Checking the Constant-Hazard Assumption

Why Check the Constant-Hazard Assumption?

An exponential model is useful when an event can occur at any time and its instantaneous risk is assumed not to change as time passes. That condition is called a constant hazard. It is a modeling assumption, not a conclusion that follows merely because the variable is a waiting time or lifetime.

For example, a hypothetical service desk might model the time until a routine request arrives, or a reliability analyst might model time until a component fails. In either case, the exponential model can be a poor description if the arrival or failure risk changes with elapsed time, calendar conditions, or the characteristics of the individuals still at risk. The practical question here is whether the proposed scenario is reasonably consistent with a constant hazard—not whether a finite data set can prove that the hazard is constant.

Definition: For a continuous waiting time or lifetime $T$ with density $f_T(t)$, survival function $S_T(t)=P(T>t)$, and $S_T(t)>0$, the hazard at time $t$ is $$ h(t)=\frac{f_T(t)}{S_T(t)}. $$ It describes the instantaneous event rate at time $t$, conditional on the event not having occurred before $t$. A constant-hazard model sets $h(t)=\lambda$ for $t\geq 0$, where $\lambda>0$ is a rate per unit of time.

Under this model, $T$ has an exponential distribution with rate $\lambda$, and its survival function is $S_T(t)=e^{-\lambda t}$. A direct practical implication is that the probability of surviving an interval depends on the interval’s length, not on how long the subject has already survived. Thus, conditional survival over successive intervals of the same length should be the same under the model.

What the Model Predicts Across Equal Intervals

Suppose an interval has length $\Delta>0$. Under a constant hazard, the conditional probability of surviving the next interval, given survival to its start, is $$ P(T>t+\Delta\mid T>t)=\frac{S_T(t+\Delta)}{S_T(t)} =\frac{e^{-\lambda(t+\Delta)}}{e^{-\lambda t}} =e^{-\lambda\Delta}. $$ The expression does not depend on $t$. This gives a useful diagnostic: compare conditional survival proportions across intervals of equal length. Similar proportions are consistent with a constant hazard; a systematic increase or decrease is a warning that the assumption may not fit.

Key diagnostic: For equal-length intervals, a constant-hazard model predicts equal conditional survival probabilities. If the conditional survival probability over an interval is $p$, the exponential rate implied by that interval is $$ \lambda=-\frac{\ln(p)}{\Delta}. $$ Rates inferred from different equal-length intervals should agree approximately if the model is suitable. This is a diagnostic, not a formal proof of the model.

The comparison must use the correct denominator. The conditional survival proportion in an interval is the number surviving that interval divided by the number at risk at its start—not the number surviving divided by the original starting count. The two proportions answer different questions.

Worked Checks

Worked Example: A Lifetime Pattern That Matches a Constant Hazard

In a hypothetical study of 1,000 identical electronic valves, 800 are still operating after one year, 640 after two years, and 512 after three years. Assess whether these observed proportions are consistent with a constant-hazard exponential model, and find the rate suggested by the data. Assume complete follow-up at each anniversary and no entry into or removal from the cohort before the measurements. This is a device-reliability example: no interest rate or mortality assumption applies. The proposed lifetime model is $T\sim\operatorname{Exponential}(\lambda)$, with time measured in years.

Method. Calculate the conditional survival proportion in each one-year interval. The numbers at risk at the starts of the three intervals are 1,000, 800, and 640.

From year 0 to year 1, the conditional survival proportion is $800/1{,}000=0.80$. From year 1 to year 2, it is $640/800=0.80$. From year 2 to year 3, it is $512/640=0.80$. The equal proportions are consistent with the constant-hazard prediction.

For a one-year interval, $p=e^{-\lambda}$, so the implied rate is $$ \lambda=-\ln(0.80)=0.2231436\text{ per year}. $$ Using the exact rate, the predicted three-year survival is $$ e^{-3[-\ln(0.80)]}=0.80^3=0.5120. $$ The observed three-year survival is $512/1{,}000=0.5120$, so it agrees exactly with this pattern.

