Variance of Log Normal Distribution

Theorem

Let $X$ be a continuous random variable with the Log Normal distribution with $\mu \in \R, \sigma \in \R_{> 0}$.

Then the variance of $X$ is given by:

$\var X = \map \exp {2 \mu + \sigma^2} \paren {\map \exp {\sigma^2} - 1}$

Proof

By Variance as Expectation of Square minus Square of Expectation:

$\var X = \expect {X^2} - \paren {\expect X}^2$

By Expectation of Log Normal Distribution:

$\expect X = \map \exp {\mu + \dfrac {\sigma^2} 2}$

From Raw Moment of Log Normal Distribution:

The $n$th raw moment $\expect {X^n}$ of $X$ is given by:

$\expect {X^n} = \map \exp {n \mu + \dfrac {\sigma^2 n^2} 2}$

Therefore, for $n = 2$ we have:

$\expect {X^2} = \map \exp {2 \mu + 2 \sigma^2}$

So:

\(\ds \var X\)

\(=\)

\(\ds \map \exp {2 \mu + 2 \sigma^2} - \paren {\map \exp {\mu + \dfrac {\sigma^2} 2} }^2\)

\(\ds \)

\(=\)

\(\ds \map \exp {2 \mu + 2 \sigma^2} - \map \exp {2 \mu + \sigma^2}\)

Power of Power

\(\ds \)

\(=\)

\(\ds \map \exp {2 \mu + \sigma^2} \paren {\map \exp {\sigma^2} - 1}\)

$\blacksquare$