BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: . For a fully continuous 20 -year term insurance policy of 100,000 on (50), you are given:
(i) Gross premiums, calculated using the equivalence principle, are payable at an annual rate of 4500
(ii) Expenses at an annual rate of \(R\) are payable continuously throughout the life of the policy
(iii) \(\quad \mu_{50+t}=0.04\), for \(t>0\)
(iv) \(\delta=0.08\)

Calculate \(R\).

Options:
A
400
B
500
C
600
D
700
E
800

Correct answer: B

Solution: By the equivalence principle,
\[
4500 \bar{a}_{x: 20 \mid}=100,000 \bar{A}_{x: 20 \mid}^{1}+R \bar{a}_{x: 20 \mid}
\]
where
\[
\begin{aligned}
& \bar{A}_{x: 201}^{1}=\frac{\mu}{\mu+\delta}\left(1-e^{-20(\mu+\delta)}\right)=\frac{0.04}{0.12}\left(1-e^{-20(0.12)}\right)=0.3031 \\
& \bar{a}_{x: \overline{201}}=\frac{1-e^{-20(\mu+\delta)}}{\mu+\delta}=\frac{1-e^{-20(0.12)}}{0.12}=7.5774
\end{aligned}
\]

Solving for \(R\), we have
\[
R=4500-100,000\left(\frac{0.3031}{7.5774}\right)=500
\]
Question: . For a fully discrete whole life insurance policy of 50,000 on (35), with premiums payable for a maximum of 10 years, you are given:
(i) Expenses of 100 are payable at the end of each year including the year of death
(ii) Mortality follows the Standard Ultimate Life Table
(iii) \(\quad i=0.05\)

Calculate the annual gross premium using the equivalence principle.

Options:
A
790
B
800
C
810
D
820
E
830

Correct answer: D

Solution: By the equivalence principle, we have
\[
G \ddot{a}_{35: \overline{10}}=50,000 A_{35}+100 a_{35}+100 A_{35}
\]
so
\[
G=\frac{50,100 A_{35}+100\left(\ddot{a}_{35}-1\right)}{\ddot{a}_{35: \overline{10}}}=\frac{50,100(0.09653)+100(17.9728)}{8.0926}=819.69
\]
Question: . For an \(n\)-year endowment insurance of 1000 on \((x)\), you are given:
(i) Death benefits are payable at the moment of death
(ii) Premiums are payable annually at the beginning of each year
(iii) Deaths are uniformly distributed over each year of age
(iv) \(\quad i=0.05\)
(v) \(\quad{ }_{n} E_{x}=0.172\)
(vi) \(\quad \bar{A}_{x: n}=0.192\)

Calculate the annual net premium for this insurance.

Options:
A
10.1
B
11.3
C
12.5
D
13.7
E
14.9

Correct answer: B

Solution: Let \(P\) be the annual net premium
\[
P=\frac{1000 \bar{A}_{x: \bar{n}}}{\ddot{a}_{x: n}}=\frac{1000(0.192)}{\ddot{a}_{x: \bar{n}}}
\]
where
\[
\begin{aligned}
& \ddot{a}_{x: \bar{n}}=\frac{1-A_{x: \bar{n}}}{d}=\frac{(1.05)}{(0.05)}\left(1-A_{x: \bar{n}}^{1}-A_{x: \bar{n}}\right) \\
& A_{x: \bar{n}}=\frac{i}{\delta}\left(A_{x: \bar{n}}^{1}\right)+{ }_{n} E_{x} \\
& \Rightarrow 0.192=\frac{0.05}{0.04879}\left(A_{x: \bar{n}}^{1}\right)+0.172 \\
& \Rightarrow A_{x: \bar{n}}^{1}=0.019516 \\
& \Rightarrow \ddot{a}_{x: \bar{n}}=\frac{1.05}{0.05}(1-0.019516-0.172)=16.978
\end{aligned}
\]

Therefore, we have
\[
P=\frac{1000(0.192)}{16.978}=11.31
\]
Question: . XYZ Insurance writes 10,000 fully discrete whole life insurance policies of 1000 on lives age 40 and an additional 10,000 fully discrete whole life policies of 1000 on lives age 80.
XYZ used the following assumptions to determine the net premiums for these policies:
(i) Mortality follows the Standard Ultimate Life Table
(ii) \(\quad i=0.05\)

During the first ten years, mortality did follow the Standard Ultimate Life Table.
Calculate the average net premium per policy in force received at the beginning of the eleventh year.

Options:
A
29
B
32
C
35
D
38
E
41

Correct answer: A

Solution: Premium at issue for (40): \(\frac{1000 A_{40}}{\ddot{a}_{40}}=\frac{121.06}{18.4578}=6.5587\)
Premium at issue for (80): \(\frac{1000 A_{80}}{\ddot{a}_{80}}=\frac{592.93}{8.5484}=69.3615\)
Lives in force after ten years:
Issued at age 40: \(10,000_{10} p_{40}=10,000 \times \frac{98,576.4}{99,338.3}=9923.30\)
Issued at age 80: \(10,000_{10} p_{80}=10,000 \times \frac{41,841.1}{75,657.2}=5530.35\)

The total number of lives after ten years is therefore: \(9923.30+5530.35=15,453.65\)
The average premium after ten years is therefore:
\(\frac{(6.5587 \times 9923.30)+(69.3615 \times 5530.35)}{15,453.65}=29.03\)
Question: . For a special fully discrete whole life insurance, you are given:
(i) The death benefit is \(1000(1.03)^{k}\) for death in policy year \(k\), for \(k=1,2,3 \ldots\)
(ii) \(q_{x}=0.05\)
(iii) \(\quad i=0.06\)
(iv) \(\quad \ddot{a}_{x+1}=7.00\)
(v) The annual net premium for this insurance at issue age \(x\) is 110

Calculate the annual net premium for this insurance at issue age \(x+1\).

Options:
A
110
B
112
C
116
D
120
E
122

Correct answer: C

Solution: Let \(P\) be the annual net premium at \(x+1\). Also, let \(A_{y}^{*}\) be the expected present value for the special insurance described in the problem issued to \((y)\).
\(P \ddot{a}_{x+1}=1000 \sum_{k=0}^{\infty}(1.03)^{k+1} v^{k+1}{ }_{k \mid} q_{x+1}=1000 A_{x+1}^{*}\)
We are given
\(110 \ddot{a}_{x}=1000 \sum_{k=0}^{\infty}(1.03)^{k+1} v^{k+1}{ }_{k \mid} q_{x}=1000 A_{x}^{*}\)
Which implies that
\(110\left(1+v p_{x} \ddot{a}_{x+1}\right)=1000\left(1.03 v q_{x}+1.03 v p_{x} A_{x+1}^{*}\right)\)
Solving for \(A_{x+1}^{*}\), we get
\(A_{x+1}^{*}=\frac{\frac{110}{1000}[1+v(0.95)(7)]-1.03 v(0.05)}{1.03 v(0.95)}=0.8141032\)
Thus, we have
\(P=\frac{1000(0.8141032)}{7}=116.3005\)
Question: . For a special fully discrete 2 -year term insurance on \((x)\), you are given:
(i) \(\quad q_{x}=0.01\)
(ii) \(\quad q_{x+1}=0.02\)
(iii) \(\quad i=0.05\)
(iv) The death benefit in the first year is 100,000
(v) Both the benefits and premiums increase by \(1 \%\) in the second year Calculate the annual net premium in the first year.

Options:
A
1410
B
1417
C
1424
D
1431
E
1438

Correct answer: B

Solution: Let \(P\) be the net premium for year 1 .
Then:
\(P+1.01 P v p_{x}=100,000 v q_{x}+(1.01)(100,000) v^{2} p_{x} q_{x+1}\)
\(P\left[1+\frac{1.01}{1.05} 0.99\right]=100,000\left(\frac{0.01}{1.05}+\frac{(1.01)(0.99)(0.02)}{(1.05)^{2}}\right) \Rightarrow P=1416.93\)
Question: . For a fully discrete, 5 -payment 10 -year term insurance of 200,000 on ( 30 ), you are given:
(i) Mortality follows the Standard Ultimate Life Table
(ii) The following expenses are incurred at the beginning of each respective year:

\begin{tabular}{lc|c|cc} 
& \multicolumn{2}{c|}{ Year 1 } & \multicolumn{2}{c}{ Years 2-10 } \\
\cline { 2 - 5 } & \begin{tabular}{c} 
Percent of \\
Premium
\end{tabular} & Per Policy & \begin{tabular}{c} 
Percent of \\
Premium
\end{tabular} & Per Policy \\
\hline Taxes & \(5 \%\) & 0 & \(5 \%\) & 0 \\
Commissions & \(30 \%\) & 0 & \(10 \%\) & 0 \\
Maintenance & \(0 \%\) & 8 & \(0 \%\) & 4 \\
\hline
\end{tabular}
(iii) \(\quad i=0.05\)
(iv) \(\quad \ddot{a}_{30: 51}=4.5431\)

Calculate the annual gross premium using the equivalence principle.

Options:
A
150
B
160
C
170
D
180
E
190

Correct answer: C

Solution: \(A P V(\) expenses \()=0.35 G+8+0.15 G a_{30: \overline{4}}+4 a_{30: \overline{9}}\)
\(=0.20 G+4+0.15 G \ddot{a}_{30: 5 \mid}+4 \ddot{a}_{30: 10 \mid}\)
\(G \ddot{a}_{30: 51}=0.20 G+4+0.15 G \ddot{a}_{30: 51}+4 \ddot{a}_{30: \overline{10} 1}+200,000 A_{30: \overline{10} 1}^{1}\)
\(G=\frac{200,000 A_{30: \overline{10} \mid}^{1}+4+4 \ddot{a}_{30: \overline{10} \mid}}{0.85 \ddot{a}_{30: 5 \mid}-0.20}\)
\(200,000 A_{30: \overline{10} \mid}^{1}=200,000\left[A_{30: \overline{10}}-{ }_{10} E_{30}\right]\)
\(=200,000(0.61447-0.61152)=590\)
\(G=\frac{590+4+4(8.0961)}{0.85(4.5431)-0.20}=171.07\)
Question: . For a special fully discrete 10 -year deferred whole life insurance of 100 on (50), you are given:
(i) Premiums are payable annually, at the beginning of each year, only during the deferral period
(ii) For deaths during the deferral period, the benefit is equal to the return of all premiums paid, without interest
(iii) \(\quad i=0.05\)
(iv) \(\quad \ddot{a}_{50}=17.0\)
(v) \(\quad \ddot{a}_{60}=15.0\)
(vi) \(\quad{ }_{10} E_{50}=0.60\)
(vii) \(\quad(I A)_{50: 101}^{1}=0.15\)

Calculate the annual net premium for this insurance.

Options:
A
1.3
B
1.6
C
1.9
D
2.2
E
2.5

Correct answer: D

Solution: Let \(P\) be the premium per 1 of insurance.
\[
\begin{aligned}
& P \ddot{a}_{50: \overline{10}}=P(I A)_{50: \overline{10}}^{1}+{ }_{10} E_{50} A_{60} \\
& \ddot{a}_{50: \overline{10}}=\ddot{a}_{50}-{ }_{10} E_{50} \ddot{a}_{60}=17.0-0.60 \times 15.0=8 \\
& A_{60}=1-d \ddot{a}_{60}=1-\left(\frac{0.05}{1.05}\right) 15=0.285714 \\
& P\left(\ddot{a}_{50: \overline{10}}-(I A)_{50: \overline{10}}^{1}\right)={ }_{10} E_{50} A_{60} \\
& P=\frac{{ }_{10} E_{50} A_{60}}{\ddot{a}_{50: \overline{10}}-(I A)_{50: \overline{10}}^{1}}=\frac{0.6 \times 0.285714}{8-0.15}=0.021838
\end{aligned}
\]
\(100 P=2.18\)
Question: . For a fully continuous whole life insurance of 100,000 on (35), you are given:
(i) The annual rate of premium is 560
(ii) Mortality follows the Standard Ultimate Life Table
(iii) Deaths are uniformly distributed over each year of age
(iv) \(\quad i=0.05\)

Calculate the \(75{ }^{\text {th }}\) percentile of the loss at issue random variable for this policy.

