Breiman et al recommend the 1-SE rule, and show that it is successful in screening out noise variables. At page no. 80 of their book, I get confuse about the '1 S.E. Rule'. $$R'[T(k_l)]\leq R'[T(k_0)]+S.E\{R'[T(k_0)]\} $$

Where $T_1, T_2, ... $ are the number of sequence of trees (number of variables) and the corresponding estimates of K-fold cross validation prediction error are $R'[T_1], R'[T_2], ... $ Then, the tree selected is $T(k_l)$, where $k_l$ is the maximum $k$ satisfying above equation. Please note that $R'[T(k_0)]=min_kR'[T_k]$

My question is that how I will calculate the $S.E\{R'[T(k_0)]\}$ ? because it is only one value, Please correct me where I am wrong.


Isn't it as simple as calculating error of mean of $R'[T_i]$ (for a given i) using each cross validation fold as an "independent" measurement? (i.e. calculating standard deviation of $R'[T_i]$ (across K folds) and then dividing by $\sqrt{K-1}$ gives a reasonable resampling-based proxy of that standard error)

  • $\begingroup$ what are $i$ and $K$? $\endgroup$ – Biostat Nov 22 '11 at 17:56
  • $\begingroup$ K is the number of resamples (which is equal to the number of folds in the case of a single replication of K-fold cross validation.) i corresponds to the model with lowest value of metric R' (k0 in your notation above). Basically, resamapling is used to estimate both R' and it's SE. $\endgroup$ – Yevgeny Nov 22 '11 at 18:05
  • $\begingroup$ Can I use the following estimate for SE? $SD(X_m)$, where $X_m=mean(R[T_i])$ and $m$ denote replication of k-fold CV. $\endgroup$ – Biostat Nov 22 '11 at 18:45
  • $\begingroup$ That is likely to cause a way too parsimonious model to be chosen (since it overestimates the standard error of R' due to the fact that you are not using error of the mean). If you are in doubt you could try to calculate SE with sqrt(k-1) and without and see what kind of models you get. $\endgroup$ – Yevgeny Nov 22 '11 at 19:07
  • 1
    $\begingroup$ By the way, any reason why you are not using R package caret that supports this feature? (and lots of other goodies) - see bottom of the page 19 of caretTrain vignette. $\endgroup$ – Yevgeny Nov 22 '11 at 19:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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