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I am using a univariate filter to reduce the number of features prior to applying a learning algorithm to a huge binary classification dataset (22510066 features x 500 examples). All the features are binary-valued (0 or 1). Most of these features are supposed to be irrelevant. I am using a $\chi^2$ test to obtain a p-value for each feature and then, based on some significance threshold, I would like to keep only the significant features.

Since I am performing so many hypothesis tests, it is necessary to correct for multiple testing. I have chosen to use a method that limits the FDR, as it seems more natural for feature selection. The features in the dataset are not independent and some of them are negatively correlated, thus I am using the Benjamini-Yekutieli method to do the correction. After computing the p-value for each feature, the minimum p-value is 2.7061601033755451e-09.

I have noticed something that I do not understand. If I set the maximum FDR ($\alpha$) to 1.0, which means that the proportion of type I errors is $\leq 1.0$, even the smallest p-value is not small enough to be considered significant. I understand that alpha is an upper bound on the FDR and that if $\alpha=1.0$, it does not mean that 100% of the tests for which the null hypothesis was rejected are type I errors, but I still don't know how to interpret this result.

Could someone please help me interpret this result? Does this mean that the Benjamini-Yekutieli procedure is not powerful enough for such a huge number of tests?

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  • $\begingroup$ Two issues: First, the paper of Benjamini and Yekuteli (2001) assumes positive regression dependency of the test statistics, not negative. Second: It would be no problem for a test to break down for $\alpha=100%$ since nobody needs this anyway. Maybe these superficial comments help. $\endgroup$ – Horst Grünbusch Apr 15 '15 at 22:01
  • $\begingroup$ Hi Horst, thanks for your reply! I am using the version of Benjamini and Yekutieli that is valid under arbitrary dependencies (see here). I agree that there is no use for a method that is valid at $\alpha=100$. I am asking this question, since at such a high value for alpha, I would have expected to have at least a few significant tests. $\endgroup$ – jamesbond Apr 15 '15 at 22:12
  • $\begingroup$ There are 122 significant tests at $\alpha=1.1$. $\endgroup$ – jamesbond Apr 15 '15 at 22:16
  • $\begingroup$ @HorstGrünbusch The Benjamini-Yekutieli paper refers to the Benjamini-Hochberg FDR adjustment as requiring either independence or positive dependence. The Benjamini-Yekutieli procedure is applicable to dependent tests (regardless of positive or negative dependence), as the last two sentences of the abstract make abundantly clear. $\endgroup$ – Alexis Apr 15 '15 at 22:17
  • $\begingroup$ jamesbond, you should be choosing FDR for values of $\alpha$ akin to what you would use for a single test (e.g. 0.05, 0.001, etc.). Also: both the B-H and the B-Y procedures are step-down procedures, meaning that the decision to reject a test depends on more than simply its adjusted p-value, but also on the ordering of its unadjusted p-value relative to other tests. $\endgroup$ – Alexis Apr 15 '15 at 22:21
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Benjamini and Yekutieli (2001) shows that that the BH procedure from Benjamini-Hochberg (1995) controls the FDR under certain dependence conditions, in particular including positive regression dependency. They proposed the modified procedure that the original poster is using, which controls the FDR under general dependency.

Since this modified procedure works so generally, it will tend to be conservative and thus not be as powerful as the original BH procedure. Given the millions of hypotheses you're testing, yes, you would need an extremely low p-value (even smaller than your 2.7e-09 value) to be considered significant using the modified procedure.

There are some suggestions at the end of this presentation regarding dependence in a multiple testing framework: http://www.stat.cmu.edu/~genovese/talks/hannover1-04.pdf

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    $\begingroup$ Nice answer. A quibble, though: it is inaccurate to say that the BY procedure is as conservative as the Bonferroni adjustment, as the BY amounts to using the BH with $\alpha/C$ where $C=\sum_{i=1}^{m}{i^{-1}}$, where $m$ is the number of comparisons. In @jamesbond 's case, $C=17.50669$, so the BY amounts to the BH for $\alpha/17.50669$ which is a far cry from the $\alpha/22510066$ for the Bonferroni. $\endgroup$ – Alexis Apr 15 '15 at 22:29
  • $\begingroup$ Hi, thanks for the comment. In general, you're right, I agree that BY is not as conservative as Bonferroni (and in this case, probably not even close). But depending on how the observed p-values are distributed, the BY procedure could reject the same number of tests as Bonferroni. $\endgroup$ – gnaist Apr 15 '15 at 22:42
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    $\begingroup$ The number of rejections is not the same thing as the Type I error rate, or the false discovery rate. :) $\endgroup$ – Alexis Apr 16 '15 at 1:09
  • $\begingroup$ Touché, touché! I removed that bit about Bonferroni from my original post. Thanks! :) $\endgroup$ – gnaist Apr 16 '15 at 1:46
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    $\begingroup$ I will have a look at the presentation! Thank you all for your answers! $\endgroup$ – jamesbond Apr 16 '15 at 2:25

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