Excess Kurtosis of Erlang Distribution

Theorem
Let $k$ be a strictly positive integer.

Let $\lambda$ be a strictly positive real number.

Let $X$ be a continuous random variable with an Erlang distribution with parameters $k$ and $\lambda$.

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


 * $\gamma_2 = \dfrac 6 k$

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 Erlang Distribution we have:


 * $\mu = \dfrac k \lambda$

By Variance of Erlang Distribution we have:


 * $\sigma = \dfrac {\sqrt k} \lambda$

So: