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

Theorem: The probability density function of the normal distribution with mean $\mu$ and variance $\sigma^2$ has two inflection points at $x = \mu - \sigma$ and $x = \mu + \sigma$, i.e. exactly one standard deviation away from the expected value.

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 three deriatives of this function are:

\[\label{eq:norm-pdf-der1} f'_X(x) = \frac{\mathrm{d}f_X(x)}{\mathrm{d}x} = \frac{1}{\sqrt{2 \pi} \sigma} \cdot \left( - \frac{x - \mu}{\sigma^2} \right) \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right]\] \[\label{eq:norm-pdf-der2} \begin{split} f''_X(x) = \frac{\mathrm{d}^2f_X(x)}{\mathrm{d}x^2} &= \frac{1}{\sqrt{2 \pi} \sigma} \cdot \left( - \frac{1}{\sigma^2} \right) \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] + \frac{1}{\sqrt{2 \pi} \sigma} \cdot \left( \frac{x-\mu}{\sigma^2} \right)^2 \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] \\ &= \frac{1}{\sqrt{2 \pi} \sigma} \cdot \left[ \left( \frac{x-\mu}{\sigma^2} \right)^2 - \frac{1}{\sigma^2} \right] \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] \end{split}\] \[\label{eq:norm-pdf-der3} \begin{split} f'''_X(x) = \frac{\mathrm{d}^3f_X(x)}{\mathrm{d}x^3} &= \frac{1}{\sqrt{2 \pi} \sigma} \cdot \left[ \frac{2}{\sigma^2} \left( \frac{x-\mu}{\sigma^2} \right) \right] \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] - \frac{1}{\sqrt{2 \pi} \sigma} \cdot \left[ \left( \frac{x-\mu}{\sigma^2} \right)^2 - \frac{1}{\sigma^2} \right] \cdot \left( \frac{x - \mu}{\sigma^2} \right) \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] \\ &= \frac{1}{\sqrt{2 \pi} \sigma} \cdot \left[ -\left( \frac{x - \mu}{\sigma^2} \right)^3 + 3 \left( \frac{x - \mu}{\sigma^4} \right) \right] \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] \; . \end{split}\]

The second derivative is zero, if and only if

\[\label{eq:norm-pdf-der2-zero} \begin{split} 0 &= \left[ \left( \frac{x-\mu}{\sigma^2} \right)^2 - \frac{1}{\sigma^2} \right] \\ 0 &= \frac{x^2}{\sigma^4} - \frac{2 \mu x}{\sigma^4} + \frac{\mu^2}{\sigma^4} - \frac{1}{\sigma^2} \\ 0 &= x^2 - 2 \mu x + (\mu^2 - \sigma^2) \\ x_{1/2} &= -\frac{-2 \mu}{2} \pm \sqrt{ \left(\frac{-2 \mu}{2}\right)^2 - (\mu^2 - \sigma^2)} \\ x_{1/2} &= \mu \pm \sqrt{ \mu^2 - \mu^2 + \sigma^2} \\ x_{1/2} &= \mu \pm \sigma \; . \end{split}\]

Since the third derivative is non-zero at this value

\[\label{eq:norm-pdf-der3-infl} \begin{split} f'''_X(\mu \pm \sigma) &= \frac{1}{\sqrt{2 \pi} \sigma} \cdot \left[ -\left( \frac{\pm \sigma}{\sigma^2} \right)^3 + 3 \left( \frac{\pm \sigma}{\sigma^4} \right) \right] \cdot \exp \left[ -\frac{1}{2} \left( \frac{\pm \sigma}{\sigma} \right)^2 \right] \\ &= \frac{1}{\sqrt{2 \pi} \sigma} \cdot \left( \pm \frac{2}{\sigma^3} \right) \cdot \exp \left( - \frac{1}{2} \right) \neq 0 \; , \end{split}\]

there are inflection points at $x_{1/2} = \mu \pm \sigma$. Because $\mu$ is the mean and $\sigma^2$ is the variance of a normal distribution, these points are exactly one standard deviation away from the mean.

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Metadata: ID: P252 | shortcut: norm-infl | author: JoramSoch | date: 2020-08-26, 12:26.