I want to show the "mirror" to the survival probability calculated with the Kaplan-Meier estimator, that is the cumulative incidence (or probability).

I would like to add the pointwise confidence interval. In my software it is provided for the Kaplan-Meier, but not for the CIF=1-KM.

I guess it's as easy as just swapping the ends and subtract from 1, that is [0.2 ; 0.6] --> [1-0.6 ; 1-0.2] --> [0.4, 0.8]

That's the Greenwood plain one. But I want also the one on the log-scale. It is asymmetric. But can I just swap the KM CI for the survival and get the CI for my cumulative incidence?

By the way, is there any proper name for the "log" CI? I mean - "a variance estimator calculated using the cumulative hazard" or maybe "Greenwood estimator on the log-scale"?

Just to make 100% clear - I have NO competing risks. This is not competing risk adjusted CIF. This is just pure 1-KM cumulative probability with just one kind of event and censoring.


1 Answer 1


The cumulative incidence function with just one event possible for each individual is, as you note, $F(t)=1-S(t)$, where $S(t)$ is the survival function over time. That's by definition.

Whether $S(t)$ is estimated by Kaplan-Meier or other methods doesn't matter. For confidence limits, you can just replace $S(t)$ in that formula with its upper confidence limit and lower confidence limit, as you suggest. Again, it doesn't matter how the confidence limits were determined or whether they are symmetric.

Klein and Moeschberger use the simple phrase "log-transformed confidence interval" on page 105, so that should do.

  • $\begingroup$ +1, but for future readers note that, in general, it is not sufficient to evaluate the transformed variable at the endpoints of a CI (though it works here). For instance, if $(-1, 2)$ is a CI for $\theta$ and $\phi = \theta^2$ it would be a mistake to assume that $(1, 4)$ is the corresponding CI for $\phi$. Instead, consider all values inside the CI to obtain $(0, 4)$. $\endgroup$
    – knrumsey
    Sep 1, 2023 at 18:10

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