BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: . You are given:
i) The following extract from a three-year select and ultimate table:

\begin{tabular}{|l|l|l|l|l|l|}
\hline [ \(x\) ] & \(q_{[x]}\) & \(q_{[x]+1}\) & \(q_{\{x\}+2}\) & \(q_{x+3}\) & \(x+3\) \\
\hline 50 & 0.020 & 0.031 & 0.043 & 0.056 & 53 \\
\hline 51 & 0.025 & 0.037 & 0.050 & 0.065 & 54 \\
\hline 52 & 0.030 & 0.043 & 0.057 & 0.072 & 55 \\
\hline 53 & 0.035 & 0.049 & 0.065 & 0.091 & 56 \\
\hline 54 & 0.040 & 0.055 & 0.076 & 0.113 & 57 \\
\hline 55 & 0.045 & 0.061 & 0.090 & 0.140 & 58 \\
\hline
\end{tabular}
ii) Mortality follows a uniform distribution of deaths over each year of age

Calculate \(1000\left({ }_{0.6 \mid 1.5} q_{[52]+1.7}\right)\).

Options:
A
91
B
92
C
93
D
94
E
95

Correct answer: C

Solution: \begin{tabular}{|c|c|c|c|c|c|}
\hline\([x]\) & \(q_{[x]}\) & \(q_{[x]+1}\) & \(q_{[x]+2}\) & \(q_{x+3}\) & \(x+3\) \\
\hline 52 & 0.030 & 0.043 & 0.057 & 0.072 & 55 \\
\hline
\end{tabular}
\(1000\left({ }_{0.6 \mid 1.5} q_{[52]+1.7}\right)=1000 \frac{l_{[52]+2.3}-l_{[52]+3.8}}{l_{[52]+1.7}}\)
\(l_{\text {[52] }}=1000\) (arbitrary, for convenience; any other value would work and give the same answer)
\(l_{[52]+1}=1000(1-0.03)=970\)
\(l_{[52]+2}=970(1-0.043)=928.29\)
\(l_{55}=928.29(1-0.057)=875.3775\)
\(l_{56}=875.3775(1-0.072)=812.3503\)
\(l_{[52]+2.3}=(0.7) l_{[52]+2}+(0.3) l_{55}=912.42\)
\(l_{[52]+3.8}=l_{55.8}=(0.2) l_{55}+(0.8) l_{56}=824.96\)
\(l_{[52]+1.7}=(0.3) l_{[52]+1}+(0.7) l_{[52]+2}=940.80\)
\(1000\left({ }_{0.6 \mid 1.5} q_{[52]+1.7}\right)=1000 \frac{912.42-824.96}{940.80}=92.96\)
Question: . For a special whole life insurance policy issued on (40), you are given:
(i) Death benefits are payable at the end of the year of death
(ii) The amount of benefit is 2 if death occurs within the first 20 years and is 1 thereafter
(iii) \(\quad Z\) is the present value random variable for the payments under this insurance
(iv) \(\quad i=0.03\)
(v)

\begin{tabular}{lll}
\hline\(x\) & \(A_{x}\) & \({ }_{20} E_{x}\) \\
\hline 40 & 0.36987 & 0.51276 \\
60 & 0.62567 & 0.17878 \\
\hline
\end{tabular}
(vi) \(\quad E\left[Z^{2}\right]=0.24954\)

Calculate the standard deviation of \(Z\).

Options:
A
0.27
B
0.32
C
0.37
D
0.42
E
0.47

Correct answer: A

Solution: \(E[Z]=2 \cdot A_{40}-{ }_{20} E_{40} A_{60}=(2)(0.36987)-(0.51276)(0.62567)=0.41892\)
\(E\left[Z^{2}\right]=0.24954\) which is given in the problem.
\(\operatorname{Var}(Z)=E\left[Z^{2}\right]-(E[Z])^{2}=0.24954-0.41892^{2}=0.07405\)
\(S D(Z)=\sqrt{0.07405}=0.27212\)

