BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: You are given:
i) The current price to buy one share of XYZ stock is 500 .
ii) The stock does not pay dividends.
iii) The continuously compounded risk-free interest rate is \(6 \%\).
iv) A European call option on one share of XYZ stock with a strike price of \(K\) that expires in one year costs 66.59.
v) A European put option on one share of XYZ stock with a strike price of \(K\) that expires in one year costs 18.64.

Using put-call parity, calculate the strike price, \(K\).

Options:
A
449
B
452
C
480
D
559
E
582

Correct answer: C

Solution: \(66.59-18.64=500-K \cdot \exp (-0.06)\) and so \(K=(500-66.59+18.64) / \exp (-0.06)=480\).
Question: The current price of a non-dividend paying stock is 40 and the continuously compounded risk-free interest rate is \(8 \%\). You are given that the price of a 35 -strike call option is 3.35 higher than the price of a 40-strike call option, where both options expire in 3 months.

Calculate the amount by which the price of an otherwise equivalent 40 -strike put option exceeds the price of an otherwise equivalent 35-strike put option.

Options:
A
1.55
B
1.65
C
1.75
D
3.25
E
3.35

Correct answer: A

Solution: Let \(C\) be the price for the 40 -strike call option. Then, \(C+3.35\) is the price for the 35 -strike call option. Similarly, let \(P\) be the price for the 40 -strike put option. Then, \(P-x\) is the price for the 35 -strike put option, where \(x\) is the desired quantity. Using put-call parity, the equations for the 35 -strike and 40 -strike options are, respectively,
\((C+3.35)+35 e^{-0.02}-40=P-x\)
\(C+40 e^{-0.02}-40=P\).
Subtracting the first equation from the second, \(5 e^{-0.02}-3.35=x, x=1.55\).
Question: Annual losses in Year 1 follow an exponential distribution with mean \(\theta\). An inflation factor of \(20 \%\) applies to all Year 2 losses.

The ordinary deductible for Year 1 is \(0.25 \theta\). The deductible is doubled in Year 2 .

Calculate the percentage increase in the loss elimination ratio from Year 1 to Year 2.

Options:
A
\(19 \%\)
B
\(28 \%\)
C
\(37 \%\)
D
\(54 \%\)
E
\(78 \%\)

Correct answer: D

Solution: For Year 1, we have: \(E(X)=\theta, E(X \wedge 0.25 \theta)=\theta\left(1-e^{-0.25 \theta / \theta}\right)=\theta\left(1-e^{-0.25}\right)\)
Therefore, \(L E R_{1}=\frac{E(X \wedge 0.25 \theta)}{E(X)}=1-e^{-0.25}=0.2212\)
For Year 2, \(E(Y)=1.20 \theta, E(Y \wedge 0.5 \theta)=1.2 E\left(X \wedge \frac{0.5 \theta}{1.20}\right)=1.2 \theta\left(1-e^{-5 / 12}\right)\)
and \(L E R_{2}=\frac{E(Y \wedge 0.5 \theta)}{E(Y)}=1-e^{-5 / 12}=0.34076\)
\(\frac{L E R_{2}}{L E R_{1}}=\frac{0.34076}{0.22120}=1.54 \Rightarrow 54 \%\) increase
Question: For a medical insurance company, you are given:
i) Losses for a new product are assumed to follow a lognormal distribution with parameters \(\mu=6\) and \(\sigma=1.5\).
ii) The new product has a per-loss deductible that results in a loss elimination ratio of 0.33 .

In a review of the business after five years of experience, it is determined that:
i) Losses for this product actually followed an exponential distribution.
ii) The initial mean for the exponential distribution is the same as the initial mean under the lognormal assumption.
iii) Since it was introduced, the expected value of a loss for this product increased at an annual compound rate of \(4 \%\).
iv) The per-loss deductible required to target the same loss elimination ratio is d.

Calculate \(d\).

Options:
A
605
B
659
C
722
D
775
E
852

Correct answer: A

Solution: Under the initial assumptions, losses, \(X\), have \(E(X)=\exp \left(6+\frac{1.5^{2}}{2}\right)=1242.65\)
After 5 years: exponential distribution with mean \(1242.65\left(1.04^{5}\right)=1511.87\) and \(\mathrm{LER}=0.33\), so: \(0.33=\frac{E(Y \wedge d)}{E(Y)}=\frac{\theta\left(1-e^{-d / \theta}\right)}{\theta}=1-e^{-d / 1511.87} \Rightarrow d=-1511.87 \ln (1-0.33)=605.47\)
Question: The distribution of the number of claims, \(N\), is a member of the ( \(a, b, 0\) ) class. You are given:
i) \(\quad P(N=k)=p_{k}\)
ii) \(\quad \frac{p_{6}}{p_{4}}=0.5\) and \(\frac{p_{5}}{p_{4}}=0.8\)

A zero-modified distribution, \(N^{M}\), associated with \(N\) has \(P\left(N^{M}=0\right)=0.1\).

Calculate \(E\left(N^{M}\right)\).

Options:
A
\(\quad 3.64\)
B
3.73
C
3.85
D
4.00
E
4.05

Correct answer: A

Solution: \(\frac{p_{k}}{p_{k-1}}=a+\frac{b}{k}=>0.8=a+\frac{b}{5}\) and \(\frac{0.5}{0.8}=0.625=a+\frac{b}{6}\)
\(=>0.8-0.625=b\left(\frac{1}{5}-\frac{1}{6}\right) \Rightarrow b=5.25\) and \(a=-0.25\)
\(\Rightarrow N\) is binomial with \(\frac{-q}{1-q}=-0.25\) and \((m+1) \frac{q}{1-q}=5.25 \Rightarrow m=20\) and \(q=0.2\)
\(=>E\left(N^{M}\right)=\left(1-p_{0}^{M}\right) \frac{m q}{1-(1-q)^{m}}=3.64\)
Question: You are given the following properties of the distribution of the annual number of claims, \(N\) :
i) \(\quad P(N=k)=p_{k}, \quad k=0,1,2, \ldots\)
ii) \(\quad p_{0}=0.45\)
iii) \(\quad \frac{p_{n}}{p_{m}}=\frac{m!}{n!} \quad\) for \(m \geq 1\) and \(n \geq 1\)

Calculate the probability that at least two claims occur during a year.

Options:
A
\(\quad 0.16\)
B
0.18
C
0.21
D
0.23
E
\(\quad 0.26\)

Correct answer: D

Solution: Members of the ( \(a, b, 1\) ) class have \(\frac{p_{k}}{p_{k-1}}=a+\frac{b}{k}\) for \(k=2,3,4, \ldots\), so point iii) implies that \(N\) is an ( \(a, b, 1\) ) distribution with \(a=0\) and \(b=\lambda=1\). This, together with point ii), implies the distribution is a zero-modified Poisson distribution with \(p_{0}^{M}=0.45\).
\(P(N \geq 2)=1-P(N \leq 1)=1-(P(N=0)+P(N=1))=1-\left(0.45+\frac{1-p_{0}^{M}}{1-P(L=0)} P(L=1)\right)\),
where \(L\) has a Poisson distribution with \(\lambda=1\) and \(P(L=0)=P(L=1)=e^{-1}\), so \(P(N \geq 2)=1-\left(0.45+\frac{0.55}{1-e^{-1}} e^{-1}\right)=0.2299\)
Question: For the workers' compensation claims of a construction company you are given:
i) The annual number of claims follows the Poisson distribution with mean 20.
ii) Claim sizes \(X\) follow the lognormal distribution with \(\mu=4.2\) and \(\sigma=1.1\).
iii) The company retains the first 500 of each claim.
iv) Annual aggregate retained claims approximately follow the normal distribution.
v) \(\quad E\left[(X \wedge 500)^{2}\right]=26,189\)

Determine the \(90^{\text {th }}\) percentile of the aggregate distribution of retained claims.

Options:
A
Less than 2900
B
At least 2900, but less than 3100
C
At least 3100, but less than 3300
D
At least 3300, but less than 3500
E
At least 3500

Correct answer: C

Solution: \[
E(X \wedge 500)=\exp \left(4.2+\frac{1.1^{2}}{2}\right) \Phi(0.73)+500(1-\Phi(1.83))=110.5
\]

Let \(S\) denote the aggregate distribution of retained claims.
\(E(S)=E(N) E(X \wedge 500)=20(110.5)=2210\)
\(\operatorname{Var}(S)=E(N) \operatorname{Var}(X \wedge 500)+E\left((X \wedge 500)^{2}\right) \operatorname{Var}(N)=20(26,189)=523,780\)
The 90th percentile of \(S\) is \(2210+1.282 \sqrt{523,780}=3137.82\)
Question: For a product liability policy, you are given:
i) The following statistics for products X and Y

\begin{tabular}{|l|l|c|c|}
\cline { 3 - 4 } \multicolumn{2}{c|}{} & Product X & Product Y \\
\hline \multirow{2}{*}{ Number of Claims } & Mean & 10 & 2 \\
\cline { 2 - 4 } & Standard Deviation & 3 & 1 \\
\hline \multirow{2}{*}{ Loss Amount } & Mean & 20 & 50 \\
\cline { 2 - 4 } & Standard Deviation & 5 & 10 \\
\hline
\end{tabular}
ii) Loss amounts and claim numbers are independent within and between products.
iii) Aggregate losses, for both products combined, approximately follow the normal distribution.

Determine the probability that aggregate losses for both products combined exceed 400.

Options:
A
Less than 0.10
B
At least 0.10 , but less than 0.15
C
At least 0.15 , but less than 0.20
D
At least 0.20 , but less than 0.25
E
At least 0.25

Correct answer: B

Solution: For Product X: aggregate losses have mean \(10 \times 20=200\) and variance \(10 \times 25+400 \times 9=3850\).
For Product Y: aggregate losses have mean \(2 \times 50=100\) and variance \(2 \times 100+2500=2700\).
Because Product X and Product Y are independent, total aggregate losses, \(S\), have mean 300 and variance 6550.
Using the normal approximation, we have:
\(P(S>400)=P\left(\frac{S-300}{\sqrt{6550}}>\frac{400-300}{\sqrt{6550}}\right)=P(Z>1.24)=0.11\)
Question: You are given:
i) The random variable \(X\) has probability density function
\[
f(x)=\alpha(1500)^{\alpha}(1500+x)^{-(\alpha+1)}, \quad \alpha>0 \text { and } x>0
\]
ii) Five sample observations are:
50
250
450
650
850

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

Options:
A
\(\quad 0.16\)
B
0.79
C
1.85
D
2.91
E
\(\quad 3.97\)

Correct answer: E

Solution: The likelihood function is \(L(\alpha)=\alpha^{5} 1500^{5 \alpha} \prod\left(1500+x_{i}\right)^{-(\alpha+1)}\)
The loglikelihood function is \(l(\alpha)=5 \ln \alpha+5 \alpha \ln 1500-(\alpha+1) \sum \ln \left(1500+x_{i}\right)\)
\(l^{\prime}(\alpha)=\frac{5}{\alpha}+5 \ln 1500-\sum \ln \left(1500+x_{i}\right)\)
Setting \(l^{\prime}(\alpha)=0\), we find: \(\frac{5}{\alpha}=-5 \ln 1500+\sum \ln \left(1500+x_{i}\right)=>\hat{\alpha}=3.974\)
Question: The following random sample of five claims is drawn from a Pareto distribution with parameters \(\alpha\) (unknown) and \(\theta=1\) :
 1.2 1.6 1.4 and two other values that exceed 2.0