Options:
A
610
B
630
C
650
D
670
E
690

Correct answer: E

Solution: \(L_{0}=100,000 v^{T}-560 \bar{a}_{\bar{T}}=\left(100,000+\frac{560}{\delta}\right) e^{-\delta T}-\frac{560}{\delta}\)
Since \(L_{0}\) is a decreasing function of \(T\), the \(75^{\text {th }}\) percentile of \(L_{0}\) is \(L_{0}(t)\) where \(t\) is such that \(\operatorname{Pr}\left[T_{35}>t\right]=0.75\).
\(\frac{l_{35+t}}{l_{35}}=0.75\)
\(l_{35+t}=0.75 l_{35}=0.75 \times 99,556.7=74,667.5\)
\(l_{81}<74,667.5<l_{80}\)
\(t=(80-35)+s\)
\(l_{80+s}=s l_{81}+(1-s) l_{80}\)
\(74,667.5=73,186.3 s+75,657.2(1-s)\)
\(s=0.40054\)
\(t=45.40054\)
\(L_{0}(45.40054)=\left(100,000+\frac{560}{\ln (1.05)}\right) e^{-45.40054 \ln (1.05)}-\frac{560}{\ln (1.05)}=689.25\)
Question: . For a 10 -year deferred whole life annuity-due with payments of 100,000 per year on ( 70 ), you are given:
(i) Annual gross premiums of \(G\) are payable for 10 years
(ii) First year expenses are \(75 \%\) of premium
(iii) Renewal expenses for years 2 and later are \(5 \%\) of premium during the premium paying period
(iv) Mortality follows the Standard Ultimate Life Table
(v) \(\quad i=0.05\)

Calculate \(G\) using the equivalence principle.

Options:
A
\(\quad 64,900\)
B
65,400
C
\(\quad 65,900\)
D
66,400
E
\(\quad 66,900\)

Correct answer: D

Solution: \(G \ddot{a}_{70: \overline{10}}=100,000_{10} E_{70} \ddot{a}_{80}+0.05 G \ddot{a}_{70: \overline{10}}+0.7 G\)
\(7.6491 G=(100,000)(0.50994)(8.5484)+0.05 G(7.6491)+0.7 G\)
\(\Rightarrow G=66,383.54\)
Question: . For a special fully discrete 5 -year deferred 3 -year term insurance of 100,000 on ( \(x\) ) you are given:
(i) There are two premium payments, each equal to \(P\). The first is paid at the beginning of the first year and the second is paid at the end of the 5 -year deferral period
(ii) The following probabilities:
(iii) \(\quad{ }_{5} p_{x}=0.95\)
(iv) \(\quad q_{x+5}=0.02, \quad q_{x+6}=0.03, \quad q_{x+7}=0.04\)
(v) \(\quad i=0.06\)

Calculate \(P\) using the equivalence principle.

Options:
A
3195
B
3345
C
3495
D
3645
E
3895

Correct answer: A

Solution: Actuarial present value of insured benefits:
\(100,000\left[\frac{0.95 \times 0.02}{1.06^{6}}+\frac{0.95 \times 0.98 \times 0.03}{1.06^{7}}+\frac{0.95 \times 0.98 \times 0.97 \times 0.04}{1.06^{8}}\right]=5,463.32\)
\(\Rightarrow P\left(1+\frac{0.95}{1.06^{5}}\right)=5,463.32 \Rightarrow P=3,195.12\)
Question: . For a special whole life insurance of 100,000 on (40), you are given:
(i) The death benefit is payable at the moment of death
(ii) Level gross premiums are payable monthly for a maximum of 20 years
(iii) Mortality follows the Standard Ultimate Life Table
(iv) \(\quad i=0.05\)
(v) Deaths are uniformly distributed over each year of age
(vi) Initial expenses are 200
(vii) Renewal expenses are \(4 \%\) of each premium including the first
(viii) Gross premiums are calculated using the equivalence principle Calculate the monthly gross premium.

