Why Three Quantities Describe Event Time
A time-to-event outcome is not simply a yes-or-no response: the time at which an event occurs matters. For example, in a hypothetical PK/PD analysis, the event might be loss of response after dosing, and the scientific question might concern the chance that a patient remains event-free through a specified time. The hazard, survival function, and cumulative hazard give three connected ways to describe that event-time distribution.
The hazard is an instantaneous rate among those still event-free. The survival function gives the probability of remaining event-free beyond a time. The cumulative hazard accumulates the hazard over time. Understanding their mathematical relationship lets us translate a modeled hazard into probabilities that are directly interpretable, while avoiding the mistake of treating a rate as a probability.
Time is measured from a defined origin, and the standard expressions below assume that individuals are event-free at time zero, so \(S(0)=1\). The event-time distribution is continuous over the time range considered, and its hazard is nonnegative and integrable. The hazard has units of inverse time, such as per day; \(H(t)\), \(S(t)\), and \(F(t)\) are dimensionless. If the distribution has point masses, such as events recorded only at discrete time points, the continuous hazard-density relationship requires a different formulation.
From Hazard to Survival
For a continuous event-time distribution, \(S(t)=\int_t^\infty f(u)\,du\), so wherever the density is defined, \(S'(t)=-f(t)\). Since \(f(t)=\lambda(t)S(t)\), survival satisfies a differential equation. This gives a direct derivation of the relationship between hazard and survival.
With \(S(0)=1\), \(\log S(0)=0\), and therefore \(S(t)=\exp[-H(t)]\). Conversely, if \(S(t)>0\) and is differentiable, then \(\lambda(t)=-S'(t)/S(t)\), and \(H(t)=-\log S(t)\). These are not separate assumptions about the same outcome: under the stated conditions, each representation determines the others.
For an interval from \(a\) to \(b\), with \(0\leq a<b\), the probability of an event in that interval among the original population is \(P(a<T\leq b)=S(a)-S(b)\). Among individuals who are event-free at \(a\), the conditional probability of an event by \(b\) is \(P(T\leq b\mid T>a)=1-S(b)/S(a)\). In hazard terms, these become \(e^{-H(a)}-e^{-H(b)}\) and \(1-e^{-[H(b)-H(a)]}\), respectively. The two probabilities answer different questions because they use different denominators.
Worked Example: Constant Hazard and Interval Risk
Suppose, hypothetically, that the event hazard is constant at \(\lambda(t)=0.0800\) per day. Find survival at days 2 and 5, the event probability by day 5, and the probability of an event between days 2 and 5 among (a) the original population and (b) those event-free at day 2.
Model and assumptions. The hazard is nonnegative, constant, and expressed per day. Thus \(H(t)=\int_0^t0.0800\,du=0.0800t\), with time in days, and \(S(t)=e^{-0.0800t}\).
At day 2, \(H(2)=0.160\), so \(S(2)=e^{-0.160}=0.8521\). At day 5, \(H(5)=0.400\), so \(S(5)=e^{-0.400}=0.6703\). The event probability by day 5 is \(1-S(5)=1-0.6703=0.3297\), or 33.0%.
For the original population, the probability of an event in days 2 through 5 is \(S(2)-S(5)=e^{-0.160}-e^{-0.400}=0.8521438-0.6703200=0.1818\), or 18.2%.
Conditional on being event-free at day 2, the probability is \(1-S(5)/S(2)=1-e^{-[H(5)-H(2)]}=1-e^{-0.240}=0.2134\), or 21.3%. This conditional percentage is larger because it is calculated within the group that remained event-free to day 2, not within the original population.
Time-Varying Hazards
When the hazard changes with time, cumulative hazard is still obtained by integration; the constant-hazard formula is a special case, not a general shortcut. If \(\lambda(t)\) is higher during some periods than others, those periods contribute more to \(H(t)\), and hence reduce survival more rapidly. Calculations may use an analytic integral or, when a model has no convenient closed-form integral, a numerical approximation.
Worked Example: A Smoothly Increasing Hazard
Consider a hypothetical hazard that increases linearly over time: \(\lambda(t)=0.0200+0.0100t\), where \(t\) is in days and the hazard is per day. Calculate the hazard at day 3, survival through day 4, and the probability of an event in days 2 through 4, both unconditionally and conditional on being event-free at day 2.
Model and assumptions. The intercept \(0.0200\) has units per day and the slope \(0.0100\) has units per day squared, so both terms in the hazard have units per day. The hazard is nonnegative for \(t\geq0\), is integrable, and the continuous-event-time relationship applies.
At day 3, \(\lambda(3)=0.0200+0.0100(3)=0.0500\) per day. Integrating gives \(H(t)=0.0200t+0.00500t^2\). Thus \(H(2)=0.0400+0.0200=0.0600\), while \(H(4)=0.0800+0.0800=0.160\).
Survival through day 4 is \(S(4)=e^{-0.160}=0.8521\), so the event probability by day 4 is \(1-0.8521=0.1479\), or 14.8%. At day 2, \(S(2)=e^{-0.0600}=0.9418\). Therefore, the unconditional probability of an event in days 2 through 4 is \(S(2)-S(4)=e^{-0.0600}-e^{-0.160}=0.9417645-0.8521438=0.08962\), or 8.96%.
