Skewness of Normal Distribution

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 1

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:

$\gamma_1 = 0$

$\blacksquare$

Proof 2

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 Normal Distribution, we have:

$\mu = \mu$

By Variance of Normal 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 Normal 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 Normal 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:

\(\ds \gamma_1\)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \frac {\paren {3 \mu \sigma^2 + \mu^3 } - 3 \mu \paren {\sigma^2 + \mu^2} + 3 \mu^2 \paren {\mu} - \mu^3} {\sigma^3}\)

\(\ds \)

\(=\)

\(\ds 0\)

$\blacksquare$