BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: . For a fully discrete whole life insurance of 1000 on ( \(x\) ), you are given:
(i) The following expenses are incurred at the beginning of each year:

\begin{tabular}{|l|c|c|}
\hline & Year 1 & Years 2+ \\
\hline Percent of premium & \(75 \%\) & \(10 \%\) \\
\hline Maintenance expenses & 10 & 2 \\
\hline
\end{tabular}
(ii) An additional expense of 20 is paid when the death benefit is paid
(iii) The gross premium is determined using the equivalence principle
(iv) \(\quad i=0.06\)
(v) \(\quad \ddot{a}_{x}=12.0\)
(vi) \({ }^{2} A_{x}=0.14\)

Calculate the variance of the loss at issue random variable.

Options:
A
14,600
B
33,100
C
51,700
D
70,300
E
\(\quad 88,900\)

Correct answer: E

Solution: 1,020 in the solution is the 1,000 death benefit plus the 20 death benefit claim expense.
\(A_{x}=1-d \ddot{a}_{x}=1-d(12.0)=0.320755\)
\(G \ddot{a}_{x}=1,020 A_{x}+0.65 G+0.10 G \ddot{a}_{x}+8+2 \ddot{a}_{x}\)
\(G=\frac{1,020 A_{x}+8+2 \ddot{a}_{x}}{\ddot{a}_{x}-0.65-0.10 \ddot{a}_{x}}=\frac{1,020(0.320755)+8+2(12.0)}{12.0-0.65-0.10(12.0)}=35.38622\)
Let \(Z=v^{K_{x}+1}\) denote the present value random variable for a whole life insurance of 1 on ( \(x\) ).
Let \(Y=\ddot{a}_{\overline{K_{x}+1}}\) denote the present value random variable for a life annuity-due of 1 on ( \(x\) ).
\[
\begin{aligned}
& L=1,020 Z+0.65 G+0.10 G Y+8+2 Y-G Y \\
& =1,020 Z+(2-0.9 G) Y+0.65 G+8 \\
& =1,020 v^{K_{x}+1}+(2-0.9 G) \frac{1-v^{K_{x}+1}}{d}+0.65 G+8 \\
& =\left(1,020+\frac{0.9 G-2}{d}\right) v^{K_{x}+1}+\frac{2-0.9 G}{d}+0.65 G+8 \\
& \begin{aligned}
\operatorname{Var}(L) & =\left[{ }^{2} A_{x}-\left(A_{x}\right)^{2}\right]\left(1,020+\frac{0.9 G-2}{d}\right)^{2} \\
& =\left(0.14-0.320755^{2}\right)\left(1,020+\frac{0.9(35.38622)-2}{d}\right)^{2} \\
& =0.037116(2,394,161) \\
& =88,861
\end{aligned}
\end{aligned}
\]
Question: . For a fully discrete whole life insurance of 10,000 on (45), you are given:
(i) Commissions are \(80 \%\) of the first year premium and \(10 \%\) of subsequent premiums. There are no other expenses
(ii) Mortality follows the Standard Ultimate Life Table
(iii) \(\quad i=0.05\)
(iv) \(\quad{ }_{0} L\) denotes the loss at issue random variable
(v) If \(T_{45}=10.5\), then \({ }_{0} L=4953\)

Calculate \(\mathrm{E}\left[{ }_{0} L\right]\).

Options:
A
-580
B
-520
C
-460
D
\(\quad-400\)
E
\(\quad-340\)

Correct answer: D

Solution: If \(T_{45}=10.5\), then \(K_{45}=10\) and \(K_{45}+1=11\).
\[
\begin{aligned}
& { }_{0} L=10,000 v^{K_{45}+1}-G(1-0.10) \ddot{a}_{\overline{K_{45}+1}}+G(0.80-0.10)=10,000 v^{11}-0.9 G \ddot{a}_{\overline{11}}+0.7 G \\
& 4953=10,000(0.58468)-0.9 G(8.72173)+0.7 G \\
& G=(5846.8-4953) /(7.14956)=125.01 \\
& E\left({ }_{0} L\right)=10,000 A_{45}-(1-0.1) G \ddot{a}_{45}+(0.8-0.1) G \\
& \quad=(10,000)(0.15161)-(0.9)(125.01)(17.8162)+(0.7)(125.01) \\
& E\left({ }_{0} L\right)=-400.87
\end{aligned}
\]
Question: . For a fully discrete whole life insurance of 1000 on \((x)\) with net premiums payable quarterly, you are given:
(i) \(\quad i=0.05\)
(ii) \(\quad \ddot{a}_{x}=3.4611\)
(iii) \(\quad P^{(W)}\) and \(P^{(U D D)}\) are the annualized net premiums calculated using the 2-term Woolhouse ( \(W\) ) and the uniform distribution of deaths (UDD) assumptions, respectively
Calculate \(\frac{P^{(U D D)}}{P^{(W)}}\).

