BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: You are given:
(i) The following are observed claim amounts:
\[
\begin{array}{lllllll}
400 & 1000 & 1600 & 3000 & 5000 & 5400 & 6200
\end{array}
\]
(ii) An exponential distribution with \(\theta=3300\) is hypothesized for the data.
(iii) The goodness of fit is to be assessed by a \(p-p\) plot and a \(D(x)\) plot.

Let \((s, t)\) be the coordinates of the \(p-p\) plot for a claim amount of 3000 .
Calculate \((s-t)-D(3000)\)

Options:
A
\(\quad-0.12\)
B
\(\quad-0.07\)
C
0.00
D
\(\quad 0.07\)
E
\(\quad 0.12\)

Correct answer: B

Solution: \(s=F_{n}(3000)=4 / 8=0.5\) because for the \(p-p\) plot the denominator is \(n+1\).
\(t=F(3000)=1-e^{-3000 / 3300}=0.59711\). For the difference plot, \(D\) uses a denominator of \(n\) and so \(D= 4 / 7-0.59711=-0.02568\) and the answer is \(0.5-0.59711+0.02568=-0.071\).
Question: You are given:
(i) In a portfolio of risks, each policyholder can have at most two claims per year.
(ii) For each year, the distribution of the number of claims is:

\begin{tabular}{|c|c|}
\hline Number of Claims & Probability \\
\hline 0 & 0.10 \\
\hline 1 & \(0.90-q\) \\
\hline 2 & \(q\) \\
\hline
\end{tabular}
(iii) The prior density is:
\[
\pi(q)=\frac{q^{2}}{0.039}, \quad 0.2<q<0.5
\]

A randomly selected policyholder had two claims in Year 1 and two claims in Year 2. For this insured, calculate the Bayesian estimate of the expected number of claims in Year 3.

Options:
A
Less than 1.30
B
At least 1.30, but less than 1.40
C
At least 1.40 , but less than 1.50
D
At least 1.50 , but less than 1.60
E
At least 1.60

Correct answer: B

Solution: \(\pi(q \mid 2,2) \propto f(2 \mid q) f(2 \mid q) \pi(q)=q(q)\left(q^{2} / 0.039\right) \propto q^{4}\). Because \(\int_{0.2}^{0.5} q^{4} d q=0.006186\),
\(\pi(q \mid 2,2)=q^{4} / 0.006186\). Given \(q\), the expected number of claims is
\(E(N \mid q)=0(0.1)+1(0.9-q)+2 q=0.9+q\). The Bayesian estimate is
\(E(N \mid 2,2)=\int_{0.2}^{0.5}(0.9+q) \frac{q^{4}}{0.006186} d q=1.319\).
Question: You are given:
(i) The number of claims follows a Poisson distribution.
(ii) Claim sizes follow a gamma distribution with parameters \(\alpha\) (unknown) and \(\theta=10,000\).
(iii) The number of claims and claim sizes are independent.
(iv) The full credibility standard has been selected so that actual aggregate losses will be within \(10 \%\) of expected aggregate losses \(95 \%\) of the time.

Using limited fluctuation (classical) credibility, calculate the expected number of claims required for full credibility.

Options:
A
Less than 400
B
At least 400 , but less than 450
C
At least 450 , but less than 500
D
At least 500
E
The value cannot be determined from the information given.

