Why a Constant-Hazard Model?
The exponential survival model is the simplest parametric description of an event time. Its central assumption is strong and easy to state: for an individual who has remained event-free up to time \(t\), the instantaneous event rate is the same at every time. This can be useful when a simple summary is needed, but it can be misleading when the event rate changes as time passes.
In survival analysis, \(T\) denotes a nonnegative event time, \(S(t)=P(T>t)\) its survival function, and \(h(t)\) its hazard function. The earlier material on deriving survival from the hazard establishes the general relationship between these functions. Here we apply that relationship to a constant hazard and show how right-censored data estimate its rate.
From Constant Hazard to Survival
The hazard at time \(t\) is the instantaneous event rate among individuals still event-free just before \(t\). In terms of the survival function and event-time density \(f(t)\), it is \(h(t)=f(t)/S(t)\). Since \(f(t)=-S'(t)\) for a continuous event time, a constant hazard \(\lambda\) gives the differential equation \(S'(t)=-\lambda S(t)\), with initial condition \(S(0)=1\).
The density follows by differentiating the survival function: \(f(t)=-S'(t)=\lambda\exp(-\lambda t)\). These expressions describe a continuous event time on \(t\geq0\); the density has units of inverse time, and the survival probability is dimensionless. The rate is not itself a probability. For example, a rate of \(0.08\) per month does not mean that exactly 8% of individuals experience the event each month; the probability depends on the time interval and who is still at risk.
A further implication is the memoryless property. For \(s,t\geq0\),
Thus, under this model, the chance of surviving a further \(t\) time units does not depend on how long the individual has already remained event-free. This is a substantive assumption, not a general property of survival data.
Worked Example: Interpreting a Specified Rate
Suppose a hypothetical device lifetime follows an exponential model with \(\lambda=0.08\) per month. Find the probability that a device lasts beyond 6 months, the probability that its event occurs between 6 and 12 months, and the median lifetime.
Survival past 6 months. Substitution into the survival function gives \(S(6)=\exp\{-(0.08)(6)\}=\exp(-0.48)\approx0.6188\). Under the model, about 61.88% of devices survive beyond 6 months.
Event between 6 and 12 months. This probability is \(P(6<T\leq12)=S(6)-S(12)\). Therefore, \(\exp(-0.48)-\exp(-0.96)\approx0.6188-0.3829=0.2359\). The modeled probability is about 23.59%.
Median lifetime. Solving \(\exp(-0.08t)=0.5\) gives \(t=\log(2)/0.08\approx8.664\) months. This is the model’s median, not a guarantee that half of any particular finite sample will fail by exactly that time.
Estimating the Rate With Right Censoring
For each of \(n\) independent individuals, let \(Y_i=\min(T_i,C_i)\) be the observed follow-up time, where \(C_i\) is the censoring time. Let \(\delta_i=1\) if the event was observed and \(\delta_i=0\) if the observation was right-censored. A censored individual is known to have survived through \(Y_i\), but their event time after that point is unknown. The likelihood contribution is \(f(Y_i)\) for an observed event and \(S(Y_i)\) for a censored observation.
For estimating \(\lambda\), this calculation assumes independent event times across individuals and noninformative right censoring: censoring does not depend on the event time in a way that supplies unmodeled information about \(\lambda\). In the usual likelihood formulation, the censoring mechanism is separate from the event-time model, so its factors do not involve \(\lambda\). If censoring is informative, treating censored follow-up as if it were independent can invalidate the estimate.
Write \(D=\sum_{i=1}^n\delta_i\) for the number of observed events and \(Y_+=\sum_{i=1}^nY_i\) for total observed follow-up time. The event and censoring contributions combine into the likelihood below.
Taking logarithms gives \(\ell(\lambda)=D\log(\lambda)-\lambda Y_+\). When \(D>0\) and \(Y_+>0\), differentiating and setting the derivative to zero yields the maximum likelihood estimate. The second derivative, \(-D/\lambda^2\), is negative, confirming a maximum.
The estimate is the number of observed events divided by total observed person-time. Its units are inverse time. Censored observations contribute follow-up time to \(Y_+\), but do not contribute an event to \(D\). This is why replacing a censored time with an assumed event time, or simply dropping censored individuals, gives the wrong likelihood calculation.
Worked Example: Estimating the Rate From Censored Follow-Up
In a hypothetical study, six independent participants are followed for a time-to-event outcome, measured in months. Their observed times and event indicators are:
| Participant | \(Y_i\), months | \(\delta_i\) | Observation |
|---|---|---|---|
| 1 | 2 | 1 | Event observed |
| 2 | 4 | 1 | Event observed |
| 3 | 5 | 0 | Right-censored |
| 4 | 7 | 1 | Event observed |
| 5 | 8 | 1 | Event observed |
| 6 | 10 | 0 | Right-censored |
There are \(D=4\) observed events, and total observed follow-up is \(Y_+=2+4+5+7+8+10=36\) person-months. Therefore,
The censored participants contribute 5 and 10 months to total follow-up, but neither contributes an event. The fitted exponential model estimates survival at 6 months as \(\exp\{-(0.1111)(6)\}\approx0.5134\). Using the unrounded rate \(4/36\) gives the same four-significant-figure result. The estimated median is \(\log(2)/(4/36)\approx6.238\) months. These are model-based estimates, and their interpretation depends on the constant-hazard and censoring assumptions.
If no events are observed, \(D=0\), the likelihood is \(\exp(-\lambda Y_+)\), which decreases as \(\lambda\) increases. Its maximum over \(\lambda\geq0\) occurs at the boundary \(\hat{\lambda}=0\). This boundary estimate does not establish that the true event rate is zero; it reflects the limited information in a sample with no observed events.
