BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: For a stop-loss insurance on a three person group:
(i) Loss amounts are independent.
(ii) The distribution of loss amount for each person is:

\begin{tabular}{|c|c|}
\hline Loss Amount & Probability \\
\hline 0 & 0.4 \\
\hline 1 & 0.3 \\
\hline 2 & 0.2 \\
\hline 3 & 0.1 \\
\hline
\end{tabular}
(iii) The stop-loss insurance has a deductible of 1 for the group. Calculate the net stop-loss premium.

Options:
A
\(\quad 2.00\)
B
2.03
C
2.06
D
2.09
E
\(\quad 2.12\)

Correct answer: C

Solution: \(X=\) losses on one life
\(E(X)=0.3(1)+0.2(2)+0.1(3)=1\)
\(S=\) total losses
\(E(S)=3 E(X)=3(1)=3\)
\(E\left[(S-1)_{+}\right]=E(S)-1\left[1-F_{S}(0)\right]=3-1\left[1-f_{S}(0)\right]=3-\left(1-0.4^{3}\right)=2.064\)

Alternatively, the expected retained payment is \(0 f_{s}(0)+1\left[1-f_{s}(0)\right]=0.936\) and the stop-loss premium is \(3-0.936=2.064\).
Question: For a discrete probability distribution, you are given the recursion relation
\[
p(k)=\frac{2}{k} p(k-1), \quad k=1,2, \ldots
\]

Calculate \(p(4)\).

Options:
A
0.07
B
0.08
C
0.09
D
0.10
E
\(\quad 0.11\)

Correct answer: C

Solution: \(p(k)=\frac{2}{k} p(k-1)=\left(0+\frac{2}{k}\right) p(k-1)\)
Thus an ( \(a, b, 0\) ) distribution with \(a=0, b=2\).
Thus, Poisson with \(\lambda=2\).
\(p(4)=\frac{e^{-2} 2^{4}}{4!}=0.09\)
Question: A company insures a fleet of vehicles. Aggregate losses have a compound Poisson distribution. The expected number of losses is 20. Loss amounts, regardless of vehicle type, have exponential distribution with \(\theta=200\).

To reduce the cost of the insurance, two modifications are to be made:
(i) A certain type of vehicle will not be insured. It is estimated that this will reduce loss frequency by \(20 \%\).
(ii) A deductible of 100 per loss will be imposed.

Calculate the expected aggregate amount paid by the insurer after the modifications.

Options:
A
1600
B
1940
C
2520
D
3200
E
3880

Correct answer: B

Solution: By the memoryless property, the distribution of amounts paid in excess of 100 is still exponential with mean 200.

With the deductible, the probability that the amount paid is 0 is \(F(100)=1-e^{-100 / 200}=0.393\).
Thus the average amount paid per loss is \((0.393)(0)+(0.607)(200)=121.4\)
The expected number of losses is \((20)(0.8)=16\).
The expected amount paid is \((16)(121.4)=1942\).
Question: The number of claims, \(N\), made on an insurance portfolio follows the following distribution:

\begin{tabular}{|c|c|}
\hline\(n\) & \(\operatorname{Pr}(N=n)\) \\
\hline 0 & 0.7 \\
\hline 2 & 0.2 \\
\hline 3 & 0.1 \\
\hline
\end{tabular}

If a claim occurs, the benefit is 0 or 10 with probability 0.8 and 0.2 , respectively.
The number of claims and the benefit for each claim are independent.
Calculate the probability that aggregate benefits will exceed expected benefits by more than 2 standard deviations.

