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  "body": "\n## Definition\n\nLet $X$ be a real [[random variable|0PNRKMRC55]] such that $M(t) = E[e^{tX}]$ is finite for all $t$ in an open interval containing $0$.\nLet $K(t) = \\log M(t)$ be its [[cumulant generating function|0N2SP6K9ZC]] on that interval.\nFor $n \\ge 1$, the *$n$-th cumulant* of $X$ is\n$$\\kappa_n(X) = K^{(n)}(0),$$\nso that $K(t) = \\sum_{n \\ge 1} \\kappa_n \\frac{t^n}{n!}$ near $0$.\n\n## Remarks\n\nWriting $\\mu = E[X]$ and $\\mu_k = E[(X - \\mu)^k]$, the first cumulants are $\\kappa_1 = \\mu$, $\\kappa_2 = \\mu_2 = \\operatorname{Var}(X)$, $\\kappa_3 = \\mu_3$ and $\\kappa_4 = \\mu_4 - 3 \\mu_2^2$.\nIf $X$ and $Y$ are independent, then $K_{X+Y} = K_X + K_Y$, so $\\kappa_n(X + Y) = \\kappa_n(X) + \\kappa_n(Y)$.\nAlso $\\kappa_n(cX) = c^n \\kappa_n(X)$ for real $c$, and $\\kappa_n(X + c) = \\kappa_n(X)$ for $n \\ge 2$.\nExample: if $X$ is normal with mean $\\mu$ and variance $\\sigma^2$, then $K(t) = \\mu t + \\sigma^2 t^2 / 2$, so $\\kappa_1 = \\mu$, $\\kappa_2 = \\sigma^2$ and $\\kappa_n = 0$ for $n \\ge 3$.\nExample: if $X$ is Poisson with mean $\\lambda$, then $K(t) = \\lambda(e^t - 1)$, so $\\kappa_n = \\lambda$ for all $n$.\nNon-example: a Cauchy variable has $E[e^{tX}] = \\infty$ for all $t \\ne 0$, so the hypothesis fails and it has no cumulants.\nWhen only $E[\\abs{X}^n] \u003c \\infty$ is known, one defines $\\kappa_1, \\dots, \\kappa_n$ instead from the expansion of $\\log E[e^{itX}]$ at $0$, as $i^k \\kappa_k$ for the coefficient of $t^k/k!$; the two definitions agree when both apply.\n",
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