Holm step-down procedure

For $m\ge 1$, fix a finite family of null hypotheses $H_{0i}$, $1\le i\le m$, before observing the data, and let $p_i:\mathcal{X}\to [0,1]$ be p-value statistics valid for their respective nulls in the sense of Null-valid p-values and familywise and false-discovery error rates. Given $\alpha \in [0,1]$ and an observation $x\in \mathcal{X}$, order the indices by the pairs $(p_i(x),i)$ in lexicographic order. This gives a unique permutation $\pi$ with $p_{\pi (1)}(x)\le \cdots \le p_{\pi (m)}(x)$; ties are ordered by increasing index.

Set $r_{\alpha }(x)=\max \{k\in \{0,\dots ,m\}:p_{\pi (j)}(x)\le \alpha \mathrm{/}(m-j+1)\text{ for every }1\le j\le k\}\mathrm{.}$ The rejected-index set is $R_{\alpha }(x)=\{\pi (j):1\le j\le r_{\alpha }(x)\}$. Reject $H_{0i}$ exactly when $i\in R_{\alpha }(x)$; equivalently, compare sequentially to $\alpha \mathrm{/}(m-j+1)$, stop at the first failed comparison, and reject only the preceding ranks.

For $m=0$, extend the procedure to the empty input by returning the empty rejected set. For $m\ge 1$, every denominator in a tested rank is positive. The ordering is a unique finite permutation, and the finite set defining $r_{\alpha }(x)$ contains $0$, so the output is well-defined.