BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: . For a fully discrete whole life insurance policy of 2000 on (45), you are given:
(i) The gross premium is calculated using the equivalence principle
(ii) Expenses, payable at the beginning of the year, are:

\begin{tabular}{|l|c|c|c|}
\hline & \% of Premium & Per 1000 & Per Policy \\
\hline First year & \(25 \%\) & 1.5 & 30 \\
\hline Renewal years & \(5 \%\) & 0.5 & 10 \\
\hline
\end{tabular}
(iii) Mortality follows the Standard Ultimate Life Table
(iv) \(\quad i=0.05\)

Calculate the expense policy value at the end of policy year 10.

Options:
A
-2
B
-8
C
\(\quad-12\)
D
-21
E
-25

Correct answer: E

Solution: Gross premium \(=G\)
\(G \ddot{a}_{45}=2000 A_{45}+\underbrace{\left(1\left(\frac{2000}{1000}\right)+20\right)}_{22}+\underbrace{\left(0.5\left(\frac{2000}{1000}\right)+10\right)}_{11} \ddot{a}_{45}+0.20 G+0.05 G \ddot{a}_{45}\)
\(\left(0.95 \ddot{a}_{45}-0.20\right) G=2000 A_{45}+22+11 \ddot{a}_{45}\)
\(G=\frac{2000 A_{45}+22+11 \ddot{a}_{45}}{0.95 \ddot{a}_{45}-0.20}=\frac{2000(0.15161)+22+11(17.8162)}{0.95(17.8162)-0.20}=31.16\)
There are two ways to proceed. The first is to calculate the gross premium policy value (with equivalence principle gross premium and original assumptions, both of which do apply here) and the net premium policy value and take the difference.
The net premium is \(\frac{2000 A_{45}}{\ddot{a}_{45}}=\frac{2000(0.15161)}{17.8162}=17.02\)
The net premium policy value is \(2000 A_{55}-17.02 \ddot{a}_{55}=2000(0.23524)-17.02(16.0599)=197.14\)

The gross premium policy value is
\(2000 A_{55}+[0.05(31.16)+0.5(2000 / 1000)+10] \ddot{a}_{55}-31.16 \ddot{a}_{55}\)
\(=2000(0.23524)+(12.56-31.16)(16.0599)=171.77\)
Expense policy value is \(171.77-197.14=-25\)
The second is to calculate the expense policy value directly based on the pattern of expenses. The first step is to determine the expense premium.

The present value of expenses is
\([0.05 G+0.5(2000 / 1000)+10] \ddot{a}_{45}+0.20 G+1.0(2000 / 1000)+20\)
\(=12.558(17.8162)+28.232=251.97\)
The expense premium is \(251.97 / 17.8162=14.14\)
The expense policy value is the expected present value of future expenses less future expense premiums, that is,
\([0.05 G+0.5(2000 / 1000)+10] \ddot{a}_{55}-14.14 \ddot{a}_{55}=-1.582(16.0599)=-25\)

There is a shortcut with the second approach based on recognizing that expenses that are level throughout create no expense policy value (the level expense premium equals the actual expenses). Therefore, the expense policy value in this case is created entirely from the extra first year expenses. They occur only at issue, so the expected present value is \(0.20(31.16)+1.0(2000 / 1000)+20=28.232\). The expense premium for those expenses is then \(28.232 / 17.8162=1.585\) and the expense policy value is the present value of future non-level expenses ( 0 ) less the present value of those future expense premiums, which is 1.585 (16.0599) = 25 for a reserve of -25 .
Question: . For a whole life insurance of 10,000 on \((x)\), you are given:
(i) Death benefits are payable at the end of the year of death
(ii) A premium of 30 is payable at the start of each month
(iii) Commissions are \(5 \%\) of each premium
(iv) Expenses of 100 are payable at the start of each year
(v) \(\quad i=0.05\)
(vi) \(\quad 1000 A_{x+10}=400\)
(vii) \({ }_{10} V\) is the gross premium policy value at the end of year 10 for this insurance Calculate \({ }_{10} V\) using the two-term Woolhouse formula for annuities.

Options:
A
950
B
980
C
1010
D
1110
E
1140

Correct answer: D

Solution: \(\ddot{a}_{x+10}=\left(1-A_{x+10}\right) / d=(1-0.4) /(0.05 / 1.05)=12.6\)
\(\ddot{a}_{x+10}^{(12)} \approx 12.6-11 / 24=12.142\)
\({ }_{10} V=10,000 A_{x+10}+100 \ddot{a}_{x+10}-12 \ddot{a}_{x+10}^{(12)}(30)(1-0.05)\)
\({ }_{10} V=10,000(0.4)+100(12.6)-12(12.142)(28.50)\)
\({ }_{10} V=1107\)
Question: . For a fully discrete whole life insurance of 1000 on a select life [70], you are given:
(i) Ultimate mortality follows the Standard Ultimate Life Table
(ii) During the three-year select period, \(q_{[x]+k}=(0.7+0.1 k) q_{x+k}, k=0,1,2\)
(iii) \(\quad i=0.05\)
(iv) The net premium for this insurance is 35.168

Calculate \({ }_{1} V\), the net premium policy value at the end of year 1 for this insurance.

Options:
A
25.25
B
27.30
C
29.85
D
31.60
E
33.35

Correct answer: C

Solution: The simplest solution is recursive:
\({ }_{0} V=0\) since the policy values are net premium policy values.
\(q_{[70]}=(0.7)(0.010413)=0.007289\)
\({ }_{1} V=\frac{(0+35.168)(1.05)-(1000)(0.007289)}{1-0.007289}=29.86\)
Prospectively, \(q_{[70]+1}=(0.8)(0.011670)=0.009336 ; \quad q_{[70]+2}=(0.9)(0.013081)=0.011773\)
\(A_{[70]+1}=(0.009336) v+(1-0.009336)(0.011773) v^{2} +(1-0.009336)(1-0.011773)(0.47580) v^{2}=0.44197\)
\(\ddot{a}_{[70]+1}=\left(1-A_{[70]+1}\right) / d=(1-0.44197) /(0.05 / 1.05)=11.7186\)
\({ }_{1} V=(1000)(0.44197)-(11.7186)(35.168)=29.85\)
Question: . For a semi-continuous 20 -year endowment insurance of 100,000 on (45), you are given:
(i) Net premiums of 253 are payable monthly
(ii) Mortality follows the Standard Ultimate Life Table
(iii) Deaths are uniformly distributed over each year of age
(iv) \(\quad i=0.05\)

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

Options:
A
\(\quad 38,100\)
B
38,300
C
38,500
D
38,700
E
\(\quad 38,900\)

Correct answer: A

Solution: Let \(P=0.00253\) be the monthly net premium per 1 of insurance.
\[
\begin{aligned}
{ }_{10} V & =100,000\left[\frac{i}{\delta} A_{55: 101}^{1}+A_{55: 101}-12 P \ddot{a}_{55: 101}^{(12)}\right] \\
& =100,000[1.02480(0.02471)+0.59342-(12)(0.00253)(7.8311)] \\
& \approx 38,100
\end{aligned}
\]

Where
\[
\begin{aligned}
A_{55: \overline{10}}^{1} & =A_{55: \overline{10}}-{ }_{10} E_{55}=0.61813-0.59342=0.02471 \\
A_{55: \overline{10}} & ={ }_{10} E_{55}=0.59342 \\
\ddot{a}_{55: \overline{10}} & =8.0192 \\
\ddot{a}_{55: \overline{10}}^{(12)} & =\alpha(12) \ddot{a}_{55: \overline{10}}-\beta(12)\left[1-{ }_{10} E_{55}\right] \\
& =1.00020(8.0192)-0.46651(1-0.59342)=7.8311
\end{aligned}
\]
Question: . For a fully discrete whole life insurance of 100,000 on (45), you are given:
(i) Mortality follows the Standard Ultimate Life Table
(ii) \(\quad i=0.05\)
(iii) Commission expenses are \(60 \%\) of the first year's gross premium and \(2 \%\) of renewal gross premiums
(iv) Administrative expenses are 500 in the first year and 50 in each renewal year
(v) All expenses are payable at the start of the year
(vi) The gross premium, calculated using the equivalence principle, is 977.60

Calculate \({ }_{5} V^{e}\), the expense policy value at the end of year 5 for this insurance.

Options:
A
\(\quad-1070\)
B
-1020
C
\(\quad-970\)
D
-920
E
-870

Correct answer: C

Solution: Use superscript \(g\) for gross premiums and gross premium policy values.
Use superscript \(n\) (representing "net") for net premiums and net premium policy values.
Use superscript \(e\) for expense premiums and expense policy values.
\(P^{g}=977.60\) (given)
\[
\begin{aligned}
P^{e} & =\frac{0.58 P^{g}+450+\left(0.02 P^{g}+50\right) \ddot{a}_{45}}{\ddot{a}_{45}} \\
& =\frac{0.58(977.60)+450+[0.02(977.60)+50] 17.8162}{17.8162}=126.64
\end{aligned}
\]

Alternatively,
\[
P^{n}=\frac{100,000 A_{45}}{\ddot{a}_{45}}=850.97 \quad P^{e}=P^{g}-P^{n}=126.63
\]
\({ }_{5} V^{e}=\left(0.02 P^{g}+50\right) \ddot{a}_{50}-P^{e} \ddot{a}_{50}=[0.02(977.60)+50](17.0245)-126.64(17.0245)=-972\)
Alternatively,
\[
\begin{aligned}
{ }_{5} V^{n} & =100,000 A_{50}-P^{n} \ddot{a}_{50} \\
& =100,000(0.18931)-850.97(17.0245)=4443.66 \\
{ }_{5} V^{g} & =100,000 A_{50}+\left(50+0.02 P^{g}-P^{g}\right) \ddot{a}_{50} \\
& =100,000(0.18931)+[50+0.02(977.60)-977.60](17.0245)=3471.93 \\
{ }_{5} V^{e} & ={ }_{5} V^{g}-{ }_{5} V^{n}=-972
\end{aligned}
\]
Question: . For a fully discrete whole life insurance of 10,000 on (45), you are given:
(i) \(\quad i=0.05\)
(ii) \({ }_{0} L\) denotes the loss at issue random variable based on the net premium
(iii) If \(K_{45}=10\), then \({ }_{0} L=4450\)
(iv) \(\quad \ddot{a}_{55}=13.4205\)

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

Options:
A
1010
B
1460
C
1820
D
2140
E
2300

Correct answer: B

Solution: \(L=10,000 v^{K_{45}+1}-P \ddot{a}_{\overline{K_{45}+1}}=10,000 v^{11}-P \ddot{a}_{\overline{11}}\)
\(4450=10,000(0.58468)-8.7217 P\)
\(P=(5,846.8-4,450) / 8.7217=160.15\)
\(A_{55}=1-d \ddot{a}_{55}=1-(0.05 / 1.05)(13.4205)=0.36093\)
\({ }_{10} V=10,000 A_{55}-P \ddot{a}_{55}=(10,000)(0.36093)-(160.15)(13.4205)=1,460\)
Question: . For a special fully discrete 25 -year endowment insurance on (44), you are given:
(i) The death benefit is \((26-k)\) for death in year \(k\), for \(k=1,2,3 \ldots 25\)
(ii) The endowment benefit in year 25 is 1
(iii) Net premiums are level
(iv) \(q_{55}=0.15\)
(v) \(\quad i=0.04\)
(vi) \({ }_{11} V\), the net premium policy value at the end of year 11, is 5.00
(vii) \({ }_{24} V\), the net premium policy value at the end of year 24 , is 0.60

Calculate \({ }_{12} V\), the net premium policy value at end of year 12 .

