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A cure model mixes an individual class with identically zero hazard function and a susceptible class with survivor function $S_{\ast }$. For $0\le \pi <1$, its representation as a proportional frailty model with $\mathbb{E}U=1$ is $U=0$ with probability $\pi$ and $U=1/(1-\pi )$ otherwise, with baseline hazard $h_0=(1-\pi )h_{\ast }$. The frailty law is a mixture of two Dirac measures, rather than an ordinary density. Its Laplace transform evaluated at $H_0$ gives the displayed survivor function. The limiting survival equals $\pi$ when $S_{\ast }(t)\to 0$; otherwise the plateau also includes susceptible individuals who never experience the event.

In a proportional frailty model, Bayes' theorem tilts the original density $g(u)$ among subjects surviving to $t$ toThis survival selection progressively favors smaller frailties.

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 $1/(1-\pi )$.

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 isLarge frailty gives earlier events on average. Normalizing $\mathbb{E}U=1$ 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, $S^{\prime }(t\mid U)=-h(t\mid U)S(t\mid U)$ givesThis 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 ${\sum }_i{h}_i(t)S_i(t)/{\sum }_i{S}_i(t)$. 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 $e^{-u{H}_0(t)}$ times the original frailty distribution.

A frailty model uses an unobserved nonnegative random hazard multiplier to represent variation in individual risk. In a proportional frailty model, conditional on $U=u$ the cumulative hazard function is $u{H}_0(t)$, giving survivor $e^{-u{H}_0(t)}$. Averaging over the frailty random variable by the law of total probability yieldsthe Laplace transform of the frailty distribution evaluated at the baseline integrated hazard. This also applies to atoms by integration against the probability measure; the printed Dirac expression is not an ordinary continuous density.

In the given two-atom case, the Dirac delta function notation assigns probabilities $1/4$ and $3/4$ to $U=4/7$ and $U=8/7$. These have mean one, fixing the frailty scale relative to the baseline hazard. With $H_0(t)=7t/4$, the two conditional integrated hazards are $t$ and $2t$. ThereforeThe frailty distribution among survivors shifts toward the smaller multiplier $4/7$ over time, reproducing the declining population hazard and limit one without changing either individual's constant conditional hazard.

A proportional frailty model specifiesIts population survival function is the Laplace transform of the frailty density,If $m=\mathbb{E}U<\infty$, replacing $U$ by $U/m$ and $h_0$ by $m{h}_0$ leaves their product, and hence the model, unchanged. The normalization $\mathbb{E}U=1$ identifies this otherwise arbitrary scale and makes $h_0(0)$ the initial population hazard when $H_0(0)=0$.

For unit-rate exponential distribution frailty, the population survival and hazard in a proportional frailty model areThe surviving frailty distribution is exponential with rate $1+H_0(t)$.