2.3 The cumulative hazard and the basic identity
Theorem 2.3.2. For a continuous survival time, \(S(t) = \exp \bigl (-H(t)\bigr )\) and \(f(t) = h(t)S(t)\). Any one of \(S\), \(f\), \(h\) and \(H\) therefore determines the others.
Proof. Since \(f = -S'\), \(h(t) = f(t)/S(t) = -\frac {d}{dt}\log S(t)\). Integrating from \(0\) to \(t\) and using \(S(0) = 1\) gives \(H(t) = -\log S(t)\), that is, \(S(t) = e^{-H(t)}\). Then \(f = hS\) by the definition of \(h\). □
The identity explains why a hazard can be any non-negative function: the only condition for a proper distribution is \(H(t) \to \infty \), so that \(S(t) \to 0\). It also gives the exponential distribution its special place: a constant hazard \(h(t) = \lambda \) gives \(H(t) = \lambda t\) and \(S(t) = e^{-\lambda t}\), and the exponential is the only continuous distribution with a constant hazard. Its conditional survival \(S(t + u)/S(t) = e^{-\lambda u} = S(u)\) does not depend on \(t\): it is memoryless, and a subject who has survived to \(t\) is as good as new. The Weibull distribution, with \(H(t) = (t/\lambda )^k\), \(S(t) = \exp (-(t/\lambda )^k)\) and \(h(t) = (k/\lambda )(t/\lambda )^{k - 1}\), has a decreasing, constant or increasing hazard as \(k < 1\), \(k = 1\) or \(k > 1\), the three curves of Figure 2.
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