Options:
A
\(\quad 3.63\)
B
3.74
C
3.88
D
3.98
E
4.09

Correct answer: E

Solution: In the final year: \(\left({ }_{24} V+P\right)(1+\mathrm{i})=b_{25}\left(q_{68}\right)+1\left(p_{68}\right)\)
Since \(b_{25}=1\), this reduces to \(\left({ }_{24} V+P\right)(1+i)=1 \Rightarrow(0.6+P)(1.04)=1 \Rightarrow P=0.36154\)
Looking back to the \(12{ }^{\text {th }}\) year: \(\left({ }_{11} V+P\right)(1+i)=b_{12}\left(q_{55}\right)+{ }_{12} V\left(p_{55}\right)\)
\(\Rightarrow(5.36154)(1.04)=14(0.15)+{ }_{12} V(0.85) \Rightarrow{ }_{12} V=4.089\)
Question: . For a fully discrete 30 -year endowment insurance of 1000 on (40), you are given:
(i) Mortality follows the Standard Ultimate Life Table
(ii) \(\quad i=0.05\)

Calculate the full preliminary term (FPT) reserve for this policy at the end of year 10.

Options:
A
180
B
185
C
190
D
195
E
200

Correct answer: A

Solution: This first solution recognizes that the full preliminary term reserve at the end of year 10 for a 30 year endowment insurance on (40) is the same as the net premium policy value at the end of year 9 for a 29 year endowment insurance on (41). Then, using superscripts of FPT for full preliminary term reserve and NLP for net premium policy value to distinguish the symbols, we have
\[
\begin{aligned}
1000_{10} V^{F P T} & =1000_{9} V^{N L P}=1000\left(A_{50: 201}-P_{41: 29} \ddot{a}_{50: 201}\right) \\
& =1000[0.38844-0.01622(12.8428)]=180
\end{aligned}
\]
or \(=1000\left(1-\frac{\ddot{a}_{50: 201}}{\ddot{a}_{41: 291}}\right)=1000\left(1-\frac{12.8428}{15.6640}\right)=180\)
where
\[
\begin{aligned}
\ddot{a}_{41: 29 l} & =\ddot{a}_{41}-{ }_{29} E_{41} \ddot{a}_{70} \\
& =18.3403-(0.2228726)(12.0083) \\
& =15.6640 \\
A_{41: 29 l} & =1-d(15.6640)=0.254095 \\
{ }_{29} E_{41} & =v^{29}\left(\frac{l_{70}}{l_{41}}\right)=(0.242946)\left(\frac{91,082.4}{99,285.9}\right)=0.2228726 \\
P_{41: 29 l} & =\frac{0.254095}{15.6640}=0.01622
\end{aligned}
\]

Alternatively, working from the definition of full preliminary term reserves as having \({ }_{1} V^{F P T}=0\) and the discussion of modified net premium reserves in the Notation and Terminology Study Note, let \(\alpha\) be the valuation premium in year 1 and \(\beta\) be the valuation premium thereafter. Then (with some of the values taken from above),
\(\alpha=1000 v q_{40}=0.5019\)
APV (valuation premiums) = APV (benefits)
\(\alpha+{ }_{1} E_{40}\left(\ddot{a}_{41: 29}\right) \beta=1000 A_{40: 30}\)
\(0.5019+0.95188(15.6640) \beta=242.37\)
\(\beta=\frac{242.37-0.5019}{14.9102}=16.22\)
Where
\({ }_{1} E_{40}=(1-0.000527) v=0.95188\)
\(A_{40: 301}=A_{40}+{ }_{20} E_{40}\left({ }_{10} E_{60}\right)\left(1-A_{70}\right)\)
\[
=0.12106+0.36663(0.57864)(1-0.42818)=0.24237
\]
\({ }_{10} V^{F P T}=1000 A_{50: 20 \mid}-\beta \ddot{a}_{50: 20 \mid}=1000(0.38844)-16.22(12.8427)=180\)
Question: . For a fully discrete whole life insurance of 100,000 on (45), you are given:
(i) The gross premium policy value at duration 5 is 5500 and at duration 6 is 7100
(ii) \(\quad q_{50}=0.009\)
(iii) \(\quad i=0.05\)
(iv) Renewal expenses at the start of each year are 50 plus \(4 \%\) of the gross premium.
(v) Claim expenses are 200.

Calculate the gross premium.

