This is an extension from my last post. It seems to be too long to discuss there.
When testing some null $H$ versus alternative $K$ by a test statistic $U(X)$, the p-value for $U$ on a sample $X$ can be defined as the infinimum of size $\alpha$ of the test rule, with varying critical value $c$ while still rejecting the null on $X$: $$p_U(X) := \sup_{F \in H} P(U(Y) \geq U(X) | Y \sim F), \quad i.e. \quad \alpha(U(X)),$$ where $\alpha(c)$ means the size of the test rule $(U, c)$.
When the permutation test with the set $G$ of permutations can be applied to the above test (the null, $U$ and $X$), the p-value of the permutation test is said to be $$ p_P(X) := \frac{\# \{\pi \in G: U(\pi X) \geq U(X)\}}{|G|} \\ \stackrel{?}{=} P(U(Y) \geq U(X) | Y \sim F_0) $$ where $F_0 \in H$ is a common distribution of $X$ and $\pi X$ under null, $\pi X$ means applying the permutation $\pi$ to the sample $X = \{x_1, ..., x_n\}$, and the part after $\stackrel{?}{=} $ is what I thought of myself.
Questions:
- I was wondering if $p_U(X) = p_P(X)$? I.e., is the p-value of the permutation test equal to the p-value based on $U(X)$, for sample $X$?
If yes, how are they equal when the null is composite?
$p_U(X)$ considers all the null distributions $F$'s one by one and then take $\sup_{F\in H}$,
$p_P(X)$ seems to equal $P(U(Y) \geq U(X) | Y \sim F_0)$ for a single null distribution $F_0$ of $X$, and doesn't consider other null distributions of $X$.
Thanks!