An alternative way to obtain the mean is \(E[Z]=2 A_{40: 20 \mid}^{1}+20 \mid A_{40}\). Had the problem asked for the evaluation of the second moment, a formula is
\(E\left[Z^{2}\right]=\left(2^{2}\right)\left({ }^{2} A_{40: 201}\right)+\left(v^{2}\right)^{20}\left({ }_{20} p_{40}\right)\left({ }^{2} A_{60}\right)\)
Question: . For a special 2 -year term insurance policy on \((x)\), you are given:
(i) Death benefits are payable at the end of the half-year of death
(ii) The amount of the death benefit is 300,000 for the first half-year and increases by 30,000 per half-year thereafter
(iii) \(q_{x}=0.16\) and \(q_{x+1}=0.23\)
(iv) \(\quad i^{(2)}=0.18\)
(v) Deaths are assumed to follow a constant force of mortality between integral ages
(vi) \(\quad Z\) is the present value random variable for this insurance

Calculate \(\operatorname{Pr}(Z>277,000)\).

Options:
A
0.08
B
0.11
C
0.14
D
0.18
E
0.21

Correct answer: D

Solution: \begin{tabular}{|l|l|}
\hline Half-year & PV of Benefit \\
\hline 1 & \(300,000 v^{0.5}=(300,000)(1.09)^{-1}=275,229\) \\
\hline 2 & \(330,000 v^{1}=(330,000)(1.09)^{-2}=277,754\) \\
\hline 3 & \(360,000 v^{1.5}=(360,000)(1.09)^{-3}=277,986\) \\
\hline 4 & \(390,000 v^{2}=(390,000)(1.09)^{-4}=276,286\) \\
\hline
\end{tabular}
\(P V>277,000\) if and only if ( \(x\) ) dies in the \(2^{\text {nd }}\) or \(3^{\text {rd }}\) half years.

Under CF assumption, \({ }_{0.5} p_{x}={ }_{0.5} p_{x+0.5}=(0.84)^{0.5}=0.9165\) and
\({ }_{0.5} p_{x+1}={ }_{0.5} p_{x+1.5}=(0.77)^{0.5}=0.8775\) Then the probability of dying in the \(2^{\text {nd }}\) or \(3^{\text {rd }}\) half-years is \(\left({ }_{0.5} p_{x}\right)\left(1-{ }_{0.5} p_{x+0.5}\right)+\left(p_{x}\right)\left(1-{ }_{0.5} p_{x+1}\right)=(0.9165)(0.0835)+(0.84)(0.1225)=0.1794\)
Question: . You are given:
(i) \(\quad q_{60}=0.01\)
(ii) Using \(i=0.05, A_{60: 3 \mid}=0.86545\)

Using \(i=0.045\) calculate \(A_{60: 31} \cdot\)

Options:
A
0.866
B
0.870
C
0.874
D
0.878
E
0.882

Correct answer: D

Solution: \[
\begin{aligned}
& A_{60: 3 \mid}=q_{60} v+\left(1-q_{60}\right) q_{60+1} v^{2}+\left(1-q_{60}\right)\left(1-q_{60+1}\right) v^{3}=0.86545 \\
& q_{60+1}=\frac{A_{60: 31}-q_{60} v-\left(1-q_{60}\right) v^{3}}{\left(1-q_{60}\right) v^{2}-\left(1-q_{60}\right) v^{3}}=\frac{0.86545-\frac{0.01}{1.05}-\frac{0.99}{1.05^{3}}}{\frac{0.99}{1.05^{2}}-\frac{0.99}{1.05^{3}}}=0.017 \text { when } v=1 / 1.05
\end{aligned}
\]

The primes indicate calculations at \(4.5 \%\) interest.
\[
\begin{aligned}
A_{60: 31}^{\prime} & =q_{60} v^{\prime}+\left(1-q_{60}\right) q_{60+1} v^{\prime 2}+\left(1-q_{60}\right)\left(1-q_{60+1}\right) v^{\prime 3} \\
& =\frac{0.01}{1.045}+\frac{0.99(0.017)}{1.045^{2}}+\frac{0.99(0.983)}{1.045^{3}} \\
& =0.87777
\end{aligned}
\]
Question: For a special increasing whole life insurance on (40), payable at the moment of death, you are given:
(i) The death benefit at time \(t\) is \(b_{t}=1+0.2 t, \quad t \geq 0\)
(ii) The interest discount factor at time \(t\) is \(v(t)=(1+0.2 t)^{-2}, \quad t \geq 0\)
(iii) \(\quad{ }_{t} p_{40} \mu_{40+t}= \begin{cases}0.025, & 0 \leq t<40 \\ 0, & \text { otherwise }\end{cases}\)
(iv) \(\quad Z\) is the present value random variable for this insurance

Calculate \(\operatorname{Var}(Z)\).

