From an Instantaneous Rate to Survival
A hazard describes how quickly events occur among individuals who have remained event-free up to a given time. It is not itself the probability of an event by that time. To obtain an event-time distribution, we need to accumulate this instantaneous rate over time. The key result is that survival is the exponential of the negative cumulative hazard.
Let \(T\) denote an event time, and let \(S(t)=P(T>t)\) be the probability of remaining event-free beyond time \(t\). For a continuous event-time distribution, the hazard \(h(t)\) is the instantaneous event rate at time \(t\), conditional on survival to that time. When the density \(f(t)\) exists and \(S(t)>0\), the hazard is \(h(t)=f(t)/S(t)\). The cumulative hazard \(H(t)\) is the accumulated hazard from time zero through time \(t\).
The formulas assume a continuous event-time distribution on the interval under consideration, with a differentiable survival function and positive survival probability there. A piecewise-defined hazard is allowed: integrate each piece over its time interval. The hazard must be nonnegative, so its cumulative integral cannot decrease and the resulting survival function cannot increase. If the hazard is integrable over every finite interval, the formula defines survival at each such time. For the event to occur eventually with probability one, the additional condition \(H(t)\to\infty\) as \(t\to\infty\) is needed.
Why the Exponential Formula Holds
For a continuous event-time distribution, \(S'(t)=-f(t)\): the rate at which the probability of remaining event-free decreases is the negative of the event-time density. Since \(h(t)=f(t)/S(t)\), it follows that
Integrate both sides from zero to \(t\). At time zero, \(S(0)=1\) for an event time that is nonnegative and has no event at time zero. Therefore,
This derivation also gives the density formula: \(f(t)=h(t)S(t)=h(t)\exp\{-H(t)\}\). Thus, knowing the hazard over time determines the survival function and density, and therefore probabilities and quantiles of the event time. For an interval \((a,b]\), the probability of an event in that interval is \(S(a)-S(b)\). Conditional on survival beyond \(a\), the probability of an event by \(b\) is \(1-S(b)/S(a)\).
Worked Example: Constant Hazard and Conditional Risk
Suppose a hypothetical event-time distribution has a constant hazard of \(0.04\) per year. Find the probability of remaining event-free beyond 5 years and the probability of an event between years 5 and 10, conditional on remaining event-free beyond year 5.
Check the assumptions. The specified hazard is nonnegative and constant, hence integrable over finite intervals. Under the continuous-event-time assumption, its cumulative hazard through \(t\) years is \(H(t)=0.04t\), with \(t\) measured in years.
Survival beyond 5 years. The cumulative hazard is \(H(5)=0.04(5)=0.20\). Thus
The probability of remaining event-free beyond 5 years is about \(0.819\), or \(81.9\%\).
Conditional event probability from year 5 to year 10. Since \(S(10)=\exp\{-0.04(10)\}=\exp(-0.40)\), the conditional probability is
Among those event-free at year 5, about \(18.1\%\) experience the event by year 10 under this distribution. The answer is not \(0.04(5)=0.20\): that product is the cumulative hazard over the interval, not the conditional event probability.
Time-Varying Hazards
When the hazard changes with time, the same result applies, but the cumulative hazard is an integral rather than a constant multiplied by time. The shape of the hazard determines how quickly cumulative hazard builds up. Because the exponential function is decreasing, larger accumulated hazard corresponds to lower survival.
Worked Example: Increasing Hazard, Survival, and Density
Consider a hypothetical event-time distribution with hazard \(h(t)=0.02+0.01t\) per year, for \(t\geq0\) measured in years. Find survival at 10 years, the density at 10 years, and the median event time.
Check the assumptions. The hazard is nonnegative for all \(t\geq0\), continuous, and integrable over every finite interval. It is therefore suitable for the continuous-event-time derivation. Its units are per year: the intercept is \(0.02\) per year and the slope \(0.01\) per year squared.
Cumulative hazard. Integrating the hazard gives
At 10 years, \(H(10)=0.02(10)+0.005(10^2)=0.20+0.50=0.70\). Hence
The probability of remaining event-free beyond 10 years is about \(49.7\%\).
Density at 10 years. The hazard there is \(h(10)=0.02+0.01(10)=0.12\) per year. Thus
This is the event-time density at 10 years, not a probability of an event at exactly 10 years. For a sufficiently short interval around 10 years, the density approximates probability per unit time.
