Consider two generic policies that have level premiums (P_1) and (P_2) and policy values (~_k V_1) and (~_k V_2), respectively. Suppose that they have the same accumulated cost of insurance, denoted as (~_k AC_x), at policy duration (k). Then, from the retrospective formula, we may express the difference in policy values as
begin{eqnarray*}
~_k V_1 – ~_k V_2 &=& (P_1-P_2) frac{sum_{h=0}^{k-1} v^h ~_h p_x }{ ~_k E_x}
= (P_1-P_2) ddot{s}_{x:overline{k|}}.
end{eqnarray*}
Here, the difference in accumulated values drops out because they are assumed equivalent. Thus, the difference in policy values is a simple function of the difference in premiums
Example 1
Suppose that Policy 1 is a traditional (n)-year endowment policy and Policy 2 is a traditional (n)-year term life policy. Then, we may write the excess policy value of the (n)-year endowment over the (n)-year term as
begin{eqnarray*}
~_k V_1 – ~_k V_2 &=& (P_{x:overline{n|}}-P_{x:overline{n|}}^{1}) ddot{s}_{x:overline{h|}}.
end{eqnarray*}
Example 2
Suppose that Policy 1 is a traditional (n)-year endowment policy and Policy 2 is a traditional (n)-pay whole life policy. Then, we may write the excess premium of the (n)-year endowment over the (n)-pay whole as
begin{eqnarray*}
P_{x:overline{n|}}- ~_n P_x &=& frac{1}{ddot{s}_{x:overline{h|}}}(~_k V_1 – ~_k V_2 ) \
&=& P_{x:overline{n|}}^{~~1}(~_k V_1 – ~_k V_2 ) .
end{eqnarray*}
In particular, if we take (k=n), then we get
begin{eqnarray*}
P_{x:overline{n|}}- ~_n P_x &=& P_{x:overline{n|}}^{~~1}(1 – A_{x+n} ) .
end{eqnarray*}
Exercise
(a) Use the same logic to check the relation
begin{eqnarray*}
P_x &=& P_{x:overline{n|}}^{1}+ P_{x:overline{n|}}^{~~1}~_n V_x ,
end{eqnarray*}
where (~_n V_x) is the policy value of a whole life policy at duration (n).
Solution
Suppose that Policy 1 is a traditional whole life policy and Policy 2 is a traditional (n)-year term life policy. Then, we may write the excess policy value of the whole life over the term policy as
begin{eqnarray*}
~_k V_1 – ~_k V_2 &=& (P_x -P_{x:overline{n|}}^{1}) ddot{s}_{x:overline{k|}}.
end{eqnarray*}
At duration (k=n), the term policy has value 0 and the right-hand side is (~_n V_x). As before, (1/ddot{s}_{x:overline{n|}} = P_{x:overline{n|}}^{~~1}), yielding the desired result.
(b) Suppose that (~_n V_x = 0.8), (P_x =0.024), and (P_{x:overline{n|}}^{~~1} = 0.02). Calculate (
P_{x:overline{n|}}^{1}).
Solution
From part (a), we have
begin{eqnarray*}
0.024 = P_x &=& P_{x:overline{n|}}^{1}+ P_{x:overline{n|}}^{~~1}~_n V_x = P_{x:overline{n|}}^{1} + (0.02)(0.8).
end{eqnarray*}
This yields (P_{x:overline{n|}}^{1} = 0.008).