Correct answer: E

Solution: \(n \lambda \geq \lambda_{0}\left[1+\left(\frac{\sigma_{Y}}{\theta_{Y}}\right)^{2}\right] ; \theta_{Y}=\alpha \theta=10,000 \alpha ; \sigma_{Y}^{2}=\alpha \theta^{2}=10^{8} \alpha\)
\(n \lambda \geq\left(\frac{1.96}{0.1}\right)^{2}\left[1+\frac{10^{8} \alpha}{10^{8} \alpha^{2}}\right]=384.16\left(1+\alpha^{-1}\right)\)
Because \(\alpha\) is needed, but not given, the answer cannot be determined from the information given.
Question: You are given:
(i) Losses follow a Burr distribution with \(\alpha=2\).
(ii) A random sample of 15 losses is:
\(\begin{array}{lllllllllllllll}195 & 255 & 270 & 280 & 350 & 360 & 365 & 380 & 415 & 450 & 490 & 550 & 575 & 590 & 615\end{array}\)
(iii) The parameters \(\gamma\) and \(\theta\) are estimated by percentile matching using the smoothed empirical estimates of the 30th and 65th percentiles.

Calculate the estimate of \(\gamma\).

Options:
A
Less than 2.9
B
At least 2.9, but less than 3.2
C
At least 3.2, but less than 3.5
D
At least 3.5 , but less than 3.8
E
At least 3.8

Correct answer: E

Solution: With \(n+1=16\), we need the \(0.3(16)=4.8\) and \(0.65(16)=10.4\) smallest observations. They are \(0.2(280)+0.8(350)=336\) and \(0.6(450)+0.4(490)=466\).
The equations to solve are:
\(0.3=1-\left(\frac{\theta^{\gamma}}{\theta^{\gamma}+336^{\gamma}}\right)^{2}\) and \(0.65=1-\left(\frac{\theta^{\gamma}}{\theta^{\gamma}+466^{\gamma}}\right)^{2}\)
\((0.7)^{-1 / 2}=1+(336 / \theta)^{\gamma}\) and \((0.35)^{-1 / 2}=1+(466 / \theta)^{\gamma}\)
\(\frac{(0.7)^{-1 / 2}-1}{(0.35)^{-1 / 2}-1}=\frac{(336 / \theta)^{\gamma}}{(466 / \theta)^{\gamma}}\)
\(0.282814=(336 / 466)^{\gamma}\)
\(\ln (0.282814)=\gamma \ln (336 / 466)\)
\(\gamma=3.8614\).
Question: You are planning a simulation to estimate the mean of a non-negative random variable. It is known that the population standard deviation is \(20 \%\) larger than the population mean.
\(M\) denotes the estimate, using the central limit theorem, of the smallest number of trials needed so that you will be at least \(95 \%\) confident that the simulated mean is within \(5 \%\) of the population mean.

\section*{Calculate \(M\).}

Options:
A
944
B
1299
C
1559
D
1844
E
2213

Correct answer: E

Solution: \(0.95=\operatorname{Pr}(0.95 \mu<\bar{X}<1.05 \mu)\)
\(\bar{X} \sim N\left(\mu, \sigma^{2} / n=1.44 \mu^{2} / n\right)\)
\(0.95=\operatorname{Pr}\left(\frac{0.95 \mu-\mu}{1.2 \mu / \sqrt{n}}<Z<\frac{1.05 \mu-\mu}{1.2 \mu / \sqrt{n}}\right)\)
\(0.95=\operatorname{Pr}(-0.05 \sqrt{n} / 1.2<Z<0.05 \sqrt{n} / 1.2)\)
\(0.05 \sqrt{n} / 1.2=1.96\)
\(n=2212.76\).
Question: You are given:
(i) The distribution of the number of claims per policy during a one-year period for 10,000 insurance policies is:

\begin{tabular}{|c|c|}
\hline Number of Claims per Policy & Number of Policies \\
\hline 0 & 5000 \\
\hline 1 & 5000 \\
\hline 2 or more & 0 \\
\hline
\end{tabular}
(ii) You fit a binomial model with parameters \(m\) and \(q\) using the method of maximum likelihood.

Calculate the maximum value of the loglikelihood function when \(m=2\).