Options:
A
0.036
B
0.038
C
0.040
D
0.042
E
0.044

Correct answer: A

Solution: \[
\begin{aligned}
\operatorname{Var}(Z) & =E\left(Z^{2}\right)-E(Z)^{2} \\
E(Z) & =E\left[(1+0.2 T)(1+0.2 T)^{-2}\right]=E\left[(1+0.2 T)^{-1}\right] \\
& =\int_{0}^{40} \frac{1}{(1+0.2 t)} f_{T}(t) d t=\frac{1}{40} \int_{0}^{40} \frac{1}{1+0.2 t} d t \\
& =\left.\frac{1}{40} \frac{1}{0.2} \ln (1+0.2 t)\right|_{0} ^{40}=\frac{1}{8} \ln (9)=0.27465 \\
E\left(Z^{2}\right) & =E\left\{(1+0.2 T)^{2}\left[(1+0.2 T)^{-2}\right]^{2}\right\}=E\left[(1+0.2 T)^{-2}\right] \\
& =\int_{0}^{40} \frac{1}{(1+0.2 t)^{2}} f_{T}(t)=\frac{1}{40} \frac{1}{0.2}\left[\frac{-1}{(1+0.2 t)}\right]_{0}^{40} \\
& =\frac{1}{8}\left(1-\frac{1}{9}\right)=\frac{1}{9}=0.11111 \\
\operatorname{Var}(Z) & =0.11111-(0.27465)^{2}=0.03568
\end{aligned}
\]
Question: . For a 30 -year term life insurance of 100,000 on (45), you are given:
(i) The death benefit is payable at the moment of death
(ii) Mortality follows the Standard Ultimate Life Table
(iii) \(\delta=0.05\)
(iv) Deaths are uniformly distributed over each year of age

Calculate the \(95{ }^{\text {th }}\) percentile of the present value of benefits random variable for this insurance.

Options:
A
\(\quad 30,200\)
B
31,200
C
35,200
D
36,200
E
37,200

Correct answer: C

Solution: The earlier the death (before year 30), the larger the loss. Since we are looking for the \(95{ }^{\text {th }}\) percentile of the present value of benefits random variable, we must find the time at which \(5 \%\) of the insureds have died. The present value of the death benefit for that insured is what is being asked for.
\[
\begin{aligned}
& l_{45}=99,033.9 \Rightarrow 0.95 l_{45}=94,082.2 \\
& l_{65}=94,579.7 \\
& l_{66}=94,020.3
\end{aligned}
\]

So, the time is between ages 65 and 66, i.e., time 20 and time 21.
\[
\begin{aligned}
& l_{65}-l_{66}=94,579.7-94,020.3=559.4 \\
& l_{65+t}-l_{66}=94,579.7-94,082.2=497.5 \\
& 497.5 / 559.4=0.8893
\end{aligned}
\]

The time just before the last \(5 \%\) of deaths is expected to occur is: \(20+0.8893=20.8893\)
The present value of death benefits at this time is:
\[
100,000 e^{-20.8893(0.05)}=35,188
\]
Question: . For a 3 -year term insurance of 1000 on (70), you are given:
(i) \(\quad q_{70+k}^{S U L T}\) is the mortality rate from the Standard Ultimate Life Table, for \(k=0,1,2\)
(ii) \(\quad q_{70+k}\) is the mortality rate used to price this insurance, for \(k=0,1,2\)
(iii) \(\quad q_{70+k}=(0.95)^{k} q_{70+k}^{\text {SULT }}\), for \(k=0,1,2\)
(iv) \(\quad i=0.05\)

Calculate the single net premium.

