Hazard and Cumulative Hazard

Why Hazard Is Not an Event Probability

The survival function describes the probability that an event time exceeds a specified time. The hazard answers a different question: among subjects who have not yet experienced the event, how rapidly are events occurring at a particular time? This distinction matters when describing event risk over time. A hazard is an instantaneous rate, not a probability that an event occurs at that exact instant.

The survival function and event-time quantiles provide probabilities over time intervals or up to a time point. Hazard and cumulative hazard give complementary descriptions: the hazard is local in time, while the cumulative hazard adds up that local rate over time. The relationships among these quantities let us translate between a rate and survival probabilities.

Definition: Let $T$ be a nonnegative, continuous event time, and let $S(t)=P(T>t)$ be its survival function. At a time $t$ with $S(t)>0$, the hazard is $$ \lambda(t)=\lim_{\Delta t\to 0^+}\frac{P(t\leq T<t+\Delta t\mid T\geq t)}{\Delta t}. $$ When $T$ has a density $f(t)$ that is continuous at $t$, this limit equals $\lambda(t)=f(t)/S(t)$. The cumulative hazard from time 0 to $t$ is $H(t)=\int_0^t \lambda(u)\,du$, when this integral exists.

The conditioning in the hazard definition is essential: it concerns individuals still event-free at time $t$. It is not the probability of an event among the original population, nor is it an unconditional event rate. For continuous event times, whether the condition is written $T\geq t$ or $T>t$ does not affect the probability at a single time point. We assume here that there is no point mass at time 0 and that the event-time distribution is continuous over the intervals considered.

Units and Interpretations

If time is measured in years, $\lambda(t)$ has units of events per person-at-risk per year, often shortened in a model description to “per year.” It is a rate, so it can be numerically greater than 1 per year without being an impossible probability. In contrast, $H(t)$ is dimensionless: integrating a rate per year over time in years cancels the time units. Neither $\lambda(t)$ nor $H(t)$ is itself an event probability.

The cumulative hazard $H(t)$ is nondecreasing because the hazard is nonnegative. Its increment between two times, $H(t_2)-H(t_1)$, summarizes the accumulated hazard over that interval. In a continuous-time model, the conditional probability of surviving the interval and the conditional probability of an event in it are determined by that increment:

P(T>t_2\mid T>t_1)=\exp\{-[H(t_2)-H(t_1)]\}, \qquad P(t_1<T\leq t_2\mid T>t_1)=1-\exp\{-[H(t_2)-H(t_1)]\}.

These formulas apply for $t_2>t_1$ when the event-time distribution is continuous and the conditioning event has positive probability. The interval event probability depends on the cumulative-hazard increment, not on the hazard at one endpoint alone. A larger increment means a smaller conditional probability of surviving the interval.

Worked Example: Constant Hazard Over Five Years

Suppose a hypothetical event-time model has a constant hazard of $0.12$ per year. Assume the model is continuous and applies from time 0. Find the cumulative hazard by year 5, survival through year 5, and the conditional event probability from year 3 to year 5 among those event-free at year 3.

Because the hazard is constant, the cumulative hazard is its value multiplied by elapsed time:

H(5)=\int_0^5 0.12\,du=0.12(5)=0.600.

Thus survival through year 5 is $\exp[-H(5)]=\exp(-0.600)\approx 0.5488$. The probability of an event by year 5 is $1-0.5488=0.4512$, or about 45.12%.

For the interval from year 3 to year 5, the cumulative-hazard increment is $0.12(5-3)=0.240$. Therefore the conditional event probability is

P(3<T\leq 5\mid T>3)=1-\exp(-0.240)\approx 0.2134.

Among subjects who are event-free at year 3, the model assigns about a 21.34% probability of an event by year 5. This is not the unconditional probability of an event between years 3 and 5 in the original population.

Deriving the Relationships

For a continuous event time with density $f(t)$, the probability of surviving beyond $t$ changes at rate $S'(t)=-f(t)$. Since $\lambda(t)=f(t)/S(t)$ wherever $S(t)>0$, the derivative of the logarithm of survival is

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

Integrating from 0 to $t$, and using $S(0)=1$ under the no-event-at-zero assumption, gives

\log S(t)-\log S(0)=-\int_0^t\lambda(u)\,du, \qquad S(t)=\exp[-H(t)].

Taking logarithms also gives $H(t)=-\log S(t)$. Subtracting this identity at $t_1$ from the identity at $t_2$ yields $S(t_2)/S(t_1)=\exp\{-[H(t_2)-H(t_1)]\}$. That ratio is the conditional survival probability from $t_1$ to $t_2$, which gives the interval probability formulas above by taking its complement.

This derivation requires a continuous event-time distribution with a well-defined hazard and survival differentiable over the interval. If event times can have point masses, such as a non-negligible probability of an event at one exact time, the continuous hazard formulas need adjustment at those times. A discrete-time hazard is defined instead as a conditional probability for an interval, not as an instantaneous rate.

Worked Example: A Hazard That Increases With Time

Suppose a hypothetical model has $\lambda(u)=0.04u$, where $u$ is measured in years. The coefficient $0.04$ has units of inverse years squared, so the hazard has units of inverse years. Calculate the cumulative hazard at 5 and 5.2 years, then find the conditional event probability from year 5 to year 5.2 for subjects event-free at year 5.

