Assume the null distribution is a <centrally symmetric probability distribution>, so $X_i$ and $-X_i$ have the same law. A <sign-flip randomization test> draws signs $s_i\in\{-1,1\}$ independently and recomputes, for example,
$$
T(s)=n\left\lVert\frac1n\sum_{i=1}^ns_iX_i\right\rVert^2.
$$
The exact p-value averages over all $2^n$ sign vectors:
$$
p=2^{-n}\#\{s:T(s)\geq T(1,\ldots,1)\}.
$$
With $B$ random sign vectors, including the observed configuration, the standard Monte Carlo version is $(1+\#\{b:T_b\geq T_0\})/(B+1)$, where $T_0$ is the observed statistic. Joint sign invariance under the null makes this finite-sample valid.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 225 / 2 / c / Solution

A probability distribution on a vector space is centrally symmetric about zero when $X$ and $-X$ have the same distribution. Independent observations from such a distribution are jointly invariant under coordinatewise sign changes.


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 Incoming links: Centrally symmetric probability distribution