# Expectation of Gaussian Distribution

## Theorem

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

Then:

$\expect X = \mu$

## Proof 1

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 the expected value of a continuous random variable:

$\ds \expect X = \int_{-\infty}^\infty x \map {f_X} x \rd x$

So:

 $\ds \expect X$ $=$ $\ds \frac 1 {\sigma \sqrt {2 \pi} } \int_{-\infty}^\infty x \map \exp {-\frac {\paren {x - \mu}^2} {2 \sigma^2} } \rd x$ $\ds$ $=$ $\ds \frac {\sqrt 2 \sigma} {\sigma \sqrt {2 \pi} } \int_{-\infty}^\infty \paren {\sqrt 2 \sigma t + \mu} \map \exp {-t^2} \rd t$ substituting $t = \dfrac {x - \mu} {\sqrt 2 \sigma}$ $\ds$ $=$ $\ds \frac 1 {\sqrt \pi} \paren {\sqrt 2 \sigma \int_{-\infty}^\infty t \map \exp {-t^2} \rd t + \mu \int_{-\infty}^\infty \map \exp {-t^2} \rd t}$ $\ds$ $=$ $\ds \frac 1 {\sqrt \pi} \paren {\sqrt 2 \sigma \intlimits {-\frac 1 2 \map \exp {-t^2} } {-\infty} \infty + \mu \sqrt \pi}$ Fundamental Theorem of Calculus, Gaussian Integral $\ds$ $=$ $\ds \frac {\mu \sqrt \pi} {\sqrt \pi}$ Exponential Tends to Zero and Infinity $\ds$ $=$ $\ds \mu$

$\blacksquare$

## Proof 2

By Moment Generating Function of Gaussian Distribution, the moment generating function of $X$ is given by:

$\map {M_X} t = \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$
$\expect X = \map {M'_X} 0$

We have:

 $\ds \map {M_X'} t$ $=$ $\ds \frac \d {\d t} \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$ $\ds$ $=$ $\ds \map {\frac \d {\d t} } {\mu t + \frac 1 2 \sigma^2 t^2} \frac \d {\map \d {\mu t + \dfrac 1 2 \sigma^2 t^2} } \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$ Chain Rule for Derivatives $\ds$ $=$ $\ds \paren {\mu + \sigma^2 t} \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$ Derivative of Power, Derivative of Exponential Function

Setting $t = 0$:

 $\ds \map {M_X'} 0$ $=$ $\ds \paren {\mu + 0\sigma^2} \map \exp {0\mu + 0 \sigma^2}$ $\ds$ $=$ $\ds \mu \exp 0$ $\ds$ $=$ $\ds \mu$ Exponential of Zero

$\blacksquare$