Options:
A
0.02
B
0.05
C
0.07
D
0.09
E
\(\quad 0.12\)

Correct answer: E

Solution: \(E(N)=0.7(0)+0.2(2)+0.1(3)=0.7\)
\(\operatorname{Var}(N)=0.7(0)+0.2(4)+0.1(9)-0.7^{2}=1.21\)
\(E(X)=0.8(0)+0.2(10)=2\)
\(\operatorname{Var}(X)=0.8(0)+0.2(100)-2^{2}=16\)
\(E(S)=E(N) E(X)=0.7(2)=1.4\)
\(\operatorname{Var}(S)=E(N) \operatorname{Var}(X)+E(X)^{2} \operatorname{Var}(N)=0.7(16)+4(1.21)=16.04\)
\(S D(S)=\sqrt{16.04}=4\)
\(\operatorname{Pr}(S>1.4+2(4)=9.4)=1-\operatorname{Pr}(S=0)=1-0.7-0.2(0.8)^{2}-0.1(0.8)^{3}=0.12\)
The last line follows because there are no possible values for \(S\) between 0 and 10 . A value of 0 can be obtained three ways: no claims, two claims both for 0 , three claims all for 0 .
Question: Annual prescription drug costs are modeled by a Pareto distribution with \(\theta=2000\) and \(\alpha=2\).

A prescription drug plan pays annual drug costs for an insured member subject to the following provisions:
(i) The insured pays \(100 \%\) of costs up to the ordinary annual deductible of 250 .
(ii) The insured then pays \(25 \%\) of the costs between 250 and 2250 .
(iii) The insured pays \(100 \%\) of the costs above 2250 until the insured has paid 3600 in total.
(iv) The insured then pays \(5 \%\) of the remaining costs.

Calculate the expected annual plan payment.

Options:
A
1120
B
1140
C
1160
D
1180
E
1200

Correct answer: C

Solution: \(E(X \wedge x)=\frac{\theta}{\alpha-1}\left[1-\left(\frac{\theta}{x+\theta}\right)^{\alpha-1}\right]=\frac{2000}{1}\left[1-\frac{2000}{x+2000}\right]=\frac{2000 x}{x+2000}\)

\begin{tabular}{c|c}
\(x\) & \(E(X \wedge x)\) \\
\hline\(\infty\) & 2000 \\
250 & 222 \\
2250 & 1059 \\
5100 & 1437
\end{tabular}
\(0.75[E(X \wedge 2250)-E(X \wedge 250)]+0.95[E(X)-E(X \wedge 5100)]\)
\(0.75(1059-222)+0.95(2000-1437)=1162.6\)
The 5100 breakpoint was determined by when the insured's share reaches 3600:
\(3600=250+0.25(2250-250)+(5100-2250)\)
Question: Two types of insurance claims are made to an insurance company. For each type, the number of claims follows a Poisson distribution and the amount of each claim is uniformly distributed as follows:

\begin{tabular}{|c|c|c|}
\hline Type of Claim & \begin{tabular}{c} 
Poisson Parameter \(\lambda\) for \\
Number of Claims in one \\
year
\end{tabular} & \begin{tabular}{c} 
Range of Each Claim \\
Amount
\end{tabular} \\
\hline I & 12 & \((0,1)\) \\
\hline II & 4 & \((0,5)\) \\
\hline
\end{tabular}

The numbers of claims of the two types are independent and the claim amounts and claim numbers are independent.

Calculate the normal approximation to the probability that the total of claim amounts in one year exceeds 18 .

Options:
A
0.37
B
0.39
C
0.41
D
0.43
E
\(\quad 0.45\)

