Raw Moment of Log Normal Distribution

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

Let $n$ be a strictly positive integer.

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


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

Proof
From the definition of the Log Normal distribution, $X$ has probability density function:


 * $\map {f_X} x = \dfrac 1 {\sigma \sqrt {2 \pi} x} \map \exp {-\dfrac {\paren {\map \ln x - \mu}^2} {2 \sigma^2} }$

where $\Img X = \R_{>0}$.

From the definition of the expected value of a continuous random variable:
 * $\ds \expect {X^n} = \int_0^\infty x^n \map {f_X} x \rd x$

Therefore:

Let:
 * $u = \dfrac {\paren {\map \ln x - \mu}} {\sqrt 2 \sigma}$

Then by Chain Rule for Derivatives, we have:
 * $\dfrac {\d u} {\d x} = \dfrac 1 {x \sqrt 2 \sigma}$

and also:
 * $\exp {\paren {\sqrt 2 \sigma nu + n \mu} } = x^n$

We can see that:
 * $u \to -\infty$ as $x \to 0$

and
 * $u \to \infty$ as $x \to \infty$

Plugging these results back into our integral above, we have: