A <proportional frailty model> specifies
$$
h(t\mid U=u)=u h_0(t),
\qquad
S(t\mid U=u)=e^{-uH_0(t)}.
$$
Its population <survival function> is the <Laplace transform> of the frailty density,
$$
\overline S(t)=\int_0^\infty e^{-uH_0(t)}g(u)\,du.
$$
If $m=\mathbb EU<\infty$, replacing $U$ by $U/m$ and $h_0$ by $m h_0$ leaves their product, and hence the model, unchanged. The normalization $\mathbb EU=1$ identifies this otherwise arbitrary scale and makes $h_0(0)$ the initial population hazard when $H_0(0)=0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 207 / 6 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-207/6/solution