Excess Kurtosis of Bernoulli Distribution

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

Then the excess kurtosis $\gamma_2$ of $X$ is given by:


 * $\gamma_2 = \dfrac {1 - 6 p q} {p q}$

where $q = 1 - p$.

Proof
From the definition of excess kurtosis, we have:


 * $\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$

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

By Expectation of Bernoulli Distribution, we have:


 * $\mu = p$

By Variance of Bernoulli Distribution, we have:


 * $\sigma = \sqrt {p q}$

So: