Why the Baseline Hazard Matters
A time-to-event model needs to represent how event risk changes with time, even when drug exposure is absent. This underlying time pattern is the baseline hazard. It is not itself a drug effect: it describes the event process at zero exposure (and, when covariates are included, at their reference values). Choosing its form determines which patterns of underlying risk the model can represent.
For example, an event might be equally likely per unit time throughout follow-up, become more likely as time passes, or have a high early risk that later declines. An overly restrictive baseline can misrepresent these patterns and, in an exposure-response analysis, make the model compensate for the error elsewhere. The goal here is to compare the exponential, Weibull, and flexible baseline hazards before adding exposure effects.
A hazard is an instantaneous rate, not the probability that an event occurs during an entire interval. For a short interval of width \(\Delta t\), the event probability conditional on remaining event-free to its start is approximately \(\lambda_0(t)\Delta t\), when the interval is sufficiently short. The cumulative hazard combines risk over time; survival follows from it through \(S_0(t)=\exp[-H_0(t)]\).
Exponential and Weibull Baselines
The exponential baseline assumes a constant event rate. It is the simplest model, but its constant hazard is a substantive assumption rather than a neutral default. The Weibull baseline adds a shape parameter, allowing hazard to increase or decrease systematically with time.
For the Weibull model, \(k=1\) gives the exponential model with rate \(1/\eta\). When \(k>1\), the hazard increases with time; when \(0<k<1\), it decreases. This change is global: a single Weibull shape cannot describe a hazard that rises early, falls later, and then rises again. Both the exponential and Weibull models impose a particular mathematical form across the full time range.
Worked Example: Estimating an Exponential Baseline
Suppose a hypothetical study follows five individuals in the absence of drug exposure. Three have events at 2, 4, and 5 months; two are censored at 3 and 6 months. Assume individuals share a constant baseline hazard, censoring is independent of event time, and event and censoring times are measured from a common origin.
For an observed event at time \(t_i\), the event-time likelihood contribution is \(\lambda_0\exp(-\lambda_0t_i)\). For a right-censored observation at \(t_i\), it is \(\exp(-\lambda_0t_i)\), the probability of remaining event-free through that time. Thus, with \(d\) events and total observed time \(T\), the likelihood is proportional to \(\lambda_0^d\exp(-\lambda_0T)\).
Here \(d=3\), and total observed time is \(T=2+4+5+3+6=20\) months. The log-likelihood, apart from a constant, is \(\ell(\lambda_0)=3\log(\lambda_0)-20\lambda_0\). Differentiating and setting the derivative to zero gives \(3/\lambda_0-20=0\), so \(\hat{\lambda}_0=3/20=0.1500\) per month. The second derivative is \(-3/\lambda_0^2<0\), confirming a maximum.
The fitted cumulative hazard at 4 months is \(\hat{H}_0(4)=0.1500(4)=0.6000\). Therefore, estimated event-free survival at 4 months is \(\hat{S}_0(4)=\exp(-0.6000)=0.5488\), or about 54.9%. This estimate depends on the constant-hazard assumption; it does not show that the underlying event rate was actually constant.
Worked Example: Reading a Weibull Baseline
Consider a hypothetical Weibull baseline with shape \(k=1.8\) and scale \(\eta=8\) months. These values describe an increasing baseline hazard because \(k>1\).
At 4 months, \(t/\eta=4/8=0.5\). The hazard is \(\lambda_0(4)=\frac{1.8}{8}(0.5)^{0.8}=0.225(0.5743)=0.1292\) per month. The cumulative hazard is \(H_0(4)=(0.5)^{1.8}=0.2872\), so survival is \(S_0(4)=\exp(-0.2872)=0.7504\).
At 8 months, the hazard is \(\lambda_0(8)=\frac{1.8}{8}(1)^{0.8}=0.2250\) per month, and survival is \(S_0(8)=\exp(-1)=0.3679\). The increase from 0.1292 to 0.2250 per month illustrates the time-varying hazard. The hazard values are instantaneous rates, whereas 0.7504 and 0.3679 are probabilities of remaining event-free through the stated times.
Flexible Baselines
A flexible baseline allows the event risk to change shape over time without requiring one constant rate or one global Weibull pattern. One common strategy is to model a transformed baseline hazard or cumulative hazard with a smooth function, such as a spline. The smoothness and number or placement of knots control how much the estimated shape can vary. Another broad strategy estimates baseline risk in time intervals; detailed piecewise-exponential construction is a separate modeling choice.
Flexibility does not mean that the hazard can be arbitrary without constraints. A hazard must be nonnegative, and a cumulative hazard must be nondecreasing. Smooth models also require choices about complexity and boundary behavior. Too little flexibility can miss meaningful changes; too much can fit random fluctuations, especially when few events occur.
Worked Example: Interpreting a Flexible Baseline
Suppose a hypothetical smooth-baseline fit to event-time data estimates cumulative baseline hazards of 0.10, 0.32, 0.70, and 1.05 at 2, 4, 6, and 8 months, respectively. These values are illustrative model estimates, not observations. Each is nonnegative, and they increase with time, as a cumulative hazard must.
The corresponding survival estimates are \(S_0(2)=\exp(-0.10)=0.9048\), \(S_0(4)=\exp(-0.32)=0.7261\), \(S_0(6)=\exp(-0.70)=0.4966\), and \(S_0(8)=\exp(-1.05)=0.3499\). For example, the model estimates about 35.0% event-free survival at 8 months.
Average hazard over an interval can be summarized by the cumulative-hazard increment divided by interval width. From 2 to 4 months this is \((0.32-0.10)/(4-2)=0.1100\) per month. From 4 to 6 months it is \((0.70-0.32)/2=0.1900\) per month, and from 6 to 8 months it is \((1.05-0.70)/2=0.1750\) per month. These averages suggest that risk does not follow a single constant rate. They are interval averages, not instantaneous hazard values at a particular time.
Choosing and Checking a Baseline
The assumptions should be considered before fitting. Exponential and Weibull models assume a common baseline form for individuals at the same modeled covariate values. Standard likelihood-based inference also relies on an appropriate event-time likelihood and censoring that is independent of the event process, at least conditional on modeled information. If dropout is related to unobserved event risk, treating it as ordinary independent right-censoring can bias estimates. The absence of drug exposure must also be meaningful for the study question; if zero exposure is outside the data-supported range, the baseline may represent an extrapolation.
Compare models using diagnostics and criteria suited to the analysis, not solely by which curve looks smoothest. Useful checks include plotting estimated survival against a nonparametric survival estimate, examining whether the fitted curve systematically misses early or late events, and inspecting uncertainty bands. A time trend in discrepancies may indicate that a constant hazard is too restrictive. A flexible curve that changes sharply where few events are observed may instead be overfitting.
Likelihood-based comparisons require care. AIC can compare fitted models on the same observations and likelihood definition, with lower values indicating a better balance of fit and parameter count. A likelihood-ratio test is appropriate for nested models under its regularity conditions; it is not automatically valid for every pair of differently parameterized baseline models. In all cases, compare predicted survival over the clinically relevant follow-up period, not just a single summary statistic.
| Baseline choice | What it allows | Main limitation |
|---|---|---|
| Exponential | Constant event rate over time | Cannot represent a changing baseline hazard |
| Weibull | Monotonically increasing or decreasing hazard, including the constant case | Imposes one global shape across follow-up |
| Flexible smooth baseline | More varied changes in risk over time | Requires complexity choices and may be unstable with sparse events |
These choices answer different questions about how much structure to impose. A simpler model may be more stable and easier to interpret when data are limited. A flexible baseline may better represent time patterns, but its apparent detail is not necessarily reliable. The next modeling step can add drug exposure to the hazard; the baseline choice should be assessed separately so that the underlying risk pattern is not confused with an exposure effect.
Common Pitfalls
- Calling the baseline hazard a probability. A hazard is a rate per unit time; survival or an interval event probability is a probability.
- Assuming exponential is neutral. A constant hazard is a strong time-pattern assumption, not merely a convenient notation.
- Interpreting Weibull shape without checking its parameterization. With the shape-scale definition used here, \(k>1\) means increasing hazard and \(k<1\) means decreasing hazard.
- Reading a flexible curve literally where events are sparse. A wiggly estimate may reflect uncertainty or excess model complexity rather than a real biological pattern.
- Attributing baseline misspecification to drug action. The baseline describes risk without exposure; inadequate baseline fit can distort a later exposure-effect estimate.
Check Your Understanding
Answer each question, then compare with the solutions below.
- An exponential baseline has rate 0.08 per day. What are its cumulative hazard and event-free survival at 5 days?
- For a Weibull baseline with \(k=0.7\), does the hazard increase, decrease, or remain constant with time? What is the hazard at \(t=\eta\)?
- A fitted flexible baseline has \(H_0(3)=0.20\) and \(H_0(6)=0.65\). Calculate survival at 6 and the average hazard from 3 to 6 time units.
- Why can a flexible baseline be less reliable when very few events occur late in follow-up?
Solutions. (1) \(H_0(5)=0.08(5)=0.4000\), so \(S_0(5)=\exp(-0.4000)=0.6703\). (2) Since \(k<1\), the Weibull hazard decreases with time; at \(t=\eta\), \(\lambda_0(\eta)=k/\eta=0.7/\eta\), in inverse time units. (3) \(S_0(6)=\exp(-0.65)=0.5220\), and the average hazard is \([0.65-0.20]/(6-3)=0.1500\) per time unit. (4) Sparse late events provide little information about the hazard shape, so a flexible fit may be uncertain or may follow random variation; late predictions should be interpreted with that limitation in mind.