BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: . The gross annual premium, \(G\), for a fully discrete 5 -year endowment insurance of 1000 issued on \((x)\) is calculated using the equivalence principle. You are given:
(i) \(\quad 1000 P_{x: 5 \mid}=187.00\)
(ii) The expense policy value at the end of the first year, \({ }_{1} V^{e}=-38.70\)
(iii) \(q_{x}=0.008\)
(iv) Expenses, payable at the beginning of the year, are:

\begin{tabular}{|l|c|c|}
\hline Year & \begin{tabular}{c} 
Percent of \\
Premium
\end{tabular} & Per Policy \\
\hline First & \(25 \%\) & 10 \\
\hline Renewal & \(5 \%\) & 5 \\
\hline
\end{tabular}
(v) \(\quad i=0.03\)

Calculate \(G\).

Options:
A
200
B
213
C
226
D
239
E
252

Correct answer: B

Solution: Since \(G\) is determined using the equivalence principle, \({ }_{0} V=0\)

Then, \({ }_{1} V^{e}=\frac{(0+\overbrace{G-187}^{P^{e}}-0.25 G-10)(1.03)}{0.992}=-38.7\)
\(\Rightarrow 0.75 G=\frac{-38.7(0.992)}{1.03}+187+10=159.72\)
\(\Rightarrow G=212.97\)
Question: . For a special fully discrete whole life insurance of 1,000 on (45), you are given:
(i) The net premiums for year \(k\) are:
\[
\left\{\begin{array}{cc}
P, & k=1,2, \ldots, 20 \\
P+W, & k=21,22, \ldots
\end{array}\right.
\]
(ii) Mortality follows the Standard Ultimate Life Table
(iii) \(\quad i=0.05\)
(iv) \({ }_{20} V\), the net premium policy value at the end of the \(20^{\text {th }}\) year, is 0

Calculate \(W\).

Options:
A
12
B
16
C
20
D
24
E
28

Correct answer: D

Solution: \({ }_{20} V=0=>1000 A_{65}=(P+W) \times \ddot{a}_{65}\)

At issue, present value of benefits must equal present value of premium, so: \(1000 A_{45}=P \ddot{a}_{45}+W_{20} E_{45} \times \ddot{a}_{65}\)
\(354.77=(P+W)(13.5498) \Rightarrow P+W=26.182674 \Rightarrow P=26.182674-W\)
\(151.61=17.8162 P+W(0.35994)(13.5498)\)
\(151.61=17.8162(26.182674-W)+W(0.35994)(13.5498)\)
\(\Rightarrow W=24.33447\)
Question: . For a fully discrete whole life insurance of \(B\) on \((x)\), you are given:
(i) Expenses, incurred at the beginning of each year, equal 30 in the first year and 5 in subsequent years
(ii) The net premium policy value at the end of year 10 is 2290
(iii) Gross premiums are calculated using the equivalence principle
(iv) \(\quad i=0.04\)
(v) \(\quad \ddot{a}_{x}=14.8\)
(vi) \(\quad \ddot{a}_{x+10}=11.4\)

Calculate \({ }_{10} V^{g}\), the gross premium policy value at the end of year 10.

Options:
A
2190
B
2210
C
2230
D
2250
E
2270

Correct answer: E

Solution: \(V_{10}=2,290=B\left(1-\frac{\ddot{a}_{x+10}}{\ddot{a}_{x}}\right)=B\left(1-\frac{11.4}{14.8}\right) \Rightarrow B=9,968.24\)
\(G \ddot{a}_{x}=25+5 \ddot{a}_{x}+B \times A_{x}\)
\(A_{x}=1-d \ddot{a}_{x}=1-\left(\frac{0.04}{1.04} \times 14.8\right)=0.430769231\)
\(G \times 14.8=25+5 \times 14.8+9,968.24 \times 0.430769231\)
\(\Rightarrow G=296.82\)
\({ }_{10} V^{g}=9,968.24 A_{x+10}+5 \ddot{a}_{x+10}-296.82 \ddot{a}_{x+10}\)
\(A_{x+10}=1-d \ddot{a}_{x+10}=1-\left(\frac{0.04}{1.04} \times 11.4\right)=0.561538462\)
\({ }_{10} V^{g}=9,968.24 \times 0.561538462+5 \times 11.4-296.82 \times 11.4\)
\(\Rightarrow{ }_{10} V^{g}=2,270.80\)
Alternatively, the expense net premium is based on the extra expenses in year 1, so \(P^{e}=(30-5) / 14.8=1.68919\)
\({ }_{10} V^{e}=0-1.68919(11.4)=-19.26\)
\({ }_{10} V^{g}={ }_{10} V^{n}+{ }_{10} V^{e}=2290-19.26=2270.74\)
Question: . Ten years ago J, then age 25, purchased a fully discrete 10-payment whole life policy of 10,000. All actuarial calculations for this policy were based on the following:
(i) Mortality follows the Standard Ultimate Life Table
(ii) \(\quad i=0.05\)
(iii) The equivalence principle

In addition:
(i) \(\quad L_{10}\) is the present value of future losses random variable at time 10.
(ii) At the end of policy year 10 , the interest rate used to calculate \(L_{10}\) is changed to \(0 \%\). Calculate the increase in \(E\left[L_{10}\right]\) that results from this change.

Options:
A
5035
B
6035
C
7035
D
8035
E
9035

Correct answer: E

Solution: \(L_{10}=10,000 A_{35}=965.30\)
\(L_{10}^{*}=10,000\)
\(L_{10}^{*}-L_{10}=10,000-965.30=9034.70\)
Question: . For a fully discrete 3 -year endowment insurance of 1000 on \((x)\), you are given:
(i) Expenses, payable at the beginning of the year, are:

\begin{tabular}{|c|c|c|}
\hline Year(s) & Percent of Premium & Per Policy \\
\hline 1 & \(20 \%\) & 15 \\
\hline 2 and 3 & \(8 \%\) & 5 \\
\hline
\end{tabular}
(ii) The expense policy value at the end of year 2 is - 23.64
(iii) The gross annual premium calculated using the equivalence principle is \(G=368.05\)
(iv) \(G=1000 P_{x: 3]}+P^{e}\), where \(P^{e}\) is the expense loading

Calculate \(P_{x: 3]} \cdot\)

Options:
A
0.290
B
0.295
C
0.300
D
0.305
E
0.310

Correct answer: E

Solution: Future expenses at \(x+2=0.08 G+5\)
Expense load at \(x+2=P^{e}\)
\(-23.64=(0.08 G+5)-P^{e}\)
\(\Rightarrow P^{e}=58.08\)
\(1000 P_{x: 3 \mid}=368.05-58.08=309.97\)
Question: For two fully continuous whole life insurance policies on ( \(x\) ), you are given:
(i)

\begin{tabular}{lccc}
\hline \hline & \begin{tabular}{c} 
Death \\
Benefit
\end{tabular} & \begin{tabular}{c} 
Annual \\
Premium Rate
\end{tabular} & \begin{tabular}{c} 
Variance of the Present \\
Value of Future Loss at \(t\)
\end{tabular} \\
\hline Policy A & 1 & 0.10 & 0.455 \\
Policy B & 2 & 0.16 & - \\
\hline \hline
\end{tabular}
(ii) \(\delta=0.06\)

Calculate the variance of the present value of future loss at \(t\) for Policy B.

Options:
A
0.9
B
1.4
C
2.0
D
2.9
E
3.4

Correct answer: B

Solution: \(L_{A}=v^{T}-0.10 \bar{a}_{\bar{T} \mid}=\left(1+\frac{10}{6}\right) v^{T}-\frac{10}{6}\)
\(\operatorname{Var}\left(L_{A}\right)=\left(1+\frac{10}{6}\right)^{2} \operatorname{Var}\left(v^{T}\right)=0.455 \Rightarrow \operatorname{Var}\left(v^{T}\right)=0.06398\)
\(L_{B}=2 v^{T}-0.16 \bar{a}_{\bar{T} \mid}=\left(2+\frac{16}{6}\right) v^{T}-\frac{16}{6}\)
\(\operatorname{Var}\left(L_{B}\right)=\left(2+\frac{16}{6}\right)^{2} \operatorname{Var}\left(v^{T}\right)=\left(2+\frac{16}{6}\right)^{2}(0.06398)=1.39\)
Question: For a special semi-continuous 20-year endowment insurance on (70), you are given:
i) The death benefit is 1000
ii) The endowment benefit is 500
iii) Mortality follows the Standard Ultimate Life Table
iv) Deaths are uniformly distributed over each year of age
v) The annual net premium is 35.26
vi) \(i=0.05\)

Calculate the net premium policy value at the end of year 10 .