Check and trap avoided. The successive conditional proportions all equal 0.80, and their product gives the overall survival, $0.80\times0.80\times0.80=0.5120$. If the rate is rounded first to $0.2231$, the three-year prediction is $e^{-3(0.2231)}\approx0.5121$, not $0.5120$. Keep the exact rate for the calculation and round only the reported result.

Worked Example: Conditional Survival Changes Between Intervals

A hypothetical reliability test begins with 60 devices. Of these, 45 survive the first interval; of those 45, 27 survive the second interval, which has the same duration. Calculate the conditional survival proportion and implied exponential rate for each interval, then assess the constant-hazard assumption. Let each interval last one week. Assume every device is observed through each interval unless it fails. No interest rate or mortality assumption applies. The proposed model is $T\sim\operatorname{Exponential}(\lambda)$, with $\lambda$ measured per week.

Method. Use the number at risk at the beginning of each interval as the denominator. The first interval begins with 60 devices, and the second begins with the 45 devices that survived the first.

The first conditional survival proportion is $$ p_1=\frac{45}{60}=0.75. $$ The second is $$ p_2=\frac{27}{45}=0.60. $$ These are not equal. The corresponding one-week rate estimates are $$ \lambda_1=-\ln(0.75)=0.2876821\text{ per week}, $$ and $$ \lambda_2=-\ln(0.60)=0.5108256\text{ per week}. $$

The second interval has lower observed conditional survival and a higher implied rate. The pattern is not consistent with equal conditional survival across equal intervals, so it raises a concern about a constant-hazard model. It does not, by itself, establish that the true hazard changes: with only 60 devices initially, random variation could produce different proportions.

Check and trap avoided. The second interval’s denominator is 45, not 60. Using $27/60=0.45$ would calculate the fraction of the original cohort that survived both intervals, not the conditional survival in the second interval. The correct second-interval proportion is $27/45=0.60$.

Worked Example: A Clear Change in a Seasonal Failure Pattern

A hypothetical operator monitors 200 identical outdoor sensors over two consecutive three-month intervals. All 200 are operating at the start. At the end of the first interval, 180 remain operating; at the end of the second, 126 remain operating. Assume complete follow-up, no replacements, and no censoring before the interval endpoints. Assess the constant-hazard assumption and compare the observed total survival with a model that applies the first interval’s rate throughout. This is a device example, so no interest rate or mortality assumption applies. The proposed lifetime distribution is exponential, with rates measured per three-month interval.

Method. Calculate conditional survival separately for the two intervals. The first interval starts with 200 sensors; the second starts with 180.

The first interval’s conditional survival is $$ p_1=\frac{180}{200}=0.90. $$ The second interval’s conditional survival is $$ p_2=\frac{126}{180}=0.70. $$ The interval-implied rates are $$ \lambda_1=-\ln(0.90)=0.1053605 $$ and $$ \lambda_2=-\ln(0.70)=0.3566749 $$ per three-month interval. The difference is substantial, so the observed pattern is not consistent with a constant hazard in these intervals.

If the first interval’s rate continued unchanged, two-interval survival would be $0.90^2=0.81$. The observed two-interval survival is $126/200=0.63$, which also signals that the first interval’s pattern does not describe the full period well. The later interval coincides with a harsher season in this hypothetical setting, a plausible reason the risk might differ over calendar time.

Check and trap avoided. The total survival of 0.63 is not the conditional survival in the second interval. The latter is $126/180=0.70$. Also, the rate units here are per three-month interval; interpreting $0.1053605$ as a rate per year would be a time-unit error.

Why Constant Hazard Produces the Exponential Model

The link between a constant hazard and exponential survival follows from the definitions. Since $h(t)=f_T(t)/S_T(t)$, and $f_T(t)=-S_T'(t)$ for a continuous lifetime, a constant hazard $\lambda$ implies the differential equation below.

-\frac{S_T'(t)}{S_T(t)}=\lambda \quad\Longrightarrow\quad \frac{d}{dt}\ln S_T(t)=-\lambda.

