I have a question about how exactly the Benjamini-Hochberg method for multiple hypothesis testing works. The Wikipedia page for False discovery rate says:

  1. For a given $\alpha$, find the largest k such that $P_{(k)}\leq {\frac {k}{m}}\alpha$
  2. Reject the null hypothesis (i.e., declare discoveries) for all $H_{(i)}$ for $i=1,\ldots ,k$

My question is, what if there's a hypothesis $H_{(i)}$ with $i < k$ but where $P_{(k)} > {\frac {k}{m}}\alpha$ ? Would this hypothesis be called significant? This is probably rare, but for instances where p-values increase with a slope less than $\frac{1}{m}$ (where $m$ is the number of hypotheses tested), it's possible I think.

  • 1
    $\begingroup$ Yes, this happens and Benjamini-Hochberg then rejects the $i$th null hypothesis. $\endgroup$
    – frank
    Jul 23, 2022 at 9:00

1 Answer 1


Assuming that you are interested in if it is possible to have $i<k$ and $p_{(i)}>\frac{k}{m}\alpha$, then I would think it is impossible because the Benjamini-Hochberg first reorders all of the p-values from smallest to largest. (I am going to assume all p-values are unique)

$$ p_1,...,p_n\rightarrow p_{(1)},...,p_{(n)} $$

So, if $p_{(k)}\leq \frac{k}{m}\alpha$, this implies that for all $i<k$, $p_{(i)}<p_{(k)}<\frac{k}{m}\alpha$.

  • $\begingroup$ Right. I understand $p_{(i)} \le p_{(k)} \le \frac{k}{m} \alpha$. just wondering about the case where $p_{(i)} \not \le \frac{i}{m} \alpha$. For example: if we had two hypotheses ($m=2$) with p-values of $0.08, 0.09$, and a FDR $\alpha=0.1$. Then the condition would fail for $i=1$ but succeed for $i=2$. I was wondering what B-H would say in situations like this. $\endgroup$ Jul 26, 2022 at 20:28

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