Consider a population containing smokers \\((s)\\) and non-smokers \\((ns)\\) where their forces of mortality are related as
\n\\begin{eqnarray*}
\n\\mu_x^{ns} = \\frac{1}{2} \\mu_x^s .
\n\\end{eqnarray*} For the non-smoking population, mortality is given as \\(l_x = 75 – x\\) \\(x\\ge 0\\). Consider two lives, a non-smoker age 65 and a smoker age 55, and assume lives are independent. Calculate \\(\\dot{e}_{65:55}\\).<\/p>\n
For smokers, the force of mortality is \\(\\mu_x^s = \\frac{2}{75-x}\\) so that the conditional survival force is
\n\\begin{eqnarray*}
\n~_t p_x^s &=& \\exp\\left( – \\int_0^t \\mu_{x+r}^s ~ dr \\right) \\\\
\n&=& \\exp \\left( -2 \\int_0^t \\frac{1}{75-x-r} ~ dr \\right) \\\\
\n&=& \\exp \\left( 2 \\ln \\left(75-x-r \\right)|_0^t \\right) \\\\
\n&=& \\exp \\left( 2 \\ln \\frac{75-x-t}{75-x} \\right) \\\\
\n&=& \\left(1- \\frac{t}{75-x} \\right)^2
\n\\end{eqnarray*}
\nThus,
\n\\begin{eqnarray*}
\n~_t p_{65:55} &=& ~_t p_{65}^{ns} ~_t p_{55}^s = \\left(1-
\n\\frac{t}{10} \\right)\\left(1- \\frac{t}{20} \\right)^2
\n\\end{eqnarray*} and
\n\\begin{eqnarray*}
\n\\dot{e}_{65:55}&=& \\int_0^{10} ~_t p_{65:55} dt \\\\
\n&=& \\int_0^{10} \\left(1- \\frac{t}{10} \\right)\\left(1- \\frac{t}{20}
\n\\right)^2 dt = 3.54 .
\n\\end{eqnarray*}\n<\/p><\/div>\n<\/div>\n
<\/p>\n
As follow-ups, note the following extensions\/observations.<\/p>\n
1) \\(\\dot{e}_{65}^{ns} =5 (=\\frac{75-65}{2} )\\).<\/p>\n
2) \\(\\dot{e}_{55}^s = \\int_0^{20} \\left(1- \\frac{t}{20} \\right)^2 dt
\n= \\frac{20}{3}\\).<\/p>\n
3)
\n\\begin{eqnarray*}
\n\\dot{e}_{\\overline{65:55}} &=& 5 + \\frac{20}{3} – 3.54 = 8.12\\\\
\n& \\neq & \\int_0^{10} ~ _t p_{\\overline{65:55}} ~dt .
\n\\end{eqnarray*}<\/p>\n