BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: You study five lives to estimate the time from the onset of a disease to death. The times to death are:
\[
\begin{array}{lllll}
2 & 3 & 3 & 3 & 7
\end{array}
\]

Using a triangular kernel with bandwidth 2 , calculate the density function estimate at 2.5 .

Options:
A
\(\quad 8 / 40\)
B
\(\quad 12 / 40\)
C
\(\quad 14 / 40\)
D
\(\quad 16 / 40\)
E
\(\quad 17 / 40\)

Correct answer: B

Solution: The kernel is a triangle with a base of 4 and a height at the middle of 0.5 (so the area is 1 ). The length of the base is twice the bandwidth. Any observation within 2 of 2.5 will contribute to the estimate. For the observation at 2 , when the triangle is centered at 2 , the height of the triangle at 2.5 is .375 (it is one-quarter the way from 2 to the end of the triangle at 4 and so the height is one-quarter the way from 0.5 to 0 ). Similarly the points at 3 are also 0.5 away and so the height of the associated triangle is also .375. Each triangle height is weighted by the empirical probability at the associated point. So the estimate at 2.5 is \((1 / 5)(3 / 8)+(3 / 5)(3 / 8)+ (1 / 5)(0)=12 / 40\).
Question: For a sample of dental claims \(x_{1}, x_{2}, \ldots, x_{10}\), you are given:
(i) \(\quad \sum x_{i}=3860\) and \(\sum x_{i}^{2}=4,574,802\)
(ii) Claims are assumed to follow a lognormal distribution with parameters \(\mu\) and \(\sigma\)
(iii) \(\quad \mu\) and \(\sigma\) are estimated using the method of moments.

Calculate \(E[X \wedge 500]\) for the fitted distribution.

Options:
A
Less than 125
B
At least 125 , but less than 175
C
At least 175 , but less than 225
D
At least 225 , but less than 275
E
At least 275

Correct answer: D

Solution: For the method of moments estimate,
\(386=\exp \left(\mu+0.5 \sigma^{2}\right), 457,480.2=\exp \left(2 \mu+2 \sigma^{2}\right)\)
\(5.9558=\mu+0.5 \sigma^{2}, 13.0335=2 \mu+2 \sigma^{2}\)
\(\hat{\mu}=5.3949, \hat{\sigma}^{2}=1.1218\).
Then
\[
\begin{aligned}
& E(X \wedge 500)=\exp [5.3949+0.5(1.1218)] \Phi\left(\frac{\ln 500-5.3949-1.1218}{\sqrt{1.1218}}\right) \\
& \quad+500\left[1-\Phi\left(\frac{\ln 500-5.3949}{\sqrt{1.1218}}\right)\right] \\
& =386 \Phi(-0.2853)+500[1-\Phi(0.7739)]=386(0.3877)+500(0.2195)=259 .
\end{aligned}
\]

Note-these calculations use exact normal probabilities. Rounding and using the normal table that accompanies the exam will produce a different numerical answer but the same letter answer.
Question: You are given:
(i) Claim counts follow a Poisson distribution with mean \(\theta\).
(ii) Claim sizes follow an exponential distribution with mean \(10 \theta\).
(iii) Claim counts and claim sizes are independent, given \(\theta\).
(iv) The prior distribution has probability density function:
\[
\pi(\theta)=\frac{5}{\theta^{6}}, \quad \theta>1
\]

Calculate Bühlmann's \(k\) for aggregate losses.

