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
\n\\begin{eqnarray*}
\n~_k V_1 – ~_k V_2 &=& (P_1-P_2) \\frac{\\sum_{h=0}^{k-1} v^h ~_h p_x }{ ~_k E_x}
\n= (P_1-P_2) \\ddot{s}_{x:\\overline{k|}}.
\n\\end{eqnarray*}
\nHere, 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<\/p>\n
Example 1<\/strong> Example 2<\/strong> Exercise<\/strong> Solution<\/em> (b) Suppose that \\(~_n V_x = 0.8\\), \\(P_x =0.024\\), and \\(P_{x:\\overline{n|}}^{~~1} = 0.02\\). Calculate \\( Solution<\/em>
\nSuppose 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
\n\\begin{eqnarray*}
\n~_k V_1 – ~_k V_2 &=& (P_{x:\\overline{n|}}-P_{x:\\overline{n|}}^{1}) \\ddot{s}_{x:\\overline{h|}}.
\n\\end{eqnarray*}<\/p>\n
\nSuppose 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
\n\\begin{eqnarray*}
\nP_{x:\\overline{n|}}- ~_n P_x &=& \\frac{1}{\\ddot{s}_{x:\\overline{h|}}}(~_k V_1 – ~_k V_2 ) \\\\
\n&=& P_{x:\\overline{n|}}^{~~1}(~_k V_1 – ~_k V_2 ) .
\n\\end{eqnarray*}
\nIn particular, if we take \\(k=n\\), then we get
\n\\begin{eqnarray*}
\nP_{x:\\overline{n|}}- ~_n P_x &=& P_{x:\\overline{n|}}^{~~1}(1 – A_{x+n} ) .
\n\\end{eqnarray*}<\/p>\n
\n(a) Use the same logic to check the relation
\n\\begin{eqnarray*}
\nP_x &=& P_{x:\\overline{n|}}^{1}+ P_{x:\\overline{n|}}^{~~1}~_n V_x ,
\n\\end{eqnarray*}
\nwhere \\(~_n V_x\\) is the policy value of a whole life policy at duration \\(n\\).<\/p>\n
\nSuppose 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
\n\\begin{eqnarray*}
\n~_k V_1 – ~_k V_2 &=& (P_x -P_{x:\\overline{n|}}^{1}) \\ddot{s}_{x:\\overline{k|}}.
\n\\end{eqnarray*}
\nAt 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. <\/p>\n
\nP_{x:\\overline{n|}}^{1}\\).<\/p>\n
\nFrom part (a), we have
\n\\begin{eqnarray*}
\n0.024 = P_x &=& P_{x:\\overline{n|}}^{1}+ P_{x:\\overline{n|}}^{~~1}~_n V_x = P_{x:\\overline{n|}}^{1} + (0.02)(0.8).
\n\\end{eqnarray*}
\nThis yields \\(P_{x:\\overline{n|}}^{1} = 0.008\\).<\/p>\n