Example 1
Consider at 25-year endowment to a select life age [30]. The insured amount is 100,000. The insurer incurs initial expenses of 2,000 plus 50% of first year premium, renewal expenses of 2.5% of premiums. Benefits are payable immediately. Use the Illustrative Mortality Table with (i=5%).
a) Provide an expression for the gross future loss random variable.
Solution
(i) The present value of benefit outgo is (100,000 v^{min(T[30],25) } ).
(ii) The present value of gross premium income is ( G ddot{a}_{overline{min(K[30],25)|}} ).
(iii) The present value of expenses is (2000+0.025 G ddot{a}_{overline{min(K[30],25)|}} + 0.475 G)
b) Calculate the expense-augment (gross) annual premium.
Solution
(i) The EPV of benefit outgo is
begin{eqnarray*}
&=& 100,000 bar{A}_{overline{[30],25|}} \
&=& 100,000 left( frac{i}{delta} A_{overline{[30],25|}}^{~~1} +A_{overline{[30],25|}}^{~~~~~1} right)\
&=& 29,873.20 .
end{eqnarray*}
(ii) The EPV of gross premium income is
begin{eqnarray*}
&=& G ddot{a}_{overline{[30],25|}} = G (14.73113) .
end{eqnarray*}
(iii) The EPV of expenses is
begin{eqnarray*}
&=& 2000 + 0.025 G ddot{a}_{[30]:overline{25|}} + 0.475 G \
&=& 2000 + (0.843278) G .
end{eqnarray*}
Now, setting (i)+(iii)=(ii), we have
begin{eqnarray*}
G &=& frac{29,873.20 +2,000}{14.73113 – 0.843278} = 2,295.05 .
end{eqnarray*}
Example 2
For a fully discrete whole life insurance of 100,000 on (x), you are given:
(i) Expenses, paid at the beginning of the year, are as follows:
| Year | Percentage of Premium Expenses | Per 1000 Expenses | Per Policy Expenses |
| First Year | 50% | 2.0 | 150 |
| 2+ | 4% | 0.5 | 25 |
(ii) (i = 0.04)
(iii) (ddot{a}_x =10.8).
Calculate the expense-loaded premium using the equivalence principle.
Solution
EPV Premium = (G ddot{a}_x = G (10.8) )
EPV Benefit = (100000 A_x)
EPV Expenses
begin{eqnarray*}
&=& 0.5 G + 200 + 150
+ (0.04 G + 50 + 25) a_x \
&=& G (0.5 +0.04 a_x) + 350 + 75 a_x = G (0.892) + 1,085 .
end{eqnarray*}
Equating EPV Premium = EPV Benefit + EPV Expenses yields
begin{eqnarray*}
10.8 G &=& 100000 A_x + 0.892 G + 1085 .
end{eqnarray*}
Solving for G yields
begin{eqnarray*}
G &=& frac{100000 A_x + 1085}{10.8 – 0.892} = frac{58462 1085}{9.908} = 6009.99 .
end{eqnarray*}
Example 3
Consider at 10-year term life policy for 100,000 to a life age (30). Benefits are payable immediately, premiums are payable at the beginning of the year. Use the Illustrative Mortality Table with (i=6% ). Expenses are according to the following schedule:
| Type | Per Policy | Per 1000 of Insurance | Percent of Premium |
| First Year | 50 | 5.00 | 82% |
| Years 2-5 | 6 | 0.50 | 14.5 |
| Years 6-10 | 6 | 0.50 | 7.0 |
| Claim Settlement | 25 | 0.10 | |
Calculate G, the gross annual premium.
Solution
begin{eqnarray*}
G ddot{a}_{30:overline{10|}}&=&
100,000 bar{A}_{30:overline{10|}}^{~1}
+ (25+100(0.10)) bar{A}_{30:overline{10|}}^{~1}
+ 550 \
&~~~+& 0.82 G + 56 a_{30:overline{9|}}
+ 0.145 G a_{30:overline{4|}}
+ 0.07 G ~_{5|5} ddot{a}_{30} .
end{eqnarray*}
Thus,
begin{eqnarray*}
G ddot{a}_{30:overline{10|}}&=&
frac{100,035 bar{A}_{30:overline{10|}}^{~1}
+ 56 ddot{a}_{30:overline{10|}} + 494}
{0.93 ddot{a}_{30:overline{10|}} – 0.075
ddot{a}_{30:overline{5|}} – 0.675} approx $385.
end{eqnarray*}