Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 207 / 7

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.

Use Bayes' theorem with the individual survivor function. Writing $z=\theta t^2/2$, the frailty distribution among survivors has densityHenceThe conditional expectation starts at one and declines toward zero when $\theta >0$. Survival favors the less frail individuals, precisely explaining why the population hazard under exponential frailty equals $h_0(t)$ times a decreasing multiplier. The original PDF asks for conditioning on the event $T>t$; the TeX's misplaced inequality is a transcription error, not conditioning on an observed exact event time.

Let population 1 have baseline $\theta t$ and population 2 baseline $\lambda \theta t$, with $\theta >0$ and $\lambda >1$, and identical unit-rate exponential frailty distributions initially. Their population hazard functions have ratio, for $t>0$,At exactly $t=0$, both population hazards are zero, so their literal ratio is $0/0$ and is undefined. The intended initial comparison is the right-hand limit,With $z=\theta t^2/2$, $dR/dz=\lambda (1-\lambda )/(1+\lambda z{)}^2<0$, so the marginal hazard ratio decreases from $\lambda$ toward one. The faster-hazard population loses highly frail individuals sooner; its survivors become less frail than the first population's survivors.

Conditional on the same fixed frailty $u>0$, the individual hazard ratio is the constant $\lambda$. Marginalizing over frailty therefore destroys proportional hazards, even though they hold at the individual level. A population-level Cox proportional-hazards model that ignores frailty would impose an incorrect constant hazard ratio here. Convergence of the ratio to one does not mean identical survival: ${\overline{S}}_2(t)/{\overline{S}}_1(t)\to 1/\lambda$.