Moment Generating Function of Gaussian Distribution

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Theorem

Let $X \sim \Gaussian \mu {\sigma^2}$ for some $\mu \in \R, \sigma \in \R_{> 0}$, where $N$ is the Gaussian distribution.

Then the moment generating function $M_X$ of $X$ is given by:

$\map {M_X} t = \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$


Proof

From the definition of the Gaussian distribution, $X$ has probability density function:

$\map {f_X} x = \dfrac 1 {\sigma \sqrt{2 \pi} } \map \exp {-\dfrac {\paren {x - \mu}^2} {2 \sigma^2} }$

From the definition of a moment generating function:

$\displaystyle \map {M_X} t = \expect { e^{t X} } = \int_{-\infty}^\infty e^{t x} \map {f_X} x \rd x$

So:

\(\displaystyle \map {M_X} t\) \(=\) \(\displaystyle \frac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \map \exp {t x - \frac {\paren {x - \mu}^2} {2 \sigma^2} } \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sqrt 2 \sigma} {\sigma \sqrt {2 \pi} } \int_{-\infty}^\infty \map \exp {\paren {\sqrt 2 \sigma u + \mu} t - u^2} \rd u\) substituting $u = \dfrac {x - \mu} {\sqrt 2 \sigma}$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\exp \mu t} {\sqrt \pi} \int_{-\infty}^\infty \map \exp {-\paren {u^2 - \sqrt 2 \sigma u t} } \rd u\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\exp \mu t} {\sqrt \pi} \int_{-\infty}^\infty \map \exp {-\paren {u - \frac {\sqrt 2} 2 \sigma t}^2 + \frac 1 2 \sigma^2 t^2} \rd u\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\map \exp {\mu t + \frac 1 2 \sigma^2 t^2} } {\sqrt \pi} \int_{-\infty}^\infty \map \exp {-v^2} \rd v\) substituting $v = u - \dfrac {\sqrt 2} 2 \sigma t$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sqrt \pi \map \exp {\mu t + \frac 1 2 \sigma^2 t^2} } {\sqrt \pi}\) Gaussian Integral
\(\displaystyle \) \(=\) \(\displaystyle \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}\)

$\blacksquare$