Options:
A
\(\quad-10,397\)
B
\(\quad-7,781\)
C
\(-7,750\)
D
\(\quad-6,931\)
E
\(-6,730\)

Correct answer: B

Solution: \(L(q)=\left[\binom{2}{0}(1-q)^{2}\right]^{5000}\left[\binom{2}{1} q(1-q)\right]^{5000}=2^{5000} q^{5000}(1-q)^{15000}\)
\(l(q)=5000 \ln (2)+5000 \ln (q)+15000 \ln (1-q)\)
\(l^{\prime}(q)=5000 q^{-1}-15000(1-q)^{-1}=0\)
\(\hat{q}=0.25\)
\(l(0.25)=5000 \ln (2)+5000 \ln (0.25)+15000 \ln (0.75)=-7780.97\).
Question: You are given:
(i) Over a three-year period, the following claim experience was observed for two insureds who own delivery vans:

\begin{tabular}{|c|l|c|c|c|}
\hline & & \multicolumn{3}{|c|}{ Year } \\
\hline Insured & & \(\mathbf{1}\) & \(\mathbf{2}\) & \(\mathbf{3}\) \\
\hline A & Number of Vehicles & 2 & 2 & 1 \\
\hline & Number of Claims & 1 & 1 & 0 \\
\hline B & Number of Vehicles & N/A & 3 & 2 \\
\hline & Number of Claims & N/A & 2 & 3 \\
\hline
\end{tabular}
(ii) The number of claims for each insured each year follows a Poisson distribution.

Calculate the semiparametric empirical Bayes estimate of the claim frequency per vehicle for Insured A in Year 4.

Options:
A
Less than 0.55
B
At least 0.55 , but less than 0.60
C
At least 0.60 , but less than 0.65
D
At least 0.65 , but less than 0.70
E
At least 0.70

Correct answer: C

Solution: The estimate of the overall mean, \(\mu\), is the sample mean, per vehicle, which is \(7 / 10=0.7\). With the Poisson assumption, this is also the estimate of \(v=\mathrm{EPV}\). The means for the two insureds are \(2 / 5=0.4\) and \(5 / 5=1.0\). The estimate of \(a\) is the usual non-parametric estimate,
\(\mathrm{VHM}=\hat{a}=\frac{5(0.4-0.7)^{2}+5(1.0-0.7)^{2}-(2-1)(0.7)}{10-\frac{1}{10}(25+25)}=0.04\)
Then, \(k=0.7 / 0.04=17.5\) and so \(Z=5 /(5+17.5)=2 / 9\). The estimate for insured A is \((2 / 9)(0.4)+(7 / 9)(0.7)=0.6333\).
Question: You are given:
(i) The number of claims incurred in a month by any insured follows a Poisson distribution with mean \(\lambda\).
(ii) The claim frequencies of different insureds are independent.
(iii) The prior distribution of \(\lambda\) is Weibull with \(\theta=0.1\) and \(\tau=2\).
(iv) Some values of the gamma function are
\[
\Gamma(0.5)=1.77245, \Gamma(1)=1, \Gamma(1.5)=0.88623, \Gamma(2)=1
\]
(v)

\begin{tabular}{|c|c|c|}
\hline Month & Number of Insureds & Number of Claims \\
\hline 1 & 100 & 10 \\
\hline 2 & 150 & 11 \\
\hline 3 & 250 & 14 \\
\hline
\end{tabular}

Calculate the Bühlmann-Straub credibility estimate of the number of claims in the next 12 months for 300 insureds.

Options:
A
Less than 255
B
At least 255 , but less than 275
C
At least 275 , but less than 295
D
At least 295 , but less than 315
E
At least 315

Correct answer: B

Solution: \(\mu(\lambda)=v(\lambda)=\lambda\)
\(\mu=v=E(\lambda)=0.1 \Gamma(1+1 / 2)=0.088623\)
\(a=\operatorname{Var}(\lambda)=(0.1)^{2} \Gamma(1+2 / 2)-0.088623^{2}=0.002146\)
\(Z=\frac{500}{500+0.088623 / 0.002146}=0.92371\).
The estimate for one insured for one month is \(0.92371(35 / 500)+0.07629(0.088623)=\) 0.07142 . For 300 insureds for 12 months it is \((300)(12)(0.07142)=257.11\).
Question: You are given:
(i) The annual number of claims for an individual risk follows a Poisson distribution with mean \(\lambda\).
(ii) For \(75 \%\) of the risks, \(\lambda=1\).
(iii) For \(25 \%\) of the risks, \(\lambda=3\).