Median event time. The median \(m\) satisfies \(S(m)=0.5\), equivalently \(H(m)=\log(2)\). Here,
The positive root is used because time cannot be negative. Substitution gives \(H(m)=\log(2)\), so \(S(m)=\exp\{-\log(2)\}=0.5\). This median is below 10 years, consistent with \(S(10)=0.4966<0.5\).
Worked Example: A Piecewise-Constant Hazard
Suppose a hypothetical event-time distribution has hazard \(0.03\) per year from time zero through year 4, and \(0.08\) per year after year 4. Find survival at year 6 and the probability of an event between years 4 and 6, conditional on being event-free at year 4.
Check the assumptions. Both hazard values are nonnegative and finite. The hazard changes at year 4 but is integrable, so cumulative hazard is obtained by integrating each segment. The single change point does not create a point mass in event time.
The cumulative hazard at year 6 is the area from zero to 4 plus the area from 4 to 6:
Therefore \(S(6)=\exp(-0.28)=0.7558\). For the conditional probability, the hazard accumulated after year 4 is \(0.08(2)=0.16\). Using the survival ratio,
Thus, about \(14.8\%\) of individuals event-free at year 4 experience the event by year 6. The pre-year-4 hazard affects how many remain event-free at year 4, but it cancels from this conditional probability.
Checks, Alternatives, and Limitations
The hazard-to-survival relation is useful because hazards often describe how risk evolves over time, while survival probabilities answer questions about remaining event-free. It also provides a direct check on a proposed pair of hazard and survival curves: wherever derivatives exist, a valid pair must satisfy \(h(t)=-S'(t)/S(t)\), and integration must recover \(S(t)=\exp\{-H(t)\}\).
A practical diagnostic is to check the cumulative hazard before interpreting the survival curve. If a computed cumulative hazard decreases, the hazard was negative somewhere or the integration was done incorrectly. If the resulting survival curve rises above 1 or increases over time, there is an algebraic or coding error: for a nonnegative hazard, \(H(t)\geq0\) and is nondecreasing, so \(0<S(t)\leq1\) and survival is nonincreasing. Check units as well. Integrating a hazard measured per year over time in days without conversion gives a meaningless exponent.
One alternative is to begin with a density and calculate survival as the remaining probability, \(S(t)=1-\int_0^t f(u)\,du\). That route is appropriate when the density is specified directly. The hazard route is more natural when the instantaneous rate is the object specified. Both routes must describe the same distribution: if a density is available, dividing it by its survival function should reproduce the stated hazard.
The derivation here assumes continuous event times. Distributions with point masses, such as a positive probability of an event at one exact time, require accounting for those jumps; the ordinary hazard-density identity and integral alone do not represent the entire distribution. Also, if \(H(t)\) approaches a finite limit as \(t\) grows without bound, then \(S(t)\) approaches a positive value. In that case, the model assigns positive probability to never experiencing the event, rather than forcing eventual occurrence.
A frequent mistake is to treat cumulative hazard as event probability. For example, a cumulative hazard of \(0.70\) does not mean that the probability of an event by that time is \(70\%\). The probability is \(1-\exp(-0.70)=0.5034\). Another mistake is to use \(h(t)\) itself as a probability: a hazard has units of inverse time, whereas a probability is dimensionless. Convert hazard to survival or an interval probability before interpreting it probabilistically.
Check Your Understanding
Work each problem using the hazard-to-survival relationship. Solutions follow the questions.
- If \(h(t)=0.05\) per year is constant, what is \(S(8)\)?
- For that same constant hazard, what is the probability of an event between years 3 and 7, conditional on survival beyond year 3?
- If \(h(t)=0.01t\) per year for \(t\geq0\), find \(H(4)\) and \(S(4)\).
- For a continuous event-time distribution, explain why a negative hazard cannot produce a valid nonincreasing survival curve through the stated relation.
- A computed cumulative hazard falls from 0.4 to 0.3 as time increases. Give one immediate check to make.
Solutions. (1) \(H(8)=0.05(8)=0.40\), so \(S(8)=\exp(-0.40)=0.6703\). (2) The hazard increment is \(0.05(7-3)=0.20\), so the conditional event probability is \(1-\exp(-0.20)=0.1813\). (3) \(H(4)=\int_0^4 0.01u\,du=0.005(4^2)=0.08\), giving \(S(4)=\exp(-0.08)=0.9231\). (4) Since \(S'(t)=-h(t)S(t)\), a negative hazard with positive survival gives \(S'(t)>0\), contradicting the fact that survival cannot increase. (5) Check the hazard values and integration limits: a cumulative hazard from a nonnegative hazard cannot decrease.