Calculating Moments Using Moment-Generating Functions

The MGF of a random variable encodes all of its moments through its derivatives at zero.

Moment-generating property: If $M_X(t) = E[e^{tX}]$ exists in a neighborhood of $t=0$, then the $n$-th moment of $X$ equals the $n$-th derivative of $M_X$ evaluated at $0$:

$$E[X^n] \;=\; M_X^{(n)}(0).$$

In particular, the mean of $X$ is just the first derivative at $0$:

$$E[X] = M_X'(0).$$

This follows from the Taylor expansion of $e^{tX}$. Taking the expectation term-by-term,

$$M_X(t) \;=\; E\!\left[\sum_{n=0}^{\infty} \frac{(tX)^n}{n!}\right] \;=\; \sum_{n=0}^{\infty} \frac{E[X^n]}{n!}\, t^n.$$

Matching this with the Taylor series of $M_X$ about $0$, the coefficient of $t^n$ is $\dfrac{M_X^{(n)}(0)}{n!}$, so $E[X^n] = M_X^{(n)}(0)$.

Quick check: if $M_X(t) = e^{t^2/2}$, then $M_X'(t) = t\,e^{t^2/2}$, so $E[X] = M_X'(0) = 0$.