Options:
A
1.000
B
1.002
C
1.004
D
1.006
E
1.008

Correct answer: B

Solution: Woolhouse: \(\quad{ }^{W} \ddot{a}_{x}^{(4)}=3.4611-\frac{3}{8}=3.0861\)
\[
{ }^{U D D} \ddot{a}_{x}^{(4)}=\alpha(4) \ddot{a}_{x}-\beta(4)
\]

UDD:
\[
\begin{aligned}
& =1.00019(3.4611)-0.38272 \\
& =3.0790
\end{aligned}
\]
and
\[
A_{x}=1-d \ddot{a}_{x}=1-(0.04762)(3.4611)=0.83518
\]
\(P^{(W)}=\frac{1000(0.83518)}{3.0861}=270.63\)
\(P^{(\text {UDD })}=\frac{1000(0.83518)}{3.0790}=271.25\)
\(\frac{P^{(U D D)}}{P^{(W)}}=\frac{271.25}{270.63}=1.0023\)
Question: . For a fully discrete 20 -year endowment insurance of 100,000 on (30), you are given:
(i) \(\quad d=0.05\)
(ii) Expenses, payable at the beginning of each year, are:

\begin{tabular}{|l|c|c|c|c|}
\hline & \multicolumn{2}{|c|}{ First Year } & \multicolumn{2}{c|}{ Renewal Years } \\
\hline & \begin{tabular}{c} 
Percent of \\
Premium
\end{tabular} & \begin{tabular}{c} 
Per \\
Policy
\end{tabular} & \begin{tabular}{c} 
Percent of \\
Premium
\end{tabular} & \begin{tabular}{c} 
Per \\
Policy
\end{tabular} \\
\hline Taxes & \(4 \%\) & 0 & \(4 \%\) & 0 \\
\hline Sales Commission & \(35 \%\) & 0 & \(2 \%\) & 0 \\
\hline Policy Maintenance & \(0 \%\) & 250 & \(0 \%\) & 50 \\
\hline
\end{tabular}
(iii) The net premium is 2143

Calculate the gross premium using the equivalence principle.

Options:
A
2410
B
2530
C
2800
D
3130
E
3280

Correct answer: A

Solution: \(P_{30: 201}=\frac{1}{\ddot{a}_{30: 201}}-d \Rightarrow \frac{2,143}{100,000}+0.05=\frac{1}{\ddot{a}_{30: 201}} \Rightarrow \ddot{a}_{30: 201}=14\)
\(A_{30: \overline{20}}=1-d \ddot{a}_{30: \overline{20}}=1-0.05(14)=0.3\)
\(G \ddot{a}_{30: \overline{20}}=100,000 A_{30: \overline{20}}+\left(200+50 \ddot{a}_{30: \overline{20}}\right)+\left(0.33 G+0.06 G \ddot{a}_{30: \overline{20}}\right)\)
\(14 G=100,000(0.3)+[200+50(14)]+(0.33 G+0.84 G)\)
\(12.83 G=30,900\)
\(G=2408\)
Question: . An insurance company sells special fully discrete two-year endowment insurance policies to smokers (S) and non-smokers (NS) age \(x\). You are given:
(i) The death benefit is 100,000 ; the maturity benefit is 30,000
(ii) The level annual premium for non-smoker policies is determined by the equivalence principle
(iii) The annual premium for smoker policies is twice the non-smoker annual premium
(iv) \(\quad \mu_{x+t}^{\mathrm{NS}}=0.1, t>0\)
(v) \(\quad q_{x+k}^{\mathrm{S}}=1.5 q_{x+k}^{\mathrm{NS}}\) for \(k=0,1\)
(vi) \(\quad i=0.08\)

Calculate the expected present value of the loss at issue random variable on a smoker policy.

Options:
A
\(-30,000\)
B
\(-29,000\)
C
\(\quad-28.000\)
D
\(\quad-27.000\)
E
-26.000

Correct answer: A

Solution: \[
q_{x}^{\mathrm{NS}}=q_{x+1}^{\mathrm{NS}}=1-e^{-0.1}=0.095
\]

Then the annual premium for the non-smoker policies is \(P^{\mathrm{NS}}\), where
\[
\begin{aligned}
P^{\mathrm{NS}}\left(1+v p_{x}^{\mathrm{NS}}\right) & =100,000 v q_{x}^{\mathrm{NS}}+100,000 v^{2} p_{x}^{\mathrm{NS}} q_{x+1}^{\mathrm{NS}}+30,000 v^{2} p_{x}^{\mathrm{NS}} p_{x+1}^{\mathrm{NS}} \\
P^{\mathrm{NS}} & =\frac{100,000(0.926)(0.095)+100,000(0.857)(0.905)(0.095)+30,000(0.857)(0.905)^{2}}{1+(0.926)(0.905)} \\
P^{\mathrm{NS}} & =20,251
\end{aligned}
\]

And then \(P^{\mathrm{S}}=40,502\).
\[
\begin{aligned}
q_{x}^{\mathrm{S}}=q_{x+1}^{\mathrm{S}}= & 1.5\left(1-e^{-0.1}\right)=0.143 \\
\operatorname{EPV}\left(L^{\mathrm{S}}\right)= & 100,000 v q_{x}^{\mathrm{S}}+100,000 v^{2} p_{x}^{\mathrm{S}} q_{x+1}^{\mathrm{S}}+30,000 v^{2} p_{x}^{\mathrm{S}} p_{x+1}^{\mathrm{S}}-P^{\mathrm{S}}-P^{\mathrm{S}} v p_{x}^{\mathrm{S}} \\
= & 100,000(0.926)(0.143)+100,000(0.857)(0.857)(0.143) \\
& \quad+30,000(0.857)(0.857)^{2}-40,502-40,502(0.926)(0.857) \\
= & -30,017
\end{aligned}
\]
Question: . For a 20 -year deferred whole life annuity-due with annual payments of 30,000 on (40), you are given:
(i) The single net premium is refunded without interest at the end of the year of death if death occurs during the deferral period
(ii) Mortality follows the Standard Ultimate Life Table
(iii) \(\quad i=0.05\)

Calculate the single net premium for this annuity.

Options:
A
162,000
B
164,000
C
165,200
D
166,400
E
168,800

Correct answer: D

Solution: \[
\begin{aligned}
P & =30,000_{20 \mid} \ddot{a}_{40}+P A_{40: 20 \mid}^{1} \\
\Rightarrow \quad P & =30,000_{20 \mid} \ddot{a}_{40} /\left(1-A_{40: 20 \mid}^{1}\right) \\
& =30,000(5.46429) /(1-0.0146346)=166,363
\end{aligned}
\]
Question: . For a fully discrete whole life insurance of 1 on (50), you are given:
(i) Expenses of 0.20 at the start of the first year and 0.01 at the start of each renewal year are incurred
(ii) Mortality follows the Standard Ultimate Life Table
(iii) \(\quad i=0.05\)
(iv) Gross premiums are determined using the equivalence principle.

Calculate the variance of \(L_{0}\), the gross loss-at-issue random variable.

Options:
A
0.023
B
0.028
C
0.033
D
0.038
E
0.043

Correct answer: C

Solution: Let \(\pi\) be the annual premium, so that \(\pi \ddot{a}_{50}=A_{50}+0.01 \ddot{a}_{50}+0.19\)
\(\Rightarrow \pi=\frac{A_{50}+0.19}{\ddot{a}_{50}}+0.01=\frac{0.18931+0.19}{17.0245}+0.01=0.03228\)
Loss at issue: \(L_{0}=v^{k+1}-(\pi-0.01) \ddot{a}_{\overline{k+1}}\left(1-v^{k+1}\right) / d+0.19\)
\[
\begin{aligned}
\Rightarrow \operatorname{Var}\left[L_{0}\right] & =\left(1+\frac{(\pi-0.01)}{d}\right)^{2}\left({ }^{2} A_{50}-A_{50}^{2}\right) \\
& =(2.15467)\left(0.05108-0.18931^{2}\right) \\
& =(2.15467)(0.015242) \\
& =0.033
\end{aligned}
\]
Question: . For a special fully discrete 3 -year term insurance on (75), you are given:
(i) The death benefit during the first two years is the sum of the net premiums paid without interest
(ii) The death benefit in the third year is 10,000
(iii)

\begin{tabular}{|c|c|}
\hline\(x\) & \(p_{x}\) \\
\hline 75 & 0.90 \\
76 & 0.88 \\
77 & 0.85 \\
\hline
\end{tabular}
(iv) \(\quad i=0.04\)

Calculate the annual net premium.

Options:
A
449
B
459
C
469
D
479
E
489

Correct answer: B

Solution: EPV \((\) premiums \()=\) EPV \((\) benefits \()\)
\(P\left(1+v p_{x}+v^{2}{ }_{2} p_{x}\right)=P\left(v q_{x}+2 v^{2} p_{x} q_{x+1}\right)+10000\left(v^{3}{ }_{2} p_{x} q_{x+2}\right)\)
\(P\left(1+\frac{0.9}{1.04}+\frac{0.9 \times 0.88}{1.04^{2}}\right)=P\left(\frac{0.1}{1.04}+\frac{2 \times 0.9 \times 0.12}{1.04^{2}}\right)+10000\left(\frac{0.9 \times 0.88 \times 0.15}{1.04^{3}}\right)\)
\(2.5976 P=0.29588 P+1056.13\)
\(P=459\)
Question: . For a special fully discrete 15 -year endowment insurance on (75), you are given:
(i) The death benefit is 1000
(ii) The endowment benefit is the sum of the net premiums paid without interest
(iii) \(\quad d=0.04\)
(iv) \(\quad A_{75: \overline{15} \mid}=0.70\)
(v) \(\quad A_{75: 15} \frac{1}{15}=0.11\)

Calculate the annual net premium.

Options:
A
80
B
90
C
100
D
110
E
120

Correct answer: C

Solution: \(P \times \ddot{a}_{75: 15 \mid}=1000\left(A_{75: 15 \mid}^{1}+15 \times P \times A_{75: 15 \mid}\right) \rightarrow P=\frac{1000 A_{75: 15 \mid}^{1}}{\ddot{a}_{75: 15 \mid}-15 \times A_{75: 15 \mid} \frac{1}{15 \mid}}\)
\(A_{75: \overline{15} \mid}^{1}=A_{75: \overline{15} \mid}-A_{75: \overline{15} \mid}=0.7-0.11=0.59\)
\(\ddot{a}_{75: \overline{15} \mid}=\frac{1-A_{75: 15 \mid}}{d}=(1-0.7) / 0.04=7.5\)
So \(P=\frac{590}{7.5-15(0.11)}=100.85\)
Question: . For a whole life insurance of 100,000 on (45) with premiums payable monthly for a period of 20 years, you are given:
(i) The death benefit is paid immediately upon death
(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 monthly net premium.

Options:
A
98
B
100
C
102
D
104
E
106

Correct answer: C

Solution: Let the monthly net premium \(=\pi\)
\[
12 \pi=\frac{100,000 \bar{A}_{45}}{\ddot{a}_{45: 20}^{(12)}} \quad \begin{aligned}
& \alpha(12)=1.00020 \\
& \beta(12)=0.46651 \\
& \frac{i}{\delta}=1.02480
\end{aligned}
\]
\[
\begin{aligned}
& 100,000 \bar{A}_{45}=100,000 \frac{i}{\delta} A_{45}=(1.02480)(15,161)=15,536.99 \\
& \begin{aligned}
\ddot{a}_{45: 20}^{(12)} & =\alpha(12) \ddot{a}_{45: 20}-\beta(12)\left(1-{ }_{20} E_{45}\right) \\
& =1.00020[12.9391]-0.46651(1-0.35994) \\
& =12.6431
\end{aligned}
\end{aligned}
\]
\[
\begin{aligned}
12 \pi & =\frac{15,536.99}{12.6431} \\
12 \pi & =1228.891 \\
\pi & =102.41
\end{aligned}
\]
Question: . For fully discrete 30 -payment whole life insurance policies on ( \(x\) ), you are given:
(i) The following expenses payable at the beginning of the year:

\begin{tabular}{|l|c|c|c|c|}
\hline & \(1^{\text {st }}\) Year & \begin{tabular}{c} 
Years \\
\(2-15\)
\end{tabular} & \begin{tabular}{c} 
Years \\
\(16-30\)
\end{tabular} & \begin{tabular}{c} 
Years 31 \\
and after
\end{tabular} \\
\hline Per policy & 60 & 30 & 30 & 30 \\
\hline Percent of premium & \(80 \%\) & \(20 \%\) & \(10 \%\) & \(0 \%\) \\
\hline
\end{tabular}
(ii) \(\quad \ddot{a}_{x}=15.3926\)
(iii) \(\quad \ddot{a}_{x: \overline{15} \mid}=10.1329\)
(iv) \(\quad \ddot{a}_{x: \overline{30}}=14.0145\)
(v) Annual gross premiums are calculated using the equivalence principle
(vi) The annual gross premium is expressed as \(k F+h\), where \(F\) is the death benefit and \(k\) and \(h\) are constants for all \(F\)

Calculate \(h\).

Options:
A
30.3
B
35.1
C
39.9
D
44.7
E
49.5

Correct answer: D

Solution: \[
\begin{aligned}
& G \ddot{a}_{x: 301}=\operatorname{APV}[\text { gross premium }]=\operatorname{APV}[\text { Benefits }+ \text { expenses }] \\
& \quad=F A_{x}+\left(30+30 \ddot{a}_{x}\right)+G\left(0.6+0.10 \ddot{a}_{x: 301}+0.10 \ddot{a}_{x: 151}\right) \\
& \begin{aligned}
G & =\frac{F A_{x}+30+30 \ddot{a}_{x}}{\ddot{a}_{x: 301}-0.6-0.1 \ddot{a}_{x: 301}-0.1 \ddot{a}_{x: 15}} \\
& =\frac{F A_{x}+30+30(15.3926)}{14.0145-0.6-0.1(14.0145)-0.1(10.1329)} \\
& =\frac{F A_{x}+491.78}{10.9998} \\
& =\frac{F A_{x}}{10.9998}+\frac{491.78}{10.9998}=\frac{F A_{x}}{10.9998}+44.71
\end{aligned} \\
& \Rightarrow h=44.71
\end{aligned}
\]
Question: . For a fully continuous whole life insurance of 1 on ( \(x\) ), you are given:
(i) \(\quad L\) is the present value of the loss at issue random variable if the premium rate is determined by the equivalence principle
(ii) \(\quad L^{*}\) is the present value of the loss at issue random variable if the premium rate is 0.06
(iii) \(\delta=0.07\)
(iv) \(\bar{A}_{x}=0.30\)
(v) \(\quad \operatorname{Var}(L)=0.18\)

Calculate \(\operatorname{Var}\left(L^{*}\right)\).

Options:
A
0.18
B
0.21
C
0.24
D
0.27
E
0.30

Correct answer: E

Solution: In general, the loss at issue random variable can be expressed as:
\(L=\bar{Z}_{x}-P \times \bar{Y}_{x}=\bar{Z}_{x}-P \times\left(\frac{1-\bar{Z}_{x}}{\delta}\right)=\bar{Z}_{x} \times\left(1+\frac{P}{\delta}\right)-\frac{P}{\delta}\)
Using actuarial equivalence to determine the premium rate:
\[
\begin{aligned}
P & =\frac{\bar{A}_{x}}{\bar{a}_{x}}=\frac{0.3}{(1-0.3) / 0.07}=0.03 \\
\operatorname{Var}(L) & =\left(1+\frac{P}{\delta}\right)^{2} \times \operatorname{Var}\left(\bar{Z}_{x}\right)=\left(1+\frac{0.03}{0.07}\right)^{2} \times \operatorname{Var}\left(\bar{Z}_{x}\right)=0.18 \\
\operatorname{Var}\left(\bar{Z}_{x}\right) & =\frac{0.18}{\left(1+\frac{0.03}{0.07}\right)^{2}}=0.088 \\
\operatorname{Var}\left(L^{*}\right) & =\left(1+\frac{P^{*}}{\delta}\right)^{2} \times \operatorname{Var}\left(\bar{Z}_{x}\right)=\left(1+\frac{0.06}{0.07}\right)^{2}(0.088)=0.304
\end{aligned}
\]
Question: . For a fully discrete 10 -year deferred whole life annuity-due of 1000 per month on (55), you are given:
(i) The premium, \(G\), will be paid annually at the beginning of each year during the deferral period
(ii) Expenses are expected to be 300 per year for all years, payable at the beginning of the year
(iii) Mortality follows the Standard Ultimate Life Table
(iv) \(\quad i=0.05\)
(v) Using the two-term Woolhouse approximation, the expected loss at issue is -800 Calculate \(G\).

Options:
A
\(\quad 12,110\)
B
12,220
C
12,330
D
12,440
E
12,550

Correct answer: C

Solution: Need EPV(Ben + Exp) - EPV(Prem) \(=-800\)
\(\operatorname{EPV}(\) Prem \()=G \ddot{a}_{55: \overline{10}}=8.0192 G\)
\(\operatorname{EPV}(B e n+\operatorname{Exp})=12,000_{10} \ddot{a}_{55}^{(12)}+300 \ddot{a}_{55}\)
\[
\begin{aligned}
& =12,000_{10} E_{55} \ddot{a}_{65}^{(12)}+300 \ddot{a}_{55} \\
& =12,000_{10} E_{55}\left(\ddot{a}_{65}-\frac{m-1}{2 m}\right)+300 \ddot{a}_{55} \\
& =12,000(0.59342)(13.5498-11 / 24)+300(16.0599) \\
& =98,042.83
\end{aligned}
\]

Therefore, \(\quad 98,042.83-8.0192 G=-800\)
\[
G=12,326
\]
Question: . For a special fully discrete whole life insurance policy of 1000 on (90), you are given:
(i) The first year premium is 0
(ii) \(P\) is the renewal premium
(iii) Mortality follows the Standard Ultimate Life Table
(iv) \(\quad i=0.05\)
(v) Premiums are calculated using the equivalence principle Calculate \(P\).

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

Correct answer: D

Solution: EPV \((\) Premiums \()=P a_{90}=P\left(\ddot{a}_{90}-1\right)=(4.1835) P\)
\(\operatorname{EPV}(\) Benefits \()=1000 A_{90}=1000(0.75317)=753.17\)
Therefore,
\(P=\frac{753.17}{4.1835}=180.03\)
Question: . For a fully discrete 5 -payment whole life insurance of 1000 on (40), you are given:
(i) Expenses incurred at the beginning of the first five policy years are as follows:

\begin{tabular}{|l|c|c|c|c|}
\hline & \multicolumn{2}{|c|}{ Year 1 } & \multicolumn{2}{c|}{ Years 2-5 } \\
\hline & \begin{tabular}{c} 
Percent \\
of premium
\end{tabular} & \begin{tabular}{c} 
Per \\
policy
\end{tabular} & \begin{tabular}{c} 
Percent \\
of premium
\end{tabular} & \begin{tabular}{c} 
Per \\
policy
\end{tabular} \\
\hline Sales Commissions & \(20 \%\) & 0 & \(5 \%\) & 0 \\
\hline Policy Maintenance & \(0 \%\) & 10 & \(0 \%\) & 5 \\
\hline
\end{tabular}
(ii) No expenses are incurred after Year 5
(iii) Mortality follows the Standard Ultimate Life Table
(iv) \(\quad i=0.05\)

Calculate the gross premium using the equivalence principle.

Options:
A
31
B
36
C
41
D
46
E
51

Correct answer: B

Solution: \(G \ddot{a}_{40: 51}=1000 A_{40}+0.15 G+0.05 G \ddot{a}_{40: 51}+5+5 \ddot{a}_{40: 51}\)
\(\ddot{a}_{40: 5 \mid}=\ddot{a}_{40}-{ }_{5} E_{40} \bullet \ddot{a}_{45}=18.4578-(0.78113)(17.8162)=4.5410\)
\(G=\frac{121.06+5+5(4.5410)}{-0.15+0.95(4.5410)}=35.73\)
Question: . (35) purchases a fully discrete whole life insurance policy of 100,000 .

You are given:
(i) The annual gross premium, calculated using the equivalence principle, is 1770
(ii) The expenses in policy year 1 are \(50 \%\) of premium and 200 per policy
(iii) The expenses in policy years 2 and later are \(10 \%\) of premium and 50 per policy
(iv) All expenses are incurred at the beginning of the policy year
(v) \(\quad i=0.035\)

Calculate \(\ddot{a}_{35}\).

Options:
A
20.0
B
20.5
C
21.0
D
21.5
E
22.0

Correct answer: B

Solution: Per equivalence Principle:
\[
\begin{aligned}
G \ddot{a}_{35} & =100,000 A_{35}+0.4 G+150+0.1 G \ddot{a}_{35}+50 \ddot{a}_{35} \\
1770 \ddot{a}_{35} & =100,000\left(1-d \ddot{a}_{35}\right)+0.4(1770)+150+0.1(1770) \ddot{a}_{35}+50 \ddot{a}_{35} \\
1770 \ddot{a}_{35} & =100,000+708+150+\ddot{a}_{35}\left(177+50-100,000\left(\frac{0.035}{1.035}\right)\right)
\end{aligned}
\]

Solving for \(\ddot{a}_{35}\), we have
\(\ddot{a}=\frac{100,858}{1770+3154.64}=\frac{100,858}{4924.64}=20.48\)
Question: . For a fully discrete whole life insurance of 100 on \((x)\), you are given:
(i) The first year expense is \(10 \%\) of the gross annual premium
(ii) Expenses in subsequent years are \(5 \%\) of the gross annual premium
(iii) The gross premium calculated using the equivalence principle is 2.338
(iv) \(\quad i=0.04\)
(v) \(\quad \ddot{a}_{x}=16.50\)
(vi) \({ }^{2} A_{x}=0.17\)

Calculate the variance of the loss at issue random variable.

Options:
A
900
B
1200
C
1500
D
1800
E
2100

Correct answer: A

Solution: The loss at issue is given by:
\[
\begin{aligned}
& L_{0}=100 v^{K+1}+0.05 G+0.05 G \ddot{a}_{\overline{K+1}}-G \ddot{a}_{\overline{K+1}} \\
& =100 v^{K+1}+0.05 G-0.95 G\left(\frac{1-v^{K+1}}{d}\right) \\
& =\left(100+\frac{0.95 G}{d}\right) v^{K+1}+0.05 G-0.95 \frac{G}{d}
\end{aligned}
\]

Thus, the variance is
\[
\begin{aligned}
& \operatorname{Var}\left(L_{0}\right)=\left[100+\frac{0.95(2.338)}{0.04 / 1.04}\right]^{2}\left({ }^{2} A_{x}-\left(A_{x}\right)^{2}\right) \\
& \quad=\left[100+\frac{0.95(2.338)}{0.04 / 1.04}\right]^{2}\left(0.17-\left(1-\frac{0.04}{1.04}(16.50)\right)^{2}\right) \\
& \quad=908.1414
\end{aligned}
\]
Question: . For a fully continuous whole life insurance policy of 100,000 on (35), you are given:
(i) The density function of the future lifetime of a newborn:
\[
f(t)= \begin{cases}0.01 e^{-0.01 t}, & 0 \leq t<70 \\ g(t), & t \geq 70\end{cases}
\]
(ii) \(\delta=0.05\)
(iii) \(\quad \bar{A}_{70}=0.51791\)

Calculate the annual net premium rate for this policy.

