Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 41 / 5 / a

A survivor function is $S(t)=\mathbb{P}(T>t)$. For a nonnegative event time it is nonincreasing and right-continuous, with values between zero and one; a proper finite event time has $S(t)\to 0$ as $t\to \infty$. Here $w(0)=1$, $w(t)>0$, $w(t)\to 0$, and $w^{\prime }(t)=-(e^{-t}+6e^{-2t})/4<0$ for $t\ge 0$. Extending it by one for $t<0$ supplies a valid continuous survivor function.

The density is $-w^{\prime }$, so the hazard function isIn particular $h(0)=7/4$ and $\boxed{\underset{t\to \infty }{\lim }h(t)=1}$. The survivor is a mixture of rate-one and rate-two exponential distributions with weights $1/4$ and $3/4$. Among long-term survivors the rate-one component dominates, explaining the limiting hazard. More generally the hazard derivative for a mixture of exponential distributions is minus the variance of the component rate among survivors; the decreasing hazard is selection, not an individual component's changing hazard.