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  "body": "\n## Statement\n\nLet $\\set{P_\\theta : \\theta \\in \\Theta}$ be a statistical model with a loss function, with [[risk function|N26TSVRH2S]] $R(\\theta, \\delta)$.\nLet $\\pi$ be a prior on $\\Theta$ such that $\\theta \\mapsto R(\\theta, \\delta)$ is measurable for every rule $\\delta$.\nThen the [[Bayes risk|VEBDNCJJTN]] of $\\pi$ is at most the [[minimax risk|8EG1M1WZ9C]]:\n$$r(\\pi) = \\inf_\\delta \\int_\\Theta R(\\theta, \\delta) \\, \\pi(d\\theta) \\le \\inf_\\delta \\sup_{\\theta \\in \\Theta} R(\\theta, \\delta).$$\n\n## Remarks\n\nFor each fixed rule $\\delta$, an average of $R(\\cdot, \\delta)$ is at most its supremum, so $\\int R(\\theta, \\delta) \\, \\pi(d\\theta) \\le \\sup_\\theta R(\\theta, \\delta)$.\nTaking the infimum over $\\delta$ on both sides gives the claim.\nHence also $\\sup_\\pi r(\\pi) \\le R^*$, where $R^*$ is the minimax risk.\n\nThis inequality is the standard way to prove lower bounds for the minimax risk.\nIf a rule $\\delta^*$ has $\\sup_\\theta R(\\theta, \\delta^*) = r(\\pi)$ for some prior $\\pi$, then $\\delta^*$ is minimax.\nFor instance, a Bayes rule with constant risk is minimax, as for the binomial proportion with the beta prior of parameters $\\sqrt{n}/2$ and $\\sqrt{n}/2$.\n\n## Explanations\n\n### L1 human\n\nThe worst case is always at least as bad as any average.\nSo the best achievable average risk, for any choice of prior, is never more than the best achievable worst-case risk.\n\n### L2 human\n\nTo show that no estimator can do better than some value in the worst case, it is enough to find a prior whose Bayes risk equals that value.\nThis is the Bayesian route to minimax lower bounds, often used with a sequence of priors whose Bayes risks increase to the candidate value.\nFor a finite parameter set with randomized rules allowed, minimax theorems show that the two sides are equal when the supremum over priors is taken.\n",
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