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

Options:
A
0.62
B
0.72
C
0.94
D
1.04
E
\(\quad 1.15\)

Correct answer: A

Solution: The likelihood function is:
\(L(\alpha)=f(1.2) f(1.6) f(1.4)[S(2.0)]^{2}=\alpha^{3}(1 / 3)^{2 \alpha} \prod\left(x_{i}+1\right)^{-(\alpha+1)}\)
The loglikelihood function is \(l(\alpha)=3 \ln \alpha+2 \alpha \ln (1 / 3)-(\alpha+1) \sum \ln \left(x_{i}+1\right)\)
\(l^{\prime}(\alpha)=\frac{3}{\alpha}+2 \ln (1 / 3)-\sum \ln \left(x_{i}+1\right)=\frac{3}{\alpha}-4.81666\)
Setting \(l^{\prime}(\alpha)=0\), we find \(\hat{\alpha}=0.623\)
Question: You are given the following observations on 185 small business policies:

\begin{tabular}{|c|c|}
\hline Number of Claims & Number of Policies \\
\hline 0 & 80 \\
\hline 1 or more & 105 \\
\hline
\end{tabular}

The number of claims per policy follows a Poisson distribution with parameter \(\lambda\).

Using the maximum likelihood estimate of \(\lambda\), determine the estimated probability of a policy having fewer than two claims.

Options:
A
0.79
B
0.84
C
0.89
D
0.95
E
\(\quad 0.98\)

Correct answer: A

Solution: The likelihood function is \(L(\lambda)=\left(e^{-\lambda}\right)^{80}\left(1-e^{-\lambda}\right)^{105}\)
The loglikelihood function is \(l(\lambda)=-80 \lambda+105 \ln \left(1-e^{-\lambda}\right)\)
Setting \(l^{\prime}(\lambda)=-80+\frac{105 e^{-\lambda}}{1-e^{-\lambda}}\) equal to 0 , we find \(e^{-\lambda}=\frac{80}{185}=>\hat{\lambda}=-\ln \frac{80}{185}=0.838329\)
The probability the number of claims, \(N\), is less than 2 is \(P(N<2)=e^{-\hat{\lambda}}+\hat{\lambda} e^{-\hat{\lambda}}=0.79495\)
Question: You are given:
i) The number of claims follows a Poisson distribution.
ii) The number of claims and claim severity are independent.
iii) The severity distribution is:

\begin{tabular}{|c|c|}
\hline Claim Size & Probability \\
\hline 5 & 0.60 \\
\hline 40 & 0.35 \\
\hline 60 & 0.05 \\
\hline
\end{tabular}

Calculate the expected number of claims needed so the total cost of claims is within 5\% of the expected with probability 0.90 .

Options:
A
511
B
726
C
1083
D
2044
E
3126

Correct answer: D

Solution: For the severity distribution, we have \(\mu_{X}=20\) and \(\sigma_{X}^{2}=355\). The standard for full credibility is \(\left(\frac{1.645}{0.05}\right)^{2}\left(1+C V_{X}^{2}\right)=\left(\frac{1.645}{0.05}\right)^{2}\left(1+\frac{355}{20^{2}}\right)=2043.05\), round up to 2044
Question: You are given:
i) Claim frequency follows a Poisson distribution.
ii) Claim frequency and claim severity are independent.
iii) The probability density function of claim severity is
\[
f(x)=\frac{3(100)^{3}}{x^{4}}, \quad x>100
\]

Calculate the expected number of claims needed so the observed total cost of claims is within \(5 \%\) of the true value with probability 0.95 .

