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  "body": "\n## Statement\n\nLet $X$ be a random variable with values in $[0, \\infty)$ and let $a \u003e 0$. Then\n$$P(X \\ge a) \\le \\frac{E[X]}{a}.$$\n\n## Remarks\n\nProof: $a \\, 1_{\\{X \\ge a\\}} \\le X$ pointwise. Take expectations: $a \\, P(X \\ge a) \\le E[X]$.\nMore generally, for each nondecreasing $g \\colon [0, \\infty) \\to [0, \\infty)$ with $g(a) \u003e 0$, $P(X \\ge a) \\le E[g(X)]/g(a)$.\nApplied to $(X - E[X])^2$, the inequality gives [[Chebyshev's inequality|02HZ1BK3MW]]. Applied to $e^{tX}$, it gives the [[Chernoff bound|014PEBGDYF]].\n\n## History\n\nThe inequality is named after Andrey Markov. It appears earlier in the work of Pafnuty Chebyshev, the teacher of Markov.\n\n## Explanations\n\n### L1 human\n\nA nonnegative quantity with a small average cannot often be large.\nIf the mean income is $30000$, then at most one person in five can have an income of $150000$ or more.\n",
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