Moment-generating function of the normal distribution

Statement

Let $\mu \in \mathbb{R}$, let $\sigma^2 > 0$ and let $X \sim \mathcal{N}(\mu, \sigma^2)$. Then for each $t \in \mathbb{R}$, $$M_X(t) = E\left[e^{tX}\right] = \exp\left(\mu t + \tfrac{1}{2} \sigma^2 t^2\right).$$

Remarks

Proof for $Z \sim \mathcal{N}(0, 1)$: complete the square, $tz - z^2/2 = t^2/2 - (z - t)^2/2$. So $E[e^{tZ}] = e^{t^2/2} \int \varphi(z - t) \, dz = e^{t^2/2}$. The general case follows from $X = \mu + \sigma Z$ with $\sigma = \sqrt{\sigma^2}$: $M_X(t) = e^{\mu t} M_Z(\sigma t)$. The moment-generating function is finite for all $t$. Its derivatives at $0$ give $E[X] = \mu$ and $E[X^2] = \mu^2 + \sigma^2$. The logarithm $\mu t + \sigma^2 t^2 / 2$ is a polynomial of degree $2$. So all cumulants of order $3$ or more are $0$.

Explanations

L1 human

The moment-generating function of a normal variable has a simple closed form. With the multiplication rule for sums, it shows quickly that a sum of independent normal variables is again normal.