= Solution
A <frailty model> represents unobserved heterogeneity in susceptibility by a positive latent <frailty random variable> $U$. In the <proportional frailty model>, conditional on $U=u$, the individual hazard is $h(t\mid u)=u h_0(t)$, where $h_0$ is a baseline <hazard function>. If $H_0(t)=\int_0^t h_0(s)\,ds$, the conditional <survivor function> is
$$
S(t\mid u)=e^{-uH_0(t)}.
$$
Large frailty gives earlier events on average. Normalizing $\mathbb EU=1$ separates the frailty scale from that of the baseline hazard.
The unheaded final request follows for an arbitrary survival mixture, not only this proportional model. Let a randomly selected individual's type be $U$, with individual survivor function $S(t\mid U)$ and hazard $h(t\mid U)$. The population survivor function is $\overline S(t)=\mathbb E[S(t\mid U)]$. Where differentiation under the expectation is justified, $S'(t\mid U)=-h(t\mid U)S(t\mid U)$ gives
$$
\boxed{\overline h(t)=-\frac{\overline S'(t)}{\overline S(t)}
=\frac{\mathbb E[h(t\mid U)S(t\mid U)]}{\mathbb E[S(t\mid U)]}
=\mathbb E[h(t\mid U)\mid T>t].}
$$
This proves the <population hazard of a survival mixture> formula: the weights are the individual probabilities of still surviving, normalized to sum or integrate to one. For a finite equally weighted population it becomes $\sum_i h_i(t)S_i(t)/\sum_iS_i(t)$. Generally it is not the unweighted average of the original individual hazards. In a <proportional frailty model>, the <frailty distribution among survivors> is proportional to $e^{-uH_0(t)}$ times the original frailty distribution.
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