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Absolute Risk Prediction · Tutorial 89 of 699

Baseline Hazard Choices For Absolute Risk

Learn how baseline cumulative-hazard estimation and parameterization affect absolute-risk predictions, and how to check that the chosen estimates and CIF calculation are consistent.

Advanced 11 min read

What You'll Learn

  • Define the Breslow and Efron baseline cumulative-hazard estimators for a cause-specific Cox model.
  • Distinguish a baseline hazard referenced to zero covariates from one referenced to centered covariates.
  • Calculate how competing-cause cumulative hazards determine event-free probability and cause-specific CIFs.
  • Explain why different baseline estimates can produce different absolute risks even when relative effects are similar.
  • Identify tie, centering, sparse-risk-set, and hazard-to-probability pitfalls.

Why the Baseline Choice Matters

A relative hazard model describes how covariates multiply a hazard, but an absolute-risk prediction also needs its baseline level. In a cause-specific Cox model, the regression coefficients describe relative differences between covariate profiles, while the estimated baseline cumulative hazards set the scale of risk over time. Consequently, two implementations can give similar relative hazard estimates yet different cumulative incidence function (CIF) predictions.

The baseline estimate is not a cosmetic detail. It depends on how tied event times are handled, on the reference point used to report the baseline, and on how estimated cumulative hazards are converted to event-free probabilities and CIFs. These choices must be compatible with one another. The earlier material on cause-specific Cox fits and absolute risk provides the model context; here the focus is the baseline component of those predictions.

Definition: For cause \(k\), a cause-specific Cox model has hazard \(\lambda_k(t\mid z)=\lambda_{k0}(t)\exp(\beta_k^\mathsf{T}z)\), where \(z\) is the covariate vector, \(\beta_k\) is the cause-specific coefficient vector, and \(\lambda_{k0}(t)\) is the baseline hazard at \(z=0\). The baseline cumulative hazard is \(H_{k0}(t)=\int_0^t\lambda_{k0}(u)\,du\). A fitted model estimates both \(\beta_k\) and \(H_{k0}(t)\).

The baseline cumulative hazard is a nondecreasing function, usually estimated as a step function with jumps at observed event times. Given an estimated coefficient vector, define the risk score for subject \(i\) as \(r_{ik}=\exp(\hat\beta_k^\mathsf{T}z_i)\). The risk set \(R(t_j)\) contains subjects still under observation and event-free just before time \(t_j\).

Breslow and Efron Baseline Estimates

For a cause-specific Cox model, the Breslow estimator assigns a jump at each cause-\(k\) event time. If \(d_{kj}\) events of cause \(k\) occur at time \(t_j\), its estimated jump is

$$ \Delta\widehat H_{k0}^{\,B}(t_j) = \frac{d_{kj}}{\displaystyle\sum_{i\in R(t_j)}r_{ik}}. $$

The denominator is the sum of the fitted relative hazards among those at risk. The cumulative estimate is the sum of these jumps up to the time of interest. The formula is the baseline counterpart of the Breslow treatment of tied event times in the partial likelihood.

When there are tied events, another common choice is the Efron approach. At a tied time, let \(D_{kj}\) be the subjects with cause-\(k\) events, and let \(E_{kj}=\sum_{i\in D_{kj}}r_{ik}\). The Efron baseline jump is

$$ \Delta\widehat H_{k0}^{\,E}(t_j) = \sum_{\ell=0}^{d_{kj}-1} \frac{1}{ \displaystyle\sum_{i\in R(t_j)}r_{ik} -\frac{\ell}{d_{kj}}E_{kj} }. $$

For a single event at a time, the two formulas coincide. With tied events, the Efron denominators remove an increasing fraction of the tied subjects’ risk scores across the tied failures. Its jump will generally differ from the Breslow jump. The coefficient estimation and baseline estimation should use a compatible tie convention; it is not sensible to treat the resulting predictions as if the convention had no effect.

Worked Example: A Tied-Time Baseline Jump

Suppose a hypothetical cause-specific Cox fit has two cause-1 events tied at time 4. There are six people at risk, and their fitted risk scores sum to 10. The two people with tied events have risk scores summing to 4. Calculate the Breslow and Efron baseline jumps at this time.

For Breslow, \(d_{1j}=2\) and the risk-score sum is 10, so

$$ \Delta\widehat H_{10}^{\,B}(4)=\frac{2}{10}=0.2000. $$

For Efron, the tied-event score sum is \(E_{1j}=4\). The two denominators are \(10\) and \(10-(1/2)(4)=8\), giving

$$ \Delta\widehat H_{10}^{\,E}(4) =\frac{1}{10}+\frac{1}{8} =0.2250. $$

The Efron jump is larger in this example because its second denominator is smaller than the full risk-score sum. This comparison uses the stated fitted scores and risk set; it does not imply that Efron must always give a larger jump. With few ties, the two estimates may be close. If ties are frequent, the choice can meaningfully affect predicted risks.

