Skewness of Bernoulli Distribution

Theorem

Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$.

Then the skewness $\gamma_1$ of $X$ is given by:

$\gamma_1 = \dfrac {1 - 2 p} {\sqrt {p q} }$

where $q = 1 - p$.

Proof

From Skewness in terms of Non-Central Moments:

$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$

where $\mu$ is the mean of $X$, and $\sigma$ the standard deviation.

We have, by Expectation of Bernoulli Distribution:

$\mu = p$

By Variance of Bernoulli Distribution, we also have:

$\var X = \sigma^2 = p \paren {1 - p}$

so:

$\sigma = \sqrt {p \paren {1 - p} }$

By Raw Moment of Bernoulli Distribution, we have:

$\expect {X^3} = p$

So:

\(\ds \gamma_1\)

\(=\)

\(\ds \frac {p - 3 p^2 \paren {1 - p} - p^3} {p^{3/2} \paren {1 - p}^{3/2} }\)

\(\ds \)

\(=\)

\(\ds \frac {p \paren {1 - p^2} - 3 p^2 \paren {1 - p} } {p^{3/2} \paren {1 - p}^{3/2} }\)

\(\ds \)

\(=\)

\(\ds \frac {p \paren {1 - p} \paren {1 + p} - 3 p^2 \paren {1 - p} } {p^{3/2} \paren {1 - p}^{3/2} }\)

Difference of Two Squares

\(\ds \)

\(=\)

\(\ds \frac {p \paren {1 - p} \paren {1 + p - 3p} } {p^{3/2} \paren {1 - p}^{3/2} }\)

\(\ds \)

\(=\)

\(\ds \frac {1 - 2 p} {\sqrt {p \paren {1 - p} } }\)

\(\ds \)

\(=\)

\(\ds \frac {1 - 2 p} {\sqrt {p q} }\)

$q = 1 - p$

$\blacksquare$