Reconstructing Survival from Cumulative Hazard

From an Instantaneous Rate to Survival

For a lifetime \(T\), the survival function \(S(t)=P(T>t)\) describes the probability that the lifetime exceeds \(t\). The hazard \(h(t)\) describes the instantaneous failure rate at time \(t\), among units that have survived up to that time. These quantities answer different questions: survival is a probability, while hazard is a rate conditional on continued survival. Integrating the hazard over time accumulates that rate, and the accumulated quantity determines survival exactly in the continuous-lifetime setting developed here.

Assume \(T\) is nonnegative, has a density \(f\), and has no atom at zero. Then \(S(0)=1\), and on times for which \(S(t)>0\), the hazard is \(h(t)=f(t)/S(t)\). We will assume the hazard is measurable and integrable on every bounded interval where it is used. These conditions allow the integral and the logarithm in the derivation below to be treated rigorously.

Cumulative hazard starts at zero and is nondecreasing because \(h(u)\geq 0\). It does not have to stay below one: it is not a probability. The key relation is that survival decays exponentially in this accumulated quantity.

Deriving Survival from Cumulative Hazard

Proof. Since \(T\) has density \(f\), its survival function is absolutely continuous on bounded time intervals, with derivative \(S'(u)=-f(u)\) for almost every \(u\). On any interval \([0,t]\) on which \(S(t)>0\), monotonicity of \(S\) implies \(S(u)\geq S(t)>0\). Thus \(\log S(u)\) is well defined throughout that interval. The chain rule for absolutely continuous functions gives $$ \frac{d}{du}\log S(u)=\frac{S'(u)}{S(u)} =-\frac{f(u)}{S(u)} =-h(u) $$ for almost every \(u\in[0,t]\). Integrating both sides from \(0\) to \(t\), and using \(S(0)=1\), yields $$ \log S(t)-\log S(0)=-\int_0^t h(u)\,du, \qquad \log S(t)=-H(t). $$ Exponentiating gives \(S(t)=\exp\{-H(t)\}\), and rearranging gives \(H(t)=-\log S(t)\). \(\square\)

The proof shows why the initial condition matters: the integration constant is \(\log S(0)=0\). If a lifetime has positive probability of being zero, this version of the identity cannot be used from time zero without accounting separately for that atom.

Recovering Hazard from Survival

The relation can also be read in the reverse direction. If a survival function is positive and differentiable, its logarithmic rate of decrease determines the hazard. Under absolute continuity, differentiability is needed only almost everywhere.

Proof. On any bounded interval where \(S\) is positive, it is bounded away from zero by its value at the right endpoint, since survival is nonincreasing. The logarithm is continuously differentiable on the range of \(S\) there, so the chain rule for absolutely continuous functions applies. It gives \(H'(t)=-S'(t)/S(t)\) almost everywhere. When \(S'(t)=-f(t)\) almost everywhere, substitution gives \(H'(t)=f(t)/S(t)=h(t)\). Since \(H(0)=-\log 1=0\), integrating its derivative recovers \(H(t)=\int_0^t h(u)\,du\). \(\square\)

Conversely, a nonnegative, measurable hazard that is integrable on bounded intervals can be used to construct a survival function by setting \(H(t)=\int_0^t h(u)\,du\) and \(S(t)=e^{-H(t)}\). This function is nonincreasing, starts at one, and is absolutely continuous on bounded intervals. Almost everywhere, \(S'(t)=-h(t)S(t)\). For this construction to describe a lifetime that is finite with probability one, it must also satisfy \(S(t)\to 0\) as \(t\to\infty\), equivalently \(H(t)\to\infty\). If \(H\) instead approaches a finite limit, the constructed survival function approaches a positive limit; it does not describe a finite lifetime with probability one.

What Cumulative Hazard Does—and Does Not—Measure

The value \(H(t)\) measures accumulated hazard, not the probability of failure by time \(t\). The probability of surviving to \(t\) is \(e^{-H(t)}\); therefore the probability of failing by \(t\) is \(1-e^{-H(t)}\). This distinction follows from the survival identity, rather than from treating hazard as a probability.

A common mistake is to add hazards over time and call the result a failure probability. For example, if \(H(t)=0.8\), the failure probability by time \(t\) is \(1-e^{-0.8}\), approximately \(0.5507\), not \(0.8\). The correction is to accumulate the hazard first and then apply the exponential relation.

The ordinary integral formula in this tutorial applies to continuous lifetimes with a density. A lifetime with atoms can have jumps in its survival function; an ordinary density-based hazard integral alone does not account for those jumps. Within the continuous setting, however, the inverse relations are exact: integrate hazard to obtain cumulative hazard, exponentiate its negative to obtain survival, or take the negative logarithm of survival to recover cumulative hazard.