I know that various questions already deal with this topic, but I don't think anyone has answered this before!

I'm reading a paper in which the authors claim to use 20-fold cross validation to estimate the bandwidth of a Gaussian KDE.

What does this mean? I've read up on cross validation for KDE bandwidth estimation, but every example I have found (including the 'standards' least squares and maximum likelihood) don't have an 'n-fold' mentioned, because they use the leave-one-out approach (hence n is always equal to the number of data points).

Thanks for any light you can shed.

  • 1
    $\begingroup$ see here: en.wikipedia.org/wiki/… k-fold (as opposed to n-fold) cross-validation saves time because now you have to run your procedure only $k$ times, here $k=20$. $\endgroup$
    – PA6OTA
    May 19, 2014 at 13:42
  • $\begingroup$ Yes, but how does that fit in with the specific case of estimating bandwidth for a KDE? Because the formulations are all set up for a leave-out-one method. My guess is we leave out 1/20 of the data points rather than one, but then does the maximum likelihood formulation change as you'd expect it to? $\endgroup$
    – Gabriel
    May 19, 2014 at 13:49
  • $\begingroup$ if you have, say, 50 points now instead of 1 point, then the likelihood is the product of 50 individual likelihoods. $\endgroup$
    – PA6OTA
    May 19, 2014 at 13:52
  • $\begingroup$ Got it, thanks - my guess is that they use this method to save time then since it's quite computationally expensive. I thought that this is what was meant, but I can't find a single reference to this approach in the context of KDEs so wanted to see if anyone else had. $\endgroup$
    – Gabriel
    May 19, 2014 at 13:55


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