= Solution
A <frailty model> represents unobserved individual heterogeneity by a <random effect> in the <hazard function>. In a <proportional frailty model>, an individual with <frailty random variable> $U=u$ has hazard $u h_0(t)$. The normalization $\mathbb E U=1$ identifies the scale of the <baseline hazard>: without it, multiplying all frailties by a constant and dividing $h_0$ by that constant would leave all individual hazards unchanged.
Let $H_0(t)=\int_0^t h_0(v)\,dv$. Conditional survival follows by integrating the <hazard function>:
$$
S(t\mid U=u)=\exp\left(-\int_0^tu h_0(v)\,dv\right)=e^{-uH_0(t)}.
$$
Taking the <expectation> over the frailty law yields
$$
\boxed{S(t)=\int_0^\infty e^{-uH_0(t)}g(u)\,du
=\widetilde g\big(H_0(t)\big),\qquad
\widetilde g(s)=\int_0^\infty e^{-su}g(u)\,du.}
$$
This is the <Laplace transform> of the frailty density evaluated at the <cumulative hazard>. More generally, the same calculation integrates against a <probability measure>, allowing atoms. It averages conditional survival rather than substituting $\mathbb E U$ into its exponent. Indeed, the <frailty distribution among survivors> is proportional to $e^{-uH_0(t)}g(u)$, giving population hazard
$$
h_{\rm pop}(t)=h_0(t)\frac{\mathbb E[Ue^{-UH_0(t)}]}{\mathbb E[e^{-UH_0(t)}]}.
$$
Survival progressively selects smaller frailties, so the population hazard need not be the baseline or even retain the individual's proportional structure.
For the <cure model>, first take $0\leq\pi<1$. The necessary mean-one frailty law and <baseline hazard> are
$$
\boxed{\Pr(U=0)=\pi,\qquad
\Pr\left(U=\frac1{1-\pi}\right)=1-\pi,
\qquad h_0(t)=(1-\pi)h_*(t).}
$$
Then $\mathbb E U=\pi\cdot0+(1-\pi)/(1-\pi)=1$. A cured person's conditional hazard is zero, while a susceptible person's conditional hazard is $h_0/(1-\pi)=h_*$. Taking $U$ to be zero or one while leaving $h_0=h_*$ would instead have mean frailty $1-\pi$ and violate the required normalization.
The frailty law is the measure
$$
g(du)=\pi\delta_0(du)+(1-\pi)\delta_{1/(1-\pi)}(du),
$$
where the $\delta$ terms are <Dirac measures>. This is not an ordinary Lebesgue density; the cure construction requires permitting a discrete frailty law. Its <Laplace transform> is
$$
\widetilde g(s)=\pi+(1-\pi)e^{-s/(1-\pi)}.
$$
Writing $H_*(t)=\int_0^t h_*(v)\,dv$, we have $H_0=(1-\pi)H_*$, and therefore
$$
\boxed{S(t)=\widetilde g(H_0(t))=\pi+(1-\pi)e^{-H_*(t)}.}
$$
If $H_*(t)\to\infty$, the susceptible class eventually experiences the event and $S(t)\to\pi$. If $H_*(\infty)<\infty$, the limiting survival is larger than $\pi$ because some susceptible subjects also never experience the event. At $\pi=1$, everyone is cured: set $h_0\equiv0$ and, for example, $U\equiv1$ to retain mean one, giving $S\equiv1$ without using the singular formula $1/(1-\pi)$.
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