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

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.