BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: The random variable \(X\) has survival function:
\[
S_{X}(x)=\frac{\theta^{4}}{\left(\theta^{2}+x^{2}\right)^{2}}
\]

Two values of \(X\) are observed to be 2 and 4. One other value exceeds 4. Calculate the maximum likelihood estimate of \(\theta\).

Options:
A
Less than 4.0
B
At least 4.0 , but less than 4.5
C
At least 4.5 , but less than 5.0
D
At least 5.0 , but less than 5.5
E
At least 5.5

Correct answer: E

Solution: \(f(x)=-S^{\prime}(x)=\frac{4 x \theta^{4}}{\left(\theta^{2}+x^{2}\right)^{3}}\)
\(L(\theta)=f(2) f(4) S(4)=\frac{4(2) \theta^{4}}{\left(\theta^{2}+2^{2}\right)^{3}} \frac{4(4) \theta^{4}}{\left(\theta^{2}+4^{2}\right)^{3}} \frac{\theta^{4}}{\left(\theta^{2}+4^{2}\right)^{2}}=\frac{128 \theta^{12}}{\left(\theta^{2}+4\right)^{3}\left(\theta^{2}+16\right)^{5}}\)
\(l(\theta)=\ln 128+12 \ln \theta-3 \ln \left(\theta^{2}+4\right)-5 \ln \left(\theta^{2}+16\right)\)
\(l^{\prime}(\theta)=\frac{12}{\theta}-\frac{6 \theta}{\theta^{2}+4}-\frac{10 \theta}{\theta^{2}+16}=0 ; 12\left(\theta^{4}+20 \theta^{2}+64\right)-6\left(\theta^{4}+16 \theta^{2}\right)-10\left(\theta^{4}+4 \theta^{2}\right)=0\)
\(0=-4 \theta^{4}+104 \theta^{2}+768=\theta^{4}-26 \theta^{2}-192\)
\(\theta^{2}=\frac{26 \pm \sqrt{26^{2}+4(192)}}{2}=32 ; \theta=5.657\)
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
\(\quad-6,730\)

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) At time 4 hours, there are 5 working light bulbs.
(ii) The 5 bulbs are observed for \(p\) more hours.
(iii) Three light bulbs burn out at times 5,9, and 13 hours, while the remaining light bulbs are still working at time \(4+p\) hours.
(iv) The distribution of failure times is uniform on ( \(0, \omega\) ).
(v) The maximum likelihood estimate of \(\omega\) is 29 .

Calculate \(p\).

Options:
A
Less than 10
B
At least 10 , but less than 12
C
At least 12 , but less than 14
D
At least 14 , but less than 16
E
At least 16

Correct answer: D

Solution: \(L(\omega)=\frac{\frac{1}{\omega} \frac{1}{\omega} \frac{1}{\omega}\left(\frac{\omega-4-p}{\omega}\right)^{2}}{\left(\frac{\omega-4}{\omega}\right)^{5}}=\frac{(\omega-4-p)^{2}}{(\omega-4)^{5}}\)
\(l(\omega)=2 \ln (\omega-4-p)-5 \ln (\omega-4), l^{\prime}(\omega)=\frac{2}{\omega-4-p}-\frac{5}{\omega-4}=0\)
\(0=l^{\prime}(29)=\frac{2}{25-p}-\frac{5}{25}=>p=15\).
The denominator in the likelihood function is \(S(4)\) to the power of five to reflect the fact that it is known that each observation is greater than 4 .
Question: A company has determined that the limited fluctuation full credibility standard is 2000 claims if:
(i) The total number of claims is to be within \(3 \%\) of the true value with probability \(p\).
(ii) The number of claims follows a Poisson distribution.

The standard is changed so that the total cost of claims is to be within \(5 \%\) of the true value with probability \(p\), where claim severity has probability density function:
\[
f(x)=\frac{1}{10,000}, \quad 0 \leq x \leq 10,000
\]

Calculate the expected number of claims necessary to obtain full credibility under the new standard using limited fluctuation credibility.

Options:
A
720
B
960
C
2160
D
2667
E
2880

Correct answer: B

Solution: For the severity distribution the mean is 5,000 and the variance is \(10,000^{2} / 12\). For credibility based on accuracy with regard to the number of claims,
\(2000=\left(\frac{z}{0.03}\right)^{2}, \quad z^{2}=1.8\)
Where \(z\) is the appropriate value from the standard normal distribution. For credibility based on accuracy with regard to the total cost of claims, the number of claims needed is
\(\frac{z^{2}}{0.05^{2}}\left(1+\frac{10000^{2} / 12}{5000^{2}}\right)=960\).
Question: For a group of policies, you are given:
(i) Losses follow the distribution function
\[
F(x)=1-\theta / x, \quad x>\theta .
\]
(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: Loss amounts have the distribution function
\[
F(x)= \begin{cases}(x / 100)^{2}, & 0 \leq x \leq 100 \\ 1, & x>100\end{cases}
\]

An insurance policy pays \(80 \%\) of the amount of the loss in excess of an ordinary deductible of 20 , subject to a maximum payment of 60 per loss.

Calculate the conditional expected claim payment, given that a payment has been made.

Options:
A
37
B
39
C
43
D
47
E
49

Correct answer: B

Solution: Pays \(80 \%\) of loss over 20, with cap of payment at 60, hence \(u=60 / 0.8+20=95\).
\[
\begin{aligned}
& E(Y \text { per loss })=\alpha[E(X \wedge 95)-E(X \wedge 20)]=0.8\left[\int_{0}^{95} S(x) d x-\int_{0}^{20} S(x) d x\right] \\
& \quad=0.8 \int_{20}^{95} S(x) d x=0.8 \int_{20}^{95}\left(1-\frac{x^{2}}{10,000}\right) d x=\left.0.8\left(x-\frac{x^{3}}{30,000}\right)\right|_{20} ^{95}=37.35 \\
& E(Y \text { per payment })=\frac{E(Y \text { per loss })}{1-F(20)}=\frac{37.35}{0.96}=38.91
\end{aligned}
\]
Question: Aggregate losses are modeled as follows:
(i) The number of losses has a Poisson distribution with \(\lambda=3\).
(ii) The amount of each loss has a Burr distribution with \(\alpha=3, \theta=2, \gamma=1\).
(iii) The number of losses and the amounts of the losses are mutually independent.

Calculate the variance of aggregate losses.

Options:
A
12
B
14
C
16
D
18
E
20

Correct answer: A

Solution: Let \(S=\) aggregate losses, \(X=\) severity
Since the frequency is Poisson,
\(\operatorname{Var}(S)=\lambda E\left(X^{2}\right)\)
\(E\left(X^{2}\right)=\frac{2^{2} \Gamma(3) \Gamma(1)}{\Gamma(3)}=4\) (table lookup)
\(\operatorname{Var}(S)=3(4)=12\)
You would get the same result if you used
\(\operatorname{Var}(S)=E(N) \operatorname{Var}(X)+\operatorname{Var}(N) E(X)^{2}\)
Question: A risk has a loss amount that has a Poisson distribution with mean 3 .

An insurance policy covers the risk with an ordinary deductible of 2 . An alternative insurance policy replaces the deductible with coinsurance \(\alpha\), which is the proportion of the loss paid by the policy, so that the expected cost remains the same.

Calculate \(\alpha\).

Options:
A
0.22
B
0.27
C
0.32
D
0.37
E
\(\quad 0.42\)

Correct answer: E

Solution: \[
\begin{aligned}
E(X & \wedge 2)=1 f(1)+2[1-F(1)]=1 f(1)+2[1-f(0)-f(1)] \\
& =1\left(3 e^{-3}\right)+2\left(1-e^{-3}-3 e^{-3}\right)=2-5 e^{-3}=1.75
\end{aligned}
\]

Cost per loss with deductible is
\(E(X)-E(X \wedge 2)=3-1.75=1.25\)
Cost per loss with coinsurance is \(\alpha E(X)=3 \alpha\)
Equating cost: \(3 \alpha=1.25 \Rightarrow \alpha=0.42\)
Question: An individual performs dangerous motorcycle jumps at extreme sports events around the world.

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

An insurance policy reimburses motorcycle repair costs subject to the following provisions:
(i) The annual ordinary deductible is 1000 .
(ii) The policyholder pays \(20 \%\) of repair costs between 1000 and 6000 each year.
(iii) The policyholder pays \(100 \%\) of the annual repair costs above 6000 until they have paid 10,000 in out-of-pocket repair costs each year.
(iv) The policyholder 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 the policyholder has paid 10,000 :
\[
\begin{aligned}
& 1000=\text { deductible } \\
& 1000=20 \% \text { of costs between } 1000 \text { and } 6000 \\
& 8000=100 \% \text { of costs between } 14,000 \text { and } 6,000
\end{aligned}
\]
\[
E(X \wedge x)=\theta\left(1-\frac{\theta}{x+\theta}\right)=\frac{5000 x}{x+5000}
\]
\[
\begin{array}{ccc}
x & & E(X \wedge x) \\
\cline { 1 - 1 } \cline { 3 - 3 } 1000 & & 833.33 \\
6000 & & 2727.27 \\
14000 & & 3684.21 \\
\infty & & 5000
\end{array}
\]
\[
\begin{aligned}
0.80 & {[E(X \wedge 6000)-E(X \wedge 1000)]+0.90[E(X)-E(X \wedge 14000)] } \\
& =0.80[2727,27-833.33]+0.90[5000-3684.21] \\
& =1515.15+1184.21=2699.36
\end{aligned}
\]
Question: For an aggregate loss distribution \(S\) :
(i) The number of claims has a negative binomial distribution with \(r=16\) and \(\beta=6\).
(ii) The claim amounts are uniformly distributed on the interval ( 0,8 ).
(iii) The number of claims and claim amounts are mutually independent.

Calculate the premium such that the probability that aggregate losses will exceed the premium is 5\% using the normal approximation for aggregate losses.

Options:
A
500
B
520
C
540
D
560
E
580

Correct answer: D

Solution: We have the following table:

\begin{tabular}{|l|l|l|l|}
\hline \multicolumn{1}{|c|}{ Item } & \multicolumn{1}{c|}{ Dist } & \multicolumn{1}{c|}{\(E()\)} & \multicolumn{1}{c|}{\(\operatorname{Var}()\)} \\
\hline Number claims & \(N B(16,6)\) & \(16(6)=96\) & \(16(6)(7)=672\) \\
\hline Claims amounts & \(U(0,8)\) & \((8-0) / 2=4\) & \((8-0)^{2} / 12=5.33\) \\
\hline Aggregate & & \(4(96)=384\) & \(96(5.33)+672\left(4^{2}\right)=11,264\) \\
\hline
\end{tabular}

Premium \(=E(S)+1.645 \sqrt{\operatorname{Var}(S)}=384+1.645 \sqrt{11,264}=559\)
Question: The random variable \(X\) represents the random loss, before any deductible is applied, covered by an insurance policy. The probability density function of \(X\) is
\[
f(x)=2 x, \quad 0<x<1
\]

Payments are made subject to a deductible, \(d\), where \(0<d<1\).
The probability that a claim payment is less than 0.5 is equal to 0.64 . Calculate the value of \(d\).

Options:
A
0.1
B
0.2
C
0.3
D
0.4
E
0.5

Correct answer: C

Solution: \(F(x)=\int_{0}^{x} 2 y d y=x^{2}\). Let \(C\) be a random claim payment. Then \(C=0\) if \(X<d\) and \(C=X-d\) if \(X \geq d\). Then,
\(P(C<0.5)=0.64\)
\(P(C \geq 0.5)=0.36\)
\(P(X-d \geq 0.5)=0.36\)
\(P(X \geq 0.5+d)=0.36\)
\(F(0.5+d)=0.64\)
\((0.5+d)^{2}=0.64\)
\(0.5+d=0.8\)
\(d=0.3\)
Question: You are given the following loss data:

\begin{tabular}{|l|c|c|}
\hline Size of Loss & \begin{tabular}{c} 
Number of \\
Claims
\end{tabular} & \begin{tabular}{c} 
Ground-Up \\
Total Losses
\end{tabular} \\
\hline \(0-99\) & 1100 & 58,500 \\
\hline \(100-249\) & 400 & 70,000 \\
\hline \(250-499\) & 300 & 120,000 \\
\hline \(500-999\) & 200 & 150,000 \\
\hline\(>999\) & 100 & 200,000 \\
\hline Total & \(\mathbf{2 1 0 0}\) & \(\mathbf{5 9 8 , 5 0 0}\) \\
\hline
\end{tabular}

Calculate the percentage reduction in loss costs by moving from a 100 deductible to a 250 deductible.

Options:
A
\(25 \%\)
B
\(27 \%\)
C
\(29 \%\)
D
\(31 \%\)
E
\(33 \%\)

Correct answer: B

Solution: With no deductible, the loss cost is proportional to 598,500 .
With a 100 deductible it is proportional (in the same proportion) to \(598,500-58,500- 1000(100)=440,000\). With a 250 deductible it is \(598,500-58,500-70,000-600(250)=\) 320,000 . The reduction is \(120,000 / 440,000=0.273=27 \%\).
Question: You are given the following earned premiums for three calendar years:

\begin{tabular}{|c|c|}
\hline Calendar Year & Earned Premium \\
\hline CY5 & 7,706 \\
\hline CY6 & 9,200 \\
\hline CY7 & 10,250 \\
\hline
\end{tabular}

All policies have a one-year term and policy issues are uniformly distributed through each year.

The following rate changes have occurred:

\begin{tabular}{|c|c|}
\hline Date & Rate Change \\
\hline July 1, CY3 & \(+7 \%\) \\
\hline Nov. 15, CY5 & \(-4 \%\) \\
\hline October 1, CY6 & \(+5 \%\) \\
\hline
\end{tabular}

Rates are currently at the level set on October 1, CY6.
Calculate the earned premium at the current rate level for CY6.

Options:
A
9300
B
9400
C
9500
D
9600
E
9700

Correct answer: C

Solution: Let \(P\) be the premium after the July 1, CY3 rate change. On November 15, CY5 the premium is \(0.96 P\) and on October 1, CY6 it becomes \(1.05(0.96) P=1.008 P\). The relevant parallelogram is:
![](https://cdn.mathpix.com/cropped/2025_10_19_d7de0f14b071bc625284g-29.jpg?height=335&width=774&top_left_y=555&top_left_x=330)

The upper left triangle for CY6 has area \((1 / 2)(7 / 8)^{2}=49 / 128\) and the lower right triangle has area \((1 / 2)(1 / 4)^{2}=4 / 128\). The weighted average is \([49+4(1.008)+75(0.960)] P / 128= 0.9768 P\). The current premium is \(9200(1.008) /(0.9768)=9494\).
Question: You use the following information to determine a rate change using the loss ratio method.
(i)

\begin{tabular}{|l|c|c|c|}
\hline \begin{tabular}{l} 
Accident \\
Year
\end{tabular} & \begin{tabular}{c} 
Earned Premium \\
at Current Rates
\end{tabular} & \begin{tabular}{c} 
Reported \\
Losses
\end{tabular} & \begin{tabular}{c} 
Weight Given to \\
Accident Year
\end{tabular} \\
\hline AY8 & 4252 & 2260 & \(40 \%\) \\
\hline AY9 & 5765 & 2610 & \(60 \%\) \\
\hline
\end{tabular}
(ii) Trend Factor: \(7 \%\) per annum effective
(iii) Loss Development Factor (to Ultimate):

AY8: 1.08
AY9: 1.18
(iv) Permissible Loss Ratio: 0.657
(v) All policies are one-year policies, issued uniformly through the year, and rates will be in effect for one year.
(vi) Proposed Effective Date: July 1, CY10

Calculate the required portfolio-wide rate change.

