Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Univariate continuous distributions ▷ Normal distribution ▷ Extreme points

Theorem: The probability density function of the normal distribution with mean $\mu$ and variance $\sigma^2$ has a maximum at $x = \mu$ and no other extrema. Consequently, the normal distribution is a unimodal probability distribution.

Proof: The probability density function of the normal distribution is:

\[\label{eq:norm-pdf} f_X(x) = \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] \; .\]

The first two deriatives of this function are:

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

The first derivative is zero, if and only if

\[\label{eq:norm-pdf-der1-zero} -x + \mu = 0 \quad \Leftrightarrow \quad x = \mu \; .\]

Since the second derivative is negative at this value

\[\label{eq:norm-pdf-der2-extr} f''_X(\mu) = -\frac{1}{\sqrt{2 \pi} \sigma^3} < 0 \; ,\]

there is a maximum at $x = \mu$. From \eqref{eq:norm-pdf-der1}, it can be seen that $f’_X(x)$ is positive for $x < \mu$ and negative for $x > \mu$. Thus, there are no further extrema and $\mathcal{N}(\mu, \sigma^2)$ is unimodal.

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Metadata: ID: P251 | shortcut: norm-extr | author: JoramSoch | date: 2020-08-25, 21:11.