Options:
A
29.05
B
29.85
C
30.65
D
31.45
E
32.25

Correct answer: B

Solution: \begin{tabular}{|l|l|l|l|l|}
\hline Time & Age & \(q_{x}^{\text {SULT }}\) & \begin{tabular}{l} 
Improvement \\
factor
\end{tabular} & \(q_{x}\) \\
\hline 0 & 70 & 0.010413 & \(100.00 \%\) & 0.010413 \\
\hline 1 & 71 & 0.011670 & \(95.00 \%\) & 0.011087 \\
\hline 2 & 72 & 0.013081 & \(90.25 \%\) & 0.011806 \\
\hline
\end{tabular}
\(v=1 / 1.05=0.952381\)
\[
\begin{aligned}
E P V & =1,000\left[0.010413 v+0.989587(0.011087) v^{2}+0.989587(0.988913)(0.011806) v^{3}\right] \\
& =29.85
\end{aligned}
\]
Question: . For a 25 -year pure endowment of 1 on ( \(x\) ), you are given:
(i) \(\quad Z\) is the present value random variable at issue of the benefit payment
(ii) \(\operatorname{Var}(Z)=0.10 E[Z]\)
(iii) \(\quad{ }_{25} p_{x}=0.57\)

Calculate the annual effective interest rate.

Options:
A
\(\quad 5.8 \%\)
B
\(\quad 6.0 \%\)
C
\(\quad 6.2 \%\)
D
\(\quad 6.4 \%\)
E
\(6.6 \%\)

Correct answer: B

Solution: \[
\begin{aligned}
& \operatorname{Var}(Z)=0.10 E[Z] \Rightarrow v_{25}^{50} p_{x}\left(1-{ }_{25} p_{x}\right)=0.10 \cdot v_{25}^{25} p_{x} \\
& \Rightarrow \frac{(1-0.57)}{(1+i)^{50}}=0.10 \times \frac{1}{(1+i)^{25}} \\
& \Rightarrow(1+i)^{25}=\frac{0.43}{0.10}=4.3 \Rightarrow i=0.06
\end{aligned}
\]
Question: . For a whole life insurance of 1000 on (50), you are given:
(i) The death benefit is payable at the end of the year of death
(ii) Mortality follows the Standard Ultimate Life Table
(iii) \(\quad i=0.04\) in the first year, and \(i=0.05\) in subsequent years

Calculate the actuarial present value of this insurance.

Options:
A
187
B
189
C
191
D
193
E
195

Correct answer: C

Solution: Let \(A_{51}^{\text {SULT }}\) designate \(A_{51}\) using the Standard Ultimate Life Table at \(5 \%\).
\[
\begin{aligned}
\text { APV (insurance) } & =1000\left(\frac{1}{1.04}\right)\left(q_{50}+p_{50} A_{51}^{\text {SULT }}\right) \\
& =1000\left(\frac{1}{1.04}\right)[0.001209+(1-0.001209)(0.19780)] \\
& =191.12
\end{aligned}
\]
Question: . You are given:
(i) \(\quad A_{35: \overline{15 \mid}}=0.39\)
(ii) \(\quad A_{35: 15 \mid}^{1} 0.25\)
(iii) \(\quad A_{35}=0.32\)

Calculate \(A_{50}\).

Options:
A
0.35
B
0.40
C
0.45
D
0.50
E
0.55

Correct answer: D

Solution: \[
\begin{aligned}
A_{35}= & A_{35: 15 \mid}^{1}+A_{35: 15 \mid} \frac{1}{50} A_{50} \\
& 0.32=0.25+0.14 A_{50} \\
& A_{50}=\frac{0.07}{0.14}=0.50
\end{aligned}
\]
Question: . You are given:
(i) \(\quad Z_{1}\) is the present value random variable for an \(n\)-year term insurance of 1000 issued to ( \(x\) )
(ii) \(\quad Z_{2}\) is the present value random variable for an \(n\)-year endowment insurance of 1000 issued to ( \(x\) )
(iii) For both \(Z_{1}\) and \(Z_{2}\) the death benefit is payable at the end of the year of death
(iv) \(E\left[Z_{1}\right]=528\)
(v) \(\operatorname{Var}\left(Z_{2}\right)=15,000\)
(vi) \(\quad A_{x: n}=0.209\)
(vii) \({ }^{2} A_{x: \frac{1}{n}}=0.136\)

Calculate \(\operatorname{Var}\left(Z_{1}\right)\).

