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  "body": "\n## Proof\n\nWe apply Markov's inequality to $e^{tX}$ for each $t \u003e 0$.\n\n1. Let $X$ be a real random variable with [[moment-generating function|0NNRNGKMFM]] $M_X(t) = E[e^{tX}] \\in (0, \\infty]$. Let $a \\in \\R$.\n2. Fix $t \u003e 0$. The map $x \\mapsto tx$ is increasing, and the [[exponential function is strictly increasing|0KN2V7840E]]. So $x \\ge a$ holds if and only if $e^{tx} \\ge e^{ta}$.\n3. Hence $\\set{X \\ge a} = \\set{e^{tX} \\ge e^{ta}}$ as events.\n4. The [[exponential function is positive|0P9BNBW2W3]], so $e^{tX}$ takes values in $[0, \\infty)$ and $e^{ta} \u003e 0$.\n5. By [[Markov's inequality|05PWWHYBX5]] for $e^{tX}$ and the level $e^{ta}$, $P(e^{tX} \\ge e^{ta}) \\le E[e^{tX}]/e^{ta}$. This holds in $[0, \\infty]$, also when $M_X(t) = \\infty$.\n6. By steps 3 and 5, $P(X \\ge a) \\le e^{-ta} M_X(t)$ for each $t \u003e 0$.\n7. So $P(X \\ge a)$ is a lower bound of the set $\\set{e^{-ta} M_X(t) : t \u003e 0}$. It is therefore at most the infimum of this set.\n8. Thus $P(X \\ge a) \\le \\inf_{t \u003e 0} e^{-ta} M_X(t)$.\n\n## Idea\n\nFor $t \u003e 0$ the map $x \\mapsto e^{tx}$ is increasing, so $X \\ge a$ exactly when $e^{tX} \\ge e^{ta}$. Markov's inequality bounds this event for every $t$. Then one picks the best $t$.\n",
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