Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Log-normal distribution ▷ Mode

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

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

Then, the mode of $X$ is

$\label{eq:lognorm-mode} \mathrm{mode}(X) = e^{\left( \mu -\sigma^2 \right)} \; .$

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

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

The first two derivatives of this function are:

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

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

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

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

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

we confirm that it is a maximum, showing that

$\label{eq:lognorm-mode-qed} \mathrm{mode}(X) = e^{\left( \mu -\sigma^2 \right)} \; .$
Sources:

Metadata: ID: P311 | shortcut: lognorm-mode | author: majapavlo | date: 2022-02-13, 10:15.