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

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- Solution b

Solution

Condition on a frailty random variable value

U = u

. Integrating the conditional hazard function gives

u e^{β z} H_{0} (t)

, so the conditional survivor function is

Pr (T > t ∣ U = u, z) = exp {- u e^{β z} H_{0} (t)} .

Average this conditional probability over the common frailty density

g

:

\overline{F}^{(z)} (t) = \int_{0}^{\infty} e^{- u e^{β z} H_{0} (t)} g (u) d u .

By definition the Laplace transform of

g

is

g (s) = \int_{0}^{\infty} e^{- s u} g (u) d u

. Hence

\overline{F}^{(z)} (t) = g (e^{β z} H_{0} (t)) .

This averages survivor functions, not hazards. Differentiating its logarithm, when justified, gives

\overline{h}^{(z)} (t) = e^{β z} h_{0} (t) \frac{\int _{0}^{\infty} u e ^{- u e^{β z} H_{0} (t)} g ( u ) d u}{\int _{0}^{\infty} e ^{- u e^{β z} H_{0} (t)} g ( u ) d u} = e^{β z} h_{0} (t) E [U ∣ T > t, z] .

Thus 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.

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