Options:
A
2200
B
2250
C
2300
D
2350
E
2400

Correct answer: A

Solution: \[
\begin{aligned}
\left({ }_{5} V+0.96 G-50\right)(1.05) & =q_{50}(100,200)+p_{50}{ }_{6} V \\
(5500+0.96 G-50)(1.05) & =(0.009)(100,200)+(1-0.009)(7100) \\
(1.05)(0.96) G+5722.5 & =7937.9 \\
(1.05)(0.96) G & =2215.4 \\
G & =2197.8
\end{aligned}
\]
Question: . For a fully discrete whole life insurance of 100 on \((x)\), you are given:
(i) \(\quad q_{x+15}=0.10\)
(ii) Deaths are uniformly distributed over each year of age
(iii) \(\quad i=0.05\)
(iv) \({ }_{t} V\) denotes the net premium policy value at time \(t\)
(v) \(\quad{ }_{16} V=49.78\)

Calculate \({ }_{15.6} V\).

Options:
A
49.7
B
50.0
C
50.3
D
50.6
E
\(\quad 50.9\)

Correct answer: E

Solution: \({ }_{15.6} V(1+i)^{0.4}={ }_{0.4} p_{x+15.6}{ }_{16} V+{ }_{0.4} q_{x+15.6} 100\)
\({ }_{15.6} V(1.05)^{0.4}=0.957447(49.78)+0.042553(100)\)
\({ }_{15.6} V=50.91\)
Question: . For a fully discrete 5 -payment whole life insurance of 1000 on (80), you are given:
(i) The gross premium is 130
(ii) \(\quad q_{80+k}=0.01(k+1), \quad k=0,1,2, \ldots, 5\)
(iii) \(\quad v=0.95\)
(iv) \(\quad 1000 A_{86}=683\)
(v) \({ }_{3} L\) is the prospective loss random variable at time 3 , based on the gross premium

Calculate \(E\left[{ }_{3} L\right]\).

Options:
A
330
B
350
C
360
D
380
E
390

Correct answer: D

Solution: APV future benefits \(=1000\left[0.04 v+0.05 \times 0.96 v^{2}+0.96 \times 0.95 \times(0.06+0.94 \times 0.683) v^{3}\right]=630.25\)
APV future premiums \(=130(1+0.96 v)=248.56\)
\(E\left[{ }_{3} L\right]=630.25-248.56=381.69\)
Question: . For a fully discrete whole life insurance of 1 on ( \(x\) ), you are given:
(i) \(\quad q_{x+10}=0.02067\)
(ii) \(\quad v^{2}=0.90703\)
(iii) \(A_{x+11}=0.52536\)
(iv) \({ }^{2} A_{x+11}=0.30783\)
(v) \({ }_{k} L\) is the prospective loss random variable at time \(k\)

Calculate \(\frac{\operatorname{Var}\left({ }_{10} L\right)}{\operatorname{Var}\left({ }_{11} L\right)}\).

Options:
A
1.006
B
1.010
C
1.014
D
1.018
E
1.022

Correct answer: D

Solution: \(\frac{V\left[{ }_{10} L\right]}{V\left[{ }_{11} L\right]}=\frac{\left(1+\frac{p}{d}\right)^{2}\left({ }^{2} A_{x+10}-A_{x+10}^{2}\right)}{\left(1+\frac{p}{d}\right)^{2}\left({ }^{2} A_{x+11}-A_{x+11}^{2}\right)}\)
\(A_{x+10}=v q_{x+10}+v p_{x+10} A_{x+11}\)
\(=(0.90703)^{1 / 2}(0.02067)+(0.90703)^{1 / 2}(1-0.02067)(0.52536)=0.50969\)
\({ }^{2} A_{x+10}=v^{2} q_{x+10}+v^{2} p_{x+10}{ }^{2} A_{x+11} =(0.90703)(0.02067)+(0.90703)(1-0.02067)(0.30783)=0.29219\)
\(\Rightarrow \frac{\operatorname{Var}\left({ }_{k} L\right)}{\operatorname{Var}\left({ }_{k+1} L\right)}=\frac{(0.29219)-(0.50969)^{2}}{(0.30783)-(0.52536)^{2}}=\frac{0.03241}{0.03183}=1.018\)
Question: . For a fully discrete whole life insurance of 1 on ( \(x\) ), you are given:
(i) The net premium policy value at the end of the first year is 0.012
(ii) \(q_{x}=0.009\)
(iii) \(\quad i=0.04\)

