Moments from the MGF¶

Theorem¶

If the MGF $M_X$ is finite on an open interval containing $0$ , then all moments of $X$ are finite and, for every nonnegative integer $n$ ,

\begin{aligned} \mathbb{E}[X^n]=M_X^{(n)}(0), \end{aligned}

where $M_X^{(n)}$ denotes the $n$ th derivative of $M_X$ . The assumption is what justifies differentiating under the expectation.

The Moment Generating Function is an exponential expectation. Taking its Higher-Order Derivative brings down a power of $X$ each time, and evaluating at zero leaves $\mathbb{E}[X^n]$ . The finiteness assumption justifies moving those derivatives under the expectation.

Interpretation¶

Differentiating the MGF at zero extracts the moments of the random variable one at a time: the first derivative gives $\mathbb{E}X$ , the second gives $\mathbb{E}[X^2]$ , and so on.

The MGF 1/(1 - t) of an Exponential(1) variable with its tangent line 1 + t and its quadratic approximation 1 + t + t squared at t = 0, whose coefficients give the first two moments.

For $X\sim\operatorname{Exp}(1)$ $Exp (1)$ , $M_X(t)=1/(1-t)$ for $t<1$ $1$ . The tangent line $1+t$ at $t=0$ has slope $M_X'(0)=\mathbb{E}X=1$ $1$ , and the parabola $1+t+t^2$ $t^{2}$ , written $1+t+\tfrac{2}{2!}t^2$ in the legend, matches the second derivative $M_X''(0)=\mathbb{E}[X^2]=2$ $2$ . Hence $\operatorname{Var}(X)=2-1^2=1$ $1$ .

The reason is the expansion $\mathrm{e}^{tX}=1+tX+t^2X^2/2!+\cdots$ $\dots$ . Taking expectations term by term gives $M_X(t)=1+\mathbb{E}[X]\,t+\mathbb{E}[X^2]\,t^2/2!+\cdots$ near $0$ , so, as for any Taylor series, $\mathbb{E}[X^n]$ is the $n$ th derivative at $0$ , equivalently $n!$ times the coefficient of $t^n$ . For $1/(1-t)=1+t+t^2+\cdots$ $\dots$ , a geometric series for $|t|<1$ $1$ , every coefficient is $1$ , so $\mathbb{E}[X^n]=n!$ $n!$ . Finiteness of $M_X$ on an open interval around $0$ is what justifies the term-by-term step; without it $M_X$ may be infinite on one side of $0$ , and then its derivatives at $0$ are not defined.

The MGF delivers raw moments; the variance and other central moments then follow, as in $\operatorname{Var}(X)=M_X''(0)-(M_X'(0))^2$ $(M_{X}^{'} (0))^{2}$ . A common slip is to forget to set $t=0$ $0$ : $M_X'(t)$ at some $t\neq0$ is not a moment of $X$ .

Key Results¶

$\mathbb{E}X=M_X'(0)$ and $\mathbb{E}[X^2]=M_X''(0)$ $M_{X}^{''} (0)$ .
$\operatorname{Var}(X)=M_X''(0)-(M_X'(0))^2$ $(M_{X}^{'} (0))^{2}$ .

Examples¶

For $M_X(t)=1-p+p\mathrm{e}^t$ $p e^{t}$ , the first two derivatives at zero both equal $p$ , giving variance $p-p^2$ $p^{2}$ .
For $M_X(t)=\lambda/(\lambda-t)$ $t)$ , the first two moments are $1/\lambda$ and $2/\lambda^2$ , giving variance $1/\lambda^2$ .

Prerequisites¶

Variance and Standard Deviation

Sources¶

Nicky van Foreest · Probability Distributions · Week 1

Symbols: lambda

Last updated: 20 October 2026 03:02