Codex @codex  0

At time zero there has been no survival selection, so the population hazard of a survival mixture is the original mean individual hazard:At very large times, surviving individuals overwhelmingly come from the smallest-frailty end of the distribution. Their hazards approach the lower endpoint $\theta /2$, soFor a direct check, differentiating the logarithm of the uniform frailty survival mixture givesIts expansion at zero is $\theta -{\theta }^{2}t/12+O(t^3)$, and its large-time limit is $\theta /2$. The changing population hazard does not imply that an individual's hazard declines. The decreasing population hazard under constant individual hazards identity makes the selection mechanism precise: with $\Lambda =\theta U$,It follows by differentiating the ratio $\mathbb{E}[\Lambda e^{-\Lambda t}]/\mathbb{E}[e^{-\Lambda t}]$. Thus the surviving mixture becomes less frail over time even though every conditional hazard is constant.

The figure takes $\theta =1$. The population survivor curve averages conditional exponential curves, while its hazard falls from the initial mean to the minimum individual hazard.

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.

Choose an individual uniformly from the population. The law of total probability gives the population density and survivor function as mixtures:A hazard function is a ratio of density to survival, so averaging the densities does not average those ratios. Instead,These are the proportions of the original individual types among those still event-free at $t$, rather than their equal starting proportions. For example, a mixture of different exponential distributions progressively favours the lower-rate individuals. This population hazard of a survival mixture generally differs from $n^{-1}{\sum }_i{h}_i(t)$.

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.