Options:
A
1444
B
1537
C
2049
D
3586
E
8196

Correct answer: C

Solution: For the severity distribution, we have
\(E(X)=\int_{100}^{\infty} 3(100)^{3} x^{-3} d x=150, E\left(X^{2}\right)=\int_{100}^{\infty} 3(100)^{3} x^{-2} d x=30,000\) and \(\sigma_{X}^{2}=30,000-150^{2}=7500\). The standard for full credibility is
\(\left(\frac{1.96}{0.05}\right)^{2}\left(1+C V_{X}^{2}\right)=\left(\frac{1.96}{0.05}\right)^{2}\left(1+\frac{7500}{150^{2}}\right)=2048.85\), round up to 2049
Question: You are given the following information on losses paid for each of AY1 through AY4:

\begin{tabular}{|l|l|l|l|l|l|l|}
\hline & & & \multicolumn{4}{|c|}{Incremental Loss Payments through Development Year} \\
\hline Accident Year & Earned Premium & Expected Loss Ratio & 0 & 1 & 2 & 3 \\
\hline AY1 & 35,500 & 0.71 & 10,500 & 7,500 & 4,800 & 1,340 \\
\hline AY2 & 31,200 & 0.73 & 13,050 & 5,025 & 1,400 & \\
\hline AY3 & \(X\) & 0.75 & 12,500 & 7,250 & & \\
\hline AY4 & \(X+5,000\) & 0.75 & 18,400 & & & \\
\hline
\end{tabular}

The estimated loss reserve using the expected loss ratio method is 43,412 .

Calculate \(X\).

Options:
A
\(\quad 48,964\)
B
49,674
C
51,875
D
\(\quad 52,174\)
E
\(\quad 54,785\)

Correct answer: A

Solution: \begin{tabular}{|l|l|l|l|}
\hline \begin{tabular}{l} 
Accident \\
Year
\end{tabular} & \begin{tabular}{l} 
Expected Ultimate \\
Loss
\end{tabular} & \begin{tabular}{l} 
Cumulative \\
Paid-to-date
\end{tabular} & Reserve \\
\hline AY1 & \(35,500 \times 0.71=25,205\) & 24,140 & 1065 \\
\hline AY2 & \(31,200 \times 0.73=22,776\) & 19,475 & 3301 \\
\hline AY3 & \(0.75 X\) & 19,750 & \(0.75 X-19,750\) \\
\hline AY4 & \(0.75(X+5000)\) & 18,400 & \(0.75(X+5000)-18,400\) \\
\hline
\end{tabular}
\(43,412=1065+3301+0.75 X-19,750+0.75(X+5000)-18,400 \Rightarrow X=48,964\)
Question: An actuary is establishing reserves for a group of policies as of December 31, CY3. You are given the following table of reserve estimates for AY1 and AY2:

\begin{tabular}{|c|r|r|r|}
\hline \multirow{2}{*}{} & \multicolumn{3}{|c|}{ Reserve estimates as of December 31, CY3 } \\
\cline { 2 - 4 } & \(R_{B F}\) & \(R_{L R}\) & \multicolumn{1}{|c|}{\(R_{C L}\)} \\
\hline AY1 & 400,000 & 250,000 & 437,500 \\
\hline AY2 & \(1,120,000\) & \(1,200,000\) & \(1,050,000\) \\
\hline
\end{tabular}
where \(R_{B F}\) is the loss reserve under the Bornhuetter-Ferguson method, \(R_{L R}\) is the loss reserve under the Expected Loss Ratio method, and \(R_{C L}\) is the loss reserve under the Chain Ladder method.

Calculate \(f_{2}\), the loss development factor from the paid-loss-development triangle at duration 2.

