The Cumulative Hazard as a Survival Summary

Why Summarize Survival With Cumulative Hazard?

Survival probability answers a direct question: what proportion of a population is expected to remain event-free beyond time $t$? The hazard describes something different: the instantaneous event rate among individuals who have remained event-free up to a given time. Integrating that rate gives the cumulative hazard, a summary of how hazard has accumulated over follow-up.

This distinction matters in survival data used in analyses involving incomplete covariates. Individuals can have the same event status at the end of observation but very different observed follow-up, and an accumulated summary of event experience can help describe that difference. The cumulative hazard is useful only when its interpretation is kept separate from the probability of having experienced an event.

Definition: Let $T$ be a nonnegative event time and let $S(t)=P(T>t)$ be its survival function. On an interval where the event-time distribution is absolutely continuous and $S(t)>0$, the hazard is $h(t)=-S'(t)/S(t)$, defined almost everywhere. The cumulative hazard from time 0 to time $t$ is $H(t)=\int_0^t h(u)\,du$. It accumulates the hazard rate over time and is dimensionless.

A hazard has units of inverse time: for example, per year. Integrating a rate measured per year over an interval measured in years yields a dimensionless cumulative hazard. Neither the hazard nor the cumulative hazard is automatically a probability. In particular, $H(t)$ can exceed 1.

Assumptions and the Survival Relationship

The familiar relationship between survival and cumulative hazard requires regularity. For the interval under consideration, assume the event-time distribution is absolutely continuous, $S(0)=1$, and $S(t)>0$. Absolute continuity ensures that changes in survival on the interval are represented by a density, rather than by a singular component invisible to an almost-everywhere derivative. Define the hazard by $h(t)=-S'(t)/S(t)$ wherever the derivative exists, and assume it is integrable on the interval.

Theorem: Under these conditions, for every $t$ in the interval, $$ S(t)=\exp\{-H(t)\} =\exp\left\{-\int_0^t h(u)\,du\right\}. $$ Equivalently, $H(t)=-\log S(t)$.

The result follows by differentiating the logarithm of survival and integrating. Since $S(t)>0$, $\log S(t)$ is defined. Absolute continuity permits the derivative and integration argument on the interval.

\frac{d}{dt}\log S(t) =\frac{S'(t)}{S(t)} =-h(t) \quad\text{almost everywhere}.

Integrating both sides from 0 to $t$ gives

\log S(t)-\log S(0) =-\int_0^t h(u)\,du =-H(t).

Because $S(0)=1$, $\log S(0)=0$, so $\log S(t)=-H(t)$. Exponentiating yields $S(t)=\exp\{-H(t)\}$. This derivation also shows why merely saying that event times are continuous is not enough: a continuous distribution can include a singular component for which the ordinary derivative does not capture all decreases in survival.

For times $0\leq a<b$ with $S(a)>0$, the probability of an event in $(a,b]$, conditional on being event-free at $a$, is

P(a<T\leq b\mid T>a) =1-\frac{S(b)}{S(a)} =1-\exp\{-[H(b)-H(a)]\}.

The condition $S(a)>0$ is essential because the conditional probability and survival ratio are undefined if nobody can remain event-free at $a$. The formula follows from $P(T>b\mid T>a)=S(b)/S(a)$ and the survival–cumulative-hazard relationship. It applies when the assumptions above hold over the interval.

Worked Interpretations

Worked Example: Accumulating a Piecewise-Constant Hazard

Suppose, in a hypothetical population, a continuous-time model specifies a hazard of $0.040$ per year from year 0 to year 2, and $0.100$ per year from year 2 to year 5. Assume the event-time distribution is absolutely continuous, no one has an event at time 0, and survival is positive through year 5. Find the cumulative hazard and survival at years 2 and 5, and the conditional event probability from year 2 through year 5.

The cumulative hazard is the area under the hazard curve. By year 2,

H(2)=(0.040\ \text{per year})(2\ \text{years})=0.0800.

Thus $S(2)=\exp(-0.0800)=0.9231$. From year 2 to year 5, the additional cumulative hazard is

(0.100\ \text{per year})(5-2\ \text{years})=0.300.

Therefore $H(5)=0.0800+0.300=0.380$, and $S(5)=\exp(-0.380)=0.6839$. The probability of an event during years 2 to 5, conditional on being event-free at year 2, is

1-\exp\{-[H(5)-H(2)]\} =1-\exp(-0.300) =0.2592.

Under this hypothetical model, 25.92% of those who are still event-free at year 2 are expected to experience the event by year 5. This is a conditional interval risk, not the fraction of the original population experiencing the event in that interval.

Worked Example: Recovering Cumulative Hazard From Survival

Suppose a hypothetical, absolutely continuous survival model has $S(4)=0.800$ and $S(8)=0.500$, with positive survival through year 8. Calculate the cumulative hazards at years 4 and 8, then find the conditional event probability from year 4 to year 8.

Using $H(t)=-\log S(t)$,

H(4)=-\log(0.800)=0.2231, \qquad H(8)=-\log(0.500)=0.6931.

The increase in cumulative hazard over the interval is $0.6931-0.2231=0.4700$. The corresponding conditional event probability is

1-\exp(-0.4700)=0.3750.

Checking directly from the survival values gives $1-S(8)/S(4)=1-0.500/0.800=0.3750$. Thus, among those event-free at year 4, 37.50% are expected to experience the event by year 8 under this model. The two calculations agree because both describe the same conditional probability.

