Hazard And Cumulative Hazard Relationships

Why Hazard and Cumulative Hazard Are Useful

A survival function describes the chance that an event time exceeds a duration. A hazard describes something different: the event rate at a particular duration among units that have lasted at least that long. This conditional, time-local view is useful when the question is not only whether events occur, but how the event rate changes as duration passes.

For example, a constant hazard means the instantaneous event rate does not depend on elapsed duration. An increasing hazard means that rate is higher at longer durations; a decreasing hazard means it is lower. The cumulative hazard connects this local rate to overall survival: it adds up hazard over time and, through an exponential transformation, determines the survival probability.

Definitions and Conditions

Let \(T\) be a nonnegative continuous event time, such as time until a hypothetical machine fails. Define the survival function by \(S(t)=P(T>t)\), and let \(f(t)\) be the probability density of \(T\). For durations where \(S(t)>0\), the hazard is the density relative to the probability of having survived to \(t\).

Definition: For a continuous event time with density \(f(t)\) and survival function \(S(t)>0\), the hazard function is \(h(t)=f(t)/S(t)\). Equivalently, it is the limiting event probability per unit time among units that have survived to \(t\): \(h(t)=\lim_{\Delta t\to 0^+}P(t<T\le t+\Delta t\mid T>t)/\Delta t\). The cumulative hazard from the origin to \(t\) is \(H(t)=\int_0^t h(u)\,du\).

The hazard is a rate, not a probability. If time is measured in months, \(h(t)\) has units of inverse months; for a small interval \(\Delta t\), the conditional probability of an event in that interval is approximately \(h(t)\Delta t\). A hazard can exceed 1 when expressed in a chosen time unit, whereas a probability cannot. The approximation is local and becomes accurate as the interval becomes short.

The definitions and relationship below use a continuous event-time distribution with no point mass at the origin, so \(S(0)=1\), and an absolutely continuous survival function over the durations considered. These conditions ensure that changes in survival are described by a density. In applications, the hazard is interpreted among units still at risk at each duration; it is not a statement that a particular unit's event is certain or that elapsed time causes the event.

Deriving the Survival–Cumulative Hazard Relationship

Because \(S(t)=P(T>t)\), its derivative is \(S'(t)=-f(t)\) wherever the density and derivative exist. From the definition \(h(t)=f(t)/S(t)\), it follows that \(S'(t)=-h(t)S(t)\). Dividing by the positive \(S(t)\) gives a differential equation for the logarithm of survival.

\frac{d}{dt}\log S(t) =\frac{S'(t)}{S(t)} =-\frac{f(t)}{S(t)} =-h(t).

Integrate both sides from 0 to \(t\). Since \(S(0)=1\), \(\log S(0)=0\), and the result is \(\log S(t)=-\int_0^t h(u)\,du=-H(t)\). Exponentiating yields the central relationship.

Theorem: For a continuous event time satisfying the conditions above, \(S(t)=\exp[-H(t)]\), where \(H(t)=\int_0^t h(u)\,du\). Conversely, \(H(t)=-\log S(t)\), and where derivatives exist, \(h(t)=-\frac{d}{dt}\log S(t)=-S'(t)/S(t)\).

The derivation also explains why cumulative hazard is not itself an event probability. It accumulates instantaneous rates on the log-survival scale. The probability of an event by \(t\) is instead the cumulative distribution \(F(t)=P(T\le t)=1-S(t)=1-\exp[-H(t)]\).

Worked Example: Constant Hazard

Suppose, hypothetically, that a device has a constant failure hazard of \(0.08\) per month. Find its cumulative hazard and probability of surviving three months, and interpret the result.

A constant hazard \(h(t)=0.08\) gives \(H(t)=\int_0^t 0.08\,du=0.08t\). At three months,

H(3)=0.08(3)=0.24, \qquad S(3)=\exp[-H(3)]=e^{-0.24}\approx 0.7866.

Thus the hypothetical failure probability by three months is \(1-0.7866=0.2134\), or about 21.34%. The hazard is a rate of \(0.08\) per month at each duration, not a failure probability of 8% over every month regardless of survival. In a short interval of length \(\Delta t\), the conditional failure probability is approximately \(0.08\Delta t\).

Conditional Survival Over an Interval

The same relationship gives survival over a future interval among units that have already survived to its start. For \(t\ge a\) with \(S(a)>0\), conditional probability rules imply \(P(T>t\mid T>a)=S(t)/S(a)\). Substituting the exponential representation of survival gives the accumulated hazard between \(a\) and \(t\).

P(T>t\mid T>a) =\frac{S(t)}{S(a)} =\exp\left[-\int_a^t h(u)\,du\right] =\exp[-\{H(t)-H(a)\}].

This result is useful for interpreting duration dependence. If the hazard changes with elapsed time, the conditional chance of surviving a future interval depends on the duration already reached through the hazard values over that interval. A constant hazard has a special feature: the conditional survival probability over an interval of a given length does not depend on its starting duration.

Worked Example: Increasing Hazard

Consider a hypothetical process with hazard \(h(t)=0.02t\) per year, where \(t\) is measured in years. Find the cumulative hazard and survival at five years, and compare the hazard at two and five years.

The coefficient \(0.02\) has units of inverse years squared, so \(h(t)\) has units of inverse years. Integrating from zero gives

H(t)=\int_0^t 0.02u\,du=0.01t^2, \qquad H(5)=0.01(5^2)=0.25.

Therefore,

S(5)=e^{-0.25}\approx 0.7788, \qquad P(T\le 5)=1-0.7788=0.2212.