Options:
A
Less than 1
B
At least 1 , but less than 2
C
At least 2, but less than 3
D
At least 3, but less than 4
E
At least 4

Correct answer: C

Solution: Let \(N\) be the Poisson claim count variable, let \(X\) be the claim size variable, and let \(S\) be the aggregate loss variable.
\(\mu(\theta)=E(S \mid \theta)=E(N \mid \theta) E(X \mid \theta)=\theta 10 \theta=10 \theta^{2}\)
\(v(\theta)=\operatorname{Var}(S \mid \theta)=E(N \mid \theta) E\left(X^{2} \mid \theta\right)=\theta 200 \theta^{2}=200 \theta^{3}\)
\(\mu=E\left(10 \theta^{2}\right)=\int_{1}^{\infty} 10 \theta^{2}\left(5 \theta^{-6}\right) d \theta=50 / 3\)
\(E P V=E\left(200 \theta^{3}\right)=\int_{1}^{\infty} 200 \theta^{3}\left(5 \theta^{-6}\right) d \theta=500\)
\(\operatorname{VHM}=\operatorname{Var}\left(10 \theta^{2}\right)=\int_{1}^{\infty}\left(10 \theta^{2}\right)^{2}\left(5 \theta^{-6}\right) d \theta-(50 / 3)^{2}=222.22\)
\(k=500 / 222.22=2.25\).
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: \[
\begin{aligned}
& \pi(p \mid 1,1,1,1,1,1,1,1) \propto \operatorname{Pr}(1,1,1,1,1,1,1,1 \mid p) \pi(p) \propto p^{8} \\
& \pi(p \mid 1,1,1,1,1,1,1,1)=\frac{p^{8}}{\int_{0}^{0.5} p^{8} d p}=\frac{p^{8}}{0.5^{9} / 9}=9\left(0.5^{-9}\right) p^{8} \\
& \operatorname{Pr}\left(X_{9}=1 \mid 1,1,1,1,1,1,1,1\right)=\int_{0}^{0.5} \operatorname{Pr}\left(X_{9}=1 \mid p\right) \pi(p \mid 1,1,1,1,1,1,1,1) d p \\
& =\int_{0}^{0.5} p 9\left(0.5^{-9}\right) p^{8} d p=9\left(0.5^{-9}\right)\left(0.5^{10}\right) / 10=0.45
\end{aligned}
\]
Question: You are given:
(i) The number of claims incurred in a month by any insured has a Poisson distribution with mean \(\lambda\).
(ii) The claim frequencies of different insureds are independent.
(iii) The prior distribution is gamma with probability density function:
\[
f(\lambda)=\frac{(100 \lambda)^{6} e^{-100 \lambda}}{120 \lambda}
\]
(iv)

\begin{tabular}{|c|c|c|}
\hline Month & Number of Insureds & Number of Claims \\
\hline 1 & 100 & 6 \\
2 & 150 & 8 \\
3 & 200 & 11 \\
4 & 300 & \(?\) \\
\hline
\end{tabular}

Calculate the Bühlmann-Straub credibility estimate of the number of claims in Month 4 .

Options:
A
16.7
B
16.9
C
17.3
D
17.6
E
18.0

Correct answer: B

Solution: From the Poisson distribution, \(\mu(\lambda)=\lambda\) and \(v(\lambda)=\lambda\). Then, \(\mu=E(\lambda)=6 / 100=0.06, E P V=E(\lambda)=0.06, \operatorname{VHM}=\operatorname{Var}(\lambda)=6 / 100^{2}=0.0006\) where the various moments are evaluated from the gamma distribution. Then, \(k=0.06 / 0.0006=100\) and \(Z=450 /(450+100)=9 / 11\) where 450 is the total number of insureds contributing experience. The credibility estimate of the expected number of claims for one insured in month 4 is \((9 / 11)(25 / 450)+(2 / 11)(0.06)=0.056364\). For 300 insureds the expected number of claims is \(300(0.056364)=16.9\).
Question: For a sample of 15 losses, you are given:
(i)

\begin{tabular}{|c|c|}
\hline Interval & \begin{tabular}{c} 
Observed Number of \\
Losses
\end{tabular} \\
\hline\((0,2]\) & 5 \\
\hline\((2,5]\) & 5 \\
\hline\((5, \infty)\) & 5 \\
\hline
\end{tabular}
(ii) Losses follow the uniform distribution on \((0, \theta)\).

