Expectation of Gaussian Distribution/Proof 1

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

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

\(\ds \)

\(=\)

\(\ds \mu\)

$\blacksquare$