Options:
A
66
B
76
C
86
D
96
E
106

Correct answer: C

Solution: \(G \ddot{a}_{40: 201}^{(12)}=100,000\left(\frac{i}{\delta}\right) A_{40}+200+0.04 G \ddot{a}_{40: 201}^{(12)}\)
\(\ddot{a}_{40: 20 \mid}^{(12)}=\alpha(12) \ddot{a}_{40: 20 \mid}-\left(1-{ }_{20} E_{40}\right) \beta(12)\)
\(=1.00020 \cdot 12.9935-(1-0.36663) \cdot 0.46651=12.700625\)
\(G=\frac{(100,000)(1.02480)(0.12106)+200}{0.96 \times 12.700625}=1033.92\)
\(\Rightarrow \mathrm{G} / 12=86.16\)
Question: . On July 15, 2017, XYZ Corp buys fully discrete whole life insurance policies of 1,000 on each of its 10,000 workers, all age 35 . It uses the death benefits to partially pay the premiums for the following year.

You are given:
(i) Mortality follows the Standard Ultimate Life Table
(ii) \(\quad i=0.05\)
(iii) The insurance is priced using the equivalence principle

Calculate XYZ Corp's expected net cash flow from these policies during July 2018.

Options:
A
\(\quad-47,000\)
B
\(-48,000\)
C
\(\quad-49,000\)
D
\(\quad-50,000\)
E
\(-51,000\)

Correct answer: A

Solution: \(1,000 P=1,000 \frac{A_{35}}{\ddot{a}_{35}}=\frac{96.53}{18.9728}=5.0878\)

Benefits paid during July 2018:
\(10,000 \times 1,000 \times q_{35}=10,000 \times 0.391=3910\)

Premiums payable during July 2018:
\(10,000 \times\left(1-q_{35}\right) \times 5.0878=9,996.09 \times 5.0878=50,858.10\)

Cash flow during July 2018:
\(3910-50,858=-46,948\)
Question: . For a special 10 -year deferred whole life annuity-due of 50,000 on (62), you are given:
(i) Level annual net premiums are payable for 10 years
(ii) A death benefit, payable at the end of the year of death, is provided only over the deferral period and is the sum of the net premiums paid without interest
(iii) \(\quad \ddot{a}_{62}=12.2758\)
(iv) \(\quad \ddot{a}_{62: 10}=7.4574\)
(v) \(\quad A_{62: 10 \mid}^{1}=0.0910\)
(vi) \(\quad \sum_{k=1}^{10} A_{62: k}^{1}=0.4891\)

Calculate the net premium for this special annuity.

Options:
A
\(\quad 34,400\)
B
\(\quad 34,500\)
C
\(\quad 34,600\)
D
\(\quad 34,700\)
E
\(\quad 34,800\)

Correct answer: D

Solution: Under the Equivalence Principle
\(P \ddot{a}_{62: \overline{10} \mid}=50,000\left(\ddot{a}_{62}-\ddot{a}_{62: 10 \mid}\right)+P\left((I A)_{62: 10 \mid}^{1}\right)\)
where \((I A)_{62: \overline{10} \mid}^{1}=11 A_{62: \overline{10} \mid}^{1}-\sum_{k=1}^{10} A_{62: k \mid}^{1}=11(0.091)-0.4891=0.5119\)
So \(\quad P=\frac{50,000\left(\ddot{a}_{62}-\ddot{a}_{62: 10}\right)}{\ddot{a}_{62: 101}-(I A)_{62: 101}^{1}}=\frac{50,000(12.2758-7.4574)}{7.4574-0.5119}=34,687\)
Question: . For a fully discrete 10 -payment whole life insurance of \(H\) on (45), you are given:
(i) Expenses payable at the beginning of each year are as follows:

\begin{tabular}{|l|c|c|c|}
\hline Expense Type & First Year & Years 2-10 & Years 11+ \\
\hline Per policy & 100 & 20 & 10 \\
\hline\(\%\) of Premium & \(105 \%\) & \(5 \%\) & \(0 \%\) \\
\hline
\end{tabular}
(ii) Mortality follows the Standard Ultimate Life Table
(iii) \(i=0.05\)
(iv) The gross annual premium, calculated using the equivalence principle, is of the form \(G=g H+f\), where \(g\) is the premium rate per 1 of insurance and \(f\) is the per policy fee Calculate \(f\).

Options:
A
42.00
B
44.20
C
46.40
D
48.60
E
50.80

Correct answer: E

Solution: \(G \ddot{a}_{45: \overline{10}}=H A_{45}+G+0.05 G \ddot{a}_{45: \overline{10}}+80+10 \ddot{a}_{45}+10 \ddot{a}_{45: \overline{10}}\)
\[
\begin{aligned}
& G=\frac{H A_{45}+80+10\left(\ddot{a}_{45}+\ddot{a}_{45: 10}\right)}{0.95 \ddot{a}_{45: 10}-1} \\
& G=\frac{H A_{45}+80+10(17.8162+8.0751)}{(0.95 \times 8.0751)-1} \\
& G=\frac{A_{45}}{(0.95 \times 8.0751)-1} H+\frac{80+10(17.8162+8.0751)}{(0.95 \times 8.0751)-1} \\
& g=\frac{A_{45}}{(0.95 \times 8.0751)-1} \\
& f=\frac{80+10(17.8162+8.0751)}{(0.95 \times 8.0751)-1}=50.80
\end{aligned}
\]
Question: . A warranty pays 2000 at the end of the year of the first failure if a washing machine fails within three years of purchase. The warranty is purchased with a single premium, \(G\), paid at the time of purchase of the washing machine.

You are given:
(i) \(10 \%\) of the washing machines that are working at the start of each year fail by the end of that year
(ii) \(\quad i=0.08\)
(iii) The sales commission is \(35 \%\) of \(G\)
(iv) \(G\) is calculated using the equivalence principle

Calculate \(G\).

Options:
A
630
B
660
C
690
D
720
E
750

Correct answer: D

Solution: \(G=0.35 G+2000\left(\frac{0.1}{1.08}+\frac{0.9 \times 0.1}{1.08^{2}}+\frac{0.9 \times 0.9 \times 0.1}{1.08^{3}}\right)\)
\(0.65 G=468.107\)
\(G=720.16\)
Question: . For a fully discrete whole life insurance of 200,000 on (45), you are given:
(i) Mortality follows the Standard Ultimate Life Table.
(ii) \(\quad i=0.05\)
(iii) The annual premium is determined using the equivalence principle.

Calculate the standard deviation of \(L_{0}\), the present value random variable for the loss at issue.

Options:
A
\(\quad 25,440\)
B
\(\quad 30,440\)
C
\(\quad 35,440\)
D
40,440
E
\(\quad 45,440\)

Correct answer: A

Solution: On a unit basis, \(\operatorname{Var}\left(L_{0}\right)=\left(1+\frac{P}{d}\right)^{2}\left[{ }^{2} A_{45}-\left(A_{45}\right)^{2}\right]=\left(1+\frac{A_{45}}{d \ddot{a}_{45}}\right)^{2}\left[{ }^{2} A_{45}-\left(A_{45}\right)^{2}\right]\)
\(=\left(\frac{d \ddot{a}_{45}+1-d \ddot{a}_{45}}{d \ddot{a}_{45}}\right)^{2}\left[{ }^{2} A_{45}-\left(A_{45}\right)^{2}\right]=\frac{{ }^{2} A_{45}-\left(A_{45}\right)^{2}}{(d \ddot{a})^{2}}\)
\(=\frac{0.03463-0.15161^{2}}{\left(\frac{0.05}{1.05} \times 17.8162\right)^{2}}=0.016178038\)