Options:
A
268
B
272
C
276
D
280
E
284

Correct answer: C

Solution: \(A_{80: \overline{10}}^{1}=A_{80: \overline{10}}-{ }_{10} E_{80}=0.67674-0.33952=0.33722\)
\(\bar{A}_{80: \overline{10}}^{1}=\frac{0.05}{\ln (1.05)} 0.33722=0.34559\)
EPV future benefits \(=1000(0.34559)+500(0.33952)=515.35\)
EPV future net premiums \(=\ddot{a}_{80: 10} P=6.7885(35.26)=239.36\)
Policy value \(=515.36-239.36=276\)
Question: An insurer issues a 30-year term insurance policy on (40). You are given:
i) Net premiums of 750 are payable quarterly
ii) The death benefit, payable at the end of the quarter of death, is 1,000,000
iii) \({ }_{t} V\) denotes the net premium policy value at time \(t, t \geq 0\)
iv) \(10.5^{V}=10,000\)
v) \(\quad q_{50}=0.01\)
vi) Mortality is uniformly distributed over each year of age
vii) \(i=0.05\)

Calculate \({ }_{10.75} V\).

Options:
A
8,360
B
8,370
C
8,380
D
8,390
E
8,400

Correct answer: D

Solution: \[
\begin{aligned}
& \left({ }_{10.5} V+P\right)(1+i)^{0.25}=1,000,000 \times{ }_{0.25} q_{50.5}+{ }_{0.25} p_{50.5} \times{ }_{50.75} V \\
& (10,000+750)(1+0.05)^{0.25}=1,000,000 \times\left(1-\frac{(1-0.75 \times 0.01)}{(1-0.5 \times 0.01)}\right)+\left(\frac{(1-0.75 \times 0.01)}{(1-0.5 \times 0.01)}\right) \times{ }_{50.75} V \\
& { }_{50.75} V=8,390
\end{aligned}
\]
Question: For a fully discrete whole life insurance of 100,000 on (60), you are given:
i) Reserves are determined using a modified net premium reserve method
ii) The modified net premium reserve at the end of year 2 is 0
iii) Valuation premiums in years 3 and later are level
iv) Mortality follows the Standard Ultimate Life Table
v) \(i=0.05\)

Calculate the valuation premium for year 5 .

Options:
A
1,950
B
2,050
C
2,120
D
2,190
E
2,290

Correct answer: D

Solution: The modified net premium reserve at duration 2 is: \({ }_{2} V^{\bmod }=100000 A_{62}-P^{\bmod } \ddot{a}_{62}\)
\[
P^{\bmod }=100000 \frac{A_{62}}{\ddot{a}_{62}}=100000\left(\frac{0.31495}{14.3861}\right)=2189
\]
Question: You are checking gross premium policy values for a fully discrete whole life insurance of 1000 on (50).
\({ }_{k} V\) denotes the gross premium policy value at the end of year \(k, k=0,1,2, \ldots\).
The valuation assumptions were intended to include:
i) There are commissions and maintenance expenses payable at the beginning of the year
ii) There are no other expenses
iii) \(\quad q_{58}=0.002736\)
iv) \(i=0.05\)

You discover that all intended assumptions were used correctly, except that calculations were based on \(q_{58}=0.003736\).
The calculated results included \({ }_{8} V=86.74\) and \({ }_{9} V=100\).
Calculate \({ }_{8} V\) using the intended value of \(q_{58}\).