Options:
A
1000
B
1110
C
1220
D
1330
E
1440

Correct answer: D

Solution: \[
\begin{aligned}
& \bar{A}_{35}=\left(1-e^{-35(\mu+\delta)}\right) \times\left(\frac{\mu}{\mu+\delta}\right)+e^{-35(\mu+\delta)} \bar{A}_{70}=0.063421+0.146257=0.209679 \\
& \bar{a}_{35}=\frac{1-\bar{A}_{35}}{\delta}=\frac{1-0.209679}{0.05}=15.80642 \\
& \bar{P}_{35}=\frac{\bar{A}_{35}}{\bar{a}_{35}}=\frac{0.209679}{15.80642}=0.0132654
\end{aligned}
\]

The annual net premium for this policy is therefore \(100,000 \times 0.0132654=1,326.54\)
Question: . An insurance company sells 15 -year pure endowments of 10,000 to 500 lives, each age \(x\), with independent future lifetimes. The single premium for each pure endowment is determined by the equivalence principle.
(i) You are given:
(ii) \(\quad i=0.03\)
(iii) \(\quad \mu_{x}(t)=0.02 t, \quad t \geq 0\)
(iv) \({ }_{0} L\) is the aggregate loss at issue random variable for these pure endowments.

Using the normal approximation without continuity correction, calculate \(\operatorname{Pr}\left({ }_{0} L>50,000\right)\).

Options:
A
0.08
B
0.13
C
0.18
D
0.23
E
0.28

Correct answer: B

Solution: The probability that the endowment payment will be made for a given contract is:
\[
\begin{aligned}
15 p_{x} & =\exp \left(-\int_{0}^{15} 0.02 t d t\right) \\
& =\exp \left(-\left.0.01 t^{2}\right|_{0} ^{15}\right) \\
& =\exp \left(-0.01(15)^{2}\right) \\
& =0.1054
\end{aligned}
\]

Because the premium is set by the equivalence principle, we have \(E\left[{ }_{0} L\right]=0\). Further,
\[
\begin{aligned}
\operatorname{Var}\left({ }_{0} L\right) & =500\left[\left(10,000 v^{15}\right)^{2}\left({ }_{15} p_{x}\right)\left(1-{ }_{15} p_{x}\right)\right] \\
& =1,942,329,000
\end{aligned}
\]

Then, using the normal approximation, the approximate probability that the aggregate losses exceed 50,000 is
\(P\left({ }_{0} L>50,000\right)=P\left(Z>\frac{50,000-0}{\sqrt{1,942,329,000}}\right)=P(Z>1.13)=0.13\)
Question: . For a fully discrete whole life insurance policy on (61), you are given:
(i) The annual gross premium using the equivalence principle is 500
(ii) Initial expenses, incurred at policy issue, are \(15 \%\) of the premium
(iii) Renewal expenses, incurred at the beginning of each year after the first, are \(3 \%\) of the premium
(iv) Mortality follows the Standard Ultimate Life Table
(v) \(\quad i=0.05\)

Calculate the amount of the death benefit.

Options:
A
\(\quad 23,300\)
B
23,400
C
\(\quad 23,500\)
D
23,600
E
\(\quad 23,700\)

Correct answer: A

Solution: Let \(B\) be the amount of death benefit.
\(\operatorname{EPV}(\) Premiums \()=500 \ddot{a}_{61}=500(14.6491)=7324.55\)
EPV(Benefits) \(=\mathrm{B} \cdot A_{61}=(0.30243) \mathrm{B}\)
\(\operatorname{EPV}(\) Expenses \()=(0.12)(500)+(0.03)(500) \ddot{a}_{61}=(0.12)(500)+(0.03)(7324.55)=279.74\)
EPV(Premiums) = EPV(Benefits) + EPV(Expenses)
\(7324.55=(0.30243) \mathrm{B}+279.74\)
\(7044.81=(0.30243) \mathrm{B}\)
\(\mathrm{B}=23,294\)
Question: . For a fully discrete whole life insurance policy of 100,000 on (35), you are given:
(i) First year commissions are \(19 \%\) of the annual gross premium
(ii) Renewal year commissions are \(4 \%\) of the annual gross premium
(iii) Mortality follows the Standard Ultimate Life Table
(iv) \(\quad i=0.05\)

Calculate the annual gross premium for this policy using the equivalence principle.

Options:
A
410
B
450
C
490
D
530
E
570

Correct answer: D

Solution: Let \(G\) be the annual gross premium. By the equivalence principle, we have
\(G \ddot{a}_{35}=100,000 A_{35}+0.15 G+0.04 G \ddot{a}_{35}\)
so that
\(G=\frac{100,000 A_{35}}{0.96 \ddot{a}_{35}-0.15}=\frac{100,000(0.09653)}{0.96(18.9728)-0.15}=534.38\)