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  "body": "\n## Statement\n\nLet $\\mathcal{X}$ be a [[topological space|0CQHFPGJ45]], and let $(\\mu_n)$ be a sequence of [[probability measures|059TJN0NZ1]] on $\\mathcal{X}$.\nAssume that $(\\mu_n)$ satisfies the [[large deviation principle|MNPWBBNCAD]] with a good [[rate function|140N2PQSTE]] $I$.\nLet $\\varphi \\colon \\mathcal{X} \\to \\R$ be a [[continuous|0X3WGX95FE]] function that is bounded above.\nThen\n$$\\lim_{n \\to \\infty} \\frac{1}{n} \\log \\int_{\\mathcal{X}} e^{n \\varphi(x)} \\, \\mu_n(dx) = \\sup_{x \\in \\mathcal{X}} \\bigl(\\varphi(x) - I(x)\\bigr).$$\n\n## Remarks\n\nThe lemma is an infinite-dimensional form of Laplace's method: the integral is dominated by the neighbourhood of the point where $\\varphi - I$ is largest.\nFor the lower bound, take $x$ with $I(x) \u003c \\infty$ and a small open neighbourhood $G$ of $x$ on which $\\varphi \u003e \\varphi(x) - \\delta$; the large deviation lower bound gives $\\mu_n(G) \\ge e^{-n(I(x) + \\delta)}$ for large $n$.\nFor the upper bound, cover a compact sublevel set of $I$ by finitely many small closed neighbourhoods.\nUse the large deviation upper bound on each, and the bound on $\\varphi$ outside.\n\nThe boundedness of $\\varphi$ from above can be replaced by the tail condition\n$$\\lim_{M \\to \\infty} \\limsup_{n \\to \\infty} \\frac{1}{n} \\log \\int_{\\set{\\varphi \\ge M}} e^{n \\varphi} \\, d\\mu_n = -\\infty.$$\nSome such condition is needed, since otherwise rare but huge values of $e^{n\\varphi}$ can dominate the integral.\n\nExample: let $\\mu_n$ be the normal law with mean $0$ and variance $1/n$, with $I(x) = x^2/2$, and let $\\varphi(x) = -\\abs{x - 1}$.\nThe lemma gives the limit $\\sup_x (-\\abs{x - 1} - x^2/2) = -1/2$, attained at $x = 1$.\n\n## Explanations\n\n### L1 human\n\nSuppose a random quantity is very unlikely to be far from typical, with a known exponential cost $I$.\nIf you reward each outcome $x$ by an exponential weight $e^{n \\varphi(x)}$, then the average reward grows or decays at the rate $\\max(\\varphi - I)$.\nThe winner is the outcome with the best balance between reward and probability cost.\n\n### L2 human\n\nVaradhan's lemma is the main tool to compute free energies and other exponential averages once a large deviation principle is known.\nIn statistical mechanics, the limit is the pressure and the maximisers of $\\varphi - I$ are the equilibrium states.\nIts converse, Bryc's theorem, recovers a large deviation principle from the limits of such exponential integrals.\n",
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