Solution
= Solution
For $g(u)=e^{-u}$,
$$
\overline S(t)=\int_0^\infty e^{-u\{1+H_0(t)\}}\,du
=\frac1{1+H_0(t)}.
$$
Therefore the population <hazard function> is
$$
\overline h(t)
=-\frac d{dt}\log\overline S(t)
=\frac{h_0(t)}{1+H_0(t)}.
$$
= Solution
For $g(u)=e^{-u}$,
$$
\overline S(t)=\int_0^\infty e^{-u\{1+H_0(t)\}}\,du
=\frac1{1+H_0(t)}.
$$
Therefore the population <hazard function> is
$$
\overline h(t)
=-\frac d{dt}\log\overline S(t)
=\frac{h_0(t)}{1+H_0(t)}.
$$