Pointwise Confidence Intervals for Cumulative Incidence

Why an Estimate Needs an Interval

An Aalen–Johansen estimate gives a cumulative incidence at a chosen time, but the estimate varies from sample to sample. A confidence interval expresses uncertainty about the population cumulative incidence, rather than treating the observed estimate as exact. In this tutorial, the target is pointwise: the cumulative incidence for one specified cause at one fixed time \(t\).

The estimate and its variance come from the Aalen–Johansen estimator and its Aalen variance estimate. The earlier material on Aalen variance gives the variance calculation; here, the focus is how to use that estimated variance to form and interpret intervals, and how to choose an interval scale.

Definition: For cause \(k\) and a fixed time \(t\), let \(F_k(t)\) be the population cumulative incidence and let \(\hat F_k(t)\) be its Aalen–Johansen estimate. Write \(\hat V_k(t)\) for the Aalen estimated variance and \(\widehat{SE}_k(t)=\sqrt{\hat V_k(t)}\). A pointwise \(100(1-\alpha)\%\) confidence interval is an interval constructed to cover \(F_k(t)\) with approximate repeated-sampling probability \(1-\alpha\), under the model and regularity conditions used to justify the procedure.

Conditions for Using the Intervals

The usual intervals rely on large-sample approximations for the Aalen–Johansen estimate. The relevant assumptions include independent subjects sampled from the population of interest, accurate classification of the event cause, and censoring that is independent of the event process in the sense required for the estimator. If censoring depends on observed covariates, estimation must account for those covariates or use an appropriate conditional independent-censoring assumption. A subject must be at risk and observable in sufficient numbers through the chosen horizon; an estimate near the end of follow-up can be based on very little information.

At an interior probability, meaning \(0<F_k(t)<1\), the estimator is approximately normal after standardization:

\frac{\hat F_k(t)-F_k(t)}{\widehat{SE}_k(t)} \ \dot{\sim}\ N(0,1).

The approximation is asymptotic, not an exact finite-sample distribution. It is most credible when there is adequate information in the relevant risk sets and the estimate is not so close to a boundary that its sampling distribution is strongly skewed. Tied event times can be accommodated in the Aalen–Johansen estimate and variance calculation; ties do not, by themselves, make a normal interval exact.

The Direct Normal Interval

Let \(z_{1-\alpha/2}\) be the \(1-\alpha/2\) quantile of the standard normal distribution. The direct, or Wald, interval adds and subtracts a normal critical value times the estimated standard error:

\hat F_k(t)\ \pm\ z_{1-\alpha/2}\widehat{SE}_k(t).

For a 95% interval, \(z_{0.975}\) is approximately 1.96. The arithmetic is simple, and the interval can be useful when the estimate is comfortably away from zero and one and the standard error is small relative to that distance. However, the formula does not enforce the fact that a cumulative incidence must lie in \([0,1]\). A direct interval can therefore extend below zero or above one. Truncating an endpoint to the probability range is sometimes reported, but that alteration does not make the interval an exact 95% procedure or repair a poor normal approximation.

Worked Example: A Direct Interval With Complete Follow-Up

Consider a hypothetical cohort of 100 independent individuals whose event status is known by a fixed horizon. By that horizon, 28 have experienced cause A, 22 have experienced a competing cause, and 50 have had neither event. There is no censoring before the horizon. The estimated cumulative incidence for cause A is therefore \(28/100=0.28\). In this simple complete-follow-up setting, the cause-A indicator has estimated variance \(0.28(1-0.28)/100\), so:

\hat F_A(t)=0.28,\qquad \widehat{SE}_A(t) =\sqrt{\frac{0.28(0.72)}{100}} =\sqrt{0.002016} \approx 0.0449.

The 95% direct interval is:

0.28\pm1.96(0.0449) =0.28\pm0.0880 =(0.192,\ 0.368).

The interval lies within \([0,1]\). Under the stated sampling model and the normal approximation, it is reasonable to report an estimated cause-A cumulative incidence of 28% by the horizon, with an approximate 95% pointwise confidence interval from 19.2% to 36.8%. The competing events are not treated as censoring in defining the cause-A outcome.

