In this policy, a benefit of 1 is payable immediately upon failure. Premiums are payable at a constant annual rate of (P) per year that is determined at contract initiation by the equivalence principle. Assume a constant force of interest and no expenses. The policy valuation is at time (t < n).
1. Express the (net) premium using standard actuarial notation.
Solution.
The net premium is (P = frac{bar{A}_{x:overline{n}|}}{bar{a}_{x:overline{n}|}}).
2. Express the loss random variable using standard interest theory symbols.
Solution.
If (T< t), then the policy is complete and we do not consider it further. If (T ge n) (exceeding the endowment period), then the loss random variable is (L_t = v^{n-t} - P bar{a}_{overline{n-t}|}), where (P) is from part (1). If (t leq T < n), then the loss random variable is (L_t = v^{T-t} - P bar{a}_{overline{T-t}|}). We summarize this as
begin{eqnarray*}
L_t &=&
left{
begin{array}{cc}
v^{T-t} - P ddot{a}_{overline{T-t}|} & t leq T < n \
v^{n-t} - P ddot{a}_{overline{n-t}|} & T ge n
end{array}
right .
end{eqnarray*}
3. Express the policy value using standard actuarial notation.
Solution.
Recall that the “(n-t)” single premium endowment issued to a life age (x+t) has loss
begin{eqnarray*}
L_t^{Insurance} &=&
left{
begin{array}{cc}
v^{T-t} & T < n \
v^{n-t} & T ge n
end{array}
right .
end{eqnarray*}
and expectation (bar{A}_{x+t:overline{n-t}|}). Similarly, an "(n-t)"temporary life annuity issued to a life age (x+t) has loss
begin{eqnarray*}
L_t^{Annuity} &=&left{
begin{array}{cc}
bar{a}_{overline{T-t}|} & T < n \
bar{a}_{overline{n-t}|} & T ge n end{array}
right .
end{eqnarray*}
and expectation (bar{a}_{x+t:overline{n-t}|}). Thus, the loss random variable is just (L_t = L_t^{Insurance} -P L_t^{Annuity}) with expectation (policy value) (_t V =bar{A}_{x+t:overline{n-t}|}-P bar{a}_{x+t:overline{n-t}|}).
4. Express the standard deviation of the loss random variable using standard actuarial notation.
Solution.
Using the relation (bar{a}_{overline{n}|} = (1-v^{n})/delta), we have
begin{eqnarray*}
L_t &=&
left{
begin{array}{cc}
left(1+frac{P}{delta}right) v^{T-t} – frac{P}{delta} & t < T < n \
left(1+frac{P}{delta}right) v^{n-t} - frac{P}{delta} & T ge n
end{array}
right .
= left(1+frac{P}{delta}right) L_t^{Insurance} - frac{P}{delta} .
end{eqnarray*}
From this, we have
begin{eqnarray*}
textrm{Var}(L_t|T>t) &=& left(1+frac{P}{delta}right)^2 textrm{Var}(L_t^{Insurance}|T > t) \
&=& left(1+frac{P}{delta}right)^2
left(~^2 bar{A}_{x+t:overline{n-t}|}- bar{A}_{x+t:overline{n-t}|}^2 right)
end{eqnarray*}
5. Suppose that (x=35), (n=20), (i=6%), (t=5), and mortality follows deMoivre’s law with limiting age (w=100). Calculate the policy value and standard deviation.
Solution.
We first compute the premium
begin{eqnarray*}
P &=& frac{bar{A}_{35:overline{20}|}}{bar{a}_{35:overline{20}|}} = frac{delta bar{A}_{35:overline{20}|}}{1- bar{A}_{35:overline{20}|}}
end{eqnarray*}
For our problem, we have
begin{eqnarray*}
bar{A}_{35:overline{20}|} &=& int_0^{20} v^t ~_t p_{35} mu_{35+t} dt + v^{20} ~_{20} p_{35} \
&=& frac{1}{65} bar{a}_{overline{20|}} + v^{20} frac{45}{65} = 0.39756 .
end{eqnarray*}
that yields
begin{eqnarray*}
P &=&frac{ln(1.06) (0.39756)}{1- 0.39756} = 0.03845.
end{eqnarray*}
We can express the policy value as
begin{eqnarray*}
V_5 =bar{A}_{40:overline{15}|}-P bar{a}_{40:overline{15}|}
end{eqnarray*}
As before,
begin{eqnarray*}
bar{A}_{40:overline{15}|} &=& frac{1}{60} bar{a}_{overline{15|}} + v^{15} frac{45}{60} = 0.479628 .
end{eqnarray*}
and
begin{eqnarray*}
bar{a}_{40:overline{15}|} &=& frac{1-bar{A}_{overline{40:15}|}}{delta}= 8.930516 ,
end{eqnarray*}
so
begin{eqnarray*}
V_5 =0.479628-(0.03845 ) (8.930516) = 0.13625 .
end{eqnarray*}
For the variance, we have
begin{eqnarray*}
textrm{Var}(L_5|T>5)
&=& left(1+frac{P}{delta}right)^2
left(~^2 bar{A}_{40:overline{15}|}- bar{A}_{40:overline{15}|}^2 right) \
&=& left(1+frac{0.03845}{ln(1.06)}right)^2
left(0.24869- 0.479628^2 right) =0.051378,
end{eqnarray*}
using (~^2 bar{A}_{40:overline{15}|}=frac{1}{60} ~^2 bar{a}_{overline{15|}} + v^{2times15} frac{45}{60}=0.24869). Thus, the standard deviation is (sqrt{0.051378}=0.22667).