# Mode of Gaussian Distribution

## Theorem

Let $\mu$ be a real number.

Let $\sigma$ be a strictly positive real number.

Let $X \sim \Gaussian \mu {\sigma^2}$ where $\Gaussian \mu {\sigma^2}$ is the Gaussian distribution with parameters $\mu$ and $\sigma^2$.

Then the mode $M$ of $X$ is equal to $\mu$.

## Proof

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

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

By the definition of a mode, $f_X$ attains its global maximum at $M$.

From Exponential is Strictly Increasing, we have that $\exp x$ is strictly increasing for all real $x$.

Therefore, to maximise $f_X$, we must pick to $x$ to maximise:

- $-\dfrac {\paren {x - \mu}^2} {2 \sigma^2}$

That is, to minimise:

- $\dfrac {\paren {x - \mu}^2} {2 \sigma^2}$

We have:

- $\dfrac {\paren {x - \mu}^2} {2 \sigma^2} \ge 0$

with equality occurring at $x = \mu$.

Therefore, $\mu$ is a mode of $X$.

As this is the only value of $x$ for which equality occurs, we have that $\mu$ is the unique mode of $X$.

$\blacksquare$