Skewness of Normal Distribution/Proof 1

Theorem

Let $X$ be a continuous random variable with a normal 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:

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

From the definition of the normal distribution, $X$ has probability density function:

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

So, from Expectation of Function of Continuous Random Variable:

$\ds \gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3} = \dfrac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \paren {\dfrac {x - \mu} \sigma}^3 \map \exp {-\dfrac { \paren {x - \mu}^2} {2 \sigma^2} } \rd x$

Making a substitution of $u = x - \mu$:

$\ds \gamma_1 = \dfrac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \paren {\dfrac u \sigma}^3 \map \exp {-\dfrac {u^2} {2 \sigma^2} } \rd u$

We have that:

$\paren {-u}^3 \map \exp {-\dfrac {\paren {-u}^2} {2 \sigma^2} } = -u^3 \map \exp {-\dfrac {u^2} {2 \sigma^2} }$

So we can see that the integrand is odd.

So, by Definite Integral of Odd Function:

$\ds \dfrac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \paren {\dfrac u \sigma}^3 \map \exp {-\dfrac {u^2} {2 \sigma^2} } \rd u = 0$

giving:

$\gamma_1 = 0$

$\blacksquare$