A Logit-Transformed Interval

A logit transformation maps probabilities strictly between zero and one to the whole real line:

g(p)=\log\!\left(\frac{p}{1-p}\right), \qquad g^{-1}(x)=\frac{1}{1+\exp(-x)}.

If \(0<\hat F_k(t)<1\), the delta method gives the estimated standard error on the logit scale. Since \(g'(p)=1/[p(1-p)]\), the transformed standard error is approximately \(\widehat{SE}_k(t)/[\hat F_k(t)\{1-\hat F_k(t)\}]\). Construct a normal interval on that scale, then apply the inverse logit to both endpoints:

\left[ g^{-1}\!\left\{g(\hat F_k(t))-z_{1-\alpha/2} \frac{\widehat{SE}_k(t)}{\hat F_k(t)\{1-\hat F_k(t)\}}\right\}, \quad g^{-1}\!\left\{g(\hat F_k(t))+z_{1-\alpha/2} \frac{\widehat{SE}_k(t)}{\hat F_k(t)\{1-\hat F_k(t)\}}\right\} \right].

Because the inverse logit always lies strictly between zero and one for finite inputs, this interval respects the probability bounds. That is a feature of its scale, not a guarantee of good coverage: the logit approximation can also be inaccurate when information is sparse or the estimate is at a boundary.

Derivation: For \(0<F_k(t)<1\), approximate \(g(\hat F_k(t))\) as normal with mean \(g(F_k(t))\). The delta method estimates its standard error by \(g'(\hat F_k(t))\widehat{SE}_k(t)\). Thus a normal interval for \(g(F_k(t))\) has endpoints \(g(\hat F_k(t))\pm z_{1-\alpha/2}g'(\hat F_k(t))\widehat{SE}_k(t)\). Since \(g\) is increasing, applying \(g^{-1}\) to both endpoints gives an interval for \(F_k(t)\) in \((0,1)\).

Worked Example: Comparing Scales Away From a Boundary

Return to the hypothetical cohort with 28 cause-A events among 100 individuals by the horizon. The direct interval was \((0.192,0.368)\). For the logit interval, the estimate on the transformed scale is:

g(0.28)=\log(0.28/0.72)\approx-0.9445, \qquad SE_g\approx\frac{0.0449}{0.28(0.72)} =\frac{0.0449}{0.2016} \approx0.2227.

The 95% interval on the logit scale is \(-0.9445\pm1.96(0.2227)\), or approximately \((-1.381,\,-0.508)\). Applying the inverse logit to each endpoint gives:

\left( \frac{1}{1+\exp(1.381)},\ \frac{1}{1+\exp(0.508)} \right) \approx(0.201,\ 0.376).

Both intervals are plausible and fairly similar for this estimate, which is not especially close to either boundary. Their difference reflects the different scales on which normality is approximated; neither interval is guaranteed to be more accurate in every finite sample.

Boundary Behavior and Sparse Information

The contrast between scales is more visible near zero or one. For example, suppose a hypothetical cohort of 40 individuals has complete event status by the horizon: two experience cause A, eight experience a competing cause, and 30 have no event. Then \(\hat F_A(t)=2/40=0.05\), and the estimated standard error in this complete-follow-up setting is:

\widehat{SE}_A(t) =\sqrt{\frac{0.05(0.95)}{40}} =\sqrt{0.0011875} \approx0.0345.

The direct 95% interval is \(0.05\pm1.96(0.0345)\), approximately \((-0.0175,\,0.1175)\). Its negative lower endpoint is impossible for a cumulative incidence. If clipped to the probability range, it becomes \((0,\,0.1175)\), but the nominal coverage of this modified procedure is not automatically 95%.

For the logit interval, \(g(0.05)=\log(0.05/0.95)\approx-2.944\), and \(SE_g\approx0.0345/[0.05(0.95)]\approx0.725\). The transformed endpoints are \(-2.944\pm1.96(0.725)\), or approximately \((-4.365,\,-1.523)\). Transforming back gives approximately \((0.0126,\,0.179)\). This avoids an impossible endpoint, but only two cause-A events provide little information; the transformed interval should not be mistaken for a remedy for sparse data.

