I have a sample of observations where about $30\%$ of the observations are right-censored. I want to fit a kernel density estimator to this sample but I have not found a standard method to do so. Is there any widely accepted methodology for fitting a KDE in the presence of censored observations?

  • $\begingroup$ Hmm, it is similar, but not quite the same situation as in zero-inflated data. In spatial statistics with truncated data you can weight points closer to the boundary, but that logic is not easily translated to censored data off-hand. $\endgroup$
    – Andy W
    Feb 9, 2016 at 14:54

2 Answers 2


Yes, it's possible. In order to do anything with censored data you need to assume at least 'censoring at random', which says that the hazard for observations censored at time $t$ is the same conditional on covariates as for uncensored observations still alive at time $t$.

The main difference from ordinary uncensored KDE is that the effective sample size decreases as more observations are censored, so that a constant bandwidth is unlikely to be optimal.

One of the earliest references (now open access) is a 1983 paper by Tanner & Wong. There are implementations such as https://rdrr.io/cran/kernhaz/man/khazard.html and https://cran.r-project.org/web/packages/muhaz/index.html


Kernel density estimates probability density function of the empirical distribution, it doesn't "know" and "care" about the underlying distribution. By "empirical" we mean here that it uses only the observed data and makes minimal distributional assumptions. Obviously, if it uses the observed data, it cannot say anything about the censored, unobserved data. If you say that the distribution is censored, it means that you should have some notion of the underlying distribution, so you should use it to define a parametric model to estimate the density.

  • $\begingroup$ You lost me at estimating the empirical distribution. We have the empirical distribution; what is there to estimate? $\endgroup$
    – Dave
    Oct 5, 2020 at 9:35
  • $\begingroup$ @Dave improved the wording. $\endgroup$
    – Tim
    Oct 5, 2020 at 9:40

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