Among those event-free at day 2, the interval probability is \(1-e^{-[H(4)-H(2)]}=1-e^{-0.100}=0.09516\), or 9.52%. The difference between 8.96% and 9.52% is not a discrepancy: the former uses the original population as its denominator, whereas the latter is conditional on survival to day 2.
A time-varying hazard can also be specified in intervals. For a piecewise-constant hazard, integrate each segment and add its contribution. At a boundary, the hazard value at the single time point does not affect the integral or event probabilities under a continuous event-time model.
Worked Example: Accumulating Piecewise-Constant Hazard
Suppose a hypothetical model has hazard \(0.0300\) per day from day 0 to day 2, \(0.0700\) per day from day 2 to day 5, and \(0.0200\) per day after day 5. Find the cumulative hazard and survival at day 5, then calculate the probability of an event in days 2 through 5 for the original population and for those event-free at day 2.
Model and assumptions. Each segment has a nonnegative constant hazard, and the distribution is continuous. The cumulative hazard is the sum of each segment's hazard multiplied by its duration.
From day 0 to day 2, the contribution is \(0.0300(2)=0.0600\). From day 2 to day 5, it is \(0.0700(3)=0.210\). Therefore, \(H(5)=0.0600+0.210=0.270\), and \(S(5)=e^{-0.270}=0.7634\). The event probability by day 5 is \(1-0.7634=0.2366\), or 23.7%.
At day 2, \(H(2)=0.0600\) and \(S(2)=e^{-0.0600}=0.9418\). The original-population probability of an event in days 2 through 5 is \(S(2)-S(5)=e^{-0.0600}-e^{-0.270}=0.9417645-0.7633795=0.1784\), or 17.8%. Conditional on being event-free at day 2, it is \(1-e^{-0.210}=0.1894\), or 18.9%.
Only the hazard during days 2 through 5 determines the conditional interval probability, because that probability conditions on being event-free at day 2. The earlier segment still matters for the unconditional probability because it affects how many members of the original population remain event-free at the interval's start.
Diagnostics, Alternatives, and Limitations
These relationships provide simple checks on a calculation or model output. A valid continuous-time hazard must be nonnegative; its cumulative hazard must start at zero and never decrease. Survival must start at one under the stated time origin, remain between zero and one, and never increase. The event probability \(1-S(t)\) must be nondecreasing. If a proposed calculation gives a negative interval event probability, a survival probability above one, or a decreasing cumulative hazard, check the integration limits, signs, units, and use of the correct interval endpoints.
For time-varying hazards, a common numerical failure is to substitute the hazard at the end of an interval for the integrated hazard over that interval. In general, \(\lambda(b)(b-a)\) is not equal to \(\int_a^b\lambda(u)\,du\). It is valid only when the hazard is constant on that interval. When numerical integration is needed, check that the grid is fine enough for the hazard's rate of change and that the integrated hazard increases as time advances.
The three quantities are complementary representations, not competing definitions of event risk. A hazard model is useful for describing how the instantaneous event rate changes with time or covariates; survival is often easiest to communicate as a probability of remaining event-free; cumulative hazard provides an additive scale for integrating hazard contributions across time. Directly reporting \(F(t)=1-S(t)\) is an alternative to reporting survival when the scientific question is framed as event probability, but it contains the same distributional information.
A hazard is not the probability of an event over an arbitrary interval. For a short interval of width \(\Delta t\), the conditional event probability is approximately \(\lambda(t)\Delta t\) when the hazard is roughly constant over that short interval; for a wider interval, the integrated hazard and exponential relationship are needed. Likewise, \(H(t)\) is not the event probability: for example, if \(H(t)=0.270\), then \(F(t)=1-e^{-0.270}=0.2366\), not 0.270.
These formulas describe a single event time under a continuous-time model. They do not by themselves address how incomplete follow-up contributes to estimation, how to choose a particular baseline hazard, or how covariates modify hazard. Also, if the hazard is defined only while survival is positive, it is not defined beyond a time at which \(S(t)=0\). These boundaries matter when applying the identities to a model or interpreting its predictions.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- If \(H(3)=0.200\), what are \(S(3)\) and \(F(3)\)?
- A constant hazard is \(0.0500\) per hour. What is the probability of an event within 4 hours?
- If \(S(1)=0.90\) and \(S(4)=0.60\), what is the unconditional probability of an event in days 1 through 4?
- Why can the conditional probability of an event in days 1 through 4 be larger than its unconditional probability?
Solutions. (1) \(S(3)=e^{-0.200}=0.8187\), and \(F(3)=1-0.8187=0.1813\). (2) The cumulative hazard is \(H(4)=0.0500(4)=0.200\) per the dimensionless integrated scale, so the probability is \(1-e^{-0.200}=0.1813\), or 18.1%. (3) The unconditional interval probability is \(S(1)-S(4)=0.90-0.60=0.30\), or 30.0%. (4) The unconditional probability uses the original population as its denominator, while the conditional probability uses only those still event-free at day 1; specifically, the conditional probability is \(1-S(4)/S(1)=1-0.60/0.90=0.3333\), or 33.3%.