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  "body": "\n## Statement\n\nLet $X$ be a real random variable, and let $h \u003e 0$ with $M_X(t) = E[e^{tX}] \u003c \\infty$ for all $t \\in (-h, h)$.\nThen $X$ has finite [[moments|0PH0VHTQS0]] of all orders, the [[moment-generating function|0NNRNGKMFM]] $M_X$ is infinitely differentiable on $(-h, h)$, and for each integer $k \\ge 0$,\n$$M_X^{(k)}(0) = E\\left[X^k\\right].$$\nMoreover, $M_X(t) = \\sum_{k=0}^{\\infty} E[X^k] \\, t^k / k!$ for $|t| \u003c h$.\n\n## Remarks\n\nThe proof differentiates under the expectation sign. The bound $|x|^k e^{t x} \\le C(e^{a x} + e^{-a x})$ for $|t| \u003c a \u003c h$ justifies this by dominated convergence.\nExample: for $X \\sim \\operatorname{Exp}(\\lambda)$, $M_X(t) = \\lambda/(\\lambda - t)$, so $M_X'(0) = 1/\\lambda$ and $M_X''(0) = 2/\\lambda^2$. So $\\operatorname{Var}(X) = 1/\\lambda^2$.\n\n## Explanations\n\n### L1 human\n\nThis result is the reason for the name: the function \"generates\" the moments.\nTo find $E[X^k]$, you take the $k$-th derivative of $M_X$ and put $t = 0$.\n",
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