Transforming Lifetimes by Cumulative Hazard

Why Transform a Lifetime by Its Cumulative Hazard?

A hazard rate describes how failure risk accumulates over time, but its shape can make direct probability calculations or simulation awkward. The cumulative hazard provides a useful scale change: under suitable conditions, applying it to a lifetime removes the original hazard shape and produces a unit exponential random variable. Conversely, an exponential draw can be mapped back to a lifetime with a specified hazard.

The distinction between a hazard and a cumulative hazard is important. A hazard rate is a local rate, while the cumulative hazard integrates that rate over time. The transformation result depends on the lifetime’s survival distribution; deriving a density from a proposed cumulative hazard requires an additional regularity condition.

Definitions and Conditions

A function is locally absolutely continuous if it is absolutely continuous on every bounded time interval in its domain. In particular, a locally absolutely continuous \(H\) is the integral of its almost-everywhere derivative on each such interval. This condition matters: merely differentiating \(1-e^{-H(t)}\) at points where \(H'\) exists does not establish that this function has a density.

For a proper lifetime, \(S(t)\) decreases to zero as time tends to infinity, or approaches zero at a finite endpoint. Consequently, \(H(t)\) tends to infinity along the lifetime’s time range. When \(T\) has a continuous distribution, its survival probability evaluated at the observed lifetime is uniform on \((0,1)\). That fact gives the exponential transformation.

The Exponential Transformation

To justify the first statement, write \(F(t)=P(T\leq t)\). Since \(F\) is continuous, \(F(T)\) is uniform on \((0,1)\). One way to see this is to fix \(0<u<1\) and consider the rightmost point \(b\) up to which \(F(t)\leq u\). Continuity gives \(F(b)=u\), and \(P(F(T)\leq u)=P(T\leq b)=F(b)=u\). Flat stretches of \(F\) do not change this probability. Thus \(S(T)=1-F(T)\) is also uniform on \((0,1)\), and for \(y\geq0\):

This is the survival function of a unit exponential random variable, proving the transformation claim. For the density statement, \(F(t)=1-S(t)=1-e^{-H(t)}\). On each bounded interval, local absolute continuity of \(H\) implies absolute continuity of \(F\); the chain rule for absolutely continuous functions gives \(F'(t)=h(t)e^{-H(t)}\) almost everywhere. Because \(F\) is absolutely continuous, this derivative is a density, not merely a formula for a derivative at selected points. \(\square\)

Generating a Lifetime from a Hazard Function

Suppose a nonnegative hazard function \(h\) is specified, with \(h\geq0\) and locally integrable. Define \(H(t)=\int_0^t h(u)\,du\). If \(H(t)\to\infty\), it covers every nonnegative cumulative-hazard value, though it need not be strictly increasing. Let \(E\sim\operatorname{Exp}(1)\) and define the generalized inverse \(H^{-1}(e)=\inf\{t\geq0:H(t)\geq e\}\).

For any \(t\geq0\), continuity and monotonicity imply \(T\leq t\) exactly when \(E\leq H(t)\). Indeed, if \(E\leq H(t)\), the set in the generalized inverse includes \(t\); conversely, if the first time \(H\) reaches \(E\) is at or before \(t\), then \(H(t)\geq E\). Therefore:

Continuity ensures that the first time \(H\) reaches \(E\) has cumulative-hazard value \(E\), so \(H(T)=E\). If \(H\) is locally absolutely continuous, differentiating the survival function \(e^{-H(t)}\) gives the density \(h(t)e^{-H(t)}\) almost everywhere. Local absolute continuity ensures that this derivative accounts for the entire distributional increase. \(\square\)

This gives a general simulation procedure: draw a unit exponential variable, then solve \(H(t)=E\), using the generalized inverse if \(H\) has flat stretches. If \(H\) has a convenient algebraic inverse, simulation is direct. Otherwise, a monotone numerical root-finding method can be used.

Why Absolute Continuity Matters

The exponential transformation itself is a statement about the survival probability evaluated at a continuous lifetime. But a hazard function specified as an ordinary rate is commonly used through an integral, \(H(t)=\int_0^t h(u)\,du\). To recover a density by differentiating \(1-e^{-H(t)}\), \(H\) must be locally absolutely continuous.

A continuous, nondecreasing \(H\) need not be absolutely continuous. For example, a Cantor-function increase on \([0,1]\) has derivative zero almost everywhere but still increases across that interval. If one used that \(H\) in \(F(t)=1-e^{-H(t)}\), the resulting distribution would have a singular increase there. The expression \(h(t)e^{-H(t)}\), with \(h=H'=0\) almost everywhere on that interval, would miss that probability. Local absolute continuity rules out this failure by ensuring the whole increase is represented by the integral of the derivative.

A related practical mistake is to treat \(h(t)\) as though it were the density. The density is \(h(t)S(t)=h(t)e^{-H(t)}\), not \(h(t)\) alone: the hazard rate is multiplied by the probability of surviving to time \(t\). The cumulative-hazard transform is useful precisely because it keeps this distinction clear while providing a standard exponential scale for calculation and simulation.