Exact finite-group permutation p-values and tests

For each $g\in G$, $x\mapsto T(g\cdot x)$ is measurable by [F1]. For measurable real functions $f,h$, $\{f>h\}={\bigcup }_{q\in \mathbb{Q}}\{f>q\}\cap \{h<q\}$, so strict comparisons are measurable; equality sets are complements of the union of the two opposite strict comparisons. Finite sums show that both rank counts are measurable. The identity belongs to $G$, so the tied count is at least one and both p-value formulas take values in $[0,1]$. For the joint tie-rank formula, partition the data space by the finitely many possible integer pairs $(N_{>},N_{=})$; on each part it is an affine function of $U$, so it is jointly measurable.

Fix $\theta \in {\mathrm{\Theta }}_0$ and let $H$ be uniform on $G$, independent of $(X,U)$. For each fixed $h\in G$, invariance gives . Averaging over the finite law of $H$ therefore gives , so replacing $X$ by $H\cdot X$ does not change either p-value distribution.

Fix $x$ and form the finite list $a_g=T(g\cdot x)$ indexed by $g\in G$, retaining repeated values and repeated orbit points. At $H=h$, the transformed list at $h\cdot x$ is $T((gh)\cdot x)$ as $g$ ranges over $G$. Right multiplication $g\mapsto gh$ is a bijection of $G$, and the observed value $T(h\cdot x)$ is the entry indexed by $h$. Thus $p_{\mathrm{o}\mathrm{r}\mathrm{b}}(h\cdot x)$ is the inclusive upper rank of a uniform index in $(a_g{)}_{g\in G}$, while $p_{\mathrm{t}\mathrm{i}\mathrm{e}}(h\cdot x,U)$ is its independently tie-randomized rank.

By [F3], conditional on each fixed $x$, the first rank in step 2.1 is superuniform and the second is uniform. Averaging over $X$ preserves these bounds. The distributional equality in step 1.2 transfers them from $H\cdot X$ back to $X$, proving both displayed claims for this $\theta$. Since $\theta$ was arbitrary in ${\mathrm{\Theta }}_0$, the claims hold under every null law. No orbit quotient or freeness assumption is used.

By [F6], $p_{\mathrm{o}\mathrm{r}\mathrm{b}}$ is a valid p-value; [F7] gives its level-$\alpha$ cutoff test when an alternative is specified. For the tied test put $N_{>}(x)=\mathrm{\#}\{g:T(g\cdot x)>T(x)\}$ and $N_{=}(x)=\mathrm{\#}\{g:T(g\cdot x)=T(x)\}$. Its conditional rejection probability is ${\varphi }_{\alpha }(x)=\min \{1,\max \{0,(\mathrm{\mid }G\mathrm{\mid }\alpha -N_{>}(x))\mathrm{/}{N}_{=}(x)\}\}$, because $p_{\mathrm{t}\mathrm{i}\mathrm{e}}(x,U)\le \alpha$ exactly when $U\le (\mathrm{\mid }G\mathrm{\mid }\alpha -N_{>}(x))\mathrm{/}{N}_{=}(x)$, with the threshold clipped to $[0,1]$. The counts are measurable by step 1.1 and $N_{=}(x)\ge 1$, so [F8] makes ${\varphi }_{\alpha }$ a randomized test. Its null rejection probability equals ${\mathbb{P}}_{\theta }(p_{\mathrm{t}\mathrm{i}\mathrm{e}}(X,U)\le \alpha )=\alpha$ by step 3.1, and [F9] identifies its size as $\alpha$ when a nonempty disjoint alternative is specified. The argument uses only a finite uniform group index and the stated uniform tie randomizer; it uses no choice axiom.