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In the method described here

http://dmkd.cs.vt.edu/papers/TKDE17.pdf

R code implementation is provided

https://github.com/MLSurvival/ESP/blob/master/ESP_TKDE2016/TKDE_code.R

the Kaplan-Meier estimator is initially estimated from data but then (from line 244) in order to estimate time different distributions are also fitted (exponential, weibull, loglogistic). Why is this part necessary? wouldn't be enough the Kaplan Meier curve fitted before?

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    $\begingroup$ Welcome to CrossValidated. Please edit your question to include summaries or excerpts from the links you posted, so users can see what you are trying to ask without having to go to an external site and sifting through a multi-page PDF document. It would also be useful for us if you copied the relevant part of the code into your question for the same reason. $\endgroup$
    – Marquis de Carabas
    Commented Dec 2, 2016 at 19:47
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    $\begingroup$ Adding excerpts or summaries is also crucial because links can disappear over time. So having critical details in the question here helps this site reach its goal of being a useful continuing repository of questions and answers on statistical issues. $\endgroup$
    – EdM
    Commented Dec 2, 2016 at 19:52
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    $\begingroup$ Cross-posted to SO :-( $\endgroup$
    – Hack-R
    Commented Dec 2, 2016 at 20:10
  • $\begingroup$ If you want to project/forecast survival curve beyond observed data then you have to use a parametric model. Kaplan Meir is a nonparametric method is no good for projection beyond observable data. $\endgroup$
    – forecaster
    Commented Jan 22, 2017 at 15:59

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There are reasons to fit both a Kaplan-Meier estimator and a parametric or semi-parameteric survival function to the same data, even if they are doing the same thing.

The first of these is as a sanity check - do different methods give you similar answers? If not, why not? This is often why the Kaplan-Meier is fitted - as a means to evaluate whether or not a parametric estimator is performing well and to provide a comparison.

Now, why you may want to do this? Sometimes a parametric version of a survival distribution is important. For example, when estimating a parameter that is going to be used in a mathematical/computational model, life is much easier if you can describe survival time parametrically. So you could fit say, an exponential distribution, and then use the KM to check and make sure that you haven't deviated too far from the actual survival distribution for the sake of convenience.

I've written papers that do exactly this.

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When you say that they "...estimate time different distributions" I think you mean to say that they test for a difference in "survival".

The appropriate non-parametric test of survival as depicted by a KM curve is either the log-rank test, or the Weibull test. You can, however, conduct a parametric test of survival using some probability models: exponential, weibull, gamma, lognormal to name a few.

This is simply making some assumptions about the nature of the survival times to get a more powerful test.

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If I understand it right this paper does 3 main things:

  1. Create a 50% dataset and a 100% dataset and use Kaplan Meier with one of them in order to label all instances (I thought it would be the 50% dataset but looking at the R code in line 132 reads the 100% dataset and calculates Kaplan Meier with it (how's that??)

https://github.com/MLSurvival/ESP/blob/master/ESP_TKDE2016/TKDE_code.R

  1. Run weka with a known classifier (for example naive bayes) in order to classify the 50% dataset (or the 100% dataset???) Should be the 100% if it is already labeled with Kaplan Meier but I'm not sure.

  2. Fit a distribution (for example exponential or weibull to calculate probabilities that then uses to calculate performance measures with ROCR package (for example to calculate prediction and performance from ROCR

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