Options:
A
\(-26 \%\)
B
\(-16 \%\)
C
\(-8 \%\)
D
\(-1 \%\)
E
\(\quad 7 \%\)

Correct answer: D

Solution: Trend periods for losses: Average accident date in experience period is July 1, AY8 and July 1, AY9, respectively. New rates will be in effect from July 1, CY10 through June 30, CY12, with an average accident date of July 1, CY11. Trend period for losses are 3 years for AY8 and 2 years for AY9. Ultimate losses trended and developed are, AY8: \(2260\left(1.07^{3}\right)(1.08)=\) 2990, and AY9: \(2610\left(1.07^{2}\right)(1.18)=3526\). Weighted average loss ratio \(=0.4(2990 / 4252)+ 0.6(3526 / 5765)=0.648\). Required portfolio-wide rate change \(=0.648 / 0.657-1=-1.4 \%\).
Question: You are given:
i) Policies are written uniformly throughout the year.
ii) Policies have a term of 6 months.
iii) The following rate changes have occurred:

\begin{tabular}{|c|l|}
\hline Date & Amount \\
\hline October 1, CY1 & \(+7 \%\) \\
\hline July 1, CY2 & \(+10 \%\) \\
\hline September 1, CY3 & \(-6 \%\) \\
\hline
\end{tabular}

Rates are currently at the September 1, CY3 level.
Calculate the on-level factor needed to adjust CY2 earned premiums to the current rate level.

Options:
A
\(\quad 0.97\)
B
0.98
C
0.99
D
1.00
E
\(\quad 1.01\)

Correct answer: E

Solution: Let \(P\) be the premium prior to the October 1, CY1 rate change. After the change, the premium is \(1.07 P\) and on July \(1, \mathrm{CY} 2\) it becomes \(1.10(1.07) P=1.177 P\). The relevant parallelogram is:
![](https://cdn.mathpix.com/cropped/2025_10_19_d7de0f14b071bc625284g-30.jpg?height=340&width=825&top_left_y=555&top_left_x=333)

The upper left triangle for CY2 has area \((1 / 2)(1 / 4)(1 / 2)=1 / 16\) and the lower right triangle has area \((1 / 2)(1 / 2)(1)=4 / 16\). The weighted average is \([1+4(1.177)+11(1.07)] P / 16= 1.092375 P\). The factor is \((1.177 \times .94) /(1.092375)=1.0128\).
Question: You are given the following information:

\begin{tabular}{|l|l|l|l|l|l|l|}
\hline \multirow[b]{2}{*}{Accident Year} & \multirow[b]{2}{*}{Earned Premium} & \multirow{2}{*}{Expected Loss Ratio} & \multicolumn{4}{|c|}{Cumulative Loss Payments through Development Month} \\
\hline & & & 12 & 24 & 36 & 48 \\
\hline AY5 & 19,000 & 0.90 & 4,850 & 9,700 & 14,100 & 16,200 \\
\hline AY6 & 20,000 & 0.85 & 5,150 & 10,300 & 14,900 & \\
\hline AY7 & 21,000 & 0.91 & 5,400 & 10,800 & & \\
\hline AY8 & 22,000 & 0.88 & 7,200 & & & \\
\hline
\end{tabular}

There is no development past 48 months.
Calculate the indicated loss reserve using the Bornhuetter-Ferguson method and volumeweighted average loss development factors.

Options:
A
22,600
B
23,400
C
24,200
D
25,300
E
\(\quad 26,200\)

Correct answer: B

Solution: The year-to-year development factors are 12-24: \(30,800 / 15,400=2\); \(24-36\) : \(29,000 / 20,000=\) 1.45; and 36-48: \(16,200 / 14,100=1.149\). Then the factor for \(24-48\) is \(1.45(1.149)=1.666\) and for \(12-48\) is \(2(1.666)=3.332\). The expected ultimate losses are AY6: \(20,000(0.85)=17,000\); AY7: \(21,000(0.91)\) = 19,110; and AY8: \(22,000(0.88)=19,360\). The B-F reserves are \(17,000(1-1 / 1.149)=2205,19,110(1-1 / 1.666)=7639\), and \(19,360(1-1 / 3.332)=13,550\). The total is 23,394.
Question: You are given:
i)

