Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Normal distribution ▷ Mode

Theorem: Let $X$ be a random variable following a normal distribution:

$\label{eq:norm} X \sim \mathcal{N}(\mu, \sigma^2) \; .$

Then, the mode of $X$ is

$\label{eq:norm-mode} \mathrm{mode}(X) = \mu \; .$

Proof: The mode is the value which maximizes the probability density function:

$\label{eq:mode} \mathrm{mode}(X) = \operatorname*{arg\,max}_x f_X(x) \; .$ $\label{eq:norm-pdf} f_X(x) = \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] \; .$

The first two deriatives of this function are:

$\label{eq:norm-pdf-der1} f'_X(x) = \frac{\mathrm{d}f_X(x)}{\mathrm{d}x} = \frac{1}{\sqrt{2 \pi} \sigma^3} \cdot (-x + \mu) \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right]$ $\label{eq:norm-pdf-der2} f''_X(x) = \frac{\mathrm{d}^2f_X(x)}{\mathrm{d}x^2} = -\frac{1}{\sqrt{2 \pi} \sigma^3} \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] + \frac{1}{\sqrt{2 \pi} \sigma^5} \cdot (-x + \mu)^2 \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] \; .$

We now calculate the root of the first derivative \eqref{eq:norm-pdf-der1}:

$\label{eq:norm-mode-s1} \begin{split} f'_X(x) = 0 &= \frac{1}{\sqrt{2 \pi} \sigma^3} \cdot (-x + \mu) \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] \\ 0 &= -x + \mu \\ x &= \mu \; . \end{split}$

By plugging this value into the second deriative \eqref{eq:norm-pdf-der2},

$\label{eq:norm-mode-s2} \begin{split} f''_X(\mu) &= -\frac{1}{\sqrt{2 \pi} \sigma^3} \cdot \exp(0) + \frac{1}{\sqrt{2 \pi} \sigma^5} \cdot (0)^2 \cdot \exp(0) \\ &= -\frac{1}{\sqrt{2 \pi} \sigma^3} < 0 \; , \end{split}$

we confirm that it is in fact a maximum which shows that

$\label{eq:norm-mode-qed} \mathrm{mode}(X) = \mu \; .$
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Metadata: ID: P17 | shortcut: norm-mode | author: JoramSoch | date: 2020-01-09, 15:58.