Cumulant

Remarks

Writing $\mu = E[X]$ 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 $\kappa_4 = \mu_4 - 3 \mu_2^2$. If $X$ and $Y$ are independent, then $K_{X+Y} = K_X + K_Y$, so $\kappa_n(X + Y) = \kappa_n(X) + \kappa_n(Y)$. Also $\kappa_n(cX) = c^n \kappa_n(X)$ for real $c$, and $\kappa_n(X + c) = \kappa_n(X)$ for $n \ge 2$. Example: 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 $n \ge 3$. Example: if $X$ is Poisson with mean $\lambda$, then $K(t) = \lambda(e^t - 1)$, so $\kappa_n = \lambda$ for all $n$. Non-example: a Cauchy variable has $E[e^{tX}] = \infty$ for all $t \ne 0$, so the hypothesis fails and it has no cumulants. When only $E[\abs{X}^n] < \infty$ 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.