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

Options:
A
17.1
B
17.6
C
18.1
D
18.6
E
19.1

Correct answer: A

Solution: \[
\begin{aligned}
& \begin{aligned}
& V_{x}=A_{x+1}-P_{x} \ddot{a}_{x+1}=1-d \ddot{a}_{x+1}-P_{x} \ddot{a}_{x+1} \\
&=1-\underbrace{\left(P_{x}+d\right)}) \\
& \ddot{a}_{x+1}=1-\ddot{a}_{x+1} / \ddot{a}_{x}
\end{aligned} \\
& \Rightarrow \ddot{a}_{x}\left(1-V_{1}\right)=\ddot{a}_{x+1}
\end{aligned}
\]

Since \(\ddot{a}_{x}=1+v p_{x} \ddot{a}_{x+1}\) substituting we get
\[
\ddot{a}_{x}\left(1-{ }_{1} V_{x}\right)=\frac{\ddot{a}_{x}-1}{v p_{x}} \Rightarrow \ddot{a}_{x}\left(1-{ }_{1} V_{x}\right) v p_{x}=\ddot{a}_{x}-1
\]

Solving for \(\ddot{a}_{x}\), we get \(\ddot{a}_{x}=\frac{1}{1-\left(1-{ }_{1} V_{x}\right) v p_{x}}=\frac{1}{1-(1-0.012)\left(\frac{1}{1.04}\right)(1-0.009)}\)
\[
\text { = } 17.07942
\]
Question: . For a fully discrete whole life insurance of 100,000 on (40) you are given:
(i) Expenses incurred at the beginning of the first year are 300 plus \(50 \%\) of the first year premium
(ii) Renewal expenses, incurred at the beginning of the year, are \(10 \%\) of each of the renewal premiums
(iii) Mortality follows the Standard Ultimate Life Table
(iv) \(\quad i=0.05\)
(v) Gross premiums are calculated using the equivalence principle

Calculate the gross premium policy value for this insurance immediately after the second premium and associated renewal expenses are paid.

Options:
A
200
B
340
C
560
D
720
E
1060

Correct answer: D

Solution: Let \(G\) be the annual gross premium.
Using the equivalence principle, \(0.90 G \ddot{a}_{40}-0.40 G=100,000 A_{40}+300\)
So \(G=\frac{100,000(0.12106)+300}{0.90(18.4578)-0.40}=765.2347\)
The gross premium policy value after the first year and immediately after the second premium and associated expenses are paid is
\(100,000 A_{41}-0.90 G\left(\ddot{a}_{41}-1\right)\)
\(=12,665-0.90(765.2347)(17.3403)\)
= 723
Question: . For a fully discrete whole life insurance of 1000 on (35), you are given:
(i) First year expenses are \(30 \%\) of the gross premium plus 300
(ii) Renewal expenses are \(4 \%\) of the gross premium plus 30
(iii) All expenses are incurred at the beginning of the policy year
(iv) Gross premiums are calculated using the equivalence principle
(v) The gross premium policy value at the end of the first policy year is \(R\)
(vi) Using the Full Preliminary Term Method, the modified net premium reserve at the end of the first policy year is \(S\)
(vii) Mortality follows the Standard Ultimate Life Table
(viii) \(i=0.05\)

Calculate \(R-S\).

Options:
A
\(\quad-280\)
B
-140
C
0
D
140
E
280

Correct answer: A

Solution: If \(G\) denotes the gross premium, then
\[
G=\frac{1000 A_{35}+30 \ddot{a}_{35}+270}{0.96 \ddot{a}_{35}-0.26}=\frac{1000(0.09653)+30(18.9728)+270}{0.96(18.9728)-0.26}=52.12
\]

So that,
\[
\begin{aligned}
& R=1000 A_{36}+(30-0.96 G) \ddot{a}_{36} \\
& \quad=1000(0.10101)+(30-0.96 \times 52.12)(18.8788)=-277.23
\end{aligned}
\]

Note that \(S=0\) as per definition of FPT reserve.
Question: . A special fully discrete 10 -payment 10 -year deferred whole life annuity-due on ( 55 ) of 1000 per year provides for a return of premiums without interest in the event of death within the first 10 years. You are given:
(i) Annual net premiums are level
(ii) Mortality follows the Standard Ultimate Life Table
(iii) \(\quad i=0.05\)
(iv) \(\quad(I A)_{55: 10}^{1}=0.14743\)

Calculate \({ }_{9} V\), the net premium policy value at the end of year 9 .

