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  "body": "\n## Statement\n\nLet $X$ be a real random variable with [[moment-generating function|0NNRNGKMFM]] $M_X(t) = E[e^{tX}]$, and let $a \\in \\mathbb{R}$. Then\n$$P(X \\ge a) \\le \\inf_{t \u003e 0} e^{-ta} M_X(t).$$\n\n## Remarks\n\nProof: for each $t \u003e 0$, $\\{X \\ge a\\} = \\{e^{tX} \\ge e^{ta}\\}$. [[Markov's inequality|05PWWHYBX5]] gives $P(X \\ge a) \\le e^{-ta} E[e^{tX}]$. Then take the infimum over $t$.\nFor a sum $S$ of independent $X_i$, $M_S = \\prod_i M_{X_i}$. So the bound becomes $\\inf_{t \u003e 0} e^{-ta} \\prod_i M_{X_i}(t)$, which decreases exponentially in many cases.\nFor $S \\sim \\operatorname{Bin}(n, p)$ with mean $\\mu = np$ and $0 \u003c \\delta$, a standard consequence is $P(S \\ge (1 + \\delta)\\mu) \\le \\left(e^{\\delta} / (1 + \\delta)^{1 + \\delta}\\right)^{\\mu}$.\n\n## History\n\nThe bound is named after Herman Chernoff.\n\n## Explanations\n\n### L1 human\n\nThe Chernoff bound gives tail estimates that decrease very fast, often exponentially.\nIt is much stronger than Chebyshev's inequality for sums of many independent terms, such as the number of heads in many tosses.\n",
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