Estimate \(\theta\) by minimizing the function \(\sum_{j=1}^{3} \frac{\left(E_{j}-O_{j}\right)^{2}}{O_{j}}\), where \(E_{j}\) is the expected number of losses in the \(j\) th interval and \(O_{j}\) is the observed number of losses in the \(j\) th interval.

Options:
A
6.0
B
6.4
C
6.8
D
7.2
E
7.6

Correct answer: E

Solution: Assume that \(\theta>5\). Then the expected counts for the three intervals are \(15(2 / \theta)=30 / \theta, 15(3 / \theta)=45 / \theta\), and \(15(\theta-5) / \theta=15-75 / \theta\) respectively. The quantity to minimize is
\(\frac{1}{5}\left[\left(30 \theta^{-1}-5\right)^{2}+\left(45 \theta^{-1}-5\right)^{2}+\left(15-75 \theta^{-1}-5\right)^{2}\right]\).
Differentiating (and ignoring the coefficient of \(1 / 5\) ) gives the equation \(-2\left(30 \theta^{-1}-5\right) 30 \theta^{-2}-2\left(45 \theta^{-1}-5\right) 45 \theta^{-2}+2\left(10-75 \theta^{-1}\right) 75 \theta^{-2}=0\). Multiplying through by \(\theta^{3}\) and dividing by 2 reduces the equation to \(-(30-5 \theta) 30-(45-5 \theta) 45+(10 \theta-75) 75=-8550+1125 \theta=0\) for a solution of \(\hat{\theta}=8550 / 1125=7.6\).
Question: You are given:
(i) Low-hazard risks have an exponential claim size distribution with mean \(\theta\).
(ii) Medium-hazard risks have an exponential claim size distribution with mean \(2 \theta\).
(iii) High-hazard risks have an exponential claim size distribution with mean \(3 \theta\).
(iv) No claims from low-hazard risks are observed.
(v) Three claims from medium-hazard risks are observed, of sizes 1,2 and 3 .
(vi) One claim from a high-hazard risk is observed, of size 15.

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

Options:
A
1
B
2
C
3
D
4
E
5

Correct answer: B

Solution: The likelihood function is \(\frac{e^{-1 /(2 \theta)}}{2 \theta} \frac{e^{-2 /(2 \theta)}}{2 \theta} \frac{e^{-3 /(2 \theta)}}{2 \theta} \frac{e^{-15 /(3 \theta)}}{3 \theta}=\frac{e^{-8 / \theta}}{24 \theta^{4}}\). The loglikelihood function is \(-\ln (24)-4 \ln (\theta)-8 / \theta\). Differentiating with respect to \(\theta\) and setting the result equal to 0 yields \(-\frac{4}{\theta}+\frac{8}{\theta^{2}}=0\) which produces \(\hat{\theta}=2\).
Question: You are given:

\begin{tabular}{|r|c|}
\hline Claim Size \((X)\) & Number of Claims \\
\hline\((0,25]\) & 25 \\
\hline\((25,50]\) & 28 \\
\hline\((50,100]\) & 15 \\
\hline\((100,200]\) & 6 \\
\hline
\end{tabular}

Assume a uniform distribution of claim sizes within each interval.
Estimate \(E\left(X^{2}\right)-E\left[(X \wedge 150)^{2}\right]\).

Options:
A
Less than 200
B
At least 200 , but less than 300
C
At least 300 , but less than 400
D
At least 400 , but less than 500
E
At least 500

Correct answer: C

Solution: In general,
\[
\begin{aligned}
& E\left(X^{2}\right)-E\left[(X \wedge 150)^{2}\right]=\int_{0}^{200} x^{2} f(x) d x-\int_{0}^{150} x^{2} f(x) d x-150^{2} \int_{150}^{200} f(x) d x \\
& =\int_{150}^{200}\left(x^{2}-150^{2}\right) f(x) d x .
\end{aligned}
\]

Assuming a uniform distribution, the density function over the interval from 100 to 200 is 6/7400 (the probability of 6/74 assigned to the interval divided by the width of the interval). The answer is
\[
\int_{150}^{200}\left(x^{2}-150^{2}\right) \frac{6}{7400} d x=\left.\left(\frac{x^{3}}{3}-150^{2} x\right) \frac{6}{7400}\right|_{150} ^{200}=337.84 .
\]
Question: In a study of claim payment times, you are given:
(i) The data were not truncated or censored.
(ii) At most one claim was paid at any one time.
(iii) The Nelson-Aalen estimate of the cumulative hazard function, \(H(t)\), immediately following the second paid claim, was 23/132.

