= Solution
Condition on a <frailty random variable> value $U=u$. Integrating the conditional <hazard function> gives $u e^{\beta z}H_0(t)$, so the conditional <survivor function> is
$$
\Pr(T>t\mid U=u,z)=\exp\{-u e^{\beta z}H_0(t)\}.
$$
Average this conditional probability over the common frailty density $g$:
$$
\overline F^{(z)}(t)
=\int_0^\infty e^{-u e^{\beta z}H_0(t)}g(u)\,du.
$$
By definition the <Laplace transform> of $g$ is $\widetilde g(s)=\int_0^\infty e^{-su}g(u)\,du$. Hence
$$
\boxed{\overline F^{(z)}(t)=\widetilde g\bigl(e^{\beta z}H_0(t)\bigr).}
$$
This averages <survivor functions>, not hazards. Differentiating its logarithm, when justified, gives
$$
\overline h^{(z)}(t)
=e^{\beta z}h_0(t)\frac{\int_0^\infty u e^{-u e^{\beta z}H_0(t)}g(u)\,du}
{\int_0^\infty e^{-u e^{\beta z}H_0(t)}g(u)\,du}
=e^{\beta z}h_0(t)\mathbb E[U\mid T>t,z].
$$
Thus the population <hazard function> uses the mean frailty among current survivors, explaining why a conditional <proportional hazards model> may cease to be proportional after frailty is integrated out.
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