Correct answer: A

Solution: \(N_{I}, N_{I I}\) denote the random variables for \# of claims for Types I and II in one year
\(X_{I}, X_{I I}\) denote the claim amount random variables for Types I and II
\(S_{I}, S_{I I}\) denote the total claim amount random variables for Types I and II
\(S=S_{I}+S_{I I}\)
\(E\left(N_{I}\right)=\operatorname{Var}\left(N_{I}\right)=12, E\left(N_{I I}\right)=\operatorname{Var}\left(N_{I I}\right)=4\)
\(E\left(X_{I}\right)=(0+1) / 2=1 / 2, \operatorname{Var}\left(X_{I}\right)=(1-0)^{2} / 12=1 / 12\)
\(E\left(X_{\text {II }}\right)=(0+5) / 2=5 / 2, \operatorname{Var}\left(X_{\text {II }}\right)=(5-0)^{2} / 12=25 / 12\)
\(E(S)=E\left(N_{I}\right) E\left(X_{I}\right)+E\left(N_{I I}\right) E\left(X_{I I}\right)=12(1 / 2)+4(5 / 2)=16\)
\(\operatorname{Var}(S)=E\left(N_{I}\right) \operatorname{Var}\left(X_{I}\right)+E\left(X_{I}\right)^{2} \operatorname{Var}\left(N_{I}\right)+E\left(N_{I I}\right) \operatorname{Var}\left(X_{I I}\right)+E\left(X_{I I}\right)^{2} \operatorname{Var}\left(N_{I I}\right) =12(1 / 12)+(1 / 2)^{2}(12)+4(25 / 12)+(5 / 2)^{2}(4)=37.33\)
\(\operatorname{Pr}(S>18)=\operatorname{Pr}\left(Z>\frac{18-16}{\sqrt{37.33}}=0.327\right)=1-\Phi(0.327)=0.37\)
Question: The number of annual losses has a Poisson distribution with a mean of 5 . The size of each loss has a Pareto distribution with \(\theta=10\) and \(\alpha=2.5\). An insurance policy for the losses has an ordinary deductible of 5 per loss.

Calculate the expected value of the aggregate annual payments for this policy.

Options:
A
8
B
13
C
18
D
23
E
28

Correct answer: C

Solution: Let \(X\) be the loss random variable. Then, \((X-5)_{+}\)is the claim random variable.
\(E(X)=\frac{10}{2.5-1}=6.667\)
\(E(X \wedge 5)=\left(\frac{10}{2.5-1}\right)\left[1-\left(\frac{10}{5+10}\right)^{2.5-1}\right]=3.038\)
\(E\left[(X-5)_{+}\right]=E(X)-E(X \wedge 5)=6.667-3.038=3.629\)

Expected aggregate claims \(=E(N) E\left[(X-5)_{+}\right]=5(3.629)=18.15\).
Question: Losses in Year 1 follow a Pareto distribution with \(\alpha=2\) and \(\theta=5\). Losses in Year 2 are uniformly \(20 \%\) higher than in Year 1. An insurance covers each loss subject to an ordinary deductible of 10 .

Calculate the Loss Elimination Ratio in Year 2.

Options:
A
0.567
B
0.625
C
0.667
D
0.750
E
0.800

Correct answer: B

Solution: A Pareto ( \(\alpha=2, \theta=5\) ) distribution with \(20 \%\) inflation becomes Pareto with \(\alpha=2, \theta=5(1.2)=6\). In Year 2
\(E(X)=\frac{6}{2-1}=6\)
\(E(X \wedge 10)=\frac{6}{2-1}\left[1-\left(\frac{6}{10+6}\right)^{2-1}\right]=3.75\)
\(E\left[(X-10)_{+}\right]_{+}=E(X)-E(X \wedge 10)=6-3.75-2.25\)
\(\mathrm{LER}=1-\frac{E\left[(X-10)_{+}\right]}{E(X)}=1-\frac{2.25}{6}=0.625\)
Question: You are given the following three observations:
\[
\begin{array}{lll}
0.74 & 0.81 & 0.95
\end{array}
\]

You fit a distribution with the following density function to the data:
\[
f(x)=(p+1) x^{p}, \quad 0<x<1, p>-1
\]

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

Options:
A
4.0
B
4.1
C
4.2
D
4.3
E
4.4

Correct answer: D

Solution: \[
\begin{aligned}
L(p) & =f(0.74) f(0.81) f(0.95)=(p+1) 0.74^{p}(p+1) 0.81^{p}(p+1) 0.95^{p} \\
& =(p+1)^{3}(0.56943)^{p} \\
l(p) & =\ln L(p)=3 \ln (p+1)+p \ln (0.56943) \\
l^{\prime}(p) & =\frac{3}{p+1}-0.563119=0 \\
p+1 & =\frac{3}{0.563119}=5.32747, p=4.32747
\end{aligned}
\]
Question: You are given:
(i) The number of claims has probability function:
\[
p(x)=\binom{m}{x} q^{x}(1-q)^{m-x}, \quad x=0,1, \ldots, m
\]
(ii) The actual number of claims must be within \(1 \%\) of the expected number of claims with probability 0.95 .
(iii) The expected number of claims for full credibility is 34,574 .

Calculate \(q\).

Options:
A
0.05
B
0.10
C
0.20
D
0.40
E
\(\quad 0.80\)

Correct answer: B

Solution: Let \(n\) be the number of observations. For full credibility, \(n=\left(\frac{1.96}{0.01}\right)^{2} \frac{m q(1-q)}{(m q)^{2}}=38,416 \frac{1-q}{m q}\).
The required expected number of claims is \(n m q=34,574=38,416 \frac{1-q}{m q} m q=38,416(1-q)\).
Then \(q=1-34,574 / 38,416=0.1\).
Question: You are given:
(i) A sample of losses is:
\(600 \quad 700\)
900
(ii) No information is available about losses of 500 or less.
(iii) Losses are assumed to follow an exponential distribution with mean \(\theta\).

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

Options:
A
233
B
400
C
500
D
733
E
1233

Correct answer: A

Solution: These observations are truncated at 500. The contribution to the likelihood function is
\(\frac{f(x)}{1-F(500)}=\frac{\theta^{-1} e^{-x / \theta}}{e^{-500 / \theta}}\). Then the likelihood function is
\(L(\theta)=\frac{\theta^{-1} e^{-600 / \theta} \theta^{-1} e^{-700 / \theta} \theta^{-1} e^{-900 / \theta}}{\left(e^{-500 / \theta}\right)^{3}}=\theta^{-3} e^{-700 / \theta}\)
\(l(\theta)=\ln L(\theta)=-3 \ln \theta-700 \theta^{-1}\)
\(l^{\prime}(\theta)=-3 \theta^{-1}+700 \theta^{-2}=0\)
\(\theta=700 / 3=233.33\).
Question: You are given:
(i) The number of claims follows a Poisson distribution with mean \(\lambda\).
(ii) Observations other than 0 and 1 have been deleted from the data.
(iii) The data contain an equal number of observations of 0 and 1 .

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

Options:
A
0.50
B
0.75
C
1.00
D
1.25
E
1.50

Correct answer: C

Solution: There are \(n / 2\) observations of \(N=0\) (given \(N=0\) or 1 ) and \(n / 2\) observations of \(N=1\) (given \(N =0\) or 1 ). The likelihood function is
\(L=\left(\frac{e^{-\lambda}}{e^{-\lambda}+\lambda e^{-\lambda}}\right)^{n / 2}\left(\frac{\lambda e^{-\lambda}}{e^{-\lambda}+\lambda e^{-\lambda}}\right)^{n / 2}=\frac{\lambda^{n / 2} e^{-n \lambda}}{\left(e^{-\lambda}+\lambda e^{-\lambda}\right)^{n}}=\frac{\lambda^{n / 2}}{(1+\lambda)^{n}}\). Taking logs, differentiating and solving provides the answer.
\(l=\ln L=(n / 2) \ln \lambda-n \ln (1+\lambda)\)
\(l^{\prime}=\frac{n}{2 \lambda}-\frac{n}{1+\lambda}=0\)
\(n(1+\lambda)-n 2 \lambda=0\)
\(1-\lambda=0, \quad \lambda=1\).
Question: For a collective risk model the number of losses, \(N\), has a Poisson distribution with \(\lambda=20\).