Reference Points and Centering

Software may report a baseline cumulative hazard for covariates set to zero, or for covariates set to a reference vector such as their sample means. These are different parameterizations of the same fitted hazard model when the linear predictor is adjusted consistently. Let \(z_{\mathrm{ref}}\) be the reference vector and let \(\widehat H_{k,\mathrm{ref}}(t)\) be the cumulative hazard for that reference. Then the hazard for profile \(z\) is represented by

$$ \widehat H_k(t\mid z) = \widehat H_{k,\mathrm{ref}}(t) \exp\!\left\{\hat\beta_k^\mathsf{T}(z-z_{\mathrm{ref}})\right\}. $$

This is algebraically equivalent to using a zero-covariate baseline, provided that \(\widehat H_{k0}(t)=\widehat H_{k,\mathrm{ref}}(t)\exp(-\hat\beta_k^\mathsf{T}z_{\mathrm{ref}})\). The reference point changes the reported baseline level, not the fitted hazard for a given profile. Errors arise when a centered baseline is combined with an uncentered linear predictor, or vice versa.

Worked Example: Applying a Centered Baseline

Suppose a hypothetical one-cause model has coefficient \(\hat\beta=0.5\) for a binary covariate \(z\). The software reports a cumulative baseline hazard of 0.2000 at a reference value \(z_{\mathrm{ref}}=1\), at a specified horizon. Find the cumulative hazard and risk for \(z=0\) and \(z=1\), using \(1-\exp(-H)\) to convert cumulative hazard to risk.

For \(z=0\), the multiplier relative to the reference is \(\exp\{0.5(0-1)\}=\exp(-0.5)=0.6065\). Thus the cumulative hazard is \(0.2000(0.6065)=0.1213\), and the risk is

$$ 1-\exp(-0.1213)\approx 0.1142. $$

For \(z=1\), the profile equals the reference, so its cumulative hazard is 0.2000 and its risk is \(1-\exp(-0.2000)=0.1813\). The result for \(z=0\) would be wrong if the centered baseline were treated as a zero-covariate baseline: that mistake would assign cumulative hazard 0.2000 to \(z=0\), rather than 0.1213.

From Cause-Specific Baselines to CIFs

For competing risks, each cause has its own baseline cumulative hazard and coefficient vector. At a profile \(z\), the fitted cause-specific cumulative hazard is \(\widehat H_k(t\mid z)=\widehat H_{k0}(t)\exp(\hat\beta_k^\mathsf{T}z)\), with the reference adjustment used if the baseline is centered. The cause-specific cumulative hazards must be combined: event-free probability depends on all causes, not just the cause whose CIF is being predicted.

One useful derivation applies when the cause-specific hazards are constant over an interval, or when their cumulative hazards over a horizon have a constant cause proportion. Write \(A_k\) for cause \(k\)'s cumulative hazard over that horizon and \(A=\sum_k A_k\). Event-free probability is \(\exp(-A)\), so total failure probability is \(1-\exp(-A)\). The fraction of the total hazard assigned to cause \(k\) is \(A_k/A\). Therefore,

$$ F_k = \frac{A_k}{A}\{1-\exp(-A)\}, \qquad A>0. $$

This formula is not a general replacement for integrating CIFs when cause proportions change over time. In general, the CIF accumulates each cause’s hazard weighted by the probability of having had no earlier event. But it gives a transparent calculation for a single interval with constant hazards and shows why changing one baseline can affect more than one cause’s CIF: it changes both the cause’s contribution and overall event-free probability.

Worked Example: Baseline Choice Changes Both CIFs

Consider a hypothetical two-cause model at one prediction horizon. Suppose two plausible baseline estimation choices, with the same relative effects for the profile, produce cumulative hazards \(A_1=0.3600\) and \(A_2=0.1125\) under choice A. Under choice B, suppose they produce \(A_1=0.3400\) and \(A_2=0.1500\). Assume cause-specific hazards are constant over the horizon, so the preceding formula applies.

Under choice A, \(A=0.3600+0.1125=0.4725\). The total event probability is

$$ 1-\exp(-0.4725)=0.3765583\ldots\approx 0.3766. $$

The cause proportions are \(0.3600/0.4725=0.7619048\) and \(0.1125/0.4725=0.2380952\). Carrying full precision through the products gives

$$ F_{1,A}=0.7619048(0.3765583\ldots)=0.2869, \qquad F_{2,A}=0.2380952(0.3765583\ldots)=0.08966\ldots\approx 0.0897. $$

Under choice B, \(A=0.3400+0.1500=0.4900\), and \(1-\exp(-0.4900)=0.3873736\ldots\). Hence

$$ F_{1,B}=\frac{0.3400}{0.4900}(0.3873736\ldots)=0.2688, \qquad F_{2,B}=\frac{0.1500}{0.4900}(0.3873736\ldots)=0.1186. $$

The CIFs differ because the baseline choices changed the cause-specific cumulative hazards. In each calculation, the CIFs sum to the total event probability up to rounding: about 0.3766 under A and 0.3874 under B. These are hypothetical model outputs, not estimates from a real study.