Calculate the Nelson-Aalen estimate of the cumulative hazard function, \(H(t)\), immediately following the fourth paid claim.

Options:
A
0.35
B
0.37
C
0.39
D
0.41
E
0.43

Correct answer: C

Solution: At the time of the second failure,
\(\hat{H}(t)=\frac{1}{n}+\frac{1}{n-1}=\frac{23}{132} \Rightarrow n=12\).
At the time of the fourth failure,
\(\hat{H}(t)=\frac{1}{12}+\frac{1}{11}+\frac{1}{10}+\frac{1}{9}=0.3854\).
Question: You are given the following information about a commercial auto liability book of business:
(i) Each insured's claim count has a Poisson distribution with mean \(\lambda\), where \(\lambda\) has a gamma distribution with \(\alpha=1.5\) and \(\theta=0.2\).
(ii) Individual claim size amounts are independent and exponentially distributed with mean 5000.
(iii) The full credibility standard is for aggregate losses to be within \(5 \%\) of the expected with probability 0.90 .

Using limited fluctuated credibility, calculate the expected number of claims required for full credibility.

Options:
A
2165
B
2381
C
3514
D
7216
E
7938

Correct answer: B

Solution: \(X\) is the random sum \(Y_{1}+Y_{2}+\cdots+Y_{N}\).
\(N\) has a negative binomial distribution with \(r=a=1.5\) and \(\beta=\theta=0.2\).
\(E(N)=r \beta=0.3, \operatorname{Var}(N)=r \beta(1+\beta)=0.36\)
\(E(Y)=5,000, \operatorname{Var}(Y)=25,000,000\)
\(E(X)=0.3(5,000)=1,500\)
\(\operatorname{Var}(X)=0.3(25,000,000)+0.36(25,000,000)=16,500,000\)

Number of exposures (insureds) required for full credibility
\(n_{\text {FULL }}=(1.645 / 0.05)^{2}\left(16,500,000 / 1,500^{2}\right)=7,937.67\).
Number of expected claims required for full credibility
\(E(N) n_{\text {FULL }}=0.3(7,937.67)=2,381\).
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) Losses follow an exponential distribution with mean \(\theta\).
(ii) A random sample of 20 losses is distributed as follows:

\begin{tabular}{|l|c|}
\hline Loss Range & Frequency \\
\hline\([0,1000]\) & 7 \\
\hline\((1000,2000]\) & 6 \\
\hline\((2000, \infty)\) & 7 \\
\hline
\end{tabular}

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

Options:
A
Less than 1950
B
At least 1950, but less than 2100
C
At least 2100 , but less than 2250
D
At least 2250 , but less than 2400
E
At least 2400

Correct answer: B

Solution: \(L=F(1000)^{7}[F(2000)-F(1000)]^{6}[1-F(2000)]^{7}\)
\(=\left(1-e^{-1000 / \theta}\right)^{7}\left(e^{-1000 / \theta}-e^{-2000 / \theta}\right)^{6}\left(e^{-2000 / \theta}\right)^{7}\)
\(=(1-p)^{7}\left(p-p^{2}\right)^{6}\left(p^{2}\right)^{7}=p^{20}(1-p)^{13}\)
where \(p=e^{-1000 / \theta}\) The maximum occurs at \(p=20 / 33\) and so \(\hat{\theta}=-1000 \ln (20 / 33)=1996.90\).
Question: You are given four classes of insureds, each of whom may have zero or one claim, with the following probabilities:

\begin{tabular}{|c|c|c|}
\hline \multirow{2}{*}{ Class } & \multicolumn{2}{|c|}{ Number of Claims } \\
\cline { 2 - 3 } & 0 & 1 \\
\hline I & 0.9 & 0.1 \\
\hline II & 0.8 & 0.2 \\
\hline III & 0.5 & 0.5 \\
\hline IV & 0.1 & 0.9 \\
\hline
\end{tabular}

A class is selected at random (with probability 0.25 ), and four insureds are selected at random from the class. The total number of claims is two.

