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  "body": "\n## Definition\n\nLet $Z = (Z_1, \\dots, Z_N)$ be a random vector, and let $S_N$ be the [[symmetric group|0RJMNEN3BJ]], acting on $\\R^N$ by $(\\pi z)_i = z_{\\pi^{-1}(i)}$.\nLet $H_0$ be a null hypothesis under which $Z$ is exchangeable, that is, $\\pi Z$ has the same law as $Z$ for every $\\pi \\in S_N$.\nLet $T \\colon \\R^N \\to \\R$ be a [[statistic|036756NFD6]].\nThe *permutation p-value* is\n$$p(Z) = \\frac{1}{N!} \\, \\#\\set{\\pi \\in S_N : T(\\pi Z) \\ge T(Z)}.$$\nThe *permutation test* at level $\\alpha$ rejects $H_0$ when $p(Z) \\le \\alpha$.\n\n## Remarks\n\nUnder $H_0$, $P(p(Z) \\le \\alpha) \\le \\alpha$ for every $\\alpha \\in [0, 1]$, so the test has level $\\alpha$, whatever the law of $Z$.\nIn the two-sample problem, $Z$ is the pooled sample $X_1, \\dots, X_m, Y_1, \\dots, Y_n$ with $N = m + n$, and $H_0$ says that all $N$ values are i.i.d.\nSince $T$ usually depends only on which values fall in the first group, one may count the $\\binom{N}{m}$ splits instead of the $N!$ permutations.\nWhen $N!$ is too large, one uses a random sample of permutations, which gives a Monte Carlo approximation of $p(Z)$.\n\nExample: $X = (3, 5)$, $Y = (1, 4)$ and $T = \\bar X - \\bar Y$, so $T = 4 - 2.5 = 1.5$.\nThe $6$ splits of $\\set{1, 3, 4, 5}$ give the $X$-pairs $\\set{1,3}, \\set{1,4}, \\set{1,5}, \\set{3,4}, \\set{3,5}, \\set{4,5}$ and the values $-2.5, -1.5, -0.5, 0.5, 1.5, 2.5$.\nTwo of them are at least $1.5$, so $p = 2/6 = 1/3$.\n\nNon-example: if under $H_0$ the two groups have the same mean but different variances, the pooled vector is not exchangeable.\nThe clause on exchangeability fails, and the test of equal means need not have level $\\alpha$.\n\nThe [[Mann–Whitney U test|CKF9KDXBP7]] is a permutation test whose statistic depends only on the ranks.\n",
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