Worked Example: A Cumulative Hazard Greater Than One

For a hypothetical population, suppose $H(15)=1.200$ by year 15 under the continuous-time assumptions above. Calculate survival and the probability of an event by year 15.

The survival probability is

S(15)=\exp(-1.200)=0.3012.

Since $P(T\leq 15)=1-S(15)$, the event probability by year 15 is

1-\exp(-1.200)=1-0.3012=0.6988.

So the model implies about 30.12% survival beyond year 15 and a 69.88% probability of an event by then. A cumulative hazard of 1.200 is not a 120% event probability; the exponential transformation is what maps cumulative hazard to survival.

How to Read and Check a Cumulative-Hazard Summary

Cumulative hazard is nondecreasing because it is the integral of a nonnegative hazard. Under the stated continuous-time assumptions, survival is nonincreasing and equals $\exp\{-H(t)\}$. These give basic checks on a reported summary: a decreasing cumulative hazard or an increasing implied survival is inconsistent with a nonnegative hazard.

The size of $H(t)$ is interpretable through survival, not as a direct event fraction. For example, a change $\Delta H=0.200$ corresponds to conditional interval event probability $1-\exp(-0.200)=0.1813$, or 18.13%, for people who were event-free at the start of that interval. If the starting survival is $S(a)$, the probability of experiencing the event in that interval measured as a fraction of the original population is $S(a)\{1-\exp(-\Delta H)\}$. The latter is smaller when fewer individuals remain event-free at the interval's start.

In survival-data work, the summary should be checked against the time scale, event definition, and follow-up represented by the data. A hazard expressed per day cannot be integrated as though it were per year without converting units. A cumulative-hazard curve that falls, or a conversion to survival that exceeds 1, signals a calculation or interpretation problem. When survival is estimated from observed data and has jumps, the smooth, absolutely continuous derivation above may not apply literally; jump-aware survival relationships are needed instead.

Quantity	Meaning	Typical units or range
$h(t)$	Instantaneous event rate among those still event-free at time $t$	Inverse time, such as per year; nonnegative
$H(t)$	Hazard accumulated from the start of follow-up to $t$	Dimensionless; nonnegative and nondecreasing
$S(t)$	Probability of remaining event-free beyond $t$	Probability from 0 to 1
$1-S(t)$	Probability of experiencing the event by $t$, in a single-event setting	Probability from 0 to 1

Common Mistakes and Limitations

Treating cumulative hazard as cumulative probability. A value such as $H(t)=1.200$ is not a 120% event risk. Convert using $S(t)=\exp\{-H(t)\}$, then use $1-S(t)$ for event probability by $t$.
Confusing conditional interval risk with risk from the start. The expression $1-\exp\{-\Delta H\}$ conditions on being event-free at the start of the interval. It is not generally the proportion of the original cohort having an event in that interval.
Applying the continuous-time identity without its assumptions. If the event-time distribution has jumps or a singular component, the ordinary hazard integral may not capture every change in survival. Check whether an absolutely continuous model is appropriate before applying the identity.
Ignoring zero survival. A conditional risk given $T>a$ is not defined when $S(a)=0$. The survival ratio requires a positive denominator.

Cumulative hazard is a compact survival summary, not a complete description of every feature of event-time data. It does not, by itself, encode the event indicator, covariate relationships, or the mechanisms that produced censoring. In settings with multiple types of event, a single-event interpretation also cannot simply be applied to each type as though the others did not occur. The event structure and analysis target determine what additional summaries are needed.

Key takeaway: Under an absolutely continuous event-time model with positive survival, cumulative hazard is the integrated hazard and $S(t)=\exp\{-H(t)\}$. Use it to translate accumulated rate into survival or conditional interval risk; do not read it as a probability.

Check Your Understanding

Answer each question, then compare with the solutions below.

A hypothetical model has $H(3)=0.400$. What is $S(3)$, approximately, and what is the event probability by year 3?
For an interval $(a,b]$, the cumulative hazard increases by $0.100$. What is the conditional event probability in the interval among individuals event-free at $a$?
Why is the formula $P(a<T\leq b\mid T>a)=1-S(b)/S(a)$ not usable when $S(a)=0$?
Name a condition needed for $S(t)=\exp\{-H(t)\}$, and explain briefly why continuity alone may not be sufficient.

Solutions. (1) $S(3)=\exp(-0.400)=0.6703$, so the event probability by year 3 is $1-0.6703=0.3297$, or 32.97%. (2) $1-\exp(-0.100)=0.09516$, so the conditional interval risk is about 9.516%. (3) The ratio has denominator zero, and conditioning on $T>a$ is undefined if that event has probability zero. (4) The event-time distribution must be absolutely continuous on the interval, with positive survival, and the hazard must be integrable. Continuity alone can allow a singular component whose survival decrease is not represented by the ordinary derivative almost everywhere.

Quantity	Meaning	Typical units or range
\(h(t)\)	Instantaneous event rate among those still event-free at time \(t\)	Inverse time, such as per year; nonnegative
\(H(t)\)	Hazard accumulated from the start of follow-up to \(t\)	Dimensionless; nonnegative and nondecreasing
\(S(t)\)	Probability of remaining event-free beyond \(t\)	Probability from 0 to 1
\(1-S(t)\)	Probability of experiencing the event by \(t\), in a single-event setting	Probability from 0 to 1