A <frailty model> introduces an unobserved positive random effect $U$ that multiplies an individual's hazard. With $H_0(t)=\int_0^th_0(s)ds$, marginal survival is
$$
\bar S(t)=\int_0^\infty e^{-uH_0(t)}g(u)du.
$$
Differentiation gives
$$
\bar h(t)=h_0(t)
\frac{\int_0^\infty u e^{-uH_0(t)}g(u)du}
{\int_0^\infty e^{-uH_0(t)}g(u)du}
=h_0(t)\mathbb E[U\mid T\geq t].
$$
Consequently $\bar h(0)=h_0(0)$ exactly when $\mathbb EU=1$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 207 / 6 / Unconditional hazard in a frailty model / Solution

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A frailty model uses an unobserved positive random effect to represent individual heterogeneity in event hazards. Conditioning on survival progressively selects smaller frailties.


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