I am using R/survival to analyze survival data from cancer patients. I have recently learned that there are many kinds of estimators for the Cox model, and I understand, that in theory, their order of accuracy is 1. exact, 2. Efron, 3. Breslow.

However, it is also possible to use 'robust' estimator of variance in R/survival, which sounds like a useful idea. However, this does not seem to work with the exact estimator. Secondly, I assume that Breslow is the standard estimator in many programs, such as Stata and SPSS, and furthermore, survdiff in the same R package calculates the log-rank test by using Breslow (without robust) by default.

So what is really the order of preference between the estimators? Efron with robust, or exact without robust? Breslow for publication technical reasons? This is very confusing for a statistician coming outside of the survival analysis community - and my guess is that most non-statisticians don't even know about the question.


2 Answers 2


If your data have few or no ties, all three methods will agree, so it doesn't matter.

If your data have many ties, the "exact" method is preferred, but it can be computationally expensive. Apparently in practice the difference is usually not that large; I recently asked Terry Thernau, the author of the R survival package, about this, and he wrote:

I have yet to find a data set, particularly a large one, where there is any statistical justification for using the discreete likelihood over the Efron approximation. I could be wrong, but await evidence.

Prof. Thernau prefers Efron to Breslow, but this probably also only makes a difference with large numbers of ties.


In the meantime, I have done some more inquiries, too. There are in fact two kinds of exact estimator: exact partial likelihood and exact marginal likelihood. Efron and Breslow are estimators of the exact marginal, whereas R's 'exact' is exact partial. With R's exact it's impossible to calculate all proportional hazards (PH) tests, so I resort to Efron.

None of this makes any difference, if you don't have ties (right?). The computational problems of exact are not that large, if you have a only a few ties.

The robust variance estimator seems to have some very strange qualities. Namely, it may in fact shrink the confidence intervals, creating false positives. That's not what you would like something "robust" to do. However, it's possible that I've got something wrong.

As a final remark, I tend to agree with Prof. Therenau. We have an obsession with everything 'exact', yet statistical models are inexact in the sense that they are almost always phenomenological, not mechanistic in the physical sense. I.e. all models are wrong, but some are useful.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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