Independent censoring is one of the fundamental assumptions in the survival analysis. However, I cannot find any test for it or any paper which discusses how real that assumption is. I would be grateful if anybody could point me to some useful references. I have found the following paper as an interesting reference but it is not freely available.

Leung, Kwan-Moon, Robert M. Elashoff, and Abdelmonem A. Afifi. "Censoring issues in survival analysis." Annual review of public health 18.1 (1997): 83-104.

Any feedback would be much appreciated. Kind regards DK

  • $\begingroup$ Kalbfleisch and Prentice (2002) The Statistical Analysis of Failure Time Data, is a classical textbook on survival analysis. $\endgroup$
    – utobi
    Nov 29, 2016 at 10:49

1 Answer 1


Independent censoring is the requirement that the hazard rate of an at-risk subject coincides with the hazard rate in the surviving population, i.e.

$$ \lim_{\Delta t \rightarrow 0} \frac{\Pr \Big( t \leq T < t + \Delta t \,\Big|\, T \geq t, \, C \geq t \Big)}{\Delta t} = \lim_{\Delta t \rightarrow 0} \frac{\Pr \Big( t \leq T < t + \Delta t \,\Big|\, T \geq t \Big)}{\Delta t} $$

This requirement essentially means that the uncensored subjects under follow-up must be representative of the surviving population; a condition that is satisfied when censoring occurs independently of the survival time (e.g., censoring due to calendar termination of the study). If there are covariates, then the independent censoring assumption is made conditional on the covariate information.

  • $\begingroup$ Thank you Ocram for your reply. Much appreciated. However, I am not sure that I can agree with you. If I am not mistaken, your equation shows that survival time does not depend on censoring time, but not the other way around. In my opinion, your equation does not show that censoring time is independent of survival time. The independent censoring means that both variables (survival time, censoring time) are independent. $\endgroup$ Nov 29, 2016 at 16:47
  • $\begingroup$ If Pr(A|B) = Pr(A) then Pr(B|A) = Pr(B). $\endgroup$
    – ocram
    Nov 30, 2016 at 6:50
  • $\begingroup$ The above equation is not Pr(A|B)=Pr(A) but Pr(X|A,B) = Pr(X|A). I argue that if Pr(X|A,B) = Pr (X|A) then it does not mean that Pr(X|B) = Pr(X|B,A) $\endgroup$ Nov 30, 2016 at 14:04
  • $\begingroup$ In your notations, Pr(X|A,B) = Pr (X|A) means that the probability to die at time t does not depend on the censoring status at time t. This is non-informative censoring. On the other hand, Pr(X|A,B) = Pr(X|B) would mean that the probability to die at time t does not depend on the surviving status at time t. Of course, the probability to die does depend on the surviving status. Am I missing something to your point?? $\endgroup$
    – ocram
    Nov 30, 2016 at 14:30
  • $\begingroup$ Thank you very much for looking at this problem with me. You have excellently explained the difference between non-informative and independent censoring in a previous post stats.stackexchange.com/questions/22497/… However, my problem is the above equation in your answer. I might be wrong, but from it I can only conclude that survival times do not depend on censoring times, but not the other way around. In short, in your answer you have italic section "a condition,,," with which I have difficulty of agreeing. $\endgroup$ Dec 1, 2016 at 5:50

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