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  "body": "\n## Statement\n\nLet $X$ be a real random variable with finite mean $\\mu$ and finite [[variance|007QED3E0Y]] $\\sigma^2$. Then for each $a \u003e 0$,\n$$P(|X - \\mu| \\ge a) \\le \\frac{\\sigma^2}{a^2}.$$\nEquivalently, if $\\sigma \u003e 0$, then $P(|X - \\mu| \\ge k\\sigma) \\le 1/k^2$ for each $k \u003e 0$.\n\n## Remarks\n\nProof: apply [[Markov's inequality|05PWWHYBX5]] to the nonnegative variable $(X - \\mu)^2$ and the level $a^2$.\nFor $k = 2$, at least $3/4$ of the mass is within two [[standard deviations|0VTN5RKNC5]] of the mean, for every distribution with finite variance.\nThe bound is sharp: some distributions reach equality for a given $k \\ge 1$.\nChebyshev's inequality gives a short proof of the weak law of large numbers for variables with finite variance.\n\n## History\n\nThe inequality is named after Pafnuty Chebyshev. Irénée-Jules Bienaymé formulated it first.\n\n## Explanations\n\n### L1 human\n\nThe variance controls how often a variable can be far from its mean.\nFor any distribution, at most $1/9$ of the mass lies three or more standard deviations away from the mean.\n",
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