Worked Example: Pointwise Intervals Are Not Joint Intervals

In the hypothetical cohort of 100 individuals from the first example, the estimated cumulative incidence for the competing cause is \(22/100=0.22\). Its standard error under the same complete-follow-up calculation is:

\widehat{SE}_B(t) =\sqrt{\frac{0.22(0.78)}{100}} =\sqrt{0.001716} \approx0.0414.

The direct 95% interval for cause B is \(0.22\pm1.96(0.0414)\), which is approximately \((0.139,\,0.301)\). Together with the cause-A interval \((0.192,\,0.368)\), this gives two separate 95% pointwise intervals at the same horizon. It does not mean there is a 95% probability that both intervals simultaneously cover their respective population cumulative incidences. A family of intervals across multiple causes or times needs a procedure designed for joint coverage if that is the inferential goal.

Choosing and Checking an Interval

The direct normal interval is transparent and often adequate for estimates well inside the probability range with reasonably stable variance estimates. The logit interval respects probability bounds and often produces an asymmetric interval, which can better reflect the constrained scale. In the special case of complete follow-up at a single horizon, a cumulative incidence for one cause is a binary outcome proportion; a binomial interval designed for proportions is another possible approach. That special-case method does not directly replace an Aalen–Johansen interval when there is right censoring and event timing matters.

Before reporting an interval, check that the chosen horizon is supported by the observed follow-up, that the risk set is not extremely small, and that the variance estimate is finite and credible. Review whether the estimate is near zero or one, whether the direct interval crosses a boundary, and whether the logit calculation is defined. If \(\hat F_k(t)=0\) or \(1\), its logit is undefined; do not insert that value into the logit formula. Boundary estimates or very sparse events call for a method appropriate to that setting and careful qualification, rather than an automatic transformation.

These are pointwise intervals: each addresses one cause at one specified time. Reading intervals at many times as though they formed a simultaneous confidence band is a common error. Likewise, separate 95% intervals for several causes do not generally provide 95% joint coverage. The interval describes uncertainty in an estimated population risk; it is not a probability statement that the fixed population parameter lies inside the particular computed interval.

Common Mistakes

Reporting impossible Wald limits without comment. The direct formula can produce values below zero or above one. Report the issue and consider a bounded-scale method; clipping alone does not establish nominal coverage.
Treating a transformation as a cure for small samples. A logit interval respects probability bounds, but its approximation can still be poor with few events or a small risk set.
Calling pointwise intervals simultaneous. A 95% interval at each separate time or cause does not imply 95% coverage for all of them together.
Interpreting confidence as a probability about the parameter. In repeated samples, the procedure has its stated approximate coverage under its assumptions; after calculation, the population cumulative incidence is fixed.

Key takeaway: Use the Aalen–Johansen estimate and its estimated standard error to construct a fixed-time interval. The direct normal interval is simple; the logit interval respects probability bounds when the estimate is strictly between zero and one. Both rely on adequate information and large-sample approximations, and both must be interpreted as pointwise.

Check Your Understanding

Answer each question, then compare your reasoning with the brief solutions.

If an estimate is \(0.40\) with standard error \(0.05\), calculate the direct 95% interval.
Why is a logit interval undefined when the estimated cumulative incidence is exactly zero?
A direct interval for a cause-specific cumulative incidence is \((-0.03,\,0.12)\). What is the problem, and why does clipping the lower endpoint not guarantee 95% coverage?
Do separate 95% pointwise intervals for two causes at one time necessarily give 95% joint coverage? Explain.

Solutions. (1) \(0.40\pm1.96(0.05)=0.40\pm0.098\), giving \((0.302,\,0.498)\). (2) The logit contains \(\log\{p/(1-p)\}\), which is not finite at \(p=0\); the transformed standard error formula also divides by \(p(1-p)\). (3) A cumulative incidence cannot be negative, so the direct normal approximation has produced an impossible endpoint. Clipping changes the procedure, and its repeated-sampling coverage is not automatically the nominal 95%. (4) No. Each interval has pointwise coverage, but their joint coverage depends on their dependence and the joint procedure; two marginal 95% intervals do not generally have 95% simultaneous coverage.