\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline \multirow{2}{*}{Accident Year} & \multicolumn{6}{|c|}{Cumulative Paid Losses through Development Year} & \multirow{2}{*}{Earned premium} \\
\hline & 0 & 1 & 2 & 3 & 4 & 5 & \\
\hline AY4 & 1,400 & 5,200 & 7,300 & 8,800 & 9,800 & 9,800 & 18,000 \\
\hline AY5 & 2,200 & 6,400 & 8,800 & 10,200 & 11,500 & & 20,000 \\
\hline AY6 & 2,500 & 7,500 & 10,700 & 12,600 & & & 25,000 \\
\hline AY7 & 2,800 & 8,700 & 12,900 & & & & 26,000 \\
\hline AY8 & 2,500 & 7,900 & & & & & 27,000 \\
\hline AY9 & 2,600 & & & & & & 28,000 \\
\hline
\end{tabular}
ii) The expected loss ratio for each Accident Year is 0.550.

Calculate the total loss reserve using the Bornhuetter-Ferguson method and three-year arithmetic average paid loss development factors.

Options:
A
21,800
B
22,500
C
23,600
D
24,700
E
\(\quad 25,400\)

Correct answer: D

Solution: The development factors are:
\[
\begin{aligned}
& 0-1: \frac{1}{3}\left(\frac{7,900}{2,500}+\frac{8,700}{2,800}+\frac{7,500}{2,500}\right)=3.089 \quad 1-2: \frac{1}{3}\left(\frac{12,900}{8,700}+\frac{10,700}{7,500}+\frac{8,800}{6,400}\right)=1.428 \\
& 2-3: \frac{1}{3}\left(\frac{12,600}{10,700}+\frac{10,200}{8,800}+\frac{8,800}{7,300}\right)=1.181 \quad 3-4: \frac{1}{2}\left(\frac{11,500}{10,200}+\frac{9,800}{8,800}\right)=1.1205 \quad 4-5: 1
\end{aligned}
\]

The cumulative factors are 0-5: 5.8372, 1-5: 1.8897, 2-5: 1.3233, 3-5: 1.1205; 4-5: 1.
The B-F reserve is \([25,000(1-1 / 1.1205)+26,000(1-1 / 1.3233)+27,000(1-1 / 1.8897)+ 28,000(1-1 / 5.8372)](0.55)=24,726\).
Question: You are given:
i) An insurance company was formed to write workers compensation business in CY1.
ii) Earned premium in CY1 was 1,000,000.
iii) Earned premium growth through CY3 has been constant at \(20 \%\) per year (compounded).
iv) The expected loss ratio for AY 1 is \(60 \%\).
v) As of December 31, CY3, the company's reserving actuary believes the expected loss ratio has increased two percentage points each accident year since the company's inception.
vi) Selected reported loss development factors are as follows:

\begin{tabular}{|l|l|}
\hline 12 to 24 months & 1.500 \\
\hline 24 to 36 months & 1.336 \\
\hline 36 to 48 months & 1.126 \\
\hline 48 to 60 months & 1.057 \\
\hline 60 to 72 months & 1.050 \\
\hline 72 to ultimate & 1.000 \\
\hline
\end{tabular}

Calculate the total IBNR reserve as of December 31, CY3 using the BornhuetterFerguson method.

Options:
A
964,000
B
966,000
C
968,000
D
970,000
E
972,000

Correct answer: E

Solution: CY1: \(\mathrm{ELR}=60 \%\), earned premium \(=1,000,000\), expected losses \(=0.6(1,000,000)=600,000\). Cumulative development factor \((\mathrm{CDF})=(1.05)(1.057)(1.126)=1.250\). Reserve using BF method \(=600,000(1-1 / 1.250)=120,000\).
CY2: ELR = 62\%, earned premium = 1,200,000, expected losses = 0.62(1,200,000) = \(744,000 . \mathrm{CDF}=(1.05)(1.057)(1.126)(1.336)=1.670\). Reserve using BF method \(=744,000(1 -1 / 1.670\) ) = 298,491.
CY3: ELR = 64\%, earned premium = 1,440,000, expected losses = 0.64(1,440,000) = 921,600. \(\mathrm{CDF}=(1.05)(1.057)(1.126)(1.336)(1.5)=2.504\). Reserve using BF method \(= 921,600(1-1 / 2.504)=553,549\). Total reserve \(=120,000+298,491+553,549=972,040\).
Question: A primary insurance company has a 100,000 retention limit. The company purchases a catastrophe reinsurance treaty, which provides the following coverage