Options:
A
1.250
B
1.500
C
1.875
D
2.150
E
2.500

Correct answer: B

Solution: \[
\begin{aligned}
& R_{B F}=\left(1-\frac{1}{f}\right) R_{L R}+\frac{1}{f} R_{C L} \\
& 400,000=\left(1-\frac{1}{f}\right) 250,000+\frac{1}{f} 437,500 \Rightarrow f=1.25=\prod_{j=3}^{\infty} f_{j} \\
& 1,120,000=\left(1-\frac{1}{f}\right) 1,200,000+\frac{1}{f} 1,050,000 \Rightarrow f=1.875=\prod_{j=2}^{\infty} f_{j} \\
& f_{2}=\frac{1.875}{1.25}=1.5
\end{aligned}
\]
Question: You use the following information to calculate projected loss costs:
i) Based on historical experience, the slope of the straight line fitted to the natural log of the loss cost is 0.182 .
ii) The loss cost per unit exposure for the two most recent experience periods is:

\begin{tabular}{|c|c|}
\hline Accident Year & Loss Cost per Unit Exposure \\
\hline AY3 & 400 \\
\hline AY4 & 450 \\
\hline
\end{tabular}
iii) The current experience period is weighted \(90 \%\) and the prior experience period is weighted \(10 \%\).

New rates for one-year policies take effect May 1, CY5 and are in effect for one year. Calculate the projected loss cost for these new rates.

Options:
A
527
B
577
C
615
D
632
E
666

Correct answer: D

Solution: The midpoint of AY3 is July 1, AY3, the midpoint of AY4 is July 1, AY4, and May 1, AY6 is the midpoint of the effective period.

AY3 lost cost projected to May 1, AY6 is \(400 \exp (0.182 \times 2.8333)=669.90\)
AY4 lost cost projected to May 1, AY6 is \(450 \exp (0.182 \times 1.8333)=628.23\)
Projected loss cost \(=(0.9)(628.23)+(0.1)(669.90)=632.40\)
Question: An insurance company uses the following information to set new rates:
i) The projected loss cost grows at an annual effective rate of \(7 \%\) per year.
ii) The table below shows loss cost data for two policy years, with corresponding weights:

\begin{tabular}{|c|c|c|}
\hline Policy Year & Loss Cost & Weight \\
\hline PY3 & 600 & \(1-p\) \\
\hline PY4 & 670 & \(p\) \\
\hline
\end{tabular}
iii) New rates will be effective on April 1, CY9. Policies are one-year policies and rates will be in effect for one year.

The projected expected loss cost is 951.

Calculate \(p\).

Options:
A
0.089
B
0.119
C
0.567
D
0.881
E
\(\quad 0.911\)

Correct answer: D

Solution: For any given policy year, the average accident date is a full year after the beginning of that year. Similarly for new rates effective on April 1, CY9, the midpoint of the effective period is April 1, CY10.

Therefore, the PY3 and PY4 loss costs will be projected 6.25 and 5.25 years, respectively.
Using the projected loss cost, we find:
\(600(1.07)^{6.25}(1-p)+670(1.07)^{5.25}(p)=951 \Rightarrow p=0.8813\)
Question: You are given the following information:
i) \(\quad\) Case reserves \(=187,047\)
ii) Losses paid-to-date \(=243,005\)
iii) Age-to-ultimate reported loss development factor \(=1.08\)
iv) IBNR using Bornhuetter-Ferguson method \(=47,387\)

Calculate the absolute difference in the IBNR based on the chain-ladder method and the expected loss ratio method.

Options:
A
12,983
B
\(\quad 43,787\)
C
\(\quad 98,101\)
D
146,754
E
175,268

Correct answer: E

Solution: First, use the Borhuetter-Ferguson method to calculate the estimated ultimate loss as \(\frac{47,387}{1-1 / 1.08}=639,724.5\). Next, calculate IBNR under the expected loss ratio method as \(639,724.5-(187,047+243,005)=209,672.5\). Then, IBNR under the chain-ladder method is \((187,047+243,005)(0.08)=34,404.16\). The absolute difference is \(209,672.5-34,404.16=175,268.34\).
Question: An insurance company sells one-year policies that have uniformly distributed effective dates.

The following rate changes have occurred:

\begin{tabular}{|c|c|}
\hline Date & Rate Change \\
\hline June 1, CY1 & \(+10 \%\) \\
\hline August 1, CY2 & \(r\) \\
\hline
\end{tabular}

Rates are currently at the level set on August 1, CY2.

The earned premium at current rates for CY2 is 1.03 times the CY2 earned premium.

Calculate \(r\).

Options:
A
\(\quad 1.7 \%\)
B
\(2.4 \%\)
C
\(3.3 \%\)
D
\(5.3 \%\)
E
\(7.5 \%\)

Correct answer: B

Solution: Let \(P\) represent the rate level before the rate change on June 1, CY1. The rate level 1.1P takes effect on June 1, CY1. The rate level \((1.1 P)(1+r)\) takes effect on August 1, CY2, so this level is the current rate level.

The parallelogram method is shown in the diagram below.
![](https://cdn.mathpix.com/cropped/2025_10_19_d7de0f14b071bc625284g-38.jpg?height=594&width=1583&top_left_y=1247&top_left_x=302)

For CY2, the average rate level for the earned exposure is
\(\frac{1}{2}\left(\frac{5}{12}\right)^{2} P+\left[1-\frac{1}{2}\left(\frac{5}{12}\right)^{2}-\frac{1}{2}\left(\frac{5}{12}\right)^{2}\right](1.1 P)+\frac{1}{2}\left(\frac{5}{12}\right)^{2}(1.1 P)(1+r)\).
\(=0.086806 P+0.909028 P+0.0954861 P(1+r)=0.9958 \overline{3} P+0.095486 \overline{1} P(1+r)\)
The ratio of the earned premium at current rates for CY2 to the CY2 earned premium, which is the on-level factor for CY2, is
\(\frac{(1.1 P)(1+r)}{0.9958 \overline{3} P+0.095486 \overline{1} P(1+r)}=\frac{1.1}{0.9958 \overline{3} /(1+r)+0.095486 \overline{1}}=1.03\)
\(=>1+r=1.02402 \Rightarrow r=2.402 \%\)
Question: Company XYZ sells dental insurance policies. You are given:
i) All policies have a one-year term.
ii) Policy effective dates are uniformly distributed.
iii) The earned premium for Calendar Year 3 is 600 .

The following rate changes have occurred:

\begin{tabular}{|c|c|}
\hline Date & Rate change \\
\hline September 1, CY1 & \(3.0 \%\) \\
\hline May 15, CY3 & \(2.0 \%\) \\
\hline August 1, CY4 & \(5.0 \%\) \\
\hline
\end{tabular}

Rates are currently at the level set on August 1, CY4.

Calculate the earned premium at current rates for CY3.

Options:
A
610
B
617
C
632
D
640
E
659

Correct answer: D

Solution: ![](https://cdn.mathpix.com/cropped/2025_10_19_d7de0f14b071bc625284g-39.jpg?height=397&width=1557&top_left_y=769&top_left_x=292)

Let \(P\) denote the rate following the Sept 1, CY1 rate change.
For CY3, the average rate level for the earned exposure is:
\(\left[1-\frac{1}{2}\left(\frac{7.5}{12}\right)^{2}\right] P+\frac{1}{2}\left(\frac{7.5}{12}\right)^{2}(1.02 P)=1.00390625 P\)
Current rates are \(1.02 * 1.05 P=1.071 P\)
CY3 earned premium at current rates \(=600 \times \frac{1.071 P}{1.00390625 P}=640.10\)
Question: A non-dividend paying stock has a current price of \(S\). The continuously compounded risk-free interest rate is \(2.75 \%\).

The price of the stock over a six-month period follows a binomial model with \(u=1.2903\) and \(d=0.7966\). A six-month European put option on the stock with a strike price of ( \(S-4.50\) ) has a price of 2.482 .

Calculate \(S\).

Options:
A
44.22
B
45.72
C
46.97
D
49.11
E
50.24

Correct answer: A

Solution: The risk-neutral probability of an increase is:
\(1-q=\frac{e^{(0.0275)(0.5)}-d}{u-d}=0.44003\)
Then:
\(2.482=e^{(-0.0275)(0.5)}((1-q) \times 0+q \times(S-4.5-0.7966 S))=>S=44.22\)
Question: You are given:
(i) The Black-Scholes-Merton framework applies.
(ii) The prices of some 6-month European options on non-dividend paying Stock Y are:

\begin{tabular}{|c|c|c|}
\hline Strike Price & Price of Call Option & Price of Put Option \\
\hline 525 & 55.92 & \(x\) \\
\hline 550 & 45.46 & 64.57 \\
\hline
\end{tabular}

The continuously compounded risk-free rate is \(3.25 \%\).

Calculate \(x\).

Options:
A
50.03
B
50.43
C
50.83
D
51.03
E
51.23

Correct answer: B

Solution: Using put-call parity and the 550 -strike options, we have:
\(45.46-64.57=S_{0}-550 e^{(-0.0325)(0.5)} \Rightarrow S_{0}=522.0247\)
Using the current stock price and put-call parity with the 525 -strike options:
\(55.92-x=522.0247-525 e^{(-0.0325)(0.5)} \Rightarrow x=50.43\)
Question: You are given:
(i) For any random variable \(X\), the risk measure is \(\rho(X)=\mu_{X}+k \sigma_{X}\).
(ii) The variables \(Y\) and \(Z\) are defined as follows:

\begin{tabular}{|c|c|}
\hline\(y\) & \(\mathrm{P}(Y=y)\) \\
\hline 0 & 0.2 \\
\hline 1 & 0.8 \\
\hline
\end{tabular}
and \(Z\) is constant with \(Z=1\).

Calculate the largest value of \(k\) such that \(\rho\) has the monotonicity property with respect to \(Y\) and \(Z\).

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

Correct answer: A

Solution: The monotonicity property states that if \(Y \leq Z\) for all possible outcomes, then \(\rho(Y) \leq \rho(Z)\). From the variable definitions, we have \(\mu_{Y}=0.8, \sigma_{Y}^{2}=0.8-0.8^{2}=0.16, \mu_{Z}=1\), and \(\sigma_{Z}^{2}=0\). Monotonicity requires \(0.8+0.4 k \leq 1\). Therefore, \(k \leq 0.5\).
Question: The price movements of a non-dividend paying stock are modeled by a binomial tree with one-year periods.

You are given:
i) The current price of the stock is 100 .
ii) \(\quad u=1.24\)
iii) The risk-neutral probability of an up move is 0.406 .
iv) The continuously compounded risk-free interest rate is \(2 \%\).

The price of a 2 -year \(K\)-strike European put option on the stock is 25.05 .

Calculate \(K\).

Options:
A
123.50
B
124.00
C
124.50
D
125.00
E
125.50

Correct answer: E

Solution: You are given \(p^{*}=0.406\) and \(u=124 / 100=1.24\) and \(r=.02\)
Therefore \(p^{*}=\left(e^{r}-d\right) /(u-d) \rightarrow d=\left(e^{r}-u p^{*}\right) /\left(1-p^{*}\right)=0.87\)
\(S_{u u}=153.76\)
\(S_{u d}=107.88\)
\(S_{d d}=75.69\)
\(P=25.05\)
\(C=e^{-2 r}\left[\operatorname{Max}\left(S_{u u}-K, 0\right)\left(p^{*}\right)^{2}+2 \operatorname{Max}\left(S_{u d}-K, 0\right) p^{*}\left(1-p^{*}\right)+\operatorname{Max}\left(S_{d d}-K, 0\right)\left(1-p^{*}\right)^{2}\right]\)
\(P=e^{-2 r}\left[\operatorname{Max}\left(K-S_{u u}, 0\right)\left(p^{*}\right)^{2}+2 \operatorname{Max}\left(K-S_{u d}, 0\right) p^{*}\left(1-p^{*}\right)+\operatorname{Max}\left(K-S_{d d}, 0\right)\left(1-p^{*}\right)^{2}\right]\)
\(K=125.50\)