Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Univariate continuous distributions ▷ Normal distribution ▷ Moment-generating function

Theorem: Let $X$ be a random variable following a normal distribution:

\[\label{eq:norm} X \sim \mathcal{N}(\mu, \sigma^2) \; .\]

Then, the moment-generating function of $X$ is

\[\label{eq:norm-mgf} M_X(t) = \exp\left[ \mu t + \frac{1}{2} \sigma^2 t^2 \right] \; .\]

Proof: The probability density function of the normal distribution is

\[\label{eq:norm-pdf} f_X(x) = \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right]\]

and the moment-generating function is defined as

\[\label{eq:mgf-var} M_X(t) = \mathrm{E} \left[ e^{tX} \right] \; .\]

Using the expected value for continuous random variables, the moment-generating function of $X$ therefore is

\[\label{eq:norm-mgf-s1} \begin{split} M_X(t) &= \int_{-\infty}^{+\infty} \exp[tx] \cdot \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] \, \mathrm{d}x \\ &= \frac{1}{\sqrt{2 \pi} \sigma} \int_{-\infty}^{+\infty} \exp\left[ tx - \frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] \, \mathrm{d}x \; . \end{split}\]

Substituting $u = (x-\mu)/(\sqrt{2}\sigma)$, i.e. $x = \sqrt{2}\sigma u + \mu$, we have

\[\label{eq:norm-mgf-s2} \begin{split} M_X(t) &= \frac{1}{\sqrt{2 \pi} \sigma} \int_{(-\infty-\mu)/(\sqrt{2}\sigma)}^{(+\infty-\mu)/(\sqrt{2}\sigma)} \exp\left[ t\left( \sqrt{2} \sigma u + \mu \right) - \frac{1}{2} \left( \frac{ \sqrt{2} \sigma u + \mu - \mu}{\sigma} \right)^2 \right] \, \mathrm{d}\left( \sqrt{2} \sigma u + \mu \right) \\ &= \frac{\sqrt{2} \sigma}{\sqrt{2 \pi} \sigma} \int_{-\infty}^{+\infty} \exp\left[ \left( \sqrt{2} \sigma u + \mu \right) t - u^2 \right] \, \mathrm{d}u \\ &= \frac{\exp(\mu t)}{\sqrt{\pi}} \int_{-\infty}^{+\infty} \exp\left[ \sqrt{2} \sigma u t - u^2 \right] \, \mathrm{d}u \\ &= \frac{\exp(\mu t)}{\sqrt{\pi}} \int_{-\infty}^{+\infty} \exp\left[ - \left( u^2 - \sqrt{2} \sigma u t \right) \right] \, \mathrm{d}u \\ &= \frac{\exp(\mu t)}{\sqrt{\pi}} \int_{-\infty}^{+\infty} \exp\left[ - \left( u - \frac{\sqrt{2}}{2} \sigma t \right)^2 + \frac{1}{2} \sigma^2 t^2 \right] \, \mathrm{d}u \\ &= \frac{\exp\left[ \mu t + \frac{1}{2} \sigma^2 t^2 \right]}{\sqrt{\pi}} \int_{-\infty}^{+\infty} \exp\left[ - \left( u - \frac{\sqrt{2}}{2} \sigma t \right)^2 \right] \, \mathrm{d}u \end{split}\]

Now substituting $v = u - \sqrt{2}/2 \, \sigma t$, i.e. $u = v + \sqrt{2}/2 \, \sigma t$, we have

\[\label{eq:norm-mgf-s3} \begin{split} M_X(t) &= \frac{\exp\left[ \mu t + \frac{1}{2} \sigma^2 t^2 \right]}{\sqrt{\pi}} \int_{-\infty - \sqrt{2}/2 \, \sigma t}^{+\infty - \sqrt{2}/2 \, \sigma t} \exp\left[ -v^2 \right] \, \mathrm{d}\left( v + \sqrt{2}/2 \, \sigma t \right) \\ &= \frac{\exp\left[ \mu t + \frac{1}{2} \sigma^2 t^2 \right]}{\sqrt{\pi}} \int_{-\infty}^{+\infty} \exp\left[ -v^2 \right] \, \mathrm{d}v \; . \end{split}\]

With the Gaussian integral

\[\label{eq:gauss} \int_{-\infty}^{+\infty} \exp\left[ -x^2 \right] \, \mathrm{d}x = \sqrt{\pi} \; ,\]

this finally becomes

\[\label{eq:norm-mgf-qed} M_X(t) = \exp\left[ \mu t + \frac{1}{2} \sigma^2 t^2 \right] \; .\]
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Metadata: ID: P71 | shortcut: norm-mgf | author: JoramSoch | date: 2020-03-03, 11:29.