 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:Taking the expectation over the frailty law yieldsThis 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 $e^{-u{H}_0(t)}g(u)$, giving population hazardSurvival 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 $0\le \pi <1$. The necessary mean-one frailty law and baseline hazard areThen $\mathbb{E}U=\pi \cdot 0+(1-\pi )/(1-\pi )=1$. A cured person's conditional hazard is zero, while a susceptible person's conditional hazard is $h_0/(1-\pi )=h_{\ast }$. Taking $U$ to be zero or one while leaving $h_0=h_{\ast }$ would instead have mean frailty $1-\pi$ and violate the required normalization.

The frailty law is the measurewhere 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 isWriting $H_{\ast }(t)={\int }_0^t{h}_{\ast }(v)dv$, we have $H_0=(1-\pi )H_{\ast }$, and thereforeIf $H_{\ast }(t)\to \infty$, the susceptible class eventually experiences the event and $S(t)\to \pi$. If $H_{\ast }(\infty )<\infty$, the limiting survival is larger than $\pi$ because some susceptible subjects also never experience the event. At $\pi =1$, everyone is cured: set $h_0\equiv 0$ and, for example, $U\equiv 1$ to retain mean one, giving $S\equiv 1$ without using the singular formula $1/(1-\pi )$.