Options:
A
85.79
B
85.88
C
85.97
D
86.06
E
86.15

Correct answer: B

Solution: There is no change to \({ }_{9} V\) because there were no errors after time 9 .
An equation for \({ }_{8} V^{\text {orig }}\) is \({ }_{8} V^{\text {orig }}+P-e=1000(0.003736) / 1.05+100(0.996264) / 1.05\)
An equation for \({ }_{8} V^{\text {now }}\) is \({ }_{8} V^{\text {now }}+P-e=1000(0.002736) / 1.05+100(0.997264) / 1.05\)
By subtraction, \({ }_{8} V^{\text {now }}-{ }_{8} V^{\text {orig }}=1000(-0.001) / 1.05+100(0.001) / 1.05=-0.86\)
\({ }_{8} V^{\text {now }}=86.74-0.86=85.88\) (B)
In those equations, \(P-e\), which cancels, represents the gross premium less all expenses. You could think of \(P\) as the gross premium less commissions, and e as the flat expenses. However you think of representing them, they cancel.
Question: For a fully discrete increasing 20-year endowment insurance on (50), you are given:
i) The level annual net premium is 5,808
ii) The net premium policy value at the end of year 15 is 130,580
iii) Mortality after age 60 follows the Standard Ultimate Life Table
iv) \(i=0.05\)

Calculate the expected present value of future death and endowment benefits at age 65.

Options:
A
156,530
B
156,570
C
156,610
D
156,650
E
156,690

Correct answer: D

Solution: \(\ddot{a}_{65: 51}=\ddot{a}_{65}-{ }_{5} E_{65} \times \ddot{a}_{70}=13.5498-0.75455(12.0083)=4.4889\)
EPV future net premiums \(=4.4889(5,808)=26,072\)
130,580 = EPV future benefits - EPV future net premiums = EPV future benefits - 26,072
EPV future benefits \(=156,652\)
Question: For a fully discrete 30 payment whole life insurance of 1000 on (50) with level annual premiums of 16, \({ }_{k} V\) denotes the gross premium policy value at the end of year \(k, k=0,1,2, \ldots\).
The original valuation assumptions include:
i) Mortality follows the Standard Ultimate Life Table
ii) \(\quad i=0.05\)
iii) Premium taxes are \(2 \%\)
iv) There are commissions and various other expenses

Using the original valuation assumptions, \({ }_{10} V=110\).
At some point prior to year 10, your jurisdiction increased the premium tax rate so the premium tax assumption increased to \(3 \%\). All other assumptions are unchanged.

Calculate the revised value of \({ }_{10} V\).

Options:
A
112
B
114
C
116
D
118
E
120

Correct answer: A

Solution: There is no change in the EPV of future benefits or of future gross premiums. Each year that a premium is due, the expense changes by ( \(3 \%-2 \%\) ) of 16 , or 0.16 . Since at the end of year 10 , the insured is 60 with 20 remaining premiums, the EPV of future expense rises by
\(0.16 \ddot{a}_{60: 201}=0.16(12.3816)=1.981\)
The new reserve is \(110+1.981=112\)
Question: An insurer issues a 20-year deferred whole life annuity due on [45]. You are given:
i) Net premiums of 20,000 are payable at the beginning of each year during the deferral period
ii) There is no benefit paid upon death during the deferral period
iii) \({ }_{t} V\) denotes the net premium policy value at time \(t, t \geq 0\)
iv) \({ }_{19} V=575,000\)
v) \(q_{[45]+18}=0.023044\)
vi) \(i=0.05\)

Calculate \({ }_{18} V\).

Options:
A
495,000
B
505,000
C
515,000
D
525,000
E
535,000

Correct answer: C

Solution: \[
\begin{aligned}
& \left({ }_{18} V+P\right)(1+i)=p_{[45]+18} \times{ }_{19} V \\
& { }_{18} V=0.976956 \times 575,000 / 1.05-20,000=515,000
\end{aligned}
\]
Question: For a fully discrete whole life insurance of 1000 on (60), you are given:
i) Reserves are determined using a modified net premium reserve method
ii) The modified reserve at the end of year 2 is 0
iii) Valuation premiums in years 3 and later are level
iv) Mortality follows the Standard Ultimate Life Table
v) \(i=0.05\)

Calculate the modified net premium reserve at the end of year 5 .

