Excess Kurtosis of Binomial Distribution

Theorem
Let $X$ be a discrete random variable with a binomial distribution with parameters $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$:

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


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

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 Binomial Distribution, we have:


 * $\mu = n p$

By Variance of Binomial Distribution, we have:


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

So:

To calculate $\gamma_2$, we must calculate $\expect {X^4}$.

We find this using the moment generating function of $X$, $M_X$.

By Moment Generating Function of Binomial Distribution, this is given by:


 * $\map {M_X} t = \paren {1 - p + p e^t}^n$

From Moment in terms of Moment Generating Function:


 * $\expect {X^4} = \map {M_X} 0$

So:

Setting $t = 0$:

Plugging this result back into our equation above: