BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: The number of auto vandalism claims reported per month at Sunny Daze Insurance Company (SDIC) has mean 110 and variance 750. Individual losses have mean 1101 and standard deviation 70. The number of claims and the amounts of individual losses are independent.

Using the normal approximation, calculate the probability that SDIC's aggregate auto vandalism losses reported for a month will be less than 100,000 .

Options:
A
0.24
B
0.31
C
0.36
D
0.39
E
0.49

Correct answer: A

Solution: \(E(S)=E(N) E(X)=110(1,101)=121,110\)
\(\operatorname{Var}(S)=E(N) \operatorname{Var}(X)+E(X)^{2} \operatorname{Var}(N)=110\left(70^{2}\right)+1101^{2}(750)=909,689,750\)
\(\operatorname{StdDev}(S)=30,161\)
\(\operatorname{Pr}(S<100,000)=\operatorname{Pr}\left(Z<\frac{100,000-121,110}{30,161}=-0.70\right)=0.242\)
where \(Z\) has standard normal distribution.
Question: Prescription drug losses, \(S\), are modeled assuming the number of claims has a geometric distribution with mean 4 , and the amount of each prescription is 40 .

Calculate \(E\left[(S-100)_{+}\right]\).

Options:
A
60
B
82
C
92
D
114
E
146

Correct answer: C

Solution: Let \(N=\) number of prescriptions then

\begin{tabular}{|c|c|c|c|}
\hline\(n\) & \(f_{N}(n)\) & \(F_{N}(n)\) & \(1-F_{N}(n)\) \\
\hline 0 & 0.2000 & 0.2000 & 0.8000 \\
1 & 0.1600 & 0.3600 & 0.6400 \\
2 & 0.1280 & 0.4880 & 0.5120 \\
3 & 0.1024 & 0.5904 & 0.4096 \\
\hline
\end{tabular}
\[
\begin{aligned}
& E(N)=4=\sum_{j=0}^{\infty}[1-F(j)] \\
& E\left[(S-80)_{+}\right]=40 E\left[(N-2)_{+}\right]=40 \sum_{j=2}^{\infty}[1-F(j)] \\
& =40\left[\sum_{j=0}^{\infty}[1-F(j)]-\sum_{j=0}^{1}[1-F(j)]\right] \\
& =40(4-1.44)=102.40 \\
& E\left[(S-120)_{+}\right]=40 E\left[(N-3)_{+}\right]=40\left[\sum_{j=0}^{\infty}[1-F(j)]-\sum_{j=0}^{2}[1-F(j)]\right]=40(4-1.952)=81.92
\end{aligned}
\]

Because no values of S between 80 and 120 are possible,
\[
E\left[(S-100)_{+}\right]=\frac{(120-100) E\left[(S-80)_{+}\right]+(100-80) E\left[(S-120)_{+}\right]}{120-80}=92.16
\]

Alternatively,
\[
E\left[(S-100)_{+}\right]=\sum_{j=0}^{\infty}(40 j-100) f_{N}(j)+100 f_{N}(0)+60 f_{N}(1)+20 f_{N}(2)
\]
(The correction terms are needed because ( \(40 j-100\) ) would be negative for \(j=0,1,2\); we need to add back the amount those terms would be negative)
\[
\begin{aligned}
& =40 \sum_{j=0}^{\infty} j f_{N}(j)-100 \sum_{j=0}^{\infty} f_{N}(j)+100(0.2)+60(0.16)+20(0.128) \\
& =40 E(N)-100+20+9.6+2.56=160-67.84=92.16
\end{aligned}
\]
Question: At the beginning of each round of a game of chance the player pays 12.5 . The player then rolls one die with outcome \(N\). The player then rolls \(N\) dice and wins an amount equal to the total of the numbers showing on the \(N\) dice. All dice have 6 sides and are fair.

Using the normal approximation, calculate the probability that a player starting with 15,000 will have at least 15,000 after 1000 rounds.

Options:
A
0.01
B
0.04
C
0.06
D
0.09
E
0.12

Correct answer: E

Solution: Method 1:
In each round,
\(N=\) result of first roll, to see how many dice you will roll
\(X=\) result of for one of the \(N\) dice you roll
\(S=\) sum of \(X\) for the \(N\) dice
\(E(X)=E(N)=3.5\)
\(\operatorname{Var}(X)=\operatorname{Var}(N)=2.9167\)
\(E(S)=E(N) E(X)=12.25\)
\(\operatorname{Var}(S)=E(N) \operatorname{Var}(X)+\operatorname{Var}(N) E(X)^{2}=3.5(2.9167)+2.9167(3.5)^{2}=45.938\)
Let \(S_{1000}\) the sum of the winnings after 1000 rounds
\(E\left(S_{1000}\right)=1000(12.25)=12,250\)
\(S D\left(S_{1000}\right)=\sqrt{1000(45.938)}=214.33\)
After 1000 rounds, you have your initial 15,000 , less payments of 12,500 , plus winnings for a total of \(2,500+S_{1000}\) Since actual possible outcomes are discrete, the solution tests for continuous outcomes greater than 15000-0.5. In this problem, that continuity correction has negligible impact.
\(\operatorname{Pr}\left(2,500+S_{1000}>14,999.5\right)=\operatorname{Pr}\left(S_{1000}>12,499.5\right) \approx \operatorname{Pr}\left(Z>\frac{12,499.5-12,250}{214.33}=1.17\right)=0.12\).
Question: A group dental policy has a negative binomial claim count distribution with mean 300 and variance 800 .

Ground-up severity is given by the following table:

\begin{tabular}{|c|c|}
\hline Severity & Probability \\
\hline 40 & 0.25 \\
\hline 80 & 0.25 \\
\hline 120 & 0.25 \\
\hline 200 & 0.25 \\
\hline
\end{tabular}

You expect severity to increase \(50 \%\) with no change in frequency. You decide to impose a per claim deductible of 100 .

Calculate the expected total claim payment after these changes.

Options:
A
Less than 18,000
B
At least 18,000 , but less than 20,000
C
At least 20,000 , but less than 22,000
D
At least 22,000 , but less than 24,000
E
At least 24,000

Correct answer: D

Solution: \begin{tabular}{cc} 
Severity & Severity \\
after increase & after increase and deductible \\
60 & 0 \\
120 & 20 \\
180 & 80 \\
300 & 200
\end{tabular}

Expected payment per loss \(=0.25(0)+0.25(20)+0.25(80)+0.25(200)=75\)
Expected payments \(=\) Expected number of losses x Expected payment per loss \(=300(75)=22,500\)
Question: For a certain company, losses follow a Poisson frequency distribution with mean 2 per year, and the amount of a loss is 1,2 , or 3 , each with probability \(1 / 3\). Loss amounts are independent of the number of losses, and of each other.

An insurance policy covers all losses in a year, subject to an annual aggregate deductible of 2.

Calculate the expected claim payments for this insurance policy.

Options:
A
\(\quad 2.00\)
B
2.36
C
2.45
D
\(\quad 2.81\)
E
\(\quad 2.96\)

Correct answer: B

Solution: Let \(S\) denote aggregate loss before deductible.
\(\mathrm{E}(S)=2(2)=4\), since mean severity is 2 .
\(f_{S}(0)=\frac{e^{-2} 2^{0}}{0!}=0.1353\), since must have 0 losses to get aggregate loss \(=0\).
\(f_{s}(1)=\frac{e^{-2} 2^{1}}{1!} \frac{1}{3}=0.0902\), since must have 1 loss whose size is 1 to get aggregate loss \(=1\).
\(E(S \wedge 2)=0 f_{S}(0)+1 f_{S}(1)+2\left[1-f_{S}(0)-f_{S}(1)\right]\)
\(=0(0.1353)+1(0.0902)+2(1-0.01353-0.0902)=1.6392\)
\(E\left[(S-2)_{+}\right]=E(S)-E(S \wedge 2)=4-1.6392=2.3608\)
Question: The random variable for a loss, \(X\), has the following characteristics:

\begin{tabular}{|r|c|c|}
\hline \multicolumn{1}{|c|}{\(x\)} & \(F(x)\) & \(E(X \wedge x)\) \\
\hline 0 & 0.0 & 0 \\
\hline 100 & 0.2 & 91 \\
\hline 200 & 0.6 & 153 \\
\hline 1000 & 1.0 & 331 \\
\hline
\end{tabular}

Calculate the mean excess loss for a deductible of 100 .

Options:
A
250
B
300
C
350
D
400
E
450

Correct answer: B

Solution: Mean excess loss \(=\frac{E(X)-E(X \wedge 100)}{1-F(100)}=\frac{331-91}{0.8}=300\)

Note that \(E(X)=E(X \wedge 1000)\) because \(F(1000)=1\).
Question: Glen is practicing his simulation skills. He generates 1000 values of the random variable \(X\) as follows:
(i) He generates a value of \(\lambda\) from the gamma distribution with \(\alpha=2\) and \(\theta=1\) (hence with mean 2 and variance 2 ).
(ii) He then generates \(x\) from the Poisson distribution with mean \(\lambda\)
(iii) He repeats the process 999 more times: first generating a value \(\lambda\), then generating \(x\) from the Poisson distribution with mean \(\lambda\).
(iv) The repetitions are mutually independent.

Calculate the expected number of times that his simulated value of \(X\) is 3 .

Options:
A
75
B
100
C
125
D
150
E
175

Correct answer: C

Solution: \(E(N)=E[E(N \mid \Lambda)]=E(\Lambda)=2\)
\(\operatorname{Var}(N)=E[\operatorname{Var}(N \mid \Lambda)]+\operatorname{Var}[E(N \mid \Lambda)]=E(\Lambda)+\operatorname{Var}(\Lambda)=2+2=4\)
The equations to solve are
mean \(=2=r \beta\)
variance \(=4-r \beta(1+\beta)\)
This implies
\(1+\beta=2 \Rightarrow \beta=1 \Rightarrow r(1)=2 \Rightarrow r=2\)
Then \(1000 p_{3}=1000 \frac{r(r+1)(r+2) \beta^{3}}{3!(1+\beta)^{r+3}}=1000 \frac{2(3)(4)\left(1^{3}\right)}{6\left(2^{5}\right)}=125\)
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}{cr} 
Loss Amount & Probability \\
0 & 0.4 \\
1 & 0.3 \\
2 & 0.2 \\
3 & 0.1
\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: 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 accidents follows a Poisson distribution with mean 12. Each accident generates 1,2 , or 3 claimants with probabilities \(1 / 2,1 / 3\), and \(1 / 6\), respectively.

Calculate the variance of the total number of claimants.

Options:
A
20
B
25
C
30
D
35
E
40

Correct answer: E

Solution: For a compound Poisson distribution, \(\operatorname{Var}(S)=E(N) E\left(X^{2}\right)\).
\(E\left(X^{2}\right)=(1 / 2)(1)+(1 / 3)(4)+(1 / 6)(9)=10 / 3\)
\(\operatorname{Var}(S)=12(10 / 3)=40\)

Alternatively, the total is the sum \(N_{1}+2 N_{2}+3 N_{3}\) where the \(N\) s are independent Poisson variables with means 6,4 , and 2 . The variance is \(6+4(4)+9(2)=40\).
Question: An insurer has excess-of-loss reinsurance on auto insurance. You are given:
(i) Total expected losses in the year 2001 are 10,000,000.
(ii) In the year 2001 individual losses have a Pareto distribution with
\[
F(x)=1-\left(\frac{2000}{x+2000}\right)^{2}, \quad x>0
\]
(iii) Reinsurance will pay the excess of each loss over 3000 .
(iv) Each year, the reinsurer is paid a ceded premium, \(C_{\text {year }}\) equal to \(110 \%\) of the expected losses covered by the reinsurance.
(v) Individual losses increase 5\% each year due to inflation.
(vi) The frequency distribution does not change.

Calculate \(C_{2002} / C_{2001}\).

Options:
A
1.04
B
1.05
C
1.06
D
1.07
E
\(\quad 1.08\)

Correct answer: E

Solution: For 2001:
\(E(X)=2000 /(2-1)=2000\)
\(E(X \wedge 3000)=\frac{2000}{1}\left(1-\frac{2000}{3000+2000}\right)=1200\)
So the fraction of the losses expected to be covered by the reinsurance is \((2000-1200) / 2000=0.4\). The expected ceded losses are \(4,000,000\) and the ceded premium is \(4,400,000\).

For 2002:
Inflation changes the scale parameter, here to \(2000(1.05)=2100\). The revised calculations are \(E(X)=2100 /(2-1)=2100\)
\(E(X \wedge 3000)=\frac{2100}{1}\left(1-\frac{2100}{3000+2100}\right)=1235\)
The revised premium is \(1.1(10,500,000)(2100-1235) / 2100=4,757,500\).
The ratio is \(4,757,500 / 4,400,000=1.08\).
Question: The number of annual losses has a Poisson distribution with a mean of 5 . The size of each loss has a two-parameter Pareto distribution with \(\theta=10\) and \(\alpha=2.5\). An insurance for the losses has an ordinary deductible of 5 per loss.

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

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 2003 follow a two-parameter Pareto distribution with \(\alpha=2\) and \(\theta=5\). Losses in 2004 are uniformly \(20 \%\) higher than in 2003. An insurance covers each loss subject to an ordinary deductible of 10 .

Calculate the Loss Elimination Ratio in 2004.

Options:
A
\(\quad 5 / 9\)
B
\(5 / 8\)
C
\(2 / 3\)
D
\(3 / 4\)
E
\(4 / 5\)

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 2004
\(E(X)=\frac{6}{2-1}=6\)
\(E(X \wedge 10)=\frac{6}{2-1}\left[1-\left(\frac{6}{10+6}\right)^{2-1}\right]=3.75\)
\(E\left[(X-10)_{+}\right]_{+}=E(X)-E(X \wedge 10)=6-3.75-2.25\)
\(\mathrm{LER}=1-\frac{E\left[(X-10)_{+}\right]}{E(X)}=1-\frac{2.25}{6}=0.625\)
Question: You are given the following random sample of 13 claim amounts:
\[
\begin{array}{lllllllllllll}
99 & 133 & 175 & 216 & 250 & 277 & 651 & 698 & 735 & 745 & 791 & 906 & 947
\end{array}
\]

Calculate the smoothed empirical estimate of the 35th percentile.

Options:
A
219.4
B
231.3
C
234.7
D
246.6
E
256.8

Correct answer: D

Solution: \(0.35(13+1)=4.9\)
\(\hat{\pi}_{0.35}=0.1(216)+0.9(250)=246.6\)
Question: A sample of claim amounts is \(\{300,600,1500\}\). By applying the deductible to this sample, the loss elimination ratio for a deductible of 100 per claim is estimated to be 0.125 .

You are given the following simulations from the sample:

\begin{tabular}{|l|l|l|l|}
\hline Simulation & \multicolumn{3}{|c|}{Claim Amounts} \\
\hline 1 & 600 & 600 & 1500 \\
\hline 2 & 1500 & 300 & 1500 \\
\hline 3 & 1500 & 300 & 600 \\
\hline 4 & 600 & 600 & 300 \\
\hline 5 & 600 & 300 & 1500 \\
\hline 6 & 600 & 600 & 1500 \\
\hline 7 & 1500 & 1500 & 1500 \\
\hline 8 & 1500 & 300 & 1500 \\
\hline 9 & 300 & 600 & 300 \\
\hline 10 & 600 & 600 & 600 \\
\hline
\end{tabular}

Calculate the bootstrap approximation to the mean square error of the estimate.

Options:
A
0.003
B
0.010
C
0.021
D
0.054
E
0.081

Correct answer: A

Solution: For each simulation, estimate the LER and then calculate the squared difference from the estimate, 0.125.

\begin{tabular}{|l|l|l|l|l|l|}
\hline Simulation & First claim & Second claim & Third claim & LER & Squared difference \\
\hline 1 & 600 & 600 & 1500 & 0.111111 & 0.000193 \\
\hline 2 & 1500 & 300 & 1500 & 0.090909 & 0.001162 \\
\hline 3 & 1500 & 300 & 600 & 0.125000 & 0.000000 \\
\hline 4 & 600 & 600 & 300 & 0.200000 & 0.005625 \\
\hline 5 & 600 & 300 & 1500 & 0.125000 & 0.000000 \\
\hline 6 & 600 & 600 & 1500 & 0.111111 & 0.000193 \\
\hline 7 & 1500 & 1500 & 1500 & 0.066667 & 0.003403 \\
\hline 8 & 1500 & 300 & 1500 & 0.090909 & 0.001162 \\
\hline 9 & 300 & 600 & 300 & 0.250000 & 0.015625 \\
\hline 10 & 600 & 600 & 600 & 0.166667 & 0.001736 \\
\hline
\end{tabular}

The last column has an average of 0.002910 , which is the bootstrap estimate.
Question: From a population having distribution function \(F\), you are given the following sample:
\[2.0,3.3,3.3,4.0,4.0,4.7,4.7,4.7\]

Calculate the kernel density estimate of \(F(4)\), using the uniform kernel with bandwidth 1.4.

Options:
A
0.31
B
0.41
C
0.50
D
0.53
E
0.63

Correct answer: D

Solution: Let \(K_{y}(x)\) be the contribution at \(x\) of the data point at \(y\). It is
\(K_{y}(x)= \begin{cases}0, & x<y-1.4 \\ \frac{x-y+1.4}{2.8}, & y-1.4 \leq x \leq y+1.4 \\ 1, & x>y+1.4 .\end{cases}\)
For the particular points, \(K_{2}(4)=1, K_{3.3}(4)=\frac{4-3.3+1.4}{2.8}=0.75, K_{4}(4)=0.5, K_{4.7}(4)=0.25\). The kernel estimate is the weighted average \(\frac{1}{8}(1)+\frac{2}{8}(0.75)+\frac{2}{8}(0.5)+\frac{3}{8}(0.25)=0.53125\).
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
\(\quad 0.05\)
B
0.10
C
0.20
D
0.40
E
\(\quad 0.80\)

Correct answer: B

Solution: The mean is \(m q\) and the variance is \(m q(1-q)\). The mean is 34,574 and so the full credibility standard requires the confidence interval to be \(\pm 345.74\) which must be 1.96 standard deviations. Thus,
\(345.74=1.96 \sqrt{m q(1-q)}=1.96 \sqrt{34,574} \sqrt{1-q}\)
\(1-q=0.9, \quad q=0.1\).
Question: You are given:
(i) Losses are uniformly distributed on \((0, \theta)\) with \(\theta>150\).
(ii) The policy limit is 150 .
(iii) A sample of payments is:
\[
14,33,72,94,120,135,150,150
\]

Calculate the estimate of \(\theta\) obtained by matching the average sample payment to the expected payment per loss.

Options:
A
192
B
196
C
200
D
204
E
208

Correct answer: E

Solution: The sample average is \((14+33+72+94+120+135+150+150) / 8=96\). The model average is \(E(X \wedge 150)=\int_{0}^{150} x \frac{1}{\theta} d x+\int_{150}^{\theta} 150 \frac{1}{\theta} d x=\frac{150^{2}}{2 \theta}+150 \frac{\theta-150}{\theta}=150-\frac{11,250}{\theta}\). The equation to solve is \(150-\frac{11,250}{\theta}=96, \frac{11,250}{\theta}=54, \quad \theta=\frac{11,250}{54}=208.3\).
Question: You are given the following data:
\[
\begin{array}{ccccccccc}
0.49 & 0.51 & 0.66 & 1.82 & 3.71 & 5.20 & 7.62 & 12.66 & 35.24
\end{array}
\]

You use the method of percentile matching at the 40th and 80th percentiles to fit an inverse Weibull distribution to these data.

Calculate the estimate of \(\theta\).

Options:
A
Less than 1.35
B
At least 1.35 , but less than 1.45
C
At least 1.45 , but less than 1.55
D
At least 1.55 , but less than 1.65
E
At least 1.65

Correct answer: D

Solution: The equations to solve are \(0.4=e^{-(\theta / 1.82)^{\tau}}, 0.8=e^{-(\theta / 12.66)^{\tau}}\). Taking logs yields \(0.91629=(\theta / 1.82)^{\tau}, 0.22314=(\theta / 12.66)^{\tau}\). Taking the ratio of the first equation to the second equation gives \(4.10635=(12.66 / 1.82)^{\tau}=6.95604^{\tau}\). Taking logs again, \(1.41253=1.93961 \tau\) and then \(\tau=0.72825\). Returning to the first (logged) equation, \(0.91629=(\theta / 1.82)^{\tau}, 0.88688=\theta / 1.82, \theta=1.614\).
Question: A loss, \(X\), follows a 2 -parameter Pareto distribution with \(\alpha=2\) and unspecified parameter \(\theta\). You are given:
\[
E[X-100 \mid X>100]=\frac{5}{3} E[X-50 \mid X>50]
\]

Calculate \(E[X-150 \mid X>150]\).

Options:
A
150
B
175
C
200
D
225
E
250

Correct answer: B

Solution: \(E[X-d \mid X>d]\) is the expected payment per payment with an ordinary deductible of \(d\) It can be evaluated (for Pareto) as
\[
\begin{aligned}
& \frac{E(X)-E(X \wedge d)}{1-F(d)}=\frac{\frac{\theta}{\alpha-1}-\frac{\theta}{\alpha-1}\left[1-\left(\frac{\theta}{d+\theta}\right)^{\alpha-1}\right]}{1-\left[1-\left(\frac{\theta}{d+\theta}\right)^{\alpha}\right]}=\frac{\frac{\theta}{\alpha-1}\left(\frac{\theta}{d+\theta}\right)^{\alpha-1}}{\left(\frac{\theta}{d+\theta}\right)^{\alpha}}=\frac{d+\theta}{\alpha-1}=d+\theta \\
& E[X-100 \mid X>100]=(5 / 3) E[X-50 \mid X>50] \\
& 100+\theta=(5 / 3)(50+\theta) \\
& 300+3 \theta=250+5 \theta, 50=2 \theta, \theta=25 \\
& E[X-150 \mid X>150]=150+\theta=150+25=175
\end{aligned}
\]
Question: The scores on the final exam in Ms. B's Latin class have a normal distribution with mean \(\theta\) and standard deviation equal to \(8 . \theta\) is a random variable with a normal distribution with mean 75 and standard deviation 6.

Each year, Ms. B chooses a student at random and pays the student 1 times the student's score. However, if the student fails the exam (score < 65), then there is no payment.

Calculate the conditional probability that the payment is less than 90 , given that there is a payment.