Options:
A
11,540
B
11,650
C
11,760
D
11,870
E
11,980

Correct answer: D

Solution: \(\pi=\frac{1000{ }_{10} \mid \ddot{a}_{55}}{\ddot{a}_{55: \overline{10}}-(I A)_{55: \overline{10}}^{1}}=\frac{1000(0.59342)(13.5498)}{8.0192-0.14743}=1021.46\)
\({ }_{9} V=1000 \quad{ }_{1} \mid \ddot{a}_{64}+10 \pi A_{64: 1]}^{1}-\pi \ddot{a}_{64: 1]}\)
\(=1000 \frac{1}{1.05}\left(\frac{94,579.7}{95,082.5}\right) 13.5498+10(1021.46) \frac{1}{1.05}(0.005288)-1021.46\)
=11,866
Question: . For two fully discrete whole life insurance policies on ( \(x\) ), you are given:
(i)

\begin{tabular}{|l|c|c|c|}
\hline & \begin{tabular}{c} 
Death \\
Benefit
\end{tabular} & \begin{tabular}{c} 
Annual Net \\
Premium
\end{tabular} & \begin{tabular}{c} 
Variance of \\
Loss at Issue
\end{tabular} \\
\hline Policy 1 & 8 & 1.250 & 20.55 \\
\hline Policy 2 & 12 & 1.875 & \(W\) \\
\hline
\end{tabular}
(ii) \(\quad i=0.06\)
(iii) The two policies are priced using the same mortality table.

Calculate \(W\).

Options:
A
30.8
B
38.5
C
46.2
D
53.9
E
61.6

Correct answer: C

Solution: \(V\left[L_{0} \# 1\right]=\left(B_{1}+\frac{P_{1}}{d}\right)^{2}\left({ }^{2} A_{x}-A_{x}^{2}\right)=20.55 \Longrightarrow\left(8+\frac{1.25(1.06)}{0.06}\right)^{2}\left({ }^{2} A_{x}-A_{x}^{2}\right)=20.55\)
\({ }^{2} A_{x}-A_{x}^{2}=\frac{20.55}{\left(8+\frac{1.25(1.06)}{0.06}\right)^{2}}=0.0227\)
\(V\left[L_{0} \# 2\right]=\left(12+\frac{1.875(1.06)}{0.06}\right)^{2}\left({ }^{2} A_{x}-A_{x}^{2}\right)=\left(12+\frac{1.875(1.06)}{0.06}\right)^{2}(0.0227)=46.24\)
Question: . For a 40 -year endowment insurance of 10,000 issued to (25), you are given:
(i) \(\quad i=0.04\)
(ii) \(\quad p_{25}=0.995\)
(iii) \(\quad \ddot{a}_{25: \overline{20}}=11.087\)
(iv) \(\quad \ddot{a}_{25: \overline{40}}=16.645\)
(v) The annual level net premium is 216
(vi) A modified net premium reserving method is used for this policy, where the valuation premiums are:
- A first year premium equal to the first year net cost of insurance,
- Level premiums of \(\beta\) for years 2 through 20, and
- Level premiums of 216 thereafter.

Calculate \(\beta\).