The standard deviation of \(L_{0}=0.127193\)
\((200,000)\left(\right.\) The standard deviation of \(\left.L_{0}\right)=25,439\)
Question: . An insurer issues a special 20 payment insurance policy on (45) with the following benefits:
- A death benefit of 1000 , payable at the end of year of death, provided death occurs before age 65; and
- An annuity benefit that pays 2500 at the start of each year, starting at age 65

You are given:
i) Level annual premiums of \(P\) are paid at the beginning of each year
ii) Premiums are calculated based on the equivalence principle
iii) Mortality follows the Standard Ultimate Life Table
iv) \(i=0.05\)

Calculate \(P\).

Options:
A
872
B
896
C
920
D
944
E
968

Correct answer: D

Solution: EPV \((\) benefits \()=1000 \times A_{45: 201}^{1}+2500 \times{ }_{201} \ddot{a}_{45}\)
\[
=1000(0.15161-0.35994 \times 0.35477)+2500 \times 0.35994 \times 13.5498=12,216.7
\]
\(\operatorname{EPV}(\) premiums \()=P \times(17.8161-0.35994 \times 13.5498)=P \times 12.9391\)
\(P=\frac{12216.7}{12.9391}=944.17\)
Question: . For a special fully discrete whole life insurance on (40), you are given:
(i) The death benefit is 50,000 in the first 20 years and 100,000 thereafter
(ii) Level net premiums of 875 are payable for 20 years
(iii) Mortality follows the Standard Ultimate Life Table
(iv) \(\quad i=0.05\)

Calculate \({ }_{10} V\), the net premium policy value at the end of year 10 for this insurance.

Options:
A
11,090
B
11,120
C
11,150
D
11,180
E
11,210

Correct answer: C

Solution: \[
\begin{aligned}
{ }_{10} V & =50,000\left(A_{50}+{ }_{10} E_{50} A_{60}\right)-(875)\left[\ddot{a}_{50: \overline{10}}\right] \\
& =50,000[0.18931+(0.60182)(0.29028)]-875[8.0550] \\
& =11,152
\end{aligned}
\]
Question: . A special fully discrete 2 -year endowment insurance with a maturity value of 2000 is issued to \((x)\). The death benefit is 2000 plus the net premium policy value at the end of the year of death. For year 2, the net premium policy value is the net premium policy value just before the maturity benefit is paid.

You are given:
(i) \(\quad i=0.10\)
(ii) \(q_{x}=0.150\) and \(q_{x+1}=0.165\)

Calculate the level annual net premium.

Options:
A
1070
B
1110
C
1150
D
1190
E
1230

Correct answer: C

Solution: \({ }_{0} V=0\)
\({ }_{2} V=2000\)
Year 1: \(\quad\left({ }_{0} V+P\right)(1+i)=q_{x}\left(2000+{ }_{1} V\right)+p_{x}{ }_{1} V\)
\[
\begin{aligned}
& P(1.1)=0.15\left(2000+{ }_{1} V\right)+0.85\left({ }_{1} V\right) \\
& 1.1 P-300={ }_{1} V
\end{aligned}
\]