Options:
A
0.77
B
0.85
C
0.88
D
0.92
E
1.00

Correct answer: D

Solution: Let \(S=\) score
\(E(S)=E[E(S \mid \theta)]=E(\theta)=75\)
\(\operatorname{Var}(S)=E[\operatorname{Var}(S \mid \theta)]+\operatorname{Var}[E(S \mid \theta)]=E\left(8^{2}\right)+\operatorname{Var}(\theta)=64+6^{2}=100\)
\(S\) is normally distributed (a normal mixture of normal distributions with constant variance is normal).
\[
\begin{gathered}
\operatorname{Pr}(\mathrm{S}<90 \mid \mathrm{S}>65)=\frac{F(90)-F(65)}{1-F(65)}=\frac{\Phi[(90-75) / 10)]-\Phi[(65-75) / 10]}{1-\Phi[(65-75) / 10]} \\
\quad \frac{\Phi(1.5)-\Phi(-1.0)}{1-\Phi(-1.0)}=\frac{0.9332-(1-0.8413)}{1-(1-0.8413)}=0.9206
\end{gathered}
\]
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) \(E[X]=70\)
(ii) \(E[X \wedge 30]=25\)
(iii) \(\quad \operatorname{Pr}(X>30)=0.75\)
(iv) \(E\left[X^{2} \mid X>30\right]=9000\)

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
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: You are given:
(i) A mortality study covers \(n\) lives.
(ii) None were censored and no two deaths occurred at the same time.
(iii) \(t_{k}=\) time of the \(k\) th death
(iv) A Nelson-Aalen estimate of the cumulative hazard rate function is \(\hat{H}\left(t_{2}\right)=\frac{39}{380}\).

Calculate the Kaplan-Meier product-limit estimate of the survival function at time \(t_{9}\).

Options:
A
Less than 0.56
B
At least 0.56 , but less than 0.58
C
At least 0.58 , but less than 0.60
D
At least 0.60 , but less than 0.62
E
At least 0.62

Correct answer: A

Solution: \(\hat{H}\left(t_{2}\right)=\frac{1}{n}+\frac{1}{n-1}=\frac{2 n-1}{n(n-1)}=\frac{39}{380} \Rightarrow 39 n^{2}-799 n+380=0 \Rightarrow n=20, n=0.487\).
Discard the non-integer solution to have \(n=20\). The Kaplan-Meier Product-Limit Estimate is:
\(\hat{S}\left(t_{9}\right)=\frac{19}{20} \frac{18}{19} \ldots \frac{11}{12}=\frac{11}{20}=0.55\).
Question: Three observed values of the random variable \(X\) are:
\[
\begin{array}{lll}
1 & 1 & 4
\end{array}
\]

You estimate the third central moment of \(X\) using the estimator:
\[
g\left(X_{1}, X_{2}, X_{3}\right)=\frac{1}{3} \sum_{i=1}^{3}\left(X_{i}-\bar{X}\right)^{3}
\]

Calculate the bootstrap estimate of the mean-squared error of \(g\).

Options:
A
Less than 3.0
B
At least 3.0 , but less than 3.5
C
At least 3.5 , but less than 4.0
D
At least 4.0 , but less than 4.5
E
At least 4.5

Correct answer: E

Solution: There are 27 possible bootstrap samples, which produce four different results. The results, their probabilities, and the values of \(g\) are:

\begin{tabular}{lll} 
Bootstrap Sample & Prob & \(g\) \\
\(1,1,1\) & \(8 / 27\) & 0 \\
\(1,1,4\) & \(12 / 27\) & 2 \\
\(1,4,4\) & \(6 / 27\) & -2 \\
\(4,4,4\) & \(1 / 27\) & 0
\end{tabular}

The third central moment of the original sample is 2 . Then,
MSE \(=\frac{8}{27}(0-2)^{2}+\frac{12}{27}(2-2)^{2}+\frac{6}{27}(-2-2)^{2}+\frac{1}{27}(0-2)^{2}=\frac{44}{9}\).