \(\alpha=1\) and \(\theta=1.2\).

Three claims are observed in Year 1, and no claims are observed in Year 2.
Using Bühlmann credibility, estimate the number of claims in Year 3.

Options:
A
1.35
B
1.36
C
1.40
D
1.41
E
1.43

Correct answer: D

Solution: \(N\) is distributed Poisson( \(\lambda\) )
\(\mu=E(\lambda)=\alpha \theta=1(1.2)=1.2\)
\(v=E(\lambda)=1.2, a=\operatorname{Var}(\lambda)=\alpha \theta^{2}=1(1.2)^{2}=1.44\)
\(k=\frac{1.2}{1.44}=\frac{5}{6}, Z=\frac{2}{2+5 / 6}=\frac{12}{17}\)
Thus, the estimate for Year 3 is \(\frac{12}{17}(1.5)+\frac{5}{17}(1.2)=1.41\).
Note that a Bayesian approach produces the same answer.
Question: You are given the following information about a credibility model:

\begin{tabular}{|c|c|c|}
\hline First Observation & Unconditional Probability & \begin{tabular}{c} 
Bayesian Estimate of \\
Second Observation
\end{tabular} \\
\hline 1 & \(1 / 3\) & 1.50 \\
\hline 2 & \(1 / 3\) & 1.50 \\
\hline 3 & \(1 / 3\) & 3.00 \\
\hline
\end{tabular}

Calculate the Bühlmann credibility estimate of the second observation, given that the first observation is 1 .

Options:
A
0.75
B
1.00
C
1.25
D
1.50
E
1.75

Correct answer: C

Solution: The Bühlmann credibility estimate is \(Z x+(1-Z) \mu\) where \(x\) is the first observation. The Bühlmann estimate is the least squares approximation to the Bayesian estimate. Therefore, \(Z\) and \(\mu\) must be selected to minimize
\(\frac{1}{3}[Z+(1-Z) \mu-1.5]^{2}+\frac{1}{3}[2 Z+(1-Z) \mu-1.5]^{2}+\frac{1}{3}[3 Z+(1-Z) \mu-3]^{2}\).
Setting partial derivatives equal to zero will give the values. However, it should be clear that \(\mu\) is the average of the Bayesian estimates, that is,
\(\mu=\frac{1}{3}(1.5+1.5+3)=2\).
The derivative with respect to \(Z\) is (deleting the coefficients of \(1 / 3\) ):
\(2(-Z+0.5)(-1)+2(0.5)(0)+2(Z-1)(1)=0\)
\(4 Z-3=0, Z=0.75\).
The answer is \(0.75(1)+0.25(2)=1.25\).
Question: You are given:
(i) A sample of claim payments is: \(\begin{array}{lllll}29 & 64 & 90 & 135 & 182\end{array}\)
(ii) Claim sizes are assumed to follow an exponential distribution.
(iii) The mean of the exponential distribution is estimated using the method of moments.

Calculate the value of the Kolmogorov-Smirnov test statistic.

Options:
A
0.14
B
0.16
C
0.19
D
0.25
E
0.27

Correct answer: E

Solution: \begin{tabular}{|c|c|c|c|c|c|}
\hline\(X\) & \(F_{n}(x)\) & \(F_{n}\left(x^{-}\right)\) & \(F_{0}(x)\) & \(\left|F_{n}(x)-F_{0}(x)\right|\) & \(\left|F_{n}\left(x^{-}\right)-F_{0}(x)\right|\) \\
\hline 29 & 0.2 & 0 & 0.252 & 0.052 & 0.252 \\
\hline 64 & 0.4 & 0.2 & 0.473 & 0.073 & 0.273 \\
\hline 90 & 0.6 & 0.4 & 0.593 & 0.007 & 0.193 \\
\hline 135 & 0.8 & 0.6 & 0.741 & 0.059 & 0.141 \\
\hline 182 & 1.00 & 0.8 & 0.838 & 0.162 & 0.038 \\
\hline
\end{tabular}
where:
\(\hat{\theta}=\bar{x}=100\) and \(F_{0}(x)=1-e^{-x / 100}\).
The maximum value from the last two columns is 0.273 .
Question: You are given:
(i) The amount of a claim, \(X\), is uniformly distributed on the interval \([0, \theta]\).
(ii) The prior density of \(\theta\) is \(\pi(\theta)=\frac{500}{\theta^{2}}, \quad \theta>500\).

Two claims, \(x_{1}=400\) and \(x_{2} 600\), are observed. You calculate the posterior distribution as:
\[
f\left(\theta \mid x_{1}, x_{2}\right)=3\left(\frac{600^{3}}{\theta^{4}}\right), \quad \theta>600
\]

Calculate the Bayesian premium, \(E\left(X_{3} \mid x_{1}, x_{2}\right)\).

Options:
A
450
B
500
C
550
D
600
E
650

Correct answer: A

Solution: \(E(X \mid \theta)=\theta / 2\)
\(E\left(X_{3} \mid 400,600\right)=\int_{600}^{\infty} E(X \mid \theta) f(\theta \mid 400,600) d \theta=\int_{600}^{\infty} \frac{\theta}{2} 3 \frac{600^{3}}{\theta^{4}} d \theta\)
\(=\left.\frac{3\left(600^{3}\right)}{2} \frac{\theta^{-2}}{-2}\right|_{600} ^{\infty}=\frac{3\left(600^{3}\right)\left(600^{-2}\right)}{4}=450\).
Question: You are given:
(i) The number of claims per auto insured follows a Poisson distribution with mean \(\lambda\).
(ii) The prior distribution for \(\lambda\) has the following probability density function:
\[
f(\lambda)=\frac{(500 \lambda)^{50} e^{-500 \lambda}}{\lambda \Gamma(50)}
\]
(iii) A company observes the following claims experience:

\begin{tabular}{|l|c|c|}
\hline & Year 1 & Year 2 \\
\hline Number of claims & 75 & 210 \\
\hline Number of autos insured & 600 & 900 \\
\hline
\end{tabular}

The company expects to insure 1100 autos in Year 3.
Calculate the Bayesian expected number of claims in Year 3.

Options:
A
178
B
184
C
193
D
209
E
224

Correct answer: B

Solution: Because the Bayes and Bühlmann results must be identical, this problem can be solved either way. For the Bühlmann approach, \(\mu(\lambda)=v(\lambda)=\lambda\). Then, noting that the prior distribution is a gamma distribution with parameters 50 and \(1 / 500\), we have:
\(\mu=v=E(\lambda)=50(1 / 500)=0.1\)
\(a=\operatorname{Var}(\lambda)=50(1 / 500)^{2}=0.0002\)
\(k=v / a=500\)
\(Z=1500 /(1500+500)=0.75\)
\(\bar{x}=(75+210) /(600+900)=0.19\)

The credibility estimate is \(0.75(0.19)+0.25(0.1)=0.1675\). For 1100 policies, the expected number of claims is \(1100(0.1675)=184.25\).

For the Bayes approach, the posterior density is proportional to (because in a given year the number of claims has a Poisson distribution with parameter \(\lambda\) times the number of policies)
\(\frac{e^{-600 \lambda}(600 \lambda)^{75}}{75!} \frac{e^{-900 \lambda}(900 \lambda)^{210}}{210!} \frac{(500 \lambda)^{50} e^{-500 \lambda}}{\lambda \Gamma(50)} \propto \lambda^{105} e^{-2000 \lambda}\) which is a gamma density with parameters 335 and 1/2000. The expected number of claims per policy is \(335 / 2000=0.1675\) and the expected number of claims in the next year is 184.25 .