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

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 mean or expected value of $X$ is

\[\mathrm{E}(X) = \exp \left( \mu + \frac{1}{2} \sigma^2 \right)\]

Proof: The expected value is the probability-weighted average over all possible values:

\[\label{eq:mean} \mathrm{E}(X) = \int_{\mathcal{X}} x \cdot f_X(x) \, \mathrm{d}x\]

With the probability density function of the log-normal distribution, this is:

\[\label{eq:lognorm-mean-s1} \begin{split} \mathrm{E}(X) &= \int_{0}^{+\infty} x \cdot \frac{1}{x\sqrt{2 \pi \sigma^2} } \cdot \exp \left[ -\frac{1}{2} \frac{\left(\ln x-\mu\right)^2}{\sigma^2} \right] \mathrm{d}x \\ &= \frac{1}{\sqrt{2 \pi \sigma^2} } \int_{0}^{+\infty} \exp \left[ -\frac{1}{2} \frac{\left(\ln x-\mu\right)^2}{\sigma^2} \right] \mathrm{d}x \end{split}\]

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

\[\label{eq:lognorm-mean-s2} \begin{split} \mathrm{E}(X) &= \frac{1}{\sqrt{2 \pi \sigma^2} } \int_{(-\infty -\mu )/ (\sigma)}^{(\ln x -\mu )/ (\sigma)} \exp \left( -\frac{1}{2} z^2 \right) \mathrm{d} \left[ \exp \left( \mu +\sigma z \right) \right] \\ &= \frac{1}{\sqrt{2 \pi \sigma^2} } \int_{-\infty}^{+\infty} \exp \left( -\frac{1}{2} z^2 \right) \sigma \exp \left( \mu +\sigma z \right) \mathrm{d}z \\ &= \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} \exp \left( -\frac{1}{2} z^2 + \sigma z + \mu \right) \mathrm{d}z \\ &= \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} \exp \left[ -\frac{1}{2} \left( z^2 - 2 \sigma z - 2 \mu \right) \right] \mathrm{d}z \end{split}\]

Now multiplying $\exp \left( \frac{1}{2} \sigma^2 \right)$ and $\exp \left( -\frac{1}{2} \sigma^2 \right)$, we have:

\[\label{eq:lognorm-mean-s3} \begin{split} \mathrm{E}(X) &= \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} \exp \left[ -\frac{1}{2} \left( z^2 - 2 \sigma z + \sigma^2 - 2 \mu - \sigma^2 \right) \right] \mathrm{d}z \\ &= \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} \exp \left[ -\frac{1}{2} \left( z^2 - 2\sigma z + \sigma^2 \right) \right] \exp \left( \mu + \frac{1}{2} \sigma^2 \right) \mathrm{d}z \\ &= \exp \left( \mu + \frac{1}{2} \sigma^2 \right) \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2 \pi} } \exp \left[ -\frac{1}{2} \left( z - \sigma \right)^2 \right] \mathrm{d}z \end{split}\]

The probability density function of a normal distribution is given by

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

and, with unit variance $\sigma^2 = 1$, this reads:

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

Using the definition of the probability density function, we get

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

and applying \eqref{eq:def-pdf} to \eqref{eq:lognorm-mean-s3}, we have:

\[\label{eq:lognorm-mean} \mathrm{E}(X) = \exp \left( \mu + \frac{1}{2} \sigma^2 \right) .\]
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Metadata: ID: P354 | shortcut: lognorm-mean | author: majapavlo | date: 2022-10-02, 09:46.