# Median of Gaussian Distribution

## 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 median of $X$ is equal to $\mu$.

## Proof

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

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

Note that $f_X$ is non-zero, sufficient to ensure a unique median.

By the definition of a median, to prove that $\mu$ is the median of $X$ we must verify:

$\ds \map \Pr {X < \mu} = \int_{-\infty}^\mu \map {f_X} x \rd x = \frac 1 2$

We have:

 $\ds \int_{-\infty}^\mu \map {f_X} x \rd x$ $=$ $\ds \frac 1 {\sigma \sqrt {2 \pi} } \int_{-\infty}^\mu \map \exp {-\frac {\paren {x - \mu}^2} {2 \sigma^2} } \rd x$ $\ds$ $=$ $\ds \frac {\sqrt 2 \sigma} {\sigma \sqrt {2 \pi} } \int_{-\infty}^{\frac {\mu - \mu} {\sqrt 2 \sigma} } \map \exp {-t^2} \rd t$ substituting $t = \dfrac {x - \mu} {\sqrt 2 \sigma}$ $\ds$ $=$ $\ds \frac 1 {\sqrt \pi} \int_{-\infty}^0 \map \exp {-t^2} \rd t$ $\ds$ $=$ $\ds \frac 1 {2 \sqrt \pi} \int_{-\infty}^\infty \map \exp {-t^2} \rd t$ Definite Integral of Even Function $\ds$ $=$ $\ds \frac {\sqrt \pi} {2 \sqrt \pi}$ Gaussian Integral $\ds$ $=$ $\ds \frac 1 2$

$\blacksquare$