BY and BH thresholds for independent valid p-values

The five masses sum to one, and the sum of the product point masses on $S^4$ is $({\sum }_{s\in S}\mu (\{s\}){)}^4=1$, so these masses define a probability law on the finite sample space. For any subsets $B_1,\dots ,B_4\subseteq S$, summing the product masses over $B_1\times \cdots \times B_4$ factors as ${\prod }_i\mu (B_i)$; thus the coordinate p-values are mutually independent. Each coordinate has support $1\mathrm{/}100,1\mathrm{/}25,13\mathrm{/}100,29\mathrm{/}100,1$ and cumulative probabilities $1\mathrm{/}100,4\mathrm{/}100,13\mathrm{/}100,29\mathrm{/}100,1$, respectively. At each support value the cumulative probability equals that value; between successive support values it stays at the preceding value, and below $1\mathrm{/}100$ it is zero. Thus $P(p_i\le u)\le u$ for every $u\in [0,1]$, so all four p-values are valid by [F2].

Fix a coordinate $i$. For $x\in S$, define $K_i(x,\cdot )$ as the law of the four-vector obtained by fixing coordinate $i$ at $x$ and drawing the other three coordinates independently from the specified marginal; for $x\mathrm{\notin }S$, use a fixed point mass. Since $S$ is finite, each evaluation is Borel measurable and each section is a probability measure. For Borel $A\subseteq [0,1]$ and $D\subseteq [0,1{]}^4$, product-mass factorization gives ${\int }_A{K}_i(x,D)P_{p_i}(dx)=P(p_i\in A,p\in D)$, so this is a conditional law by [F4]. If $D$ is coordinatewise increasing and $x\le x^{\mathrm{\prime }}$ in $S$, then fixing any same remaining triple makes the vector at $x$ coordinatewise at most the vector at $x^{\mathrm{\prime }}$, so its indicator of $D$ is no larger. Averaging over the same three-coordinate product law yields $K_i(x,D)\le K_i(x^{\mathrm{\prime }},D)$. This holds for every $i$, and [F3] gives PRDS on the all-true-null set.

The event that the four coordinates take the four distinct values $1\mathrm{/}100,1\mathrm{/}25,13\mathrm{/}100,29\mathrm{/}100$ in any order has probability $4!(1\mathrm{/}100)(3\mathrm{/}100)(9\mathrm{/}100)(16\mathrm{/}100)>0$. On this event $H_4=25\mathrm{/}12$. BH's critical values are $1\mathrm{/}20,1\mathrm{/}10,3\mathrm{/}20,1\mathrm{/}5$, so the first three ordered p-values pass and the fourth does not, giving $R_{BH}=3$. BY's critical values are $3\mathrm{/}125,6\mathrm{/}125,9\mathrm{/}125,12\mathrm{/}125$, so ranks one and two pass and rank three does not, giving $R_{BY}=2$.

Here $m_0=m=4$. By [F8] and [F9], each rule has FDR at most $q=1\mathrm{/}5$ under this model; step 1.3 gives a positive-probability event on which their rejection counts differ.