A randomly selected risk had \(r\) claims in Year 1. The Bayesian estimate of this risk's expected number of claims in Year 2 is 2.98 .

Calculate the Bühlmann credibility estimate of the expected number of claims for this risk in Year 2.

Options:
A
Less than 1.9
B
At least 1.9, but less than 2.3
C
At least 2.3, but less than 2.7
D
At least 2.7, but less than 3.1
E
At least 3.1

Correct answer: E

Solution: \[
\begin{aligned}
\operatorname{Pr}(\lambda & \left.=1 \mid X_{1}=r\right)=\frac{\operatorname{Pr}\left(X_{1}=r \mid \lambda=1\right) \operatorname{Pr}(\lambda=1)}{\operatorname{Pr}\left(X_{1}=r \mid \lambda=1\right) \operatorname{Pr}(\lambda=1)+\operatorname{Pr}\left(X_{1}=r \mid \lambda=3\right) \operatorname{Pr}(\lambda=3)} \\
& =\frac{\frac{e^{-1}}{r!}(0.75)}{\frac{e^{-1}}{r!}(0.75)+\frac{e^{-3} 3^{r}}{r!}(0.25)}=\frac{0.2759}{0.2759+0.1245\left(3^{r}\right)} .
\end{aligned}
\]

Then,
\(2.98=\frac{0.2759}{0.2759+0.1245\left(3^{r}\right)}(1)+\frac{0.1245\left(3^{r}\right)}{0.2759+0.1245\left(3^{r}\right)}(3)=\frac{0.2759+0.3735\left(3^{r}\right)}{0.2759+0.1245\left(3^{r}\right)}\).
Rearrange to obtain
\(0.82218+0.037103\left(3^{r}\right)=0.2759+0.03735\left(3^{r}\right)\)
\(0.54628=0.00025\left(3^{r}\right)\)
\(r=7\).
Because the risks are Poisson:
\(\mu=v=E(\lambda)=0.75(1)+0.25(3)=1.5\)
\(a=\operatorname{Var}(\lambda)=0.75(1)+0.25(9)-2.25=0.75\)
\(Z=\frac{1}{1+1.5 / 0.75}=1 / 3\)
and the estimate is \((1 / 3)(7)+(2 / 3)(1.5)=3.33\).
Question: You are given the following ages at time of death for 10 individuals:
\[
\begin{array}{llllllllll}
25 & 30 & 35 & 35 & 37 & 39 & 45 & 47 & 49 & 55
\end{array}
\]

Using a uniform kernel with bandwidth \(b=10\), calculate the kernel density estimate of the probability of survival to age 40.

Options:
A
0.377
B
\(\quad 0.400\)
C
0.417
D
0.439
E
0.485

Correct answer: E

Solution: The uniform kernel spreads the probability of 0.1 to 10 units each side of an observation. So the observation at 25 contributes a density of 0.005 from 15 to 35 , contributing nothing to survival past age 40 . The same applies to the point at 30 . For the next 7 points: 35 contributes probability from 25 to 45 for \(5(0.005)=0.025\) above age 40 . 35 contributes probability from 25 to 45 for \(5(0.005)=0.025\) above age 40 . 37 contributes probability from 27 to 47 for \(7(0.005)=0.035\) above age 40 . 39 contributes probability from 29 to 49 for \(9(0.005)=0.045\) above age 40 . 45 contributes probability from 35 to 55 for \(15(0.005)=0.075\) above age 40 . 47 contributes probability from 37 to 57 for \(17(0.005)=0.085\) above age 40 . 49 contributes probability from 39 to 59 for \(19(0.005)=0.095\) above age 40. The observation at 55 contributes all 0.1 of probability. The total is 0.485 .
Question: Three individual policyholders have the following claim amounts over four years:

\begin{tabular}{|c|c|c|c|c|}
\hline Policyholder & Year 1 & Year 2 & Year 3 & Year 4 \\
\hline X & 2 & 3 & 3 & 4 \\
\hline Y & 5 & 5 & 4 & 6 \\
\hline Z & 5 & 5 & 3 & 3 \\
\hline
\end{tabular}

Using the nonparametric empirical Bayes procedure, calculate the estimated variance of the hypothetical means.

Options:
A
Less than 0.40
B
At least 0.40 , but less than 0.60
C
At least 0.60 , but less than 0.80
D
At least 0.80 , but less than 1.00
E
At least 1.00

Correct answer: C

Solution: The sample means are 3,5 , and 4 and the overall mean is 4 . Then,
\(\hat{v}=\frac{1+0+0+1+0+0+1+1+1+1+1+1}{3(4-1)}=\frac{8}{9}\)
\(\hat{a}=\frac{(3-4)^{2}+(5-4)^{2}+(4-4)^{2}}{3-1}-\frac{8 / 9}{4}=\frac{7}{9}=0.78\).
Question: For a group of policies, you are given:
(i) Losses follow the distribution function
\[
F(x)=1-\theta / x, \quad x>0 .
\]
(ii) A sample of 20 losses resulted in the following:

\begin{tabular}{|c|c|}
\hline Interval & Number of Losses \\
\hline\((0,10]\) & 9 \\
\hline\((10,25]\) & 6 \\
\hline\((25, \infty)\) & 5 \\
\hline
\end{tabular}

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

Options:
A
5.00
B
5.50
C
5.75
D
6.00
E
6.25

Correct answer: B

Solution: \(L(\theta)=\left(1-\frac{\theta}{10}\right)^{9}\left(\frac{\theta}{10}-\frac{\theta}{25}\right)^{6}\left(\frac{\theta}{25}\right)^{5} \propto(10-\theta)^{9} \theta^{11}\)
\(l(\theta)=9 \ln (10-\theta)+11 \ln (\theta)\)
\(l^{\prime}(\theta)=-\frac{9}{10-\theta}+\frac{11}{\theta}=0\)
\(11(10-\theta)=9 \theta\)
\(110=20 \theta\)
\(\theta=110 / 20=5.5\).
Question: You are given:
(i) Loss payments for a group health policy follow an exponential distribution with unknown mean.
(ii) A sample of losses is:
\[
\begin{array}{llllll}
100 & 200 & 400 & 800 & 1400 & 3100
\end{array}
\]

Using the delta method, calculate the approximation of the variance of the maximum likelihood estimator of \(S(1500)\).

Options:
A
0.019
B
0.025
C
0.032
D
0.039
E
0.045

Correct answer: A

Solution: The maximum likelihood estimate is \(\hat{\theta}=\bar{x}=1000\). The quantity to be estimated is \(S(\theta)=\exp (-1500 / \theta)\) and \(S^{\prime}(\theta)=1500 \theta^{-2} \exp (-1500 / \theta)\). For the delta method,
\(\operatorname{Var}[S(\hat{\theta})] \cong\left[S^{\prime}(\hat{\theta})\right]^{2} \operatorname{Var}(\hat{\theta})=\left[1500(1000)^{-2} \exp (-1500 / 1000)\right]^{2}\left(1000^{2} / 6\right)=0.01867\).
This is based on \(\operatorname{Var}(\hat{\theta})=\operatorname{Var}(\bar{X})=\operatorname{Var}(X) / n=\theta^{2} / n\).
Question: You are given:
(i) A random sample of payments from a portfolio of policies resulted in the following:

\begin{tabular}{|c|c|}
\hline Interval & Number of Policies \\
\hline\((0,50]\) & 36 \\
\hline\((50,150]\) & \(x\) \\
\hline\((150,250]\) & \(y\) \\
\hline\((250,500]\) & 84 \\
\hline\((500,1000]\) & 80 \\
\hline\((1000, \infty)\) & 0 \\
\hline Total & \(n\) \\
\hline
\end{tabular}
(ii) Two values of the ogive constructed from the data in (i) are:
\[
F_{n}(90)=0.21, F_{n}(210)=0.51
\]

Calculate \(x\).

Options:
A
120
B
145
C
170
D
195
E
220

Correct answer: A

Solution: Based on the information given
\(0.21=\frac{36}{n}+\frac{0.4 x}{n}\)
\(0.51=\frac{36}{n}+\frac{x}{n}+\frac{0.6 y}{n}\)
\(n=200+x+y\).
Then,
\(0.21(200+x+y)=36+0.4 x\)
\(0.51(200+x+y)=36+x+0.6 y\)
and these linear equations can be solved for \(x=119.37\).
Question: Michael is a professional stuntman who performs dangerous motorcycle jumps at extreme sports events around the world.

The annual cost of repairs to his motorcycle is modeled by a two parameter Pareto distribution with \(\theta=5000\) and \(\alpha=2\).

An insurance reimburses Michael's motorcycle repair costs subject to the following provisions:
(i) Michael pays an annual ordinary deductible of 1000 each year.
(ii) Michael pays \(20 \%\) of repair costs between 1000 and 6000 each year.
(iii) Michael pays \(100 \%\) of the annual repair costs above 6000 until Michael has paid 10,000 in out-of-pocket repair costs each year.
(iv) Michael pays \(10 \%\) of the remaining repair costs each year.

Calculate the expected annual insurance reimbursement.

Options:
A
2300
B
2500
C
2700
D
2900
E
3100

Correct answer: C

Solution: Insurance pays \(80 \%\) of the portion of annual claim between 6,000 and 1,000 , and \(90 \%\) of the portion of annual claims over 14,000.

The 14,000 breakpoint is where Michael reaches 10,000 that he has paid:
1000 = deductible
\(1000=20 \%\) of costs between 1000 and 6000
\(8000=100 \%\) of costs between 14,000 and 6,000
\(E(X \wedge x)=\theta\left(1-\frac{\theta}{x+\theta}\right)=\frac{5000 x}{x+5000}\)

\begin{tabular}{ccc}
\(x\) & & \(E(X \wedge x)\) \\
\cline { 1 - 1 } \cline { 3 - 3 } 1000 & & 833.33 \\
6000 & & 2727.27 \\
14000 & & 3684.21 \\
\(\infty\) & & 5000
\end{tabular}
\(0.80[E(X \wedge 6000)-E(X \wedge 1000)]+0.90[E(X)-\mathrm{E}(X \wedge 14000)]\)
\(=0.80[2727,27-833.33]+0.90[5000-3684.21]\)
\(=1515.15+1184.21=2699.36\)
Question: A random variable \(X\) has a two-point mixture distribution with pdf \(f(x)=\frac{1}{8} e^{-x / 2}+\frac{1}{4} e^{-x / 3}, \quad x \geq 0\).
You are to simulate one value, \(x\), from this distribution using uniform random numbers 0.2 and 0.6 . Use the value 0.2 and the inversion method to simulate \(J\) where \(J=1\) refers to the first random variable in the mixture and \(J=2\) refers to the second random variable. Then use 0.6 and the inversion method to simulate a value from \(X\).

Calculate the value of \(x\).

