You have calculated the expected present value of a last survivor life insurance of 1 on (x) and (y). You assumed:
- The death benefit is payable immediately on the second death.
- The future lifetimes of (x) and (y) are independent, and each life has a constant force of mortality with (mu=0.06).
- (delta = 0.05).
Your supervisor points out that these are not independent future lifetimes. Each mortality assumption is correct, but each includes a common shock component with constant force 0.02.
Calculate the increase in the expected present value over what you originally calculated.
– Soln
+ Soln
Solution.We wish to calculate (bar{A}_{overline{xy}}^{CS} – bar{A}_{overline{xy}}^{IND}). Under the independence model, we have
begin{eqnarray*}
bar{A}_x &=& int_0^{infty} e^{-delta t} ~_t p_x mu_{x+t} dt =int_0^{infty} e^{-delta t} e^{-mu t} mu dt = frac{mu}{mu+delta} .
end{eqnarray*}
Thus,
begin{eqnarray*}
bar{A}_x = frac{0.06}{0.06+0.05} = frac{6}{11} = bar{A}_y .
end{eqnarray*}
Further,
begin{eqnarray*}
bar{A}_{xy} &=& int_0^{infty} e^{-delta t} ~_t p_{xy} mu_{xy}(t) dt =int_0^{infty} e^{-delta t} e^{-2mu t} 2mu dt = frac{2mu}{2mu+delta} ,
end{eqnarray*}
so that ( bar{A}_{xy} = frac{0.12}{0.17} = frac{12}{17} ) and
begin{eqnarray*}
bar{A}_{overline{xy}}^{IND} = bar{A}_x +bar{A}_y-bar{A}_{xy} = 2 frac{6}{11} -frac{12}{17} = 0.385 .
end{eqnarray*}
For the common shock model, the original force of mortality was 0.06, so the new force of mortality is 0.04. Thus,
begin{eqnarray*}
bar{A}_x &=& int_0^{infty} e^{-delta t} ~_t p_x mu_{x+t} dt =int_0^{infty} e^{-0.05t} e^{-0.04 t} e^{-0.02 t}mu dt = frac{6}{11},
end{eqnarray*}
as before. The same is true for (bar{A}_y ). The joint force of mortality is (mu_{xy}(t) = 0.04+0.04+0.02 = 0.10). Thus, the joint survival function is
(~_t p_{xy} e^{-0.10 t}). This yields
begin{eqnarray*}
bar{A}_{xy}^{CS} = frac{mu_{xy}}{mu_{xy}+delta} =frac{0.10}{0.10+0.05} = frac{2}{3},
end{eqnarray*}
and
begin{eqnarray*}
bar{A}_{overline{xy}}^{CS} = bar{A}_x +bar{A}_y-bar{A}_{xy}^{CS} = 2 frac{6}{11} -frac{2}{3} = 0.42423 .
end{eqnarray*}
Thus, the increase is
begin{eqnarray*}
bar{A}_{overline{xy}}^{CS} – bar{A}_{overline{xy}}^{IND} = 0.42423 – 0.385 = 0.0392.
end{eqnarray*}
begin{eqnarray*}
bar{A}_x &=& int_0^{infty} e^{-delta t} ~_t p_x mu_{x+t} dt =int_0^{infty} e^{-delta t} e^{-mu t} mu dt = frac{mu}{mu+delta} .
end{eqnarray*}
Thus,
begin{eqnarray*}
bar{A}_x = frac{0.06}{0.06+0.05} = frac{6}{11} = bar{A}_y .
end{eqnarray*}
Further,
begin{eqnarray*}
bar{A}_{xy} &=& int_0^{infty} e^{-delta t} ~_t p_{xy} mu_{xy}(t) dt =int_0^{infty} e^{-delta t} e^{-2mu t} 2mu dt = frac{2mu}{2mu+delta} ,
end{eqnarray*}
so that ( bar{A}_{xy} = frac{0.12}{0.17} = frac{12}{17} ) and
begin{eqnarray*}
bar{A}_{overline{xy}}^{IND} = bar{A}_x +bar{A}_y-bar{A}_{xy} = 2 frac{6}{11} -frac{12}{17} = 0.385 .
end{eqnarray*}
For the common shock model, the original force of mortality was 0.06, so the new force of mortality is 0.04. Thus,
begin{eqnarray*}
bar{A}_x &=& int_0^{infty} e^{-delta t} ~_t p_x mu_{x+t} dt =int_0^{infty} e^{-0.05t} e^{-0.04 t} e^{-0.02 t}mu dt = frac{6}{11},
end{eqnarray*}
as before. The same is true for (bar{A}_y ). The joint force of mortality is (mu_{xy}(t) = 0.04+0.04+0.02 = 0.10). Thus, the joint survival function is
(~_t p_{xy} e^{-0.10 t}). This yields
begin{eqnarray*}
bar{A}_{xy}^{CS} = frac{mu_{xy}}{mu_{xy}+delta} =frac{0.10}{0.10+0.05} = frac{2}{3},
end{eqnarray*}
and
begin{eqnarray*}
bar{A}_{overline{xy}}^{CS} = bar{A}_x +bar{A}_y-bar{A}_{xy}^{CS} = 2 frac{6}{11} -frac{2}{3} = 0.42423 .
end{eqnarray*}
Thus, the increase is
begin{eqnarray*}
bar{A}_{overline{xy}}^{CS} – bar{A}_{overline{xy}}^{IND} = 0.42423 – 0.385 = 0.0392.
end{eqnarray*}