Since the KM estimator does not allow for any other state other than the event of interest is it safe to assume that $S(t=\infty) = 0$ ? Or can it be that $S(t=\infty) > 0$
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2 Answers
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When different from zero at the largest event time (i.e. when the largest observation is censored), the Kaplan-Meier estimator is usually undefined from that point on. There exist methods for completing the Kaplan-Meier estimator (for example, see here). In any case, the underlying survival function decreases to zero, $S(\infty) = 0$ even if $\hat{S}(\text{largest event time}) \neq 0$. Models that allow for $S(\infty) \neq 0$ are called cure models.
EDIT related to your comment below
The first sentence of this paper, which shows how to fit a cure model, says "In survival analysis, it is usually assumed that if complete follow-up were possible for all individuals, each would eventually experience the event of interest." That is, $S(\infty) = 0$, by assumption. Almost any other paper on cure models starts in the same way.
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$\begingroup$ Thanks. I am trying to find a citable reference for $S(\infty) = 0$ . Do you happen to know any? $\endgroup$– ECIIJan 3, 2014 at 13:41
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The Kaplan-Meier method computes the actual observed percent survival at each time a subject dies in your experiment. It describes your data, taking into account censoring. No theoretical model. No extrapolation.
Any assumption about survival at infinite times needs to be based on a model, so is beyond Kaplan and Meier.
Of course, it is hard to imagine any model of survival that doesn't end up at zero survival at long time points :)
answered Jan 3, 2014 at 18:08
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$\begingroup$ Thanks, my question was more theoretical on actuarial survival analysis without competing risk rather than focused on the KM estimator. BTW your book is one of my favorite (I read the first edition when I was a student). Pity to see GP6 now has....wait for it... pie charts. Just joking. $\endgroup$– ECIIJan 5, 2014 at 14:13