Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Univariate continuous distributions ▷ Log-normal distribution ▷ Variance

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

\[X \sim \ln \mathcal{N}(\mu, \sigma^2) .\]

Then, the variance of $X$ is

\[\label{eq:lognorm-var} \mathrm{Var}(X) = \exp \left(2\mu +2\sigma^2\right) - \exp \left(2\mu + \sigma^2\right) \; .\]

Proof: The variance of a random variable is defined as

\[\label{eq:var} \mathrm{Var}(X) = \mathrm{E}\left[ (X-\mathrm{E}(X))^2 \right]\]

which, partitioned into expected values, reads:

\[\label{eq:var2} \mathrm{Var}(X) = \mathrm{E}\left[ X^2 \right] - \mathrm{E}\left[ X \right]^2 \; .\]

The expected value of the log-normal distribution is:

\[\label{eq:lognorm-mean-ref} \mathrm{E}[X] = \exp \left( \mu + \frac{1}{2} \sigma^2 \right)\]

The second moment $\mathrm{E}[X^2]$ can be derived as follows:

\[\label{eq:second-moment} \begin{split} \mathrm{E} [X^2] &= \int_{- \infty}^{+\infty} x^2 \cdot f_\mathrm{X}(x) \, \mathrm{d}x \\ &= \int_{0}^{+\infty} x^2 \cdot \frac{1}{x\sqrt{2 \pi \sigma^2} } \cdot \exp \left[ -\frac{1}{2} \frac{\left(\ln x-\mu\right)^2}{\sigma^2} \right] \mathrm{d}x \\ &= \frac{1}{\sqrt{2 \pi \sigma^2} } \int_{0}^{+\infty} x \cdot \exp \left[ -\frac{1}{2} \frac{\left(\ln x-\mu\right)^2}{\sigma^2} \right]\mathrm{d}x \end{split}\]

Substituting $z = \frac{\ln x -\mu}{\sigma}$, i.e. $x = \exp \left( \mu + \sigma z \right )$, we have:

\[\label{eq:second-moment-2} \begin{split} \mathrm{E} [X^2] &= \frac{1}{\sqrt{2 \pi \sigma^2} } \int_{(-\infty -\mu )/ (\sigma)}^{(\ln x -\mu )/ (\sigma)} \exp \left( \mu +\sigma z \right) \exp \left( -\frac{1}{2} z^2 \right) \mathrm{d} \left[ \exp \left( \mu +\sigma z \right) \right] \\ &= \frac{1}{\sqrt{2 \pi \sigma^2} } \int_{-\infty}^{+\infty} \exp \left( -\frac{1}{2} z^2 \right) \sigma \exp \left( 2\mu + 2 \sigma z \right) \mathrm{d}z \\ &= \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} \exp \left[ -\frac{1}{2} \left( z^2 - 4 \sigma z - 4 \mu \right) \right] \mathrm{d}z \end{split}\]

Now multiplying by $\exp \left( 2 \sigma^2 \right)$ and $\exp \left(- 2 \sigma^2 \right)$, this becomes:

\[\label{eq:second-moment-3} \begin{split} \mathrm{E} [X^2] &= \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} \exp \left[ -\frac{1}{2} \left( z^2 - 4 \sigma z + 4 \sigma^2 -4 \sigma^2 - 4 \mu \right) \right] \mathrm{d}z \\ &= \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} \exp \left[ -\frac{1}{2} \left( z^2 - 4\sigma z + 4\sigma^2 \right) \right] \exp \left( 2 \sigma^2 +2 \mu \right) \mathrm{d}z \\ &= \exp \left( 2 \sigma^2 +2 \mu \right) \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2 \pi} } \exp \left[ -\frac{1}{2} \left( z - 2 \sigma \right)^2 \right] \mathrm{d}z \end{split}\]

The probability density function of a normal distribution is given by

\[f_X(x) = \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right]\]

and, with $\mu = 2 \sigma$ and unit variance, this reads:

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

Using the definition of the probability density function, we get

\[\label{eq:def-pdf} \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2 \pi}} \cdot \exp \left[ -\frac{1}{2} \left({x - 2 \sigma} \right)^2 \right] \mathrm{d}x = 1\]

and applying \eqref{eq:def-pdf} to \eqref{eq:second-moment-3}, we have:

\[\label{eq:second-moment-4} \mathrm{E}[X]^2 = \exp \left( 2 \sigma^2 +2 \mu \right) \; .\]

Finally, plugging \eqref{eq:second-moment-4} and \eqref{eq:lognorm-mean-ref} into \eqref{eq:var2}, we have:

\[\label{eq:lognorm-var-2} \begin{split} \mathrm{Var}(X) &= \mathrm{E}\left[ X^2 \right] - \mathrm{E}\left[ X \right]^2 \\ &= \exp \left(2\sigma^2 + 2\mu \right) - \left[ \exp \left(\mu + \frac{1}{2} \sigma^2 \right) \right]^2 \\ &= \exp \left(2\sigma^2 + 2\mu \right) - \exp \left(2\mu + \sigma^2\right) \; . \end{split}\]
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Metadata: ID: P355 | shortcut: lognorm-var | author: majapavlo | date: 2022-10-02, 09:02.