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proof_id |
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true |
Maja Pavlovic |
Queen Mary University London |
m.pavlovic@se22.qmul.ac.uk |
2022-10-02 02:46:00 -0700 |
Mean of the log-normal distribution |
Probability Distributions |
Univariate continuous distributions |
Log-normal distribution |
Mean |
| authors |
year |
title |
in |
pages |
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Taboga, Marco |
2022 |
Log-normal distribution |
Lectures on probability theory and mathematical statistics |
retrieved on 2022-10-01 |
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P354 |
lognorm-mean |
majapavlo |
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:
$$ \label{eq:lognorm-mean}
\mathrm{E}(X) = \exp \left( \mu + \frac{1}{2} \sigma^2 \right) .
$$