Raw Moment of Log Normal Distribution
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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:
\(\ds \expect {X^n}\) | \(=\) | \(\ds \int_0^\infty x^n \dfrac 1 {\sigma \sqrt {2 \pi} x} \map \exp {-\dfrac {\paren {\map \ln x - \mu}^2} {2 \sigma^2} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\sigma \sqrt {2 \pi} } \int_0^\infty x^{n - 1} \map \exp {-\dfrac {\paren {\map \ln x - \mu}^2} {2 \sigma^2} } \rd x\) | rewriting |
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:
\(\ds \expect {X^n}\) | \(=\) | \(\ds \dfrac {\exp {\paren {n \mu} } } {\sqrt \pi} \int_{-\infty}^\infty \exp {\paren {-u^2 + \sqrt 2 \sigma nu} } \rd u\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\exp {\paren {n \mu} } } {\sqrt \pi} \int_{-\infty}^\infty \exp {\paren {-u^2 + \sqrt 2 \sigma nu - \dfrac {\sigma^2 n^2 } 2 + \dfrac {\sigma^2 n^2} 2} } \rd u\) | adding $0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\exp {\paren {n \mu + \dfrac {\sigma^2 n^2} 2} } } {\sqrt \pi} \int_{-\infty}^\infty \exp {\paren {-u^2 + \sqrt 2 \sigma nu - \dfrac {\sigma^2 n^2} 2} } \rd u\) | Linear Combination of Definite Integrals | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\exp {\paren {n \mu + \dfrac {\sigma^2 n^2} 2} } } {\sqrt \pi} \int_{-\infty}^\infty \exp {\paren {-\paren {u - \dfrac {\sqrt 2 \sigma n} 2}^2} } \rd u\) | Square of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\exp {\paren {n \mu + \dfrac {\sigma^2 n^2} 2} } } {\sqrt \pi} \int_{-\infty}^\infty \exp {\paren {-y^2} } \rd y\) | setting $y = u - \dfrac {\sqrt 2 \sigma n} 2$ and hence $\d y = \d u$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\exp {\paren {n \mu + \dfrac {\sigma^2 n^2} 2} } } {\sqrt \pi} \sqrt \pi\) | Gaussian Integral | |||||||||||
\(\ds \) | \(=\) | \(\ds \exp {\paren {n \mu + \dfrac {\sigma^2 n^2} 2} }\) |
$\blacksquare$