Suppose that (T(x)) is exponentially distributed with force of mortality (mu_x(t)=mu_1), (T(y)) is exponentially distributed with force of mortality (mu_y(t)=mu_2), and that (T(x)) and (T(y)) are independent. Then, for (x), we have (~_t p_x = exp(-mu_1 t)) and
begin{eqnarray*}
dot{e}_x &=& E ~ T(x) = int_0^{infty} ~_t p_x ~ dt =
frac{1}{mu_1} .
end{eqnarray*} Similarly, for the joint-life status, we have
(~_t p_{xy} = ~_t p_{xy} times ~_t p_{xy} = exp(-(mu_1+mu_2) t)) which is an exponential distribution with force of mortality (mu_1+mu_2). This yields
begin{eqnarray*}
dot{e}_{xy}=frac{1}{mu_1+mu_2} .
end{eqnarray*} Now, for the last-survivor status, we have
begin{eqnarray*}
1 – ~_t p_{overline{xy}} &=& F_{T(overline{xy})}(t) = Prleft(
max(T(x),T(y)) leq t right) \
&=& Prleft( T(x) leq t right) times Prleft( T(y) leq t right) \
&=& left( 1-exp(-mu_1 t) right) times left( 1-exp(-mu_2 t)right)
end{eqnarray*} and
begin{eqnarray*}
dot{e}_{overline{xy}} &=& dot{e}_{x}+dot{e}_{y}-dot{e}_{xy}\
&=& frac{1}{mu_1} +frac{1}{mu_2} -frac{1}{mu_1+mu_2} .
end{eqnarray*}
Follow-up
Suppose in addition that (mu_1 = 0.02) and (mu_2 = 0.015). Then,
begin{eqnarray*}
dot{e}_x = frac{1}{mu_1} = 50,~~~~dot{e}_y = frac{1}{mu_2} =
66.67, ~~~dot{e}_{xy} = frac{1}{mu_1+mu_2} = 28.57 .
end{eqnarray*} Further,
begin{eqnarray*}
dot{e}_{overline{xy}} &=& dot{e}_{x}+dot{e}_{y}-dot{e}_{xy}
&=& 50 + 66.67 -28.57 = 88.1 .
end{eqnarray*} Moreover, we have
begin{eqnarray*}
Cov(T(overline{xy}),T(xy)) &=& mathrm{E~} left{T(overline{xy})times T(xy)right} – mathrm{E~} T(overline{xy}) times
mathrm{E~}T(xy)\
&=& mathrm{E~} left{max(T(x),T(y)) times min (T(x),T(y))right} – dot{e}_{overline{xy}} times dot{e}_{xy}\
&=& mathrm{E~}left{ T(x) times T(y)right}
– dot{e}_{overline{xy}} times dot{e}_{xy}\
&=& dot{e}_x dot{e}_y – dot{e}_{overline{xy}} times
dot{e}_{xy} \
&=& 50 times 66.67 – 88.1 times 28.57 = 816.48 .
end{eqnarray*}