Options:
A
143,400
B
177,500
C
211,200
D
245,300
E
279,300

Correct answer: A

Solution: \[
\begin{aligned}
& \begin{array}{r}
\operatorname{Var}\left(Z_{2}\right)=(1000)^{2}\left[{ }^{2} A_{x: n \mid}-\left(A_{x: n \mid}\right)^{2}\right]=15,000 \\
=(1000)^{2}\left({ }^{2} A_{x: n \mid}^{1}+{ }^{2} A_{x: n}\right)-(1000)^{2}\left[A_{x: n}^{1}+A_{x n}^{1}\right]^{2}
\end{array} \\
& =(1000)^{2} A_{x: n}^{1}+(1000)^{2} A_{x: n}^{1}-(1000)^{2}\left(A_{x: n}^{1}\right)^{2}-(1000)^{2}\left(A_{x} \frac{1}{n}\right)^{2} \\
& \quad-2(1000)^{2}\left(A_{x: n}^{1}\right)\left(A_{x \bar{n}}^{1}\right) \\
& \begin{array}{r}
(1000)^{2}\left[{ }^{2} A_{x: n}^{1}-\left(A_{x: n}^{1}\right)^{2}\right]+\left(1000^{2} A_{x: n}^{1}\right)-\left(1000 A_{x: n}\right)^{2} \\
-\left(1000 A_{x: n}^{1}\right)^{2}-(2)\left(1000 A_{x: n}^{1}\right)\left(1000 A_{x} \frac{1}{n}\right)
\end{array} \\
& =V\left(Z_{1}\right)+(1000)\left(1000^{2} A_{x: \frac{1}{n}}\right)-\left(1000 A_{x} \frac{1}{n}\right)^{2}-\left(1000 A_{x: n}\right)^{2} \\
& \quad-(2)\left(1000 A_{x: n}^{1}\right)\left(1000 A_{x} \frac{1}{n}\right) \\
& 15,000=\operatorname{Var}\left(Z_{1}\right)+(1000)(136)-(209)^{2}-2(528)(209)
\end{aligned}
\]

Therefore, \(\operatorname{Var}\left(Z_{1}\right)=15,000-136,000+43,681+220,704=143,385\).
Question: . For three fully discrete insurance products on the same ( \(x\) ), you are given:
(i) \(\quad Z_{1}\) is the present value random variable for a 20-year term insurance of 50
(ii) \(\quad Z_{2}\) is the present value random variable for a 20-year deferred whole life insurance of 100
(iii) \(\quad Z_{3}\) is the present value random variable for a whole life insurance of 100 .
(iv) \(E\left[Z_{1}\right]=1.65\) and \(E\left[Z_{2}\right]=10.75\)
(v) \(\operatorname{Var}\left(Z_{1}\right)=46.75\) and \(\operatorname{Var}\left(Z_{2}\right)=50.78\)

Calculate \(\operatorname{Var}\left(Z_{3}\right)\).

Options:
A
62
B
109
C
167
D
202
E
238

Correct answer: C

Solution: \[
Z_{3}=2 Z_{1}+Z_{2} \text { so that } \operatorname{Var}\left(Z_{3}\right)=4 \operatorname{Var}\left(Z_{1}\right)+\operatorname{Var}\left(Z_{2}\right)+4 \operatorname{Cov}\left(Z_{1}, Z_{2}\right)
\]
\[
\begin{aligned}
& \text { where } \operatorname{Cov}\left(Z_{1}, Z_{2}\right)=\underbrace{E\left[Z_{1} Z_{2}\right]}_{=0}-E\left[Z_{1}\right] E[ \\
& \begin{aligned}
\operatorname{Var}\left(Z_{3}\right) & =4(46.75)+50.78-4(1.65)(10.75) \\
& =166.83
\end{aligned}
\end{aligned}
\]
Question: . For a 2 -year deferred, 2 -year term insurance of 2000 on [65], you are given:
(i) The following select and ultimate mortality table with a 3 -year select period:

\begin{tabular}{|c|c|c|c|c|c|}
\hline\(x\) & \(q_{[x]}\) & \(q_{[x]+1}\) & \(q_{[x]+2}\) & \(q_{x+3}\) & \(x+3\) \\
\hline 65 & 0.08 & 0.10 & 0.12 & 0.14 & 68 \\
\hline 66 & 0.09 & 0.11 & 0.13 & 0.15 & 69 \\
\hline 67 & 0.10 & 0.12 & 0.14 & 0.16 & 70 \\
\hline 68 & 0.11 & 0.13 & 0.15 & 0.17 & 71 \\
\hline 69 & 0.12 & 0.14 & 0.16 & 0.18 & 72 \\
\hline
\end{tabular}
(ii) \(\quad i=0.04\)
(iii) The death benefit is payable at the end of the year of death Calculate the actuarial present value of this insurance.

