Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 46 / 5

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.

The population survivor function is the probability that a randomly selected person, averaging over their unobserved frailty, has not experienced the event by time $t$. Here the conditional hazard $U\theta$ is constant, so the conditional survival is $e^{-U\theta t}$. For $\theta >0$, the uniform frailty density equals one on its interval, giving the uniform frailty survival mixtureAt $t=0$ its continuous extension is $\overline{S}(0)=1$, since the numerator is $\theta t+O(t^2)$. This is the Laplace transform of the uniform frailty density at argument $\theta t$. It is not $e^{-\theta t}$: averaging individual survivor functions does not equal exponentiating the mean hazard. If $\theta =0$, every individual hazard is zero and $\overline{S}(t)=1$.

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.