Options:
A
58
B
69
C
79
D
90
E
99

Correct answer: A

Solution: The policy value at duration 2 is \({ }_{2} V^{\text {mod }}=1000 A_{62}-\left(P^{\text {mod }}\right)\left(\ddot{a}_{62}\right)=0\)
\(P^{\mathrm{mod}}=1000 A_{62} / \ddot{a}_{62}=1000(0.31495) / 14.3861=21.89\)
\({ }_{5} V^{\text {mod }}=1000 A_{65}-\left(P^{\text {mod }}\right)\left(\ddot{a}_{65}\right)=1000(0.35477)-21.89(13.5498)=58\)
Question: For a special fully discrete 20-year term insurance of 100,000 with a return of premium endowment benefit on (70), you are given:
i) If the insured survives the 20-year term, all gross premiums are returned with interest at \(5 \%\) at the end of 20 years
ii) The gross premium is 6,750
iii) Renewal commissions are \(5 \%\) of premium
iv) \({ }_{5} L\) is the gross future loss random variable at time 5, given the insured is alive at time 5
v) Mortality follows the Standard Ultimate Life Table
iv) \(\quad i=0.05\)

Calculate the \(90^{\text {th }}\) percentile of \({ }_{5} L\).

Options:
A
\(\quad 24,170\)
B
40,430
C
49,200
D
58,370
E
\(\quad 78,170\)

Correct answer: C

Solution: Note that the premium refund benefit does not affect the \(90 \%\) percentile because the benefit is largely offset by the accumulated premiums.

The highest values of \({ }_{5} L\) correspond to the earliest deaths. So, we want the first \(10 \%\) of deaths. \((0.90)\left(l_{75}\right)=76,683.15\) which is between \(l_{79}\) and \(l_{80}\). So, we want deaths occurring between 79 and 80 . Collect 5 premiums (net of commissions), pay death benefit at end of 5 years.
\(\mathrm{v}^{5}=0.78353\) annuity due \(=((1-0.78353) / 0.05) 1.05=4.54587\)
\({ }_{5} L=100,000(0.78353)-4.54587(6,750)(0.95)=49,203\)

\section*{ALL QUESTIONS FROM CHAPTER 18 HAVE BEEN DELETED}
Question: . A company offers a modern insurance contract that has the following features:
i) A benefit linked to the performance of an investment fund
ii) A guaranteed minimum return on the premiums paid
iii) A guaranteed minimum death benefit payable if the insured dies before the contract matures
iv) A contract term of 7 years

Determine the type of contract offered.

Options:
A
Term life insurance
B
Universal life insurance
C
Variable annuity
D
Endowment insurance
E
Reversionary annuity

Correct answer: C

Solution: Key is that the benefit is linked to the performance of an investment fund. That is true only for C.
Question: . The present value random variable for an insurance policy on ( \(x\) ) is expressed as:
\(Z= \begin{cases}0, & \text { if } T_{x} \leq 10 \\ v^{T_{x}}, & \text { if } 10<T_{x} \leq 20 \\ 2 v^{T_{x}}, & \text { if } 20<T_{x} \leq 30 \\ 0, & \text { thereafter }\end{cases}\)
Determine which of the following is a correct expression for \(E[Z]\).

Options:
A
\(\quad{ }_{10} \bar{A}_{x}+{ }_{20}\left|\bar{A}_{x}-{ }_{30}\right| \bar{A}_{x}\)
B
\(\bar{A}_{x}+{ }_{20} E_{x} \bar{A}_{x+20}-2{ }_{30} E_{x} \bar{A}_{x+30}\)
C
\({ }_{10} E_{x} \bar{A}_{x}+{ }_{20} E_{x} \bar{A}_{x+20}-2{ }_{30} E_{x} \bar{A}_{x+30}\)
D
\({ }_{10} E_{x} \bar{A}_{x+10}+{ }_{20} E_{x} \bar{A}_{x+20}-2{ }_{30} E_{x} \bar{A}_{x+30}\)
E
\({ }_{10} E_{x}\left[\bar{A}_{x+10}+{ }_{10} E_{x+10} \bar{A}_{x+20}-{ }_{10} E_{x+20} \bar{A}_{x+30}\right]\)

Correct answer: D

Solution: Drawing the benefit payment pattern:
![](https://cdn.mathpix.com/cropped/2025_10_19_8157dbad7f49bfd0f1c9g-21.jpg?height=443&width=757&top_left_y=501&top_left_x=376)
\(E[Z]={ }_{10} E_{x} \times \bar{A}_{x+10}+{ }_{20} E_{x} \times \bar{A}_{x+20}-2{ }_{30} E_{x} \times \bar{A}_{x+30}\)