# How does thresholding work in Benjamini-Hochberg hypothesis testing?

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.

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

Assuming that you are interested in if it is possible to have $$i and $$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, $$p_{(i)}.
• 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. Jul 26, 2022 at 20:28