Assessing Whether a Constant Hazard Is Plausible
A fitted rate is easy to calculate, but the data must also be examined for evidence against a constant hazard. A useful first check is to compare the fitted curve \(\exp(-\hat{\lambda}t)\) with the Kaplan–Meier estimate. Under the exponential model, \(\log S(t)=-\lambda t\), so a plot of the log of estimated survival against time should be approximately linear, with slope \(-\lambda\). Kaplan–Meier estimates are stepwise and uncertain in the tail, so departures should be judged alongside the number at risk and the amount of censoring.
Another descriptive check compares event rates over time intervals. Within an interval, a crude rate is the number of observed events divided by total person-time at risk in that interval. Strongly different rates can signal changing hazards, although random variation, sparse events, and selection of who remains at risk complicate interpretation. A single interval comparison is a diagnostic, not a substitute for careful model assessment.
Worked Example: A Descriptive Check of Constancy
Suppose a hypothetical follow-up dataset is summarized into two time periods. During months 0–5, there are 40 events over 400 person-months at risk. During months 5–10, there are 10 events over 500 person-months at risk. Compare the crude rates.
For the first period, the rate is \(40/400=0.1000\) per month. For the second, it is \(10/500=0.0200\) per month. The ratio of the second rate to the first is \(0.0200/0.1000=0.2000\); equivalently, the first rate is five times the second. This pattern is a substantial descriptive departure from constancy.
For a rough sense of sampling uncertainty, a large-count Poisson approximation gives the standard error of the log rate ratio as \(\sqrt{1/10+1/40}=\sqrt{0.125}\approx0.3536\). A rough 95% interval for the second-to-first rate ratio is \(\exp\{\log(0.2)\pm1.96(0.3536)\}\), or approximately \((0.100,0.400)\). This interval suggests the later rate is lower, but it is only an approximate diagnostic: it does not account for every feature of time-to-event sampling or dependence between interval counts. The pattern warrants closer investigation rather than mechanically accepting a constant rate.
Alternatives, Failure Modes, and Limits
The exponential model is attractive because one parameter determines the entire survival curve and the likelihood estimate has a simple form. When its assumptions are reasonable, this simplicity can support concise summaries and extrapolation. When the hazard changes over time, however, a constant rate can conceal important patterns and produce misleading survival predictions.
| Approach | What it provides | Trade-off relative to an exponential model |
|---|---|---|
| Exponential model | A fully specified survival curve with one constant rate | Simple, but requires a constant hazard |
| Kaplan–Meier estimator | A nonparametric estimate of survival over observed follow-up | More flexible, but does not impose a constant hazard or by itself provide a parametric extrapolation |
| Cox proportional hazards model | A regression approach that does not specify the baseline hazard’s form | More flexible about baseline shape, but it still has its own proportional-hazards assumptions and does not produce an exponential baseline by default |
Diagnostics should be interpreted with care. A curved log-survival pattern or changing interval rates may indicate nonconstant hazard, but sparse data can make real departures hard to see. Conversely, random fluctuations in a small sample do not prove the model wrong. Extensive censoring, few observed events, and follow-up concentrated in a narrow time range limit what can be learned about the hazard shape.
The constant-hazard assumption is also a population-level restriction: it does not automatically account for meaningful differences among individuals. If event rates differ across groups or covariate levels, a single pooled rate may be an inadequate summary. Finally, extrapolation beyond observed follow-up depends heavily on the assumed exponential shape; a close fit over a short observed period is not proof that the same hazard will continue indefinitely.
- Mistake: Count only observed events and divide by the number of participants. Correction: The rate estimate uses total observed person-time in the denominator, \(D/Y_+\).
- Mistake: Treat a censored time as an event time. Correction: A censored observation contributes survival through its observed time, not an event contribution.
- Mistake: Interpret \(\lambda\) as a fixed probability of failure per time unit. Correction: It is a constant instantaneous rate; interval probabilities follow from the survival function.
- Mistake: Regard a fitted exponential curve as evidence that the hazard truly is constant. Correction: The curve is conditional on the model; examine the observed survival pattern and time-specific rates.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- If \(\lambda=0.05\) per week, what is the modeled probability of surviving beyond 10 weeks?
- Five individuals contribute 3, 4, 6, 7, and 10 months of follow-up. Events are observed for the first, third, and fifth individuals; the others are censored. Find the exponential maximum likelihood estimate and state its units.
- Why does a right-censored observation contribute to total person-time but not to the event count?
- A plot of log estimated survival against time is distinctly curved over a period with many observed events. What does this suggest, and what caution is needed?
- State one important difference between an exponential model and the Kaplan–Meier estimator.
Solutions. (1) \(S(10)=\exp\{-(0.05)(10)\}=\exp(-0.5)\approx0.6065\), so the modeled survival probability is 0.6065. (2) There are \(D=3\) events and \(Y_+=3+4+6+7+10=30\) person-months, giving \(\hat{\lambda}=3/30=0.1000\) per month. (3) The individual is known to have remained event-free through the censoring time, so that time supplies exposure; no event was observed, so it does not add to \(D\). (4) Curvature suggests that a constant hazard may not describe the observed pattern well. However, uncertainty, censoring, and the number at risk should also be considered; the plot is a diagnostic, not conclusive proof. (5) The exponential model imposes a constant hazard and yields a fully specified parametric survival curve; the Kaplan–Meier estimator estimates survival without imposing that shape over observed follow-up.