Hazard Functions and Reliability Event Counts

Hazard, Survival, and Event Counts

A hazard rate describes how quickly failure risk accrues among units that have survived so far. A time-varying hazard can be converted into a survival probability by accumulating that rate over time. The same exponential calculation also appears in a nonhomogeneous Poisson process: its cumulative intensity determines the probability of no events and the distribution of the total event count. The formulas look alike, but the models answer different questions. A single lifetime ends at its first failure; a recurrent-event process can record multiple events.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \(T\) be a nonnegative lifetime, with survival function \(S(t)=\mathbb{P}(T>t)\). Suppose \(T\) has a density \(f\), and \(S(t)>0\) on the time interval under consideration. Its hazard rate is \(h(t)=f(t)/S(t)\), where the density and hazard may be specified up to changes on sets of measure zero. Informally, \(h(t)\) is the instantaneous failure rate conditional on surviving to time \(t\): when the relevant conditional probability is defined, \(h(t)\) is the limit of \(\mathbb{P}(t<T\le t+\Delta\mid T>t)/\Delta\) as \(\Delta\) decreases to zero.

The cumulative hazard is not itself a probability: it can exceed \(1\). Instead, it is the negative logarithm of survival. The formula requires an absolutely continuous lifetime with positive survival on the interval. If a lifetime has an atom, such as a positive probability of failure at an exact time, the density-based hazard formula needs adjustment; it should not be applied across that atom as if the distribution were continuous.

Proof. For times where \(S(t)>0\), absolute continuity gives \(S'(t)=-f(t)\) almost everywhere. Thus \(S'(t)/S(t)=-f(t)/S(t)=-h(t)\) almost everywhere. Integrating from \(0\) to \(t\), using \(S(0)=1\) when there is no failure at time zero, gives \(\log S(t)-\log S(0)=-\int_0^t h(u)\,du\). Since \(\log S(0)=0\), exponentiating yields \(S(t)=\exp(-H(t))\). The failure probability is \(\mathbb{P}(T\le t)=1-S(t)\), which gives the stated expression. \(\square\)

For an interval \((a,b]\), conditional failure risk among units still surviving at \(a\) follows by taking a survival ratio. Provided \(S(a)>0\),

The integrated hazard over the interval, not the hazard at its endpoint, determines this conditional probability.

Weibull Hazards

For a Weibull lifetime with shape \(\alpha>0\) and scale \(\beta>0\), the survival function and hazard are

Integrating the hazard gives \(H(t)=(t/\beta)^\alpha\). If \(\alpha>1\), the hazard increases with time; if \(\alpha=1\), it is constant; and if \(0<\alpha<1\), it decreases for \(t>0\). In the last case it may be unbounded near zero, but its integral over a finite interval remains finite. The survival formula still applies.

From Hazard to a Nonhomogeneous Poisson Count

Now consider a different model. Let \((N(t))_{t\ge0}\) be a counting process on \((\Omega,\mathcal{F},\mathbb{P})\), with state space the nonnegative integers. A nonhomogeneous Poisson process with deterministic intensity \(\lambda(t)\ge0\) starts at \(N(0)=0\), has independent increments, and has \(N(b)-N(a)\) Poisson distributed with mean \(\int_a^b\lambda(u)\,du\) for \(0\le a<b\), assuming the intensity is integrable on finite intervals. Write \(\Lambda(t)=\int_0^t\lambda(u)\,du\) for its cumulative intensity.

Proof. By the definition of the process, \(N(t)-N(0)\) has a Poisson distribution with mean \(\int_0^t\lambda(u)\,du=\Lambda(t)\). Since \(N(0)=0\), \(N(t)\) has that same distribution. A Poisson random variable with mean \(\Lambda(t)\) has probability \(e^{-\Lambda(t)}\) of equaling zero and expectation \(\Lambda(t)\). On \((a,b]\), the increment is Poisson with mean \(\int_a^b\lambda(u)\,du\); the probability of at least one event is one minus its zero-count probability. \(\square\)

When \(\lambda(t)=h(t)\), the no-event probability through time \(t\) is \(e^{-H(t)}\), the same formula as lifetime survival. This is a useful link, not an identity between the models. In a lifetime model, \(T>t\) means that one unit has not yet failed. In a recurrent-event model, \(N(t)\) may be \(0,1,2,\ldots\); its count distribution describes events under a Poisson-process assumption. The interpretation of the intensity must come from the application.

Piecewise-Constant Hazards

A piecewise-constant hazard is convenient when the rate changes at known times. If the hazard is \(c_r\) on each interval \((a_{r-1},a_r]\), the cumulative hazard through an endpoint is the sum of each interval’s rate multiplied by its duration. The same integration gives cumulative intensity for a Poisson count model with that time-varying rate. Care is needed at change points: each rate applies only over its own interval.

Common Pitfall and Interpretation

A common error is to treat cumulative hazard \(H(t)\) as a failure probability. For the piecewise example, \(H(5)=2\); using \(2\) as a probability is impossible because probabilities cannot exceed \(1\). The correct lifetime failure probability is \(1-e^{-2}\), while the survival probability is \(e^{-2}\). The exponential transformation matters even when \(H(t)\) is less than \(1\): in general, \(1-e^{-H(t)}\) is not equal to \(H(t)\).

Another error is to apply the single-lifetime interpretation to a recurrent-event count without stating a model. A survival function concerns the first failure of one unit. A nonhomogeneous Poisson process permits repeated events and additionally assumes independent increments with Poisson-distributed interval counts. If the application instead has dependence between failures, repair effects, or a changing population at risk, the Poisson count formula may not describe it. The shared cumulative-rate calculation is valid only after the relevant model and its assumptions have been specified.