Constant Hazard

Constant Hazard, the Exponential Assumption and Cycle Probabilities in Economic Models

Under a constant hazard, a person who has stayed event free for five years faces the same risk per unit of time as someone at the start of follow-up. In a health economic model that single assumption fixes the survival curve as exponential and lets one rate be turned into a probability for any cycle length, with mean survival equal to the reciprocal of the rate. These results follow from the hazard rate, and the assumption enters both Markov models and survival extrapolation, where Technical Support Document 14 (TSD 14) of the NICE Decision Support Unit sets out how to test it.

What a flat hazard implies for survival

The hazard is the event rate at each instant among people still at risk. When it does not change with time, every other survival quantity follows from one parameter, written $\lambda$, and the time-to-event distribution is exponential. The NIST/SEMATECH Engineering Statistics Handbook gives the standard results below.

$$h(t) = \lambda, \qquad H(t) = \lambda t, \qquad S(t) = \exp(-\lambda t), \qquad F(t) = 1 - \exp(-\lambda t)$$

where $h(t)$ is the hazard at time $t$, $\lambda$ is the constant hazard in events per unit of time (for example per person-year), $H(t)$ is the cumulative hazard, $S(t)$ is the probability of remaining event free beyond $t$ (the survival function), and $F(t)$ is the probability of having had the event by $t$. The same handbook gives the mean and median time to the event as follows.

$$\text{mean} = \frac{1}{\lambda}, \qquad \text{median} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda}$$

where $\ln$ is the natural logarithm. NIST adds that the exponential is the only distribution with a constant failure rate, so a constant hazard and an exponential distribution describe the same model.

Two consequences follow. The probability of surviving a further interval of length $\Delta$, given survival to time $t$, is $S(t+\Delta)/S(t) = \exp(-\lambda \Delta)$, which does not depend on $t$, so the process has no memory of time already at risk. And a constant hazard does not mean a constant number of events, because the same rate applies to a shrinking group of survivors, so events per year fall over time.

Where constant hazards enter health economic models

The assumption is built into any Markov model whose transition probabilities do not change from cycle to cycle, and into the exponential model used to extrapolate trial survival. Piecewise approaches reuse it to approximate a changing hazard.

TSD 14, written by Latimer and last updated in March 2013, reviewed the survival methods in 45 NICE cancer appraisals completed by December 2009. Of these, 23 (51%) used a Weibull model and 20 (44%) an exponential model. The document notes that an exponential model was often used when Markov models had transition probabilities that were not time dependent, and when, without patient-level data, analysts converted a reported median into a mean under an exponential assumption. TSD 14 treats that conversion as a reasonable use of medians where evidence is lacking, unlike entering a median directly as the measure of survival in the economic model.

TSD 14 calls the exponential the simplest parametric model, with one parameter and a proportional hazards form, so a treatment effect is a hazard ratio. It warns that if the exponential is used, it matters whether the hazard is likely to remain constant over an entire lifetime, the usual horizon of a cost-effectiveness model.

A piecewise model keeps the constant-hazard arithmetic while letting the rate change between periods. TSD 14 describes piecewise constant models, with exponential models fitted to different time periods, as under-used and useful when hazards vary within follow-up, but less useful for the extrapolated portion, where hazards are not observed. TSD 21 describes the Liverpool approach, which joins an exponential tail to the Kaplan-Meier estimator.

Converting a constant rate into cycle probabilities

A cohort model needs the probability that a person in a state at the start of a cycle has the event by its end. With a constant hazard it depends only on the rate and the cycle length. Gidwani and Russell give the conversion below.

$$r = \frac{-\ln(1 - p)}{t}, \qquad p = 1 - \exp(-r t)$$

where $r$ is the constant rate, $p$ is the probability of the event over a period of length $t$, and $t$ is expressed in the same time unit as the rate. A rate has time at risk in its denominator and no upper limit, while a probability lies between 0 and 1. The rate-to-probability conversion page covers the formula; both directions assume the rate is constant within the period converted.

Chhatwal, Jayasuriya and Elbasha record the warning, which they attribute to Sonnenberg and Beck and to Miller and Homan, against dividing a probability to obtain one for a shorter cycle; for a monthly cycle, those authors recommended converting the annual probability to a rate and then to a monthly probability. The divided figure understates the per-cycle risk, and the error compounds over the cycles of a model.

Competing exits from the same state

The single-event formula is only safe when one transition can leave a state. When a state has two or more exits, each with its own constant cause-specific hazard, the exits compete, because a person who leaves by one route cannot then leave by another. For a model in which at most one transition can happen in a cycle, the standard competing risks result is as follows.

$$\Lambda = \sum_{j} \lambda_j, \qquad p_{\text{stay}} = \exp(-\Lambda \Delta), \qquad p_j = \frac{\lambda_j}{\Lambda}\bigl(1 - \exp(-\Lambda \Delta)\bigr)$$

where $\lambda_j$ is the constant hazard of exit $j$, $\Lambda$ is the total hazard of leaving the state, $\Delta$ is the cycle length, $p_{\text{stay}}$ is the probability of remaining in the state and $p_j$ is the probability of leaving by exit $j$. Gidwani and Russell state that the two-transition formulas do not give correct probabilities when three or more transitions can occur in a cycle, a common situation. Chhatwal and colleagues show that the usual conversion yields a different Markov chain in such models and work on the whole transition matrix instead; the competing risks and transition probability pages cover this.

Checking whether the hazard is constant

The assumption can be examined against trial data. TSD 14 states that exponential models are only suitable if the observed hazard is approximately constant and non-zero, and recommends log-cumulative hazard plots, which show $\ln\{-\ln S(t)\}$ against $\ln t$, as the first step in choosing a survival model. Under a constant hazard this plot is a straight line with gradient 1, since $\ln H(t) = \ln \lambda + \ln t$.

TSD 14 also treats the exponential as a special case of the Weibull model with shape parameter $\gamma = 1$ (above 1 the hazard rises, below 1 it falls) and of the Gompertz model with $\theta = 0$.

Fit within follow-up does not settle the question. TSD 14 notes that these tests address the internal validity of fitted models, not their external validity, and that the extrapolated portion of a survival curve generally has a large effect on estimated mean survival. In the pemetrexed appraisal for lung cancer (TA181), the assessment group saw each arm's cumulative hazard rise linearly some time after treatment ended and fitted an exponential model to the tail; TSD 14 cautions that this convergence to a constant rate may not hold in other disease areas. The simulation study in TSD 21 found that standard parametric models gave low bias for restricted mean survival time within trial follow-up, except the exponential and in some scenarios the Gompertz, and that bias could become substantial when extrapolating to a lifetime horizon.

Worked example from a reported median to a transition row

The numbers below are illustrative and do not come from a real trial. A published study reports a median progression-free survival of 18 months, and no patient-level data are available. The modeller assumes a constant hazard of leaving the progression-free state.

1. Recover the rate from the median. With the median in years, 18 months is 1.5 years, so the rate is as follows.

$$\lambda = \frac{\ln 2}{\text{median}} = \frac{0.6931}{1.5} = 0.4621 \text{ per year}$$

where $\lambda$ is the implied constant hazard. Mean time in the state is ${1/0.4621 = 2.164}$ years, about 26 months, which exceeds the median because the exponential curve has a long right tail.

2. Read survival at whole years. $S(1) = \exp(-0.4621) = 0.6300$ and $S(3) = \exp(-1.3863) = 0.25$, so a quarter of the cohort is still progression free after two medians. The chance of surviving the next year is 0.6300 whether the cohort is at year 0 or year 3, the no-memory property in numbers.

3. Convert to annual and monthly cycle probabilities. The annual probability is ${1 - 0.6300 = 0.3700}$. The monthly probability is given below.

$$p_{\text{month}} = 1 - \exp\!\left(-\frac{0.4621}{12}\right) = 1 - 0.96222 = 0.03778$$

where $p_{\text{month}}$ is the probability of leaving the state in a one-month cycle. Twelve monthly cycles give $(1 - 0.03778)^{12} = 0.6300$, matching the annual figure exactly. Dividing the annual probability by 12 instead gives 0.0308, and twelve cycles of that leave 0.6867 progression free, overstating the time spent in the state.

