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  "body": "\n## Proof\n\nWe apply Markov's inequality to the squared deviation.\n\n1. Let $X$ be a real random variable with finite mean $\\mu$ and finite [[variance|007QED3E0Y]] $\\sigma^2$. Let $a \u003e 0$.\n2. Let $Y = (X - \\mu)^2$. Then $Y$ takes values in $[0, \\infty)$, and $E[Y] = \\sigma^2$ by the definition of the variance.\n3. For $u \\ge 0$ and $a \u003e 0$, $u \\ge a$ holds if and only if $u^2 \\ge a^2$, since squaring is increasing on $[0, \\infty)$.\n4. Apply step 3 to $u = \\abs{X - \\mu}$. This gives the equality of events $\\set{\\abs{X - \\mu} \\ge a} = \\set{Y \\ge a^2}$.\n5. By [[Markov's inequality|05PWWHYBX5]] for $Y$ and the level $a^2 \u003e 0$, $P(Y \\ge a^2) \\le E[Y]/a^2 = \\sigma^2/a^2$.\n6. By steps 4 and 5, $P(\\abs{X - \\mu} \\ge a) \\le \\sigma^2/a^2$.\n7. Now let $\\sigma \u003e 0$ be the [[standard deviation|0VTN5RKNC5]] and let $k \u003e 0$. Then $a = k\\sigma \u003e 0$.\n8. Step 6 with this $a$ gives $P(\\abs{X - \\mu} \\ge k\\sigma) \\le \\sigma^2/(k^2\\sigma^2) = 1/k^2$.\n\n## Idea\n\nA large deviation $\\abs{X - \\mu} \\ge a$ is the same event as a large squared deviation $(X - \\mu)^2 \\ge a^2$. Markov's inequality bounds the second event by the mean of the squared deviation, which is the variance.\n",
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