Options:
A
140
B
170
C
200
D
230
E
260

Correct answer: D

Solution: We have Present Value of Modified Premiums = Present Value of level net premiums
\(v q_{x}+\beta\left(\ddot{a}_{25: \overline{20}}-1\right)+P \cdot{ }_{20} E_{25} \cdot \ddot{a}_{45: 20 \mid}=P \ddot{a}_{25: \overline{40} \mid}\)
\(\Rightarrow \beta=\frac{P\left(\ddot{a}_{25: \overline{40}}\right)-P \cdot{ }_{20} E_{25} \cdot \ddot{a}_{45: 201}-v q_{x}}{\ddot{a}_{25: 201}-1}=\frac{P \ddot{a}_{25: 201}-v q_{x}}{\ddot{a}_{25: 201}-1}\)
We are given that \(P=0.0216\)
\(\Rightarrow \beta=\frac{0.0216(11.087)-(1.04)^{-1}(0.005)}{11.087-1}=0.023265\)
For insurance of 10,000, \(\beta=233\).
Question: . For a fully discrete whole life insurance policy of \(1,000,000\) on (50), you are given:
(i) The annual gross premium, calculated using the equivalence principle, is 11,800
(ii) Mortality follows the Standard Ultimate Life Table
(iii) \(\quad i=0.05\)

Calculate the expense loading, \(P^{e}\), for this policy.

Options:
A
480
B
580
C
680
D
780
E
880

Correct answer: C

Solution: \(P^{g}=P^{n}+P^{e} \quad\) where \(P^{e}\) is the expense loading
\(P^{n}=1,000,000 \frac{A_{50}}{\ddot{a}_{50}}=1,000,000\left(\frac{0.18931}{17.0245}\right)=11,119.86\)
\(P^{e}=P^{g}-P^{n}=11,800-11,120=680\)
Question: . For a fully discrete whole life insurance policy of 100,000 on [55], a professional skydiver, you are given:
(i) Level premiums are paid annually
(ii) Mortality follows a 2-year select and ultimate table
(iii) \(\quad i=0.04\)
(iv) The following table of values for \(A_{[x]+t}\) :

\begin{tabular}{|c|c|c|c|}
\hline\(x\) & \(A_{[x]}\) & \(A_{[x]+1}\) & \(A_{x+2}\) \\
\hline 55 & 0.23 & 0.24 & 0.25 \\
\hline 56 & 0.25 & 0.26 & 0.27 \\
\hline 57 & 0.27 & 0.28 & 0.29 \\
\hline 58 & 0.29 & 0.30 & 0.31 \\
\hline
\end{tabular}

Calculate the Full Preliminary Term reserve at time 3.

Options:
A
2700
B
3950
C
5200
D
6450
E
7800

Correct answer: B

Solution: \({ }_{3} V^{F P T}=100,000 A_{[55]+3}-100,000 P_{[55]+1} \ddot{a}_{[55]+3}\)
\(=100,000 A_{58}-100,000 \frac{A_{55]+1}}{\ddot{a}_{[55]+1}} \ddot{a}_{58}\)
\(=100,000\left(0.27-\frac{0.24}{\frac{1-0.24}{d}} \cdot \frac{1-0.27}{d}\right)\)
= 3947.37
Question: . For a special fully discrete 2 -year endowment insurance on \((x)\), you are given:
(i) The death benefit for year \(k\) is \(25,000 k\) plus the net premium policy value at the end of year \(k\), for \(k=1,2\). For year 2, this net premium policy value is the net premium policy value just before the maturity benefit is paid
(ii) The maturity benefit is 50,000
(iii) \(\quad p_{x}=p_{x+1}=0.85\)
(iv) \(\quad i=0.05\)
(v) \(\quad P\) is the level annual net premium

Calculate \(P\).

Options:
A
\(\quad 27,650\)
B
27,960
C
\(\quad 28,200\)
D
\(\quad 28,540\)
E
28,730

Correct answer: D

Solution: \({ }_{1} V=\left({ }_{0} V+P\right)(1+i)-\left(25,000+{ }_{1} V-{ }_{1} V\right) q_{x}=P(1+i)-(25,000) q_{x}\)
\({ }_{2} V=\left({ }_{1} V+P\right)(1+i)-\left(50,000+{ }_{2} V-{ }_{2} V\right) q_{x+1}=50,000\)
\(\left(\left(P(1+i)-25,000 q_{x}\right)+P\right)(1+i)-50,000 q_{x+1}=50,000\)
\(((P(1.05)-25,000(0.15))+P)(1.05)-50,000(0.15)=50,000\)
Solving for \(P\), we get
\(P=\frac{61,437.50}{2.1525}=28,542.39\)