Integrating from 0 to $t$, and using $S_T(0)=1$, gives $\ln S_T(t)=-\lambda t$, hence $S_T(t)=e^{-\lambda t}$. Therefore the conditional survival over any interval of length $\Delta$ is $e^{-\lambda\Delta}$, independent of the interval’s starting time. Conversely, the exponential survival function has hazard $f_T(t)/S_T(t)=\lambda$, so the two descriptions are equivalent for this continuous model.

When the Exponential Model May Be Inappropriate

A constant hazard is implausible when the event mechanism changes systematically with elapsed time or conditions. Examples include components that wear out as they age, a short-lived early defect risk followed by a stable period, or arrivals that become more or less frequent during a working day. In such cases, equal-length intervals may have different conditional event or survival proportions.

Other settings can create apparent changes even when no individual’s underlying risk changes. For instance, a group may contain subjects with different constant hazards. The higher-risk subjects may fail earlier, leaving a lower-risk group among those who remain. The aggregate conditional survival can then change over time. A comparison of aggregate proportions is a diagnostic of the proposed model for the group; it does not identify the cause of a discrepancy.

Observed proportions also vary from sample to sample. Small differences between intervals do not automatically invalidate a model, especially with small numbers at risk. Conversely, matching proportions at a few observation times does not prove the hazard is constant between those times. A useful assessment asks whether the differences are large, systematic, and plausible given the sample sizes and how the observations were collected.

State the proposed model.
Identify the event, the time unit, and the assumption that its hazard is constant over the period being assessed.

Compare like intervals.
For intervals of equal length, divide the number surviving an interval by the number at risk at its start. Do not compare an interval proportion with a cumulative proportion.

Interpret differences with care.
Unequal observed proportions challenge the model but may reflect sampling variability, changing conditions, different risk groups, or incomplete follow-up.

With censored observations, the simple ratio of endpoint counts may not represent conditional survival correctly: some subjects may leave observation before the interval ends. The complete-follow-up examples above avoid that issue. In an applied analysis, the observation and censoring process must be accounted for before interpreting interval comparisons.

Key takeaway: The exponential model predicts the same conditional survival probability over every interval of a given length. Compare conditional proportions using the correct at-risk denominators, check that time units agree, and treat the result as evidence about model suitability rather than proof.

Check Your Understanding

For each problem, compare conditional survival over intervals of equal length where possible. Treat all scenarios as hypothetical.

A group of 1,000 components has 900 operating after three months and 810 after six months. Assuming complete follow-up and no removals, calculate the two conditional survival proportions and the exponential rate per month suggested by either interval. (Answer: Both proportions are $0.9000$; $\lambda=-\ln(0.90)/3=0.0351$ per month.)
Of 80 devices at risk, 64 survive a two-week interval. Of those 64, 40 survive the next two weeks. Find both conditional proportions and their implied rates per two-week interval. Is the pattern exactly consistent with constant hazard? (Answer: $0.8000$ and $0.6250$; rates $0.2231$ and $0.4700$ per two-week interval; no.)
A hypothetical lifetime study records 300 items at the start, 240 surviving after one year, and 180 surviving after two years. Find the conditional survival in each year and state what the comparison suggests. Assume complete follow-up at each anniversary. (Answer: $240/300=0.8000$, then $180/240=0.7500$; the observed pattern is not exactly consistent with constant hazard.)
A service receives 500 requests during one four-hour observation period and 400 during the next four-hour period. Explain why these totals alone are insufficient to compare conditional survival or establish a constant hazard for waiting time. (Answer: They are counts of arrivals, not numbers surviving an at-risk interval for a defined waiting-time cohort; the cohort, risk sets, and event-time information are needed.)
A cohort has conditional survival 0.95 over each of two consecutive six-month intervals. Find the implied rate per year and the predicted one-year survival under a constant-hazard model. (Answer: $\lambda=-\ln(0.95)/0.5=0.1026$ per year; one-year survival $0.95^2=0.9025$.)

What You'll Learn