Options:
A
260
B
290
C
350
D
370
E
410

Correct answer: C

Solution: \[
\begin{aligned}
{ }_{2 \mid 2} A_{65}= & \underbrace{v^{3}}_{\text {payment year 3 }} \underbrace{{ }_{2} p_{[65]}}_{\text {Lives 2 years }} \times \underbrace{q_{[65]+2}}_{\text {Die year 3 }} \\
& +\underbrace{v^{4}}_{\text {payment year 4 }} \underbrace{{ }_{3} p_{[65]}}_{\text {Lives 3 years }} \times \underbrace{q_{65+3}}_{\text {Die year 4 }} \\
= & \left(\frac{1}{1.04}\right)^{3}(0.92)(0.9)(0.12) \\
& +\left(\frac{1}{1.04}\right)^{4}(0.92)(0.9)(0.88)(0.14) \\
= & 0.088+0.087=0.176
\end{aligned}
\]

The actuarial present value of this insurance is therefore \(2000 \times 0.176=352\).
Question: . A fund is established for the benefit of 400 workers all age 60 with independent future lifetimes. When they reach age 85 , the fund will be dissolved and distributed to the survivors.

The fund will earn interest at a rate of \(5 \%\) per year.
The initial fund balance, \(F\), is determined so that the probability that the fund will pay at least 5000 to each survivor is \(86 \%\), using the normal approximation.

Mortality follows the Standard Ultimate Life Table.
Calculate \(F\).

Options:
A
350,000
B
360,000
C
370,000
D
380,000
E
390,000

Correct answer: E

Solution: Out of 400 lives initially, we expect \(400_{25} p_{60}=400 \frac{l_{85}}{l_{60}}=400\left(\frac{61,184.9}{96,634.1}\right)=253.26\) survivors
The standard deviation of the number of survivors is \(\sqrt{400_{25} p_{60}\left(1-{ }_{25} p_{60}\right)}=9.639\)

To ensure \(86 \%\) funding, using the normal distribution table, we plan for \(253.26+1.08(9.639)=263.67\)

The initial fund must therefore be \(F=(264)(5000)\left(\frac{1}{1.05}\right)^{25}=389,800\).
Question: . For a special whole life insurance on ( \(x\) ), you are given:
(i) Death benefits are payable at the moment of death
(ii) The death benefit at time \(t\) is \(b_{t}=e^{0.02 t}\), for \(t \geq 0\)
(iii) \(\quad \mu_{x+t}=0.04\), for \(t \geq 0\)
(iv) \(\delta=0.06\)
(v) \(\quad Z\) is the present value at issue random variable for this insurance

Calculate \(\operatorname{Var}(Z)\).

Options:
A
0.020
B
0.036
C
0.052
D
0.068
E
0.083

Correct answer: E

Solution: \(E[Z]=\int_{0}^{\infty} b_{t} \cdot v^{t} \cdot{ }_{t} p_{x} \cdot \mu_{x+t} d t=\int_{0}^{\infty} e^{0.02 t} \cdot e^{-0.06 t} \cdot e^{-0.04 t} \cdot 0.04 d t\)
\(=0.04 \int_{0}^{\infty} e^{-0.08 t} d t=\frac{0.04}{0.08}=\frac{1}{2}\)
\(E\left[Z^{2}\right]=\int_{0}^{\infty}\left(b_{t} \cdot v^{t}\right)^{2}{ }_{t} p_{x} \cdot \mu_{x+t} d t=\int_{0}^{\infty}\left(e^{0.04 t}\right)\left(e^{-0.12 t}\right)\left(0.04 e^{-0.04}\right) d t=\frac{0.04}{0.12}=\frac{1}{3}\)
\(\operatorname{Var}[Z]=\frac{1}{3}-\left(\frac{1}{2}\right)^{2}=\frac{1}{12}=0.0833\)
Question: . You are given the following extract of ultimate mortality rates from a two-year select and ultimate mortality table:

\begin{tabular}{|c|c|}
\hline\(x\) & \(q_{x}\) \\
\hline 50 & 0.045 \\
\hline 51 & 0.050 \\
\hline 52 & 0.055 \\
\hline 53 & 0.060 \\
\hline
\end{tabular}