The common distribution of the individual losses has the following characteristics:
(i) \(\quad E[X]=70\)
(ii) \(\quad E[X \wedge 30]=25\)
(iii) \(\quad \operatorname{Pr}(X>30)=0.75\)
(iv) \(\quad E\left[X^{2} \mid X>30\right]=9000\)

An insurance covers aggregate losses subject to an ordinary deductible of 30 per loss. Calculate the variance of the aggregate payments of the insurance.

Options:
A
\(\quad 54,000\)
B
\(\quad 67,500\)
C
81,000
D
94,500
E
108,000

Correct answer: B

Solution: Losses in excess of the deductible occur at a Poisson rate of \(\lambda^{*}=[1-F(30)] \lambda=0.75(20)=15\) The expected payment squared per payment is
\[
\begin{aligned}
& E\left[(X-30)^{2} \mid X>30\right)=E\left(X^{2}-60 X+900 \mid X>30\right) \\
& \quad E\left[X^{2}-60(X-30)-900 \mid X>30\right] \\
& \quad E\left[X^{2} \mid X>30\right]-60 \frac{E(X)-E(X \wedge 30)}{1-F(30)}-900 \\
& \quad=9000-60 \frac{70-25}{0.75}-900=4500
\end{aligned}
\]

The variance of \(S\) is the expected number of payments times the second moment, 15(4500)= 67,500.
Question: For a collective risk model:
(i) The number of losses has a Poisson distribution with \(\lambda=2\).
(ii) The common distribution of the individual losses is:

\begin{tabular}{ccc}
\(x\) & & \(f_{X}(x)\) \\
\cline { 1 - 1 } \cline { 3 - 3 } 1 & & 0.6 \\
2 & & 0.4
\end{tabular}

An insurance covers aggregate losses subject to a deductible of 3 . Calculate the expected aggregate payments of the insurance.

Options:
A
\(\quad 0.74\)
B
0.79
C
0.84
D
0.89
E
\(\quad 0.94\)

Correct answer: A

Solution: \[
\begin{aligned}
& E\left[(S-3)_{+}\right]=E(S)-3+3 f_{S}(0)+2 f_{S}(1)+1 f_{S}(2) \\
& E(S)=2[0.6+2(0.4)]=2.8 \\
& f_{S}(0)=e^{-2}, f_{S}(1)=e^{-2}(2)(0.6)=1.2 e^{-2} \\
& f_{S}(2)=e^{-2}(2)(0.4)+\frac{e^{-2} 2^{2}}{2!}(0.6)^{2}=1.52 e^{-2} \\
& E\left[(S-3)_{+}\right]=2.8-3+3 e^{-2}+2\left(1.2 e^{-2}\right)+1\left(1.52 e^{-2}\right)=-0.2+6.92 e^{-2}=0.7365
\end{aligned}
\]
Question: A discrete probability distribution has the following properties:
(i) \(\quad p_{k}=c\left(1+\frac{1}{k}\right) p_{k-1}\) for \(k=1,2, \ldots\)
(ii) \(\quad p_{0}=0.5\)

Calculate \(c\).

Options:
A
\(\quad 0.06\)
B
0.13
C
0.29
D
0.35
E
\(\quad 0.40\)

Correct answer: C

Solution: Write (i) as \(\frac{p_{k}}{p_{k-1}}=c+\frac{c}{k}\)
This is an ( \(a, b, 0\) ) distribution with \(a=b=c\).
Which?
1. If Poisson, \(a=0\), so \(b=0\) and thus \(p_{0}=0.5\) and \(p_{1}=p_{2}=\cdots=0\). The probabilities do not sum to 1 and so not Poisson.
2. If Geometric, \(b=0\), so \(a=0\). By same reasoning as \#1, not Geometric.
3. If binomial, \(a\) and \(b\) have opposite signs. But here \(a=b\), so not binomial.
4. Thus negative binomial.
\[
\begin{aligned}
& 1=\frac{a}{b}=\frac{\beta /(1+\beta)}{(r-1) \beta /(1-\beta)}=\frac{1}{r-1} \text { so, } r=2 \\
& p_{0}=0.5=(1+\beta)^{-r}=(1+\beta)^{-2} \Rightarrow \beta=\sqrt{2}-1=0.414 \\
& c=a=\beta /(1+\beta)=0.29
\end{aligned}
\]
Question: The repair costs for boats in a marina have the following characteristics:

\begin{tabular}{|l|l|l|l|l|}
\hline Boat type & Number of boats & Probability that repair is needed & Mean of repair cost given a repair & Variance of repair cost given a repair \\
\hline Power boats & 100 & 0.3 & 300 & 10,000 \\
\hline Sailboats & 300 & 0.1 & 1000 & 400,000 \\
\hline Luxury yachts & 50 & 0.6 & 5000 & 2,000,000 \\
\hline
\end{tabular}

At most one repair is required per boat each year. Repair incidence and cost are mutually independent.

The marina budgets an amount, \(Y\), equal to the aggregate mean repair costs plus the standard deviation of the aggregate repair costs.

Calculate \(Y\).

Options:
A
200,000
B
210,000
C
220,000
D
230,000
E
240,000

Correct answer: B

Solution: The number of repairs for each boat type has a binomial distribution.
For power boats:
\(E(S)=100(0.3)(300)=9,000\),
\(\operatorname{Var}(S)=100(0.3) 10,000)+100(0.3)(0.7)\left(300^{2}\right)=2,190,000\)
For sail boats:
\(E(S)=300(0.1)(1,000)=30,000\),
\(\operatorname{Var}(S)=300(0.1)(400,000)+300(0.1)(0.9)\left(1,000^{2}\right)=39,000,000\)
For luxury yachts:
\(E(S)=50(0.6)(5,000)=150,000\),
\(\operatorname{Var}(S)=50(0.6)(0.4)(2,000,000)+50(0.6)(0.4)\left(5,000^{2}\right)=360,000,000\)
The sums are 189,000 expected and a variance of 401,190,000 for a standard deviation of 20,030 . The mean plus standard deviation is 209,030 .
Question: For an insurance policy:
(i) Losses can be 100,200 or 300 with respective probabilities \(0.2,0.2\), and 0.6 .
(ii) The policy has an ordinary deductible of 150 per loss.
(iii) \(\quad Y^{P}\) is the claim payment per payment random variable.

Calculate \(\operatorname{Var}\left(X^{P}\right)\).

Options:
A
1500
B
1875
C
2250
D
2625
E
3000

Correct answer: B

Solution: \(S_{X}(150)=1-0.2=0.8\)
\(f_{Y^{P}}(y)=\frac{f_{X}(y+150)}{S_{X}(150)}, f_{Y^{P}}(50)=\frac{0.2}{0.8}=0.25, f_{Y^{P}}(150)=\frac{0.6}{0.8}=0.75\)
\(E\left(Y^{P}\right)=0.25(50)+0.75(150)=125\)
\(E\left[\left(Y^{P}\right)^{2}\right]=0.25\left(50^{2}\right)+0.75\left(150^{2}\right)=17,500\)
\(\operatorname{Var}\left(Y^{P}\right)=17,500-125^{2}=1,875\)
Question: The distribution of a loss, \(X\), is a two-point mixture:
(i) With probability \(0.8, X\) has a Pareto distribution with \(\alpha=2\) and \(\theta=100\).
(ii) With probability 0.2, \(X\) has a Pareto distribution with with \(\alpha=4\) and \(\theta=3000\). Calculate \(\mathrm{P}(X \leq 200)\).