Probability Conversion, Diagnostics, and Alternatives

A cumulative hazard is not itself an event probability. For one cause, \(1-\exp\{-H(t)\}\) converts cumulative hazard to risk under the corresponding survival model. With competing causes, the correct calculation must also account for the other causes. For example, applying \(1-\exp\{-H_k(t\mid z)\}\) to each cause separately generally produces net risks, not competing-risk CIFs, and those values need not sum to the overall probability of failure.

A practical implementation must also choose how to handle event-time jumps. A product-integral calculation based on conditional event probabilities updates survival and CIFs by multiplying the event-free probability by the probability of no event at each step. When model-based cumulative-hazard increments are used, some prediction procedures instead use an exponential conversion over each interval; under constant cause-specific hazards within an interval, the conditional event probability is \(1-\exp(-\sum_k\Delta H_k)\), allocated across causes in proportion to their increments. These approaches agree to first order for small increments, but need not agree for large jumps. State the convention and use it consistently; do not treat a hazard increment as a probability without justification.

The Aalen-Johansen estimator is a nonparametric alternative based on observed risk sets and event counts. It does not require fitting cause-specific Cox relative effects, but it does not provide the same covariate-adjusted model predictions. Fine-Gray models use a subdistribution-hazard formulation rather than combining cause-specific hazard predictions in this way. These approaches answer related prediction questions, but their baseline quantities should not be substituted into a cause-specific Cox calculation.

Several checks help reveal implementation problems:

  • Confirm that each cause’s denominator uses the appropriate risk set and fitted risk scores, and that the chosen tie method matches the model fit.
  • Check that the baseline reference point and the profile’s linear predictor use the same centering convention.
  • Inspect jumps and risk-set sizes. Large jumps or very small late risk sets can make absolute predictions sensitive to individual event times and modeling choices.
  • Verify that predicted event-free probability and all CIFs are nonnegative and that their sum is consistent with the total failure probability under the chosen calculation.
  • Compare predictions across reasonable baseline and tie-handling choices, especially when ties are common or the prediction horizon is late in follow-up.

These diagnostics do not establish that the model is well calibrated or that its assumptions hold. The Cox model still relies on proportional cause-specific hazards and suitable handling of censoring; baseline estimation cannot repair violations. Nor does a close match between two baseline methods guarantee accurate predictions outside the observed support of follow-up.

Common pitfall: Treating a software-reported centered baseline as if it were the baseline at zero covariates, then multiplying it by \(\exp(\hat\beta^\mathsf{T}z)\), changes the fitted hazards. Use the centered form \(\widehat H_{\mathrm{ref}}\exp\{\hat\beta^\mathsf{T}(z-z_{\mathrm{ref}})\}\), or convert the baseline to the zero-covariate scale first. A second frequent error is converting each cause’s cumulative hazard to a separate net risk and calling the results CIFs; competing-risk CIFs require accounting for event-free probability across all causes.
Key takeaway: Absolute-risk predictions depend on the estimated baseline cumulative hazards as well as the relative effects. Choose a tie convention, align the baseline reference with the linear predictor, and combine all causes using a stated hazard-to-probability calculation.

Check Your Understanding

Answer each question, then compare with the brief solutions.

  1. At a tied event time, two events occur and the total risk-score sum is 12. The tied-event risk scores sum to 4. What are the Breslow and Efron baseline jumps?
  2. A baseline is reported at \(z_{\mathrm{ref}}=2\). What covariate contrast belongs in the linear predictor when predicting at \(z=1\)?
  3. Why is \(1-\exp\{-H_1(t)\}\) generally not the cause-1 CIF when cause 2 is also possible?
  4. For constant cause-specific hazards over a horizon, \(A_1=0.20\) and \(A_2=0.10\). Calculate the total event probability and both CIFs.

Solutions. (1) Breslow gives \(2/12=0.1667\). Efron gives \(1/12+1/(12-2)=0.1833\). (2) Use \(z-z_{\mathrm{ref}}=1-2=-1\), so the multiplier is \(\exp(-\hat\beta)\). (3) The cause-1 net-risk calculation ignores the chance that cause 2 occurs first; the CIF weights cause-1 hazard by event-free probability across both causes. (4) \(A=0.30\), so total event probability is \(1-\exp(-0.30)=0.2592\). The cause proportions are \(2/3\) and \(1/3\), giving \(F_1=(2/3)(0.2592)=0.1728\) and \(F_2=(1/3)(0.2592)=0.08639\). Their sum is 0.2592, as required.

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