Year 2: \(\quad\left({ }_{1} V+P\right)(1+i)=q_{x+1}\left(2000+{ }_{2} V\right)+p_{x+1}(2000)\)
\[
\begin{aligned}
& (1.1 P-300+P)(1.1)=0.165(2000+2000)+0.835(2000) \\
& 2.31 P-330=2330
\end{aligned}
\]
\[
P=\frac{2330+330}{2.31}=1152
\]
Question: . For a whole life insurance of 1000 with semi-annual premiums on (80), you are given:
(i) A gross premium of 60 is payable every 6 months starting at age 80
(ii) Commissions of \(10 \%\) are paid each time a premium is paid
(iii) Death benefits are paid at the end of the quarter of death
(iv) \({ }_{t} V\) denotes the gross premium policy value at time \(t, t \geq 0\)
(v) \(\quad{ }_{10.75} V=753.72\)
(vi)

\begin{tabular}{|c|c|}
\hline\(t\) & \(l_{90+t}\) \\
\hline 0 & 1000 \\
\hline 0.25 & 898 \\
\hline 0.50 & 800 \\
\hline 0.75 & 706 \\
\hline
\end{tabular}
(vii) \(\quad i^{(4)}=0.08\)

Calculate \({ }_{10.25} V\).

Options:
A
680
B
690
C
700
D
710
E
730

Correct answer: E

Solution: \(i^{(4)}=0.08\) means an interest rate of \(j=0.02\) per quarter. This problem can be done with two quarterly recursions or a single calculation.

Using two recursions:
\({ }_{10.75} V=\frac{\left[{ }_{10.5} V+60(1-0.1)\right](1.02)-\frac{800-706}{800}(1000)}{\frac{706}{800}}\)
\(753.72=\frac{\left[{ }_{10.5} V+54\right](1.02)-117.50}{0.8825} \Rightarrow_{10.5} V=713.31\)
\({ }_{10.5} V=\frac{[10.25 V](1.02)-\frac{898-800}{898}(1000)}{\frac{800}{898}} \Rightarrow 713.31=\frac{\left[{ }_{10.25} V\right](1.02)-109.13}{0.8909}\)
\({ }_{10.25} V=730.02\)

Using a single step, \({ }_{10.25} V\) is the \(E P V\) of cash flows through time 10.75 plus \({ }_{0.5} E_{80.25}\) times the \(E P V\) of cash flows thereafter (that is, \({ }_{10.75} V\) ).
\({ }_{10.25} V=(1000)\left[\frac{898-800}{898(1.02)}+\frac{800-706}{898(1.02)^{2}}\right]-(60)(1-0.1)\left[\frac{800}{898(1.02)}\right]+\left[\frac{706}{898(1.02)^{2}}\right](753.72)=730\)
Question: . For a special fully discrete whole life insurance on (40), you are given:
(i) The death benefit is 1000 during the first 11 years and 5000 thereafter
(ii) Expenses, payable at the beginning of the year, are 100 in year 1 and 10 in years 2 and later
(iii) \(\quad \pi\) is the level annual premium, determined using the equivalence principle
(iv) \(G=1.02 \times \pi\) is the level annual gross premium
(v) Mortality follows the Standard Ultimate Life Table
(vi) \(\quad i=0.05\)
(vii) \(\quad{ }_{11} E_{40}=0.57949\)

Calculate the gross premium policy value at the end of year 1 for this insurance.

Options:
A
-82
B
-74
C
\(\quad-66\)
D
-58
E
\(\quad-50\)

Correct answer: B

Solution: EPV of benefits at issue \(=1000 A_{40}+4_{11} E_{40}\left(1000 A_{51}\right)\)
\[
=121.06+(4)(0.57949)(197.80)=579.55
\]

EPV of expenses at issue \(=100+10\left(\ddot{a}_{40}-1\right)=100+10(17.4578)=274.58\)
\(\pi=(579.55+274.58) / \ddot{a}_{40}=854.13 / 18.4578=46.27\)
\(G=1.02 \pi=47.20\)
EPV of benefits at time \(1=1000 A_{41}+4_{10} E_{41} \times 1000 A_{51}\)
\[
=126.65+(4)(0.60879)(197.80)=608.32
\]

EPV of expenses at time \(1=10\left(\ddot{a}_{41}\right)=10(18.3403)=183.40\)
Gross Prem Policy Value \(=608.32+183.40-G \ddot{a}_{41}=791.72-47.20(18.3403)=-73.94\)