The select mortality rates satisfy the following:
- \(q_{[x]}=0.7 q_{x}\)
- \(q_{[x]+1}=0.8 q_{x+1}\)

You are also given that \(i=0.04\).
Calculate \(A_{[50]: 3]}^{1}\)

Options:
A
0.08
B
0.09
C
0.10
D
0.11
E
0.12

Correct answer: D

Solution: \(A_{[50]: 31}^{1}=v q_{[50]}+v^{2} p_{[50]} q_{[50]+1}+v^{3} p_{[50]} p_{[50]+1} q_{52}\)
where: \(v=\frac{1}{1.04}\)
\(q_{[50]}=0.7(0.045)=0.0315\)
\(p_{[50]}=1-q_{[50]}=0.9685\)
\(q_{[50]+1}=0.8(0.050)=0.040\)
\(p_{[50]+1}=1-q_{[50]+1}=0.960\)
\(q_{52}=0.055\)

So: \(A_{[50]: 31}^{1}=0.1116\)
Question: . For a special whole life policy on (48), you are given:
(i) The policy pays 5000 if the insured's death is before the median curtate future lifetime at issue and 10,000 if death is after the median curtate future lifetime at issue
(ii) Mortality follows the Standard Ultimate Life Table
(iii) Death benefits are paid at the end of the year of death
(iv) \(\quad i=0.05\)

Calculate the actuarial present value of benefits for this policy.

Options:
A
1130
B
1160
C
1190
D
1220
E
1250

Correct answer: A

Solution: The median of \(K_{48}\) is the integer \(m\) for which
\[
P\left(K_{48}<m\right) \leq 0.5 \text { and } P\left(K_{48}>m\right) \leq 0.5
\]

This is equivalent to finding \(m\) for which
\(\frac{l_{48+m}}{l_{48}} \geq 0.5\) and \(\frac{l_{48+m+1}}{l_{48}} \leq 0.5\).
Based on the SULT and \(l_{48}(0.5)=(98,783.9)(0.5)=49,391.95\), we have \(m=40\) since \(l_{88} \geq 49,391.95\) and \(l_{89} \leq 49,391.95\).

So: \(A P V=5000 A_{48}+5000_{40} E_{48} A_{88}=5000 A_{48}+5000{ }_{20} E_{48}{ }_{20} E_{68} \cdot A_{88}\)
\(=5000(0.17330)+5000(0.35370)(0.20343)(0.72349)=1126.79\)
Question: . You are given that \(T\), the time to first failure of an industrial robot, has a density \(f(t)\) given by
\[
f(t)= \begin{cases}0.1, & 0 \leq t<2 \\ 0.4 t^{-2}, & 2 \leq t<10\end{cases}
\]
with \(f(t)\) undetermined on \([10, \infty)\).
Consider a supplemental warranty on this robot that pays 100,000 at the time \(T\) of its first failure if \(2 \leq T \leq 10\), with no benefits payable otherwise.

You are also given that \(\delta=5 \%\).
Calculate the \(90^{\text {th }}\) percentile of the present value of the future benefits under this warranty.

Options:
A
\(\quad 82,000\)
B
84,000
C
87,000
D
91,000
E
95,000

Correct answer: A

Solution: The present value random variable \(\mathrm{PV}=1,000,000 e^{-0.05 T}, 2 \leq T \leq 10\) is a decreasing function of \(T\) so that its \(90^{\text {th }}\) percentile is
\(1,000,000 e^{-0.05 p}\) where \(\boldsymbol{p}\) is the solution to \(\int_{2}^{p} 0.4 t^{-2} d t=0.10\).
\[
\begin{aligned}
& \int_{2}^{p} 0.4 t^{-2} d t=-\left.0.4\left(\frac{t^{-1}}{-1}\right)\right|_{2} ^{p}=0.4\left(\frac{1}{2}-\frac{1}{p}\right)=0.10 \\
& p=4 \\
& 1,000,000 e^{-0.05 \times 4}=81,873.08
\end{aligned}
\]
Question: . (80) purchases a whole life insurance policy of 100,000 .

