10
$\begingroup$

Can someone help me find a way to estimate the variance of the Kaplan-Meier estimate with dependent observations? Specifically, I have failure time data from patients with several different observations for each patient (and different patients may have different number of observations). The observations for different patients are assumed to be independent but observations from the same patient are expected to be dependent.

I was suggested two publications, "Kaplan-Meier Analysis of Dental Implant Survival: A Strategy for Estimating Survival with Clustered Observations" and "The Kaplan-Meier Estimate for Dependent Failure Time Observations", the second of which is cited by the first.

However I was unable to make sense of these. The first has errors and the second seems far more rigorous but the equation for the variance does not make sense (double integration on a 1-form).

Improve this question
$\endgroup$
6
  • $\begingroup$ How does the dependence manifest itself in your case? Can you describe your situation in more detail? $\endgroup$
    – cardinal
    Dec 21, 2011 at 15:27
  • $\begingroup$ @cardinal Of course, should have remembered that. Added details to the question now. $\endgroup$
    – Anton
    Dec 21, 2011 at 16:05
  • $\begingroup$ So, do the failures happen one after another, i.e., similar to a renewal process, or do they happen simultaneously? For the latter, I'm imagining something like time to organ failure (each organ being considered individually) for a terminally ill patient (forgive the imagery), whereas the former might be something like, time between epileptic seizures. $\endgroup$
    – cardinal
    Dec 21, 2011 at 17:24
  • $\begingroup$ @cardinal They are individual and may or may not happen simultaneously. $\endgroup$
    – Anton
    Dec 22, 2011 at 10:20
  • $\begingroup$ @Anton if the events concern different outcomes, then you simply produce Kaplan Meier curves for each outcome. Or are you interesting in estimating a multidimensional KM that can covary for specific outcomes at different points in time? $\endgroup$
    – AdamO
    Feb 1, 2018 at 19:32

1 Answer 1

Reset to default
1
$\begingroup$

In this situation, multiple events of the same type can occur in parallel within the same group. For the first reference, grouping is of multiple implanted teeth within the same individual. For the second reference, by Ying and Wei, litters are the groups, and a multiple event is more than 1 member of a litter developing a tumor.

This within-group dependence can be taken into account with an infinitesimal jackknife variance estimator, which can be thought of as related to removing one group at a time from the model.* In the R survival package, that's done for a simple Kaplan-Meier estimate by specifying an id variable for each group.

The data used in the example by Ying and Wei turn out to be a subset of the rats data included in the survival package (untreated females). The standard errors (SE) with the jackknife are very close to those reported by Ying and Wei.

time KM estimate SE, uncorrected SE, Ying and Wei SE, jackknife
70 0.9190 0.028 0.026 0.0265
80 0.8733 0.034 0.032 0.0320
90 0.8227 0.041 0.046 0.0466
100 0.8074 0.043 0.047 0.0477

Code that does this in R:

> library(survival)
> fitJackknife <- survfit(Surv(time,status)~1,data=rats, subset=rx==0&sex=="f",id=litter)
> summary(fitJackknife,times=c(70,80,90,100))
Call: survfit(formula = Surv(time, status) ~ 1, data = rats, subset = rx == 
    0 & sex == "f", id = litter)

 time n.risk n.event survival std.err lower 95% CI upper 95% CI
   70     88       8    0.919  0.0265        0.868        0.972
   80     71       4    0.873  0.0320        0.813        0.938
   90     63       4    0.823  0.0466        0.736        0.919
  100     50       1    0.807  0.0477        0.719        0.907

*The infinitesimal jackknife is a limiting case of the original jackknife, in which one observation at a time is removed (weight = 0) while others are maintained (weights = 1). It's the liming situation as weights approach 0. Therneau and Grambsch explain in Section 7.2 how that's implemented to get variance estimates in survival models.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.