4. Split the exit between competing events. Suppose, separately and for illustration, that a state has two exits with constant annual hazards of 0.30 to progression and 0.10 to death, and the model uses annual cycles. The total hazard is 0.40, so $p_{\text{stay}} = \exp(-0.40) = 0.6703$ and the probability of any exit is 0.3297. Allocated in proportion to the hazards, progression receives ${0.75 \times 0.3297 = 0.2473}$ and death ${0.25 \times 0.3297 = 0.0824}$; the row sums to 1.0000.

5. See what separate conversion would do. Converting each hazard on its own gives ${1 - \exp(-0.30) = 0.2592}$ and ${1 - \exp(-0.10) = 0.0952}$. Added to the correct stay probability of 0.6703, the row sums to 1.0247, which is impossible; taking the stay probability as one minus the two exits instead gives about 0.646, understating time in the state.

The mean of 26 months, which drives life-years and costs, rests entirely on the hazard staying flat throughout, which a single reported median cannot test.

Constant hazard, constant hazard ratio and other hazard shapes

A constant hazard concerns one group's risk over time; a constant hazard ratio, the proportional hazards assumption, concerns the ratio of two groups' hazards and can hold when both rise or fall. Exponential arms with different rates satisfy both, but proportional hazards on their own imply nothing about whether either arm's hazard is flat.

A time-varying hazard changes over follow-up, and a bathtub hazard falls after a risky procedure, stays low and then rises with age. NIST notes that any failure rate curve can be approximated by constant rates over short periods, which is the logic of the piecewise model, and that the exponential describes the time between events while the Poisson distribution describes their number in a period.

Limitations and common misreadings

Misreadings of a constant hazard fall into two groups, one treating the rate as something it is not and the other stretching the assumption beyond the evidence. The first group biases results in a known direction; the second can move them either way.

Using a rate as if it were a probability overstates risk, since $\lambda\Delta$ exceeds ${1 - \exp(-\lambda\Delta)}$ for any positive rate (in the example, using the monthly rate of 0.0385 as a probability leaves 0.6242 progression free at 12 months instead of 0.6300). Dividing an annual probability by the number of cycles understates risk, as step 3 shows. Converting competing hazards one at a time gives rows that do not sum to 1. Assuming a lifetime constant hazard from a short trial matters most, because the tail governs mean survival and fit within follow-up does not establish it.

All-cause death rates in a life table change with age, so one constant mortality rate over a lifetime horizon misstates late survival; background mortality is normally taken from age-specific rates as the cohort ages. A constant progression hazard also says nothing about whether costs, utilities or the treatment effect stay constant.

Sources

• Chhatwal J, Jayasuriya S, Elbasha EH. Changing cycle lengths in state-transition models: challenges and solutions. Medical Decision Making. 2016;36(8):952-964. https://doi.org/10.1177/0272989X16656165 (author manuscript: https://pmc.ncbi.nlm.nih.gov/articles/PMC5045797/)
• Collett D. Modelling Survival Data in Medical Research. 3rd ed. Boca Raton: CRC Press; 2015.
• Gidwani R, Russell LB. Estimating transition probabilities from published evidence: a tutorial for decision modelers. PharmacoEconomics. 2020;38(11):1153-1164. https://doi.org/10.1007/s40273-020-00937-z (https://pmc.ncbi.nlm.nih.gov/articles/PMC7426391/)
• Latimer N. NICE DSU Technical Support Document 14: Survival analysis for economic evaluations alongside clinical trials, extrapolation with patient-level data. Sheffield: Decision Support Unit, ScHARR, University of Sheffield; 2013 (first published 2011). Sections 2.1 to 2.3, 2.7, 3.2, 3.5, 4.1.2, 4.1.5 and 4.6. https://doi.org/10.15131/shef.data.33745843
• NIST/SEMATECH. e-Handbook of Statistical Methods. Section 8.1.6.1, Exponential. Gaithersburg, MD: National Institute of Standards and Technology; accessed 9 October 2026. https://www.itl.nist.gov/div898/handbook/apr/section1/apr161.htm
• Rutherford MJ, Lambert PC, Sweeting MJ, Pennington B, Crowther MJ, Abrams KR, Latimer NR. NICE DSU Technical Support Document 21: Flexible methods for survival analysis. Sheffield: Decision Support Unit, ScHARR, University of Sheffield; 2020 (updated March 2022). Sections 3.5 and 5.3.1. https://doi.org/10.15131/shef.data.33849118