![\"F1WholeLife\"](\"http:\/\/www.ssc.wisc.edu\/~jfrees\/wp-content\/uploads\/2015\/01\/F1WholeLIfeeps.jpg\")
<\/a><\/figcaption><\/figure>\n
Suppose that mortality follows deMoivre’s law with limiting age \\(w=100\\). Let \\(i=6\\%\\) and \\(x=35\\). Then,
\n\\begin{eqnarray*}
\n\\bar{A}_{35} = \\int_0^{65} v^t ~_t p_{35} \\mu_{35+t}~ dt = \\frac{1}{65} \\int_0^{65} v^t dt = \\frac{1}{65} \\bar{a}_{\\overline{65}|}
\n\\end{eqnarray*}
\nand
\n\\begin{eqnarray*}
\nP^n = \\frac{\\bar{A}_{35}}{\\bar{a}_{35}} =\\frac{\\delta \\bar{A}_{35}}{1-\\bar{A}_{35}} =
\n\\frac{\\delta \\bar{a}_{\\overline{65}|} \/65}{1-\\bar{a}_{\\overline{65}|} \/65} = 0.020266.
\n\\end{eqnarray*}
\nSimilarly, we have \\(\\bar{A}_{35+t}=\\bar{a}_{\\overline{65-t}|}\/(65-t)\\) and
\n\\begin{eqnarray*}
\n\\bar{a}_{35+t}= \\frac{1-\\bar{A}_{35+t}}{\\delta} =\\frac{1-\\bar{a}_{\\overline{65-t}|}\/(65-t)}{\\delta}
\n\\end{eqnarray*}
\nThe policy value may be written as
\n\\begin{eqnarray*}
\n_t V^n = \\frac{\\bar{a}_{\\overline{65-t}|}}{65-t} – (0.020266)\\frac{65-t-\\bar{a}_{\\overline{65-t}|}}{\\delta(65-t)}.
\n\\end{eqnarray*}
\nand its associated variability is
\n\\begin{eqnarray*}
\n\\textrm{Var}(L_t^n|T>t)&=& \\left(1+\\frac{0.020266}{\\delta}\\right)^2 \\left( \\frac{~^2 \\bar{a}_{\\overline{65-t}|}}{65-t} – \\left[\\frac{\\bar{a}_{\\overline{65-t}|}}{65-t}\\right]^2 \\right) \\\\
\n&=&
\n1.813821 \\left( \\frac{~^2 \\bar{a}_{\\overline{65-t}|}}{65-t} – \\left[\\frac{\\bar{a}_{\\overline{65-t}|}}{65-t}\\right]^2 \\right)
\n\\end{eqnarray*}
\nusing \\(\\delta = \\ln(1.06)\\).<\/p>\n
Figure 2 summarizes these calculations. Note that the policy value is 0 at \\(t=0\\). At \\(t=65\\), the policy value becomes 1. Also at \\(t=65\\), there is no uncertainty about the value of the policy and so the standard deviation becomes 0.<\/p>\n<\/a>
<\/figcaption><\/figure>\n