Skewness of Gaussian Distribution/Proof 2

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Theorem

Let $X$ be a continuous random variable with a Gaussian distribution with parameters $\mu$ and $\sigma^2$ for some $\mu \in \R$ and $\sigma \in \R_{> 0}$.

Then the skewness $\gamma_1$ of $X$ is equal to $0$.


Proof

From the definition of skewness, we have:

$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$

where:

$\mu$ is the expectation of $X$.
$\sigma$ is the standard deviation of $X$.

By Expectation of Gaussian Distribution, we have:

$\mu = \mu$

By Variance of Gaussian Distribution, we have:

$\sigma = \sigma$

So:

\(\ds \gamma_1\) \(=\) \(\ds \frac {\expect {X^3 - 3 X^2 \mu + 3 X \mu^2 - \mu^3} } {\sigma^3}\) Cube of Difference
\(\ds \) \(=\) \(\ds \frac {\expect {X^3} - 3 \mu \expect {X^2} + 3 \mu^2 \expect X - \mu^3} {\sigma^3}\) Expectation is Linear
\(\ds \) \(=\) \(\ds \frac {\expect {X^3} - 3 \mu \paren {\sigma^2 + \mu^2} + 3 \mu^2 \paren \mu - \mu^3} {\sigma^3}\) Variance of Gaussian Distribution


To calculate $\gamma_1$, we must calculate $\expect {X^3}$.

From Moment in terms of Moment Generating Function:

$\expect {X^n} = \map { {M_X}^{\paren n} } 0$

where $M_X$ is the moment generating function of $X$.


From Moment Generating Function of Gaussian Distribution: Third Moment:

$\map { {M_X}'''} t = \paren {3 \sigma^2 \paren {\mu + \sigma^2 t} + \paren {\mu + \sigma^2 t}^3} \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$


Setting $t = 0$:

\(\ds \expect {X^3}\) \(=\) \(\ds \paren {3 \sigma^2 \paren {\mu + \sigma^2 0} + \paren {\mu + \sigma^2 0}^3} \map \exp {\mu 0 + \dfrac 1 2 \sigma^2 0^2}\)
\(\ds \) \(=\) \(\ds 3 \mu \sigma^2 + \mu^3\) Exponential of Zero

So:

\(\ds \gamma_1\) \(=\) \(\ds \frac {\expect {X^3} - 3 \mu \paren {\sigma^2 + \mu^2} + 3 \mu^2 \paren {\mu} - \mu^3} {\sigma^3}\)
\(\ds \) \(=\) \(\ds \frac {\paren {3 \mu \sigma^2 + \mu^3 } - 3 \mu \paren {\sigma^2 + \mu^2} + 3 \mu^2 \paren {\mu} - \mu^3} {\sigma^3}\)
\(\ds \) \(=\) \(\ds 0\)

$\blacksquare$