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 $X$.
$\blacksquare$