Excess Kurtosis of Gamma Distribution

Theorem
Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.

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


 * $\gamma_2 = \dfrac 6 \alpha$

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 