Options:
A
\(\quad 0.76\)
B
0.79
C
0.82
D
0.85
E
\(\quad 0.88\)

Correct answer: A

Solution: \(F(200)=0.8\left[1-\left(\frac{100}{200+100}\right)^{2}\right]+0.2\left[1-\left(\frac{3000}{3000+200}\right)^{4}\right]=0.7566\)
Question: You are given:
(i) The number of claims follows a negative binomial distribution with parameters \(r\) and \(\beta=3\).
(ii) Claim severity has the following distribution:

\begin{tabular}{|c|c|}
\hline Claim Size & Probability \\
\hline 1 & 0.4 \\
\hline 10 & 0.4 \\
\hline 100 & 0.2 \\
\hline
\end{tabular}
(iii) The number of claims is independent of the severity of claims.

Calculate the expected number of claims needed for aggregate losses to be within \(10 \%\) of expected aggregate losses with \(95 \%\) probability.

Options:
A
Less than 1200
B
At least 1200 , but less than 1600
C
At least 1600 , but less than 2000
D
At least 2000, but less than 2400
E
At least 2400

Correct answer: E

Solution: For claim severity,
\(\mu_{S}=1(0.4)+10(0.4)+100(0.2)=24.4\),
\(\sigma_{S}^{2}=1^{2}(0.4)+10^{2}(0.4)+100^{2}(0.2)-24.4^{2}=1,445.04\).
For claim frequency,
\(\mu_{F}=r \beta=3 r, \quad \sigma_{F}^{2}=r \beta(1+\beta)=12 r\),
For aggregate losses,
\(\mu=\mu_{S} \mu_{F}=24.4(3 r)=73.2 r\),
\(\sigma^{2}=\mu_{S}^{2} \sigma_{F}^{2}+\sigma_{S}^{2} \mu_{F}=24.4^{2}(12 r)+1,445.04(3 r)=11,479.44 r\).
For the given probability and tolerance, \(\lambda_{0}=(1.96 / 0.1)^{2}=384.16\).
The number of observations needed is
\(\lambda_{0} \sigma^{2} / \mu^{2}=384.16(11,479.44 r) /(73.2 r)^{2}=823.02 / r\).
The average observation produces \(3 r\) claims and so the required number of claims is \((823.02 / r)(3 r)=2,469\).
Question: You are given the following 20 bodily injury losses before the deductible is applied:

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

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

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

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

Correct answer: C

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

The derivative is \(14 / \alpha-4.5327\) and setting it equal to zero gives \(\hat{\alpha}=3.089\).
Question: Personal auto property damage claims in a certain region are known to follow the Weibull distribution:
\[
F(x)=1-\exp \left[-\left(\frac{x}{\theta}\right)^{0.2}\right], \quad x>0
\]

A sample of four claims is:
\[
\begin{array}{llll}
130 & 240 & 300 & 540
\end{array}
\]

The values of two additional claims are known to exceed 1000.
Calculate the maximum likelihood estimate of \(\theta\).

Options:
A
Less than 300
B
At least 300, but less than 1200
C
At least 1200 , but less than 2100
D
At least 2100, but less than 3000
E
At least 3000

