In a <proportional frailty model>, <Bayes theorem> tilts the original density $g(u)$ among subjects surviving to $t$ to
$$
g(u\mid T>t)=\frac{e^{-uH_0(t)}g(u)}{\int_0^\infty e^{-vH_0(t)}g(v)\,dv}.
$$
This survival selection progressively favors smaller frailties.

Frailty distribution among survivors

A <proportional frailty model> specifies
$$
h(t\mid U=u)=u h_0(t),
\qquad
S(t\mid U=u)=e^{-uH_0(t)}.
$$
Its population <survival function> is the <Laplace transform> of the frailty density,
$$
\overline S(t)=\int_0^\infty e^{-uH_0(t)}g(u)\,du.
$$
If $m=\mathbb EU<\infty$, replacing $U$ by $U/m$ and $h_0$ by $m h_0$ leaves their product, and hence the model, unchanged. The normalization $\mathbb EU=1$ identifies this otherwise arbitrary scale and makes $h_0(0)$ the initial population hazard when $H_0(0)=0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 207 / 6 / Solution

For unit-rate <exponential distribution> frailty, the population survival and hazard in a <proportional frailty model> are
$$
\overline S(t)=\frac1{1+H_0(t)},
\qquad
\overline h(t)=\frac{h_0(t)}{1+H_0(t)}.
$$
The surviving frailty distribution is exponential with rate $1+H_0(t)$.

Population hazard under exponential frailty

A proportional frailty model has conditional hazard $h(t\mid U=u)=u h_0(t)$. Its conditional survival function is $e^{-uH_0(t)}$, so its population survival function is the <Laplace transform> of the frailty distribution evaluated at $H_0(t)$.

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 Incoming links: Proportional frailty model