Integrating the hazard gives $H(t)=\int_0^t0.04u\,du=0.02t^2$. Thus $H(5)=0.02(25)=0.500$, and $H(5.2)=0.02(27.04)=0.5408$. The increment is $0.5408-0.500=0.0408$. Equivalently, evaluating the integral directly:

H(5.2)-H(5)=\int_5^{5.2}0.04u\,du =0.02(5.2^2-5^2) =0.02(27.04-25) =0.0408.

The conditional survival probability over this interval is $\exp(-0.0408)\approx 0.96002$. Therefore the conditional event probability is $1-\exp(-0.0408)\approx 0.03998$, or about 3.998%. This is a short-interval probability conditional on being event-free at year 5. The calculation uses the cumulative-hazard increment, not simply the hazard value at year 5.

Worked Example: Recovering Hazard From a Survival Model

Suppose a hypothetical model specifies survival, for time $t$ in years, as $S(t)=\exp(-0.03t-0.002t^2)$. Find the cumulative hazard and hazard, then calculate the conditional event probability from year 4 to year 6.

Taking the negative logarithm of the specified survival function gives $H(t)=0.03t+0.002t^2$. Differentiating the cumulative hazard gives $\lambda(t)=0.03+0.004t$ per year. The coefficients have units $0.03$ per year and $0.002$ per year squared, respectively, so each term in $H(t)$ is dimensionless.

At year 4, $H(4)=0.03(4)+0.002(16)=0.152$. At year 6, $H(6)=0.03(6)+0.002(36)=0.252$. Hence the interval increment is $0.252-0.152=0.100$, and

P(4<T\leq 6\mid T>4)=1-\exp(-0.100)\approx 0.09516.

The conditional event probability over these two years is approximately 9.516%. As a check, integrating the hazard directly gives $0.03(6-4)+0.002(6^2-4^2)=0.060+0.040=0.100$, the same increment.

Choosing and Checking a Summary

Hazard, cumulative hazard, and survival are related but answer distinct questions. The following comparison helps prevent them from being used interchangeably.

Quantity	What it describes	Units
Hazard $\lambda(t)$	Instantaneous event rate among subjects still event-free at $t$	Inverse time
Cumulative hazard $H(t)$	Accumulated hazard from the time origin through $t$	Dimensionless
Survival $S(t)$	Probability of remaining event-free beyond $t$	Probability

Use survival when the question concerns the probability of remaining event-free beyond a time. Use the hazard when the question concerns how the instantaneous event rate changes among those still event-free. Use cumulative hazard to summarize the accumulated rate and to obtain interval probabilities through its increments.

For a proposed continuous-time model, useful checks follow directly from the definitions. The hazard should be nonnegative and have inverse-time units. The cumulative hazard should start at zero under the stated time-origin assumption, be nondecreasing, and have no units. The survival function should lie between zero and one and be nonincreasing. Over any interval, the conditional survival probability and conditional event probability should sum to one.

In applications, an estimated hazard can be unstable, especially where few subjects remain at risk. A sharply jagged hazard estimate may reflect sparse information or smoothing choices rather than genuine rapid changes in event rate. The cumulative hazard is an integrated summary and may appear smoother, but it can still inherit estimation error. These quantities describe event-time patterns; by themselves they do not explain why an event occurs or establish a causal effect.

Common Mistakes and Limitations

Treating a hazard as a probability. A hazard of $0.12$ per year is not automatically a 12% annual event probability. For a constant hazard over one year, the probability is $1-\exp(-0.12)$, approximately 11.31%.
Using the hazard at one time for a whole interval. If the hazard varies, integrate it over the interval to obtain the cumulative-hazard increment. Do not substitute one endpoint value for that integral without a justified approximation.
Calling cumulative hazard a probability. $H(t)$ can exceed 1. The corresponding survival probability is $\exp[-H(t)]$, not $1-H(t)$ in general.
Ignoring who is at risk. Hazard is conditional on being event-free at the time in question. An interval event probability conditional on survival to its start is not an unconditional probability for the original population.

The continuous-time relationships presented here do not directly describe event distributions with point masses. They also rely on the specified time origin and event-free conditioning; changing the origin or conditioning population changes the quantity being described. In data analysis, conclusions depend on whether the model and the available follow-up adequately represent the underlying event process.

Key takeaway: The hazard is an instantaneous conditional rate, the cumulative hazard is its integral, and the survival probability is the exponential of the negative cumulative hazard. For an interval, use the cumulative-hazard increment to calculate conditional survival or event probability.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

A constant hazard is $0.20$ per year. What is the cumulative hazard over 2 years, and what is the conditional event probability over that interval?
Why can a hazard value greater than 1 per year still be valid?
If $H(3)=0.4$ and $H(5)=0.7$, what is the conditional event probability from year 3 to year 5 among subjects event-free at year 3?
For a continuous event-time model, how do you recover the hazard from a differentiable cumulative hazard?

Solutions. (1) The increment is $0.20(2)=0.400$. The conditional event probability is $1-\exp(-0.400)\approx 0.3297$, or 32.97%. (2) A hazard is a rate with inverse-time units, not a probability, so it is not restricted to be at most 1. (3) The increment is $0.7-0.4=0.300$, giving $1-\exp(-0.300)\approx 0.2592$, or 25.92%. (4) Differentiate: $\lambda(t)=dH(t)/dt$, wherever the derivative exists and the continuous-time model applies.

Quantity	What it describes	Units
Hazard \(\lambda(t)\)	Instantaneous event rate among subjects still event-free at \(t\)	Inverse time
Cumulative hazard \(H(t)\)	Accumulated hazard from the time origin through \(t\)	Dimensionless
Survival \(S(t)\)	Probability of remaining event-free beyond \(t\)	Probability

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