Moments are derivatives of the moment-generating function at zero

Statement

Let $X$ be a real random variable, and let $h > 0$ with $M_X(t) = E[e^{tX}] < \infty$ for all $t \in (-h, h)$. Then $X$ has finite moments of all orders, the moment-generating function $M_X$ is infinitely differentiable on $(-h, h)$, and for each integer $k \ge 0$, $$M_X^{(k)}(0) = E\left[X^k\right].$$ Moreover, $M_X(t) = \sum_{k=0}^{\infty} E[X^k] \, t^k / k!$ for $|t| < h$.

Remarks

The proof differentiates under the expectation sign. The bound $|x|^k e^{t x} \le C(e^{a x} + e^{-a x})$ for $|t| < a < h$ justifies this by dominated convergence. Example: 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 $\operatorname{Var}(X) = 1/\lambda^2$.

Explanations

L1 human

This result is the reason for the name: the function "generates" the moments. To find $E[X^k]$, you take the $k$-th derivative of $M_X$ and put $t = 0$.