Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 44 / 2

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.

For ${\sigma }^2>0$, a gamma distribution with unit mean and variance ${\sigma }^2$ has shape and rate $\psi =1/{\sigma }^2$. Substituting its Laplace transform into the preceding result givesIts negative logarithmic derivative is the population hazard function:Consequently, at times with $h_0(t)>0$, the gamma frailty hazard ratio isAt a time when both hazards vanish this formula is a continuous extension; the literal ratio $0/0$ is undefined.

Write $a=e^{\beta }>0$ and $x={\sigma }^2{H}_0(t)\ge 0$. Two useful identities areFor small cumulative exposure $x$, $r=a\{1+(1-a)x+O(x^2)\}$, so it starts at the conditional hazard ratio $e^{\beta }$. If $\beta >0$, it decreases towards one; if $\beta <0$, it increases towards one; and if $\beta =0$, it equals one throughout. It stays between $e^{\beta }$ and one and never crosses one.

For fixed ${\sigma }^2>0$, the late-time limit is one provided $H_0(t)\to \infty$. If instead the integrated hazard tends to a finite value $K$, the limiting hazard ratio is $a(1+{\sigma }^{2}K)/(1+a{\sigma }^{2}K)$, which need not be one. Elapsed time alone does not imply unbounded exposure.

For small ${\sigma }^2$ at a fixed time, $r\to e^{\beta }$. At ${\sigma }^2=0$ the frailty degenerates to $U=1$, giving an ordinary proportional hazards model. For large ${\sigma }^2$ at a fixed time with $H_0(t)>0$, $r\to 1$; at time zero, however, it remains $e^{\beta }$. Thus that variance limit is not uniform near zero. Larger heterogeneity accelerates the movement of the population hazard ratio towards one. The mechanism is survival selection: frailer individuals leave the population sooner, particularly in the higher-hazard group, while each individual's conditional multiplier remains fixed.