Correct answer: E

Solution: The density function is \(f(x)=\frac{0.2 x^{-0.8}}{\theta^{0.2}} e^{-(x / \theta)^{0.2}}\). The likelihood function is
\[
\begin{aligned}
L(\theta) & =f(130) f(240) f(300) f(540)[1-F(1000)]^{2} \\
& =\frac{0.2(130)^{-0.8}}{\theta^{0.2}} e^{-(130 / \theta)^{0.2}} \frac{0.2(240)^{-0.8}}{\theta^{0.2}} e^{-(240 / \theta)^{0.2}} \frac{0.2(300)^{-0.8}}{\theta^{0.2}} e^{-(300 / \theta)^{0.2}} \frac{0.2(540)^{-0.8}}{\theta^{0.2}} e^{-(540 / \theta)^{0.2}} e^{-(1000 / \theta)^{0.2}} e^{-(1000 / \theta)^{0.2}} \\
& \propto \theta^{-0.8} e^{-\theta^{-0.2}\left(130^{0.2}+240^{0.2}+300^{0.2}+540^{0.2}+1000^{0.2}+1000^{0.2}\right)} \\
l(\theta) & =-0.8 \ln (\theta)-\theta^{-0.2}\left(130^{0.2}+240^{0.2}+300^{0.2}+540^{0.2}+1000^{0.2}+1000^{0.2}\right) \\
& =-0.8 \ln (\theta)-20.2505 \theta^{-0.2} \\
l^{\prime}(\theta) & =-0.8 \theta^{-1}+0.2(20.2505) \theta^{-1.2}=0 \\
\theta^{-0.2} & =0.197526, \quad \hat{\theta}=3,325.67
\end{aligned}
\]
Question: In Year 1 a risk has a Pareto distribution with \(\alpha=2\) and \(\theta=3000\). In Year 2 losses inflate by \(20 \%\).

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

The risk is reinsured with a deductible that stays the same in each year. \(R_{i}\), the reinsurance premium in year \(i\), equals 1.1 times the expected reinsured claims.
\(\frac{R_{1}}{P_{1}}=0.55\)
Calculate \(\frac{R_{2}}{P_{2}}\).

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

Correct answer: D

Solution: For any deductible \(d\) and the given severity distribution
\(E\left[(X-d)_{+}\right]=E(X)-E(X \wedge d)=3000-3000\left(1-\frac{3000}{3000+d}\right)=3000\left(\frac{3000}{3000+d}\right)=\frac{9,000,000}{3000+d}\)
So \(P_{1}=1.2 \frac{9,000,000}{3000+600}=3000\)
Let \(r\) denote the reinsurer's deductible relative to insured losses. Thus, the reinsurer's deductible is \(600+r\) relative to losses. Thus
\(R_{1}=1.1\left(\frac{9,000,000}{3000+600+r}\right)=0.55 P_{1}=0.55(3000)=1650 \Rightarrow r=2400\)
In Year 2, after \(20 \%\) inflation, losses will have a Pareto distribution with \(\alpha=2\) and \(\theta=1.2(3000)=3600\). The general formula for expected claims with a deductible of \(d\) is
\(E\left[(X-d)_{+}\right]=3600\left(\frac{3600}{3600+d}\right)=\frac{12,960,000}{3600+d}\)
\(P_{2}=1.2 \frac{12,960,000}{3000+600}=3703, R_{2}=1.1 \frac{12,960,000}{3000+600+2400}=2160, \frac{R_{2}}{P_{2}}=\frac{2160}{3703}=0.583\)
Question: For an insurance:
(i) The number of losses per year has a Poisson distribution with \(\lambda=10\).
(ii) Loss amounts are uniformly distributed on ( 0,10 ).
(iii) Loss amounts and the number of losses are mutually independent.
(iv) There is an ordinary deductible of 4 per loss.

Calculate the variance of aggregate payments in a year.

Options:
A
36
B
48
C
72
D
96
E
120

Correct answer: C

Solution: Since loss amounts are uniform on ( 0,10 ), \(40 \%\) of losses are below the deductible (4), and \(60 \%\) are above. Thus, claims occur at a Poisson rate \(\lambda^{*}=0.6(10)=6\).

Since loss amounts were uniform on ( 0,10 ), claims are uniform on ( 0,6 ).
Let \(N=\) number of claims; \(X=\) claim amount; \(S=\) aggregate claims.
\(E(N)=\operatorname{Var}(N)=\lambda^{*}=6\)
\(E(X)=(6-0) / 2=3\)
\(\operatorname{Var}(X)=(6-0)^{2} / 12=3\)
\(\operatorname{Var}(S)=E(N) \operatorname{Var}(X)+E(X)^{2} \operatorname{Var}(N)=6(3)+3^{2}(6)=72\)