You are given:
(i) The policy is priced with a select period of one year
(ii) The select mortality rate equals \(80 \%\) of the mortality rate from the Standard Ultimate Life Table
(iii) Ultimate mortality follows the Standard Ultimate Life Table
(iv) \(\quad i=0.05\)

Calculate the actuarial present value of the death benefits for this insurance

Options:
A
\(\quad 58,950\)
B
59,050
C
\(\quad 59,150\)
D
59,250
E
59,350

Correct answer: B

Solution: Superscript SULT refers to values from the SULT. Values without superscripts refer to this select life.
\[
\begin{aligned}
q_{80} & =0.8 q_{80}^{S U L T}=0.0261264 \Rightarrow p_{80}=0.9738736 \\
A_{80} & =v q_{80}+v p_{80} A_{81}^{S U L T} \\
& =(1.05)^{-1}(0.0261264)+(1.05)^{-1}(0.9738736)(0.60984)=0.59051
\end{aligned}
\]
\(100,000 A_{80}=59,051\)
Question: . For a special fully continuous whole life insurance on ( \(x\) ), you are given:
i) \(\quad \mu_{x+t}=0.03, t \geq 0\)
ii) \(\delta=0.06\)
iii) The death benefit at time \(t\) is \(b_{t}=e^{0.05 t}, t \geq 0\)
iv) \(\quad Z\) is the present value random variable at issue for this insurance

Calculate \(\operatorname{Var}(Z)\).

Options:
A
0.0300
B
0.0325
C
0.0350
D
0.0375
E
0.0400

Correct answer: D

Solution: \[
\begin{aligned}
E(Z) & =\int_{0}^{\infty} p_{x} \times \mu \times e^{0.05 \times t} e^{-\delta \times t} d t \\
& =\int_{0}^{\infty} e^{-0.03 \times t} \times 0.03 \times e^{0.05 \times t} e^{-0.06 \times t} d t \\
& =\frac{0.03}{0.04} \times\left. e^{-0.04 t}\right|_{0} ^{\infty}=0.75 \\
E\left(Z^{2}\right) & =\int_{0}^{\infty} p_{x} \times \mu \times e^{0.05 \times 2 \times t} e^{-\delta \times 2 \times t} d t \\
& =\int_{0}^{\infty} e^{-0.03 \times t} \times 0.03 \times e^{0.05 \times 2 \times t} e^{-0.06 \times 2 \times t} d t \\
& =\frac{0.03}{0.04} \times\left. e^{-0.05 t}\right|_{0} ^{\infty}=0.6 \\
\operatorname{Var}(Z) & =E\left(Z^{2}\right)-[E(Z)]^{2}=0.6-0.75^{2}=0.375
\end{aligned}
\]
Question: . For a two-year term insurance of 1 on ( \(x\) ) payable at the moment of death, you are given:
i) \(q_{x}=0.04\)
ii) \(\quad q_{x+1}=0.06\)
iii) Deaths are uniformly distributed over each year of age
iv) \(i=0.04\)
v) \(\quad Z\) is the present value random variable for this insurance

Calculate Var[Z].

Options:
A
0.065
B
0.069
C
0.073
D
0.077
E
0.081

Correct answer: E

Solution: \[
\begin{aligned}
& E(Z)=\frac{i}{\delta}\left[v q_{x}+v^{2} p_{x} q_{x+1}\right]=\frac{0.04}{\ln (1.04)}\left[\frac{1}{1.04}(0.04)+\frac{1}{1.04^{2}}(0.96)(0.06)\right]=0.09353831 \\
& E\left(Z^{2}\right)=\frac{(1+i)^{2}-1}{2 \delta}\left[v^{2} q_{x}+v^{4} p_{x} q_{x+1}\right]=\frac{\left(1.04^{2}-1\right)}{2 \ln (1.04)}\left[\frac{1}{1.04^{2}}(0.04)+\frac{1}{1.04^{4}}(0.96)(0.06)\right]=0.08969072 \\
& \operatorname{Var}(Z)=E\left(Z^{2}\right)-[E(Z)]^{2}=0.08969072-(0.09353831)^{2}=0.0809413
\end{aligned}
\]