Index: The Book of Statistical Proofs ▷ General Theorems ▷ Probability theory ▷ Further moments ▷ Moment in terms of moment-generating function

Theorem: Let $X$ be a scalar random variable with the moment-generating function $M_X(t)$. Then, the $n$-th raw moment of $X$ can be calculated from the moment-generating function via

\[\label{eq:mom-mgf} \mathrm{E}(X^n) = M_X^{(n)}(0)\]

where $n$ is a positive integer and $M_X^{(n)}(t)$ is the $n$-th derivative of $M_X(t)$.

Proof: Using the definition of the moment-generating function, we can write:

\[\label{eq:mom-mgf-s1} M_X^{(n)}(t) = \frac{\mathrm{d}^n}{\mathrm{d}t^n} \mathrm{E}(e^{tX}) \; .\]

Using the power series expansion of the exponential function

\[\label{eq:exp-ps} e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \; ,\]

equation \eqref{eq:mom-mgf-s1} becomes

\[\label{eq:mom-mgf-s2} M_X^{(n)}(t) = \frac{\mathrm{d}^n}{\mathrm{d}t^n} \mathrm{E}\left( \sum_{m=0}^\infty \frac{t^m X^m}{m!} \right) \; .\]

Because the expected value is a linear operator, we have:

\[\label{eq:mom-mgf-s3} \begin{split} M_X^{(n)}(t) &= \frac{\mathrm{d}^n}{\mathrm{d}t^n} \sum_{m=0}^\infty \mathrm{E}\left( \frac{t^m X^m}{m!} \right) \\ &= \sum_{m=0}^\infty \frac{\mathrm{d}^n}{\mathrm{d}t^n} \frac{t^m}{m!} \mathrm{E}\left( X^m \right) \; . \end{split}\]

Using the $n$-th derivative of the $m$-th power

\[\label{eq:dndx-xm} \frac{\mathrm{d}^n}{\mathrm{d}x^n} x^m = \left\{ \begin{array}{rl} m^{\underline{n}} \, x^{m-n} \; , & \text{if} \; n \leq m \\ 0 \; , & \text{if} \; n > m \; . \end{array} \right.\]

with the falling factorial

\[\label{eq:fact-fall} m^{\underline{n}} = \prod_{i=0}^{n-1} (m-i) = \frac{m!}{(m-n)!} \; ,\]

equation \eqref{eq:mom-mgf-s3} becomes

\[\label{eq:mom-mgf-s4} \begin{split} M_X^{(n)}(t) &= \sum_{m=n}^\infty \frac{m^{\underline{n}} \, t^{m-n}}{m!} \mathrm{E}\left( X^m \right) \\ &\overset{\eqref{eq:fact-fall}}{=} \sum_{m=n}^\infty \frac{m! \, t^{m-n}}{(m-n)! \, m!} \mathrm{E}\left( X^m \right) \\ &= \sum_{m=n}^\infty \frac{t^{m-n}}{(m-n)!} \mathrm{E}\left( X^m \right) \\ &= \frac{t^{n-n}}{(n-n)!} \mathrm{E}\left( X^n \right) + \sum_{m=n+1}^\infty \frac{t^{m-n}}{(m-n)!} \mathrm{E}\left( X^m \right) \\ &= \frac{t^0}{0!} \, \mathrm{E}\left( X^n \right) + \sum_{m=n+1}^\infty \frac{t^{m-n}}{(m-n)!} \mathrm{E}\left( X^m \right) \\ &= \mathrm{E}\left( X^n \right) + \sum_{m=n+1}^\infty \frac{t^{m-n}}{(m-n)!} \mathrm{E}\left( X^m \right) \; . \end{split}\]

Setting $t = 0$ in \eqref{eq:mom-mgf-s4} yields

\[\label{eq:mom-mgf-s5} \begin{split} M_X^{(n)}(0) &= \mathrm{E}\left( X^n \right) + \sum_{m=n+1}^\infty \frac{0^{m-n}}{(m-n)!} \mathrm{E}\left( X^m \right) \\ &= \mathrm{E}\left( X^n \right) \end{split}\]

which conforms to equation \eqref{eq:mom-mgf}.

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Metadata: ID: P153 | shortcut: mom-mgf | author: JoramSoch | date: 2020-08-19, 07:51.