Layer 1: 85\% of 100,000 excess of 100,000
Layer 2: \(\quad 90 \%\) of 100,000 excess of 200,000
Layer 3: \(95 \%\) of 300,000 excess of 300,000
The primary insurance company experiences a catastrophe loss of 450,000 .
Calculate the total loss retained by the primary insurance company.

Options:
A
100,000
B
112,500
C
125,000
D
132,500
E
150,000

Correct answer: D

Solution: The retained loss is \(100,000+0.15(100,000)+0.10(100,000)+0.05(150,000)=132,500\).
Question: A primary liability insurer has a book of business with the following characteristics:
- All policies have a policy limit of 500,000
- The expected loss ratio is \(60 \%\) on premiums of \(4,000,000\)

A reinsurer provides an excess of loss treaty for the layer 300,000 in excess of 100,000.
The following table of increased limits factors is available:

\begin{tabular}{|l|l|}
\hline Limit & ILF \\
\hline 100,000 & 1.00 \\
\hline 200,000 & 1.25 \\
\hline 300,000 & 1.45 \\
\hline 400,000 & 1.60 \\
\hline 500,000 & 1.70 \\
\hline
\end{tabular}

Calculate the reinsurer's expected losses for this coverage (answer to the nearest 000s).

Options:
A
840,000
B
847,000
C
850,000
D
862,000
E
871,000

Correct answer: B

Solution: The expected losses for the primary insurer are \(0.6(4,000,000)=2,400,000\). The expected proportion of losses in the treaty layer is ( \(1.6 / 1.7-1 / 1.7=0.352941\) ). The expected cost is \(0.352941(2,400,000)=847,058\).
The relative cost of the layer can be derived using formulas from Loss Models as follows:
\(\frac{E(X \wedge 400,000)-E(X \wedge 100,000)}{E(X \wedge 500,000)}\)
\(=\frac{E(X \wedge 400,000) / E(X \wedge 100,000)-E(X \wedge 100,000) / E(X \wedge 100,000)}{E(X \wedge 500,000) / E(X \wedge 100,000)}\)
\(=\frac{\operatorname{ILF}(400,000)-\operatorname{ILF}(100,000)}{\operatorname{ILF}(500,000)}=\frac{1.60-1.00}{1.70}=0.352941\)
Question: Company XYZ sells homeowners insurance policies. You are given:
i) The loss costs by accident year are:

\begin{tabular}{|c|c|}
\hline Accident Year & Loss Cost \\
\hline AY1 & 1300 \\
\hline AY2 & 1150 \\
\hline AY3 & 1550 \\
\hline AY4 & 1800 \\
\hline
\end{tabular}
ii) The slope of the straight line fitted to the natural \(\log\) of the loss costs is 0.1275.
iii) AY4 experience is weighted \(80 \%\) and AY3 experience is weighted \(20 \%\) for rate development.

New rates take effect November 1, CY5 for one-year policies and will be in effect for one year.

Calculate the expected loss cost for these new rates.

Options:
A
2124
B
2217
C
2264
D
2381
E
2413

Correct answer: E

Solution: Policies sold from November 1, CY5 to November 1, CY6 will be in effect through November 1, CY 7 and thus have an average accident date of November 1, CY6. For losses in AY4 the projection is 2.333 years and the projected cost is \(1800 e^{0.1275(2.333)}=2423.58\). For losses in AY3 the projection is 3.333 years and the projected cost is \(1550 e^{0.1275(3.333)}=2370.76\). The projected loss cost is the weighted average, \((0.8)(2423.58)+ (0.2)(2370.76)=2413.02=2413\).