Options:
A
0.45
B
1.02
C
1.53
D
1.83
E
\(\quad 2,75\)

Correct answer: D

Solution: The mixture can be written as
\[
f(x)=\frac{1}{4} \frac{1}{2} e^{-x / 2}+\frac{3}{4} \frac{1}{3} e^{-x / 3}, x \geq 0
\]
and so \(\operatorname{Pr}(J=1)=0.25\). Because \(0.2<0.25\), the observation is from the first distribution in the mixture. Then by inversion,
\(1-e^{-x / 2}=0.6 \Rightarrow x=-2 \ln (0.4)=1.83\).
Question: Four of XYZ's policyholders have the following history.

\begin{tabular}{|l|l|l|l|l|}
\hline Policy number & Date of birth & Date policy issued & Status on 12-31-2012 & Status on policy anniversary in 2013 \\
\hline 1 & 4-1-1930 & 3-1-1980 & Died 10-1-2012 & Died 10-1-2012 \\
\hline 2 & 7-1-1930 & 10-1-1990 & Insured & Surrendered 5-1-2013 \\
\hline 3 & 8-1-1930 & 1-1-2000 & Insured & Insured \\
\hline 4 & 9-1-1930 & 8-1-2000 & Surrendered 11-1-2012 & Surrendered 11-1-2012 \\
\hline
\end{tabular}

XYZ conducts a mortality study from anniversaries in 2012 through anniversaries in 2013 using insuring ages.

XYZ assigns insuring ages based on the age at the nearest birthday when the policy is issued.

Calculate the absolute value of the difference between the estimated mortality probabilities at age 82 , using the exact and actuarial methods.

Options:
A
0.031
B
0.035
C
0.039
D
0.042
E
0.045

Correct answer: C

Solution: Policy 1 turns 82 on 3-1-2012 and has 7 months exact exposure (time to death) and 12 months actuarial exposure (deaths exposed to next birthday).

Policy 2 turns 82 on 10-1-2012 and has 7 months exposure by both methods.
Policy 3 turns 82 on 1-1-2013 and has no exposure.
Policy 4 turns 82 on 8-1-2012 and has 3 months exposure by both methods.
The difference of the probabilities is \(\left|\left[1-e^{-1 /(17 / 12)}\right]-[1 /(22 / 12)]\right|=0.039\).
Question: A mortality study of insured lives is being conducted. Insuring ages are employed and the study runs from anniversaries in 2010 through anniversaries in 2012. Assume that events that can occur between anniversaries do so uniformly. The following was observed:

There are 10,000 lives at age 60 when first observed in 2010. Between then and their next anniversary 100 die and 800 surrender their policy. In 2011, 1000 new policies are sold at age 61 . For those insured at their age 61 anniversary in 2011, 120 die and 700 surrender before their 2012 anniversary.

There are 8000 lives at age 61 when first observed in 2010. Between then and their next anniversary 80 die and 700 surrender their policy. In 2011, 800 new policies are sold at age 62 . For those insured at their age 62 anniversary in 2011, 90 die and 600 surrender before their 2012 anniversary.

Calculate the actuarial estimate of the mortality probability at age 61 .

Options:
A
\(\quad 0.0112\)
B
0.0113
C
0.0114
D
0.0115
E
\(\quad 0.0116\)

Correct answer: D

Solution: \(P_{61}=(10,000-100-800+1000)+8000=18,100\)
\(d_{61}=120+80=200\)
\(w_{61}=700+700=1400\)
\(n_{61}=0\)
\(q=\frac{200}{18,100+(0-1400) / 2}=\frac{200}{17,400}=0.0115\).
Question: A mortality study of insured lives is being conducted. Insuring ages are employed and the study runs from anniversaries in 2010 through anniversaries in 2012. Assume that events that can occur between anniversaries do so uniformly. The following was observed:

There are 10,000 lives at age 60 when first observed in 2010. Between then and their next anniversary 100 die and 800 surrender their policy. In 2011, 1000 new policies are sold at age 61 . For those insured at their age 61 anniversary in 2011, 120 die and 700 surrender before their 2012 anniversary.