At two years the hazard is \(h(2)=0.02(2)=0.04\) per year; at five years it is \(h(5)=0.02(5)=0.10\) per year. The hazard rises with elapsed duration, so this is positive duration dependence in the descriptive sense: among units still at risk, the instantaneous event rate is higher at longer durations. It does not by itself explain why the rate rises or establish a causal effect of time.

Constant, Increasing, and Decreasing Hazards

A hazard's pattern describes how the conditional event rate varies with duration. For a differentiable hazard, the sign of \(h'(t)\) indicates its local direction of change: a positive value indicates a locally increasing hazard, and a negative value indicates a locally decreasing hazard. A zero derivative at a point means there is no first-order change at that point; the hazard is locally constant only if its derivative is zero throughout a neighborhood.

The broad patterns have direct cumulative-hazard implications. A constant hazard accumulates at a constant rate, so \(H(t)\) is linear in \(t\). An increasing hazard makes \(H(t)\) rise at an increasing rate; a decreasing hazard makes \(H(t)\) rise at a decreasing rate, while it remains nondecreasing as long as \(h(t)\ge 0\). The relationships are summarized here.

Hazard pattern	Meaning among units still at risk	Implication for cumulative hazard
Constant	The instantaneous event rate is unchanged with duration.	\(H(t)\) increases linearly.
Increasing	The instantaneous event rate is higher at longer durations.	The slope of \(H(t)\) increases.
Decreasing	The instantaneous event rate is lower at longer durations.	The slope of \(H(t)\) decreases, though \(H(t)\) does not decrease.

Worked Example: Decreasing Hazard

Suppose, hypothetically, that an event process has hazard \(h(t)=0.04+0.08e^{-0.3t}\) per month, with \(t\) measured in months. Find the cumulative hazard and survival at four months, and check whether the hazard is decreasing.

Integrating each term gives \(H(t)=0.04t+\frac{0.08}{0.3}(1-e^{-0.3t})\). At four months, \(e^{-1.2}\approx0.3012\), so

H(4)=0.04(4)+\frac{0.08}{0.3}(1-e^{-1.2}) \approx 0.1600+0.1863=0.3463.

It follows that

S(4)=e^{-0.346348\ldots}\approx 0.7073, \qquad P(T\le 4)=1-0.7073=0.2927.

The hazard at four months is \(h(4)=0.04+0.08e^{-1.2}\approx0.0641\) per month, compared with \(h(0)=0.12\) per month. Differentiating gives \(h'(t)=-0.024e^{-0.3t}<0\) for all \(t\ge0\), so the hazard is decreasing throughout this range. It approaches \(0.04\) per month rather than zero; the decreasing pattern does not mean the event rate vanishes.

Reading Hazard Patterns Carefully

Duration dependence is a description of conditional rates, not necessarily a mechanism. An increasing hazard may arise because risk rises with exposure time, but it may also reflect changes in who remains at risk: units with lower underlying risk can survive longer, leaving a group with a different composition. A decreasing hazard can likewise reflect mechanisms, selection among survivors, or both. The hazard pattern alone does not distinguish these explanations.

Nor should the shape of a hazard be confused with the direction of survival. Whenever \(h(t)\ge0\), cumulative hazard \(H(t)\) is nondecreasing and survival \(S(t)=e^{-H(t)}\) is nonincreasing. A decreasing hazard therefore does not mean that survival is increasing; it means that survival continues to fall, but the conditional event rate becomes smaller as duration grows.

The total amount of cumulative hazard also matters. If \(H(t)\) grows without bound as \(t\) grows, then \(S(t)\) tends to zero. If instead \(H(t)\) approaches a finite limit \(H_\infty\), the formula gives \(\lim_{t\to\infty}S(t)=e^{-H_\infty}>0\). In a model that allows individuals never to experience the event, this positive limiting survival represents a nonzero probability of no event. A decreasing hazard does not automatically imply such a limit; the integral of the hazard determines it.

Common Mistakes

Treating a hazard as a probability. A hazard is an instantaneous rate with units of inverse time. For a short interval, the conditional probability is approximately the rate times the interval length; the rate itself is not the probability of an event over a whole period.
Calling cumulative hazard a cumulative probability. \(H(t)\) can exceed 1. The event probability by \(t\) is \(1-e^{-H(t)}\), not \(H(t)\).
Assuming falling hazard means rising survival. If the hazard remains nonnegative, \(H(t)\) still increases and survival still decreases. Only its rate of decline changes.
Overinterpreting a zero derivative. \(h'(t)=0\) at one point means no first-order change at that point, not that the hazard is constant nearby. Local constancy requires the derivative to be zero throughout a neighborhood.
Reading a duration pattern as a causal conclusion. A rising or falling hazard describes event rates among survivors at different durations. It does not, by itself, identify why those rates differ.

Key takeaway: The hazard is the instantaneous event rate conditional on survival to a duration; the cumulative hazard integrates that rate. Under the stated continuous-time conditions, \(S(t)=e^{-H(t)}\). The hazard's shape describes how conditional event rates vary with duration, while cumulative hazard determines the resulting survival.

Check Your Understanding

Answer each question using the definitions and relationships developed above.

A hypothetical process has constant hazard \(0.05\) per week. Write its cumulative hazard and survival function, and find the probability of surviving four weeks.
Explain why a hazard of \(1.2\) per month is not an invalid probability, and state what units it has.
If \(H(3)=0.4\), calculate the survival probability at duration 3 and the event probability by duration 3.
A differentiable hazard has \(h'(t)=0\) at one duration. What can be concluded about its first-order change there, and what additional condition would establish local constancy?
Give the conditional survival probability from duration \(a\) to duration \(t\) in terms of \(H\), and explain why a decreasing hazard does not mean survival is increasing.