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  "body": "\n## Problem\n\nLet $X_1, \\dots, X_m$ be i.i.d. [[normal|07RXBVTVZB]] $\\mathcal{N}(\\mu_1, \\sigma_1^2)$ and let $Y_1, \\dots, Y_n$ be i.i.d. $\\mathcal{N}(\\mu_2, \\sigma_2^2)$, independent of the $X_i$.\nAll four parameters are unknown, and $\\sigma_1^2 = \\sigma_2^2$ is not assumed.\nThe problem asks for a [[test|087P03NEND]] of $H_0 \\colon \\mu_1 = \\mu_2$ whose type I error probability is exactly $\\alpha$ for all values of $\\sigma_1^2$ and $\\sigma_2^2$.\nEquivalently, it asks for a [[confidence interval|0K4NMMJBG2]] for $\\mu_1 - \\mu_2$ with exact coverage $1 - \\alpha$ for all values of the variances.\n\n## Remarks\n\nThe natural statistic is $(\\bar X - \\bar Y)/\\sqrt{S_1^2/m + S_2^2/n}$, where $S_1^2$ and $S_2^2$ are the sample variances.\nIts law under $H_0$ depends on the unknown ratio $\\sigma_1^2/\\sigma_2^2$, which is the source of the difficulty.\nIf the ratio $\\sigma_1^2/\\sigma_2^2$ is known, a two-sample [[Student's t-test|0HYP606RYF]] solves the problem exactly.\nWelch's test compares the statistic with a t-distribution whose degrees of freedom are estimated from the data; its level is close to $\\alpha$ but not exactly $\\alpha$ in general.\nFisher proposed a fiducial solution, and its level as a frequentist test is not exactly $\\alpha$ either.\n\n## Explanations\n\n### L1 human\n\nTwo groups are measured, and one wants to know whether their true means differ.\nWhen both groups have the same spread, the classical t-test gives an exact answer.\nWhen the spreads may differ and are unknown, the exact answer becomes surprisingly hard, and this is the Behrens–Fisher problem.\nIn practice, people use Welch's approximate test, which works well for most sample sizes.\n",
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      "first_stated": "1929",
      "posed_by": "Walter-Ulrich Behrens"
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