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Let a frailty random variable have a gamma distribution with mean one and variance $v>0$. Its shape and rate are both $1/v$. For conditional hazard function $U{a}^z{h}_0(t)$, $z\in \{0,1\}$, averaging the conditional survivor function gives $(1+v{a}^z{H}_0(t){)}^{-1/v}$. Differentiating its logarithm yields the population hazard function $a^z{h}_0(t)/(1+v{a}^z{H}_0(t))$ and hence the displayed hazard ratio. Writing $x=v{H}_0(t)$ gives $r-1=(a-1)/(1+ax)$. Therefore the population hazard ratio moves monotonically from $a$ towards one as exposure $x$ increases, even though the individual conditional hazard ratio remains $a$. The limit one requires $H_0(t)\to \infty$ for fixed positive $v$, or $v\to \infty$ for fixed positive $H_0(t)$; it is not a consequence of elapsed time alone. This is survival selection through preferential removal of larger frailties.

In a frailty model, an individual's unobserved susceptibility to an event is represented by a nonnegative frailty random variable multiplying that individual's hazard function. Conditional on its value, subjects with the same measured covariates can therefore have different event rates. The frailty is a persistent latent characteristic of the individual, not newly sampled noise at each time.

Here the conditional hazard function is $u{e}^{\beta z}{h}_0(t)$, with baseline hazard $h_0$. A multiplier $u>1$ raises susceptibility and $u<1$ lowers it. Normalization $\mathbb{E}U=1$ fixes the frailty scale: without it, multiplying every frailty by a constant and dividing the baseline hazard by that constant would give the same individual hazards. The variance of $U$ measures unobserved heterogeneity. An atom at $U=0$, if allowed, would give a permanently zero conditional hazard.

The hazard ratio between groups is $e^{\beta }$ for individuals compared at the same frailty. It need not equal the population hazard ratio, because survival selection changes the frailty distribution among those still alive or event-free.

Condition on a frailty random variable value $U=u$. Integrating the conditional hazard function gives $u{e}^{\beta z}{H}_0(t)$, so the conditional survivor function isAverage this conditional probability over the common frailty density $g$:By definition the Laplace transform of $g$ is $\tilde{g}(s)={\int }_0^{\infty }{e}^{-su}g(u)du$. HenceThis averages survivor functions, not hazards. Differentiating its logarithm, when justified, givesThus the population hazard function uses the mean frailty among current survivors, explaining why a conditional proportional hazards model may cease to be proportional after frailty is integrated out.

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 frailty model represents unobserved heterogeneity using a nonnegative frailty random variable $U$ that multiplies an individual's baseline hazard. With $H_0(t)={\int }_0^t{h}_0(s)ds$, the individual and population survivor functions areThus the population survivor function is the Laplace transform of the frailty distribution at $H_0(t)$; at points where differentiation is justified, $\overline{h}(t)=h_0(t)\mathbb{E}[U\mid T>t]$. The last identity follows by differentiating the integral and dividing $-{\overline{S}}^{\prime }$ by $\overline{S}$.

Here $g(u)=e^{-u}$ for $u\ge 0$, $\theta >0$, and $H_0(t)=\theta t^2/2$. Direct integration givesThis is population hazard under exponential frailty. For every fixed $u>0$ the individual hazard function $u\theta t$ increases linearly. In contrast,so the population hazard increases up to $t=\sqrt{2/\theta }$, then decreases, tending to zero like $2/t$. More vulnerable individuals leave the risk set earlier, leaving survivors with smaller frailties. This population decline is selection, not a decline in any fixed individual's hazard. If $\theta =0$, survival is identically one and both hazards vanish; the positive-rate comparison is then inapplicable.

A frailty random variable is an unobserved positive multiplicative risk factor. A proportional frailty model hasThe scale of $U$ is not separately identifiable from $h_0$: multiplying $U$ by a constant and dividing $h_0$ by it leaves the model unchanged. We may therefore normalize $\mathbb{E}U=1$, which lets $h_0$ represent the mean initial hazard multiplier and makes relative frailty interpretable.

If $U\sim \mathrm{Exponential}(1)$ and $h_0(t)=\lambda$, then the Laplace transform of $U$ givesThus $h(0)=\lambda$, while $h(t)\to 0$ as $t\to \infty$. Survivors become increasingly enriched for low-frailty individuals.

Let the frailty random variable be uniform on $[a,b]$ with $0<a<b$, and let the conditional individual hazard function be the constant $U\theta$, $\theta >0$. Averaging conditional exponential survivor functions gives the displayed population survivor function for $t>0$, with value one at zero. Its population hazard of a survival mixture isIt starts at $\theta (a+b)/2$ and tends to $a\theta$: the frailty distribution among survivors progressively concentrates near its smallest frailty. Although each individual's hazard is constant, the population hazard decreases through survival selection.