There are 8000 lives at age 61 when first observed in 2010. Between then and their next anniversary 80 die and 700 surrender their policy. In 2011, 800 new policies are sold at age 62 . For those insured at their age 62 anniversary in 2011, 90 die and 600 surrender before their 2012 anniversary.

Using the exact exposure method, calculate the estimate of the mortality probability at age 61.

Options:
A
\(\quad 0.0112\)
B
0.0113
C
0.0114
D
0.0115
E
0.0116

Correct answer: D

Solution: \(P_{61}=(10,000-100-800+1000)+8000=18,100\)
\(d_{61}=120+80=200\)
\(w_{61}=700+700=1400\)
\(n_{61}=0\)
\(q=1-\exp \left(\frac{200}{18,100+(0-1400-200) / 2}\right)=1-e^{-200 / 17,300}=0.0115\).
Question: An insurance policy is available with deductibles of 500 or 750 , and policy limits of 1500 or 20,000 . The underlying loss distribution is assumed to be independent of deductible and policy limit.

You are given the following claim counts:

\begin{tabular}{|l|l|l|l|}
\hline \multirow[b]{2}{*}{Loss Range} & \multicolumn{2}{|c|}{Deductible} & \multirow[b]{2}{*}{Total} \\
\hline & 500 & 750 & \\
\hline 500-750 & 8 & & 8 \\
\hline 750-1000 & 10 & 10 & 20 \\
\hline 1000-1500 & 12 & 15 & 27 \\
\hline 1500-5000 & 14 & 18 & 32 \\
\hline 5000-10,000 & 10 & 17 & 27 \\
\hline 10,000-20,000 & 0 & 0 & 0 \\
\hline At 1500 limit & 6 & 0 & 6 \\
\hline At 20,000 limit & 0 & 0 & 0 \\
\hline Total & 60 & 60 & 120 \\
\hline
\end{tabular}

Calculate the probability that a loss from a policy with a 750 deductible will exceed 5000, using the Kaplan-Meier approximation with all the given data.

Options:
A
0.27
B
0.29
C
0.33
D
0.36
E
0.37

Correct answer: A

Solution: \section*{Solution:}

Conditional survival function that loss is over 750

Interval 750-1000, risk set \(=120-8=112\), losses \(=10+10=20\)
\(\hat{p}_{j}=\frac{112-20}{112}=0.8214\)
Interval 1000-1500, risk set \(=112-20=92\), losses \(=12+15=27\)
\(\hat{p}_{j}=\frac{92-27}{92}=0.7065\)
Interval 1500-5000, risk set \(=92-27-6=59\), losses \(=14+18=32\)
\(\hat{p}_{j}=\frac{59-32}{59}=0.4576\)
Probability \(=(0.8214)(0.7065)(0.4576)=0.2656=0.27\)
Question: Five models are fitted to a sample of \(n=260\) observations with the following results:

\begin{tabular}{|c|c|c|}
\hline Model & Number of Parameters & Loglikelihood \\
\hline I & 1 & -414 \\
\hline II & 2 & -412 \\
\hline III & 3 & -411 \\
\hline IV & 4 & -409 \\
\hline V & 6 & -409 \\
\hline
\end{tabular}

(This question was formerly Question 266.) Determine the model favored by the Schwarz Bayesian criterion.

Options:
A
I
B
II
C
III
D
IV
E
\(\quad \mathrm{V}\)

Correct answer: A

Solution: (This question was formerly Question 266.) The deduction to get the SBC is \((r / 2) \ln (n)=(r / 2) \ln (260)=2.78 r\) where \(r\) is the number of parameters. The SBC values are then \(-416.78,-417.56,-419.34,-420.12\), and -425.68 . The largest value is the first one, so model I is to be selected.