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  "body": "\n## Proof\n\nWe expand $e^{tX}$ in the exponential series, bound the tail, and then differentiate the resulting power series.\n\n1. Let $X$ be a real random variable and let $h \u003e 0$ with $M_X(t) = E[e^{tX}] \u003c \\infty$ for all $t \\in (-h, h)$. Here $M_X$ is the [[moment-generating function|0NNRNGKMFM]].\n2. We use two properties of the [[expected value|0DCT0QZVQ1]]. If $0 \\le U \\le W$ then $E[U] \\le E[W]$. If $Y$ is integrable then $\\abs{E[Y]} \\le E[\\abs{Y}]$.\n3. Let $0 \\le s \u003c h$. One of $e^{sX}$ and $e^{-sX}$ equals $e^{s\\abs{X}}$, and the other is positive. So $e^{s\\abs{X}} \\le e^{sX} + e^{-sX}$.\n4. By steps 2 and 3, $E[e^{s\\abs{X}}] \\le M_X(s) + M_X(-s) \u003c \\infty$.\n5. Fix $k \\ge 0$ and $s \\in (0, h)$. By the [[exponential series|06JZGMYTYD]], whose terms are nonnegative for a nonnegative argument, $e^{s\\abs{x}} \\ge (s\\abs{x})^k / k!$.\n6. So $\\abs{X}^k \\le k! \\, s^{-k} e^{s\\abs{X}}$. By steps 2 and 4, $E[\\abs{X}^k] \u003c \\infty$. Thus $X$ has finite [[moments|0PH0VHTQS0]] of all orders.\n7. Fix $t$ with $\\abs{t} \u003c h$. Choose $s$ with $\\abs{t} \u003c s \u003c h$, and let $r = \\abs{t}/s \\in [0, 1)$.\n8. For $N \\ge 0$ and real $y$, let $P_N(y) = \\sum_{k=0}^{N} y^k / k!$. By the exponential series, $\\abs{e^y - P_N(y)} \\le \\sum_{k \u003e N} \\abs{y}^k / k!$.\n9. For $k \\ge N + 1$, $\\abs{tX}^k = r^k \\abs{sX}^k \\le r^{N+1} \\abs{sX}^k$, since $r \u003c 1$.\n10. By steps 8 and 9, $\\abs{e^{tX} - P_N(tX)} \\le r^{N+1} \\sum_{k \u003e N} \\abs{sX}^k / k! \\le r^{N+1} e^{s\\abs{X}}$.\n11. By step 6 and [[linearity of expectation|0P7PTC8DDE]], $P_N(tX)$ is integrable and $E[P_N(tX)] = \\sum_{k=0}^{N} E[X^k] \\, t^k / k!$.\n12. The variable $e^{tX}$ is integrable, since $M_X(t) \u003c \\infty$. By steps 2, 10 and 11,\n$$\\abs{M_X(t) - \\sum_{k=0}^{N} \\frac{E[X^k]}{k!} t^k} \\le r^{N+1} E[e^{s\\abs{X}}].$$\n13. By step 4, the right side is a finite constant times $r^{N+1}$, which tends to $0$ as $N \\to \\infty$.\n14. So $M_X(t) = \\sum_{k=0}^{\\infty} E[X^k] \\, t^k / k!$ for every $t \\in (-h, h)$.\n15. Let $a_k = E[X^k]/k!$. The [[power series|0VP56B4EMX]] $\\sum a_k t^k$ converges at each $t \\in (-h, h)$. So its terms $a_k t^k$ are bounded for each such $t$.\n16. By the definition of the [[radius of convergence|0660QX1VPR]], the radius of $\\sum a_k t^k$ is at least $\\abs{t}$ for each $t \\in (-h, h)$. So it is at least $h$.\n17. [[Power series are differentiable inside the radius of convergence|002X7EK91H]], and the derived series has the same radius. By induction on $j$, $M_X$ is $j$ times differentiable on $(-h, h)$ for each $j$.\n18. The same induction gives, for $\\abs{t} \u003c h$,\n$$M_X^{(j)}(t) = \\sum_{k=j}^{\\infty} \\frac{k!}{(k - j)!} \\, a_k t^{k-j}.$$\n19. At $t = 0$, only the term $k = j$ survives. So $M_X^{(j)}(0) = j! \\, a_j = E[X^j]$ for each integer $j \\ge 0$.\n20. Together with step 6 and step 14, this proves all claims.\n\n## Idea\n\nNear $0$ the variable $e^{s\\abs{X}}$ is integrable, and it dominates every power $\\abs{X}^k$ and every tail of the series of $e^{tX}$ for $\\abs{t} \u003c s$. So the expectation can be taken term by term. Then $M_X$ is a power series with coefficients $E[X^k]/k!$, and its derivatives at $0$ read off these coefficients.\n",
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