If five insureds are selected at random from the same class, estimate the total number of claims using Bühlmann-Straub credibility.

Options:
A
2.0
B
2.2
C
2.4
D
2.6
E
\(\quad 2.8\)

Correct answer: C

Solution: The four classes have means \(0.1,0.2,0.5\), and 0.9 respectively and variances \(0.09,0.16,0.25\), and 0.09 respectively.

Then,
\(\mu=0.25(0.1+0.2+0.5+0.9)=0.425\)
\(v=0.25(0.09+0.16+0.25+0.09)=0.1475\)
\(a=0.25(0.010 .04+0.25+0.81)-0.425^{2}=0.096875\)
\(k=0.1475 / 0.096875=1.52258\)
\(Z=\frac{4}{4+1.52258}=0.7243\)
The estimate is \(5[0.7243(2 / 4)+0.2757(0.425)]=2.40\).
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
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
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 group of 10 claims:

\begin{tabular}{|c|c|c|}
\hline \begin{tabular}{c} 
Claim Size \\
Interval
\end{tabular} & \begin{tabular}{c} 
Number of Claims \\
in Interval
\end{tabular} & \begin{tabular}{c} 
Number of Claims \\
Censored in Interval
\end{tabular} \\
\hline\((0-15,000]\) & 1 & 2 \\
\hline\((15,000-30,000]\) & 1 & 2 \\
\hline\((30,000-45,000]\) & 4 & 0 \\
\hline
\end{tabular}

Assume that claim sizes and censorship points are uniformly distributed within each interval.
Estimate, using large data set methodology and actuarial exposures, the probability that a claim exceeds 30,000.

Options:
A
0.67
B
0.70
C
0.74
D
0.77
E
0.80

Correct answer: C

Solution: \begin{tabular}{|l|l|l|l|l|l|}
\hline\(j\) & \(P_{j}\) & \(n_{j}\) & \(d_{j}\) & \(w_{j}\) & \(e_{j}\) \\
\hline 0 & 10 & 0 & 1 & 2 & \(10-2 / 2=9.0\) \\
\hline 1 & 7 & 0 & 1 & 2 & \(7-2 / 2=6.0\) \\
\hline 2 & 4 & 0 & 4 & 0 & \(4-0 / 2=4.0\) \\
\hline
\end{tabular}
\(\operatorname{Pr}(X>30,000)=p_{0} p_{1}=\frac{9-1}{9} \frac{6-1}{6}=\frac{20}{27}=0.741\).

\section*{Question \#74}

\section*{Original version DELETED}
Question: You are given:
(i) Claim amounts follow a shifted exponential distribution with probability density function:
\[
f(x)=\frac{1}{\theta} e^{-(x-\delta) / \theta}, \quad \delta<x<\infty
\]
(ii) A random sample of claim amounts \(X_{1}, X_{2}, \ldots, X_{10}\) :
\[
\begin{array}{llllllllll}
5 & 5 & 5 & 6 & 8 & 9 & 11 & 12 & 16 & 23
\end{array}
\]
(iii) \(\quad \sum X_{i}=100\) and \(\sum X_{i}^{2}=1306\)

Estimate \(\delta\) using the method of moments.

Options:
A
3.0
B
3.5
C
4.0
D
4.5
E
5.0

Correct answer: D

Solution: \(E(X)=\int_{\delta}^{\infty} \frac{x}{\theta} e^{-(x-\delta) / \theta} d \theta=\int_{0}^{\infty} \frac{y+\delta}{\theta} e^{-y / \theta} d y=\theta+\delta\)
\(E\left(X^{2}\right)=\int_{\delta}^{\infty} \frac{x^{2}}{\theta} e^{-(x-\delta) / \theta} d \theta=\int_{0}^{\infty} \frac{y^{2}+2 y \delta+\delta^{2}}{\theta} e^{-y / \theta} d y=2 \theta^{2}+2 \theta \delta+\delta^{2}\)
Both derivations use the substitution \(y=x-\delta\) and then recognize that the various integrals are requesting moments from an ordinary exponential distribution. The method of moments solves the two equations
\(\theta+\delta=10\)
\(2 \theta^{2}+2 \theta \delta+\delta^{2}=130.6\)
\(\hat{\delta}=4.468\).
It is faster to do the problem if it is noted that \(X=Y+\delta\) where \(Y\) has an ordinary exponential distribution. Then \(E(X)=E(Y)+\delta=\theta+\delta, \operatorname{Var}(X)=\operatorname{Var}(Y)=\theta^{2}\).
Question: \(N\) is the random variable for the number of accidents in a single year. \(N\) follows the distribution:
\[
\operatorname{Pr}(N=n)=0.9(0.1)^{n-1}, \quad n=1,2, \ldots
\]
\(X_{i}\) is the random variable for the claim amount of the \(i\) th accident. \(X_{i}\) follows the distribution:
\[
g\left(x_{i}\right)=0.01 e^{-0.01 x_{i}}, \quad x_{i}>0, i=1,2, \ldots
\]

Let \(U\) and \(V_{1}, V_{2}, \ldots\) be independent random variables following the uniform distribution on \((0,1)\). You use the inversion method with \(U\) to simulate \(N\) and \(V_{i}\) to simulate \(X_{i}\). You are given the following random numbers for the first simulation:

\begin{tabular}{|c|c|c|c|c|}
\hline\(u\) & \(v_{1}\) & \(v_{2}\) & \(v_{3}\) & \(v_{4}\) \\
\hline 0.05 & 0.30 & 0.22 & 0.52 & 0.46 \\
\hline
\end{tabular}

Calculate the total amount of claims during the year for the first simulation.

Options:
A
0
B
36
C
72
D
108
E
144

Correct answer: B

Solution: For the simulated value of \(N\), note that \(f(1)=F(1)=0.9\) and because \(0.05<0.9, n=1\). For the single value of \(x\) and recognizing that it has an exponential distribution,
\(0.30=1-e^{-0.01 x}\)
\(-0.01 x=\ln (1-0.30)=-0.3567\)
\(x-35.67\).
The amount of total claims during the year is 35.67 .
Question: Actuaries have modeled auto windshield claim frequencies. They have concluded that the number of windshield claims filed per year per driver follows the Poisson distribution with parameter \(\lambda\), where \(\lambda\) follows the gamma distribution with mean 3 and variance 3 .

Calculate the probability that a driver selected at random will file no more than 1 windshield claim next year.

Options:
A
0.15
B
0.19
C
0.20
D
0.24
E
0.31

Correct answer: E

Solution: The distribution of claims (a gamma mixture of Poissons) is negative binomial.
\(E(N)=E_{\Lambda}[E(N \mid \Lambda)]=E_{\Lambda}(\Lambda)=3\)
\(\operatorname{Var}(N)=E_{\Lambda}[\operatorname{Var}(N \mid \Lambda)]+\operatorname{Var}_{\Lambda}[E(N \mid \Lambda)]=E_{\Lambda}(\Lambda)+\operatorname{Var}_{\Lambda}(\Lambda)=6\)
\(r \beta=3\)
\(r \beta(1+\beta)=6\)
\((1+\beta)=6 / 3=2 ; \beta=1\)
\(r \beta=3 ; r=3\)
\(p_{0}=(1+\beta)^{-r}=0.125\)
\(p_{1}=\frac{r \beta}{(1+\beta)^{r+1}}=0.1875\)
\(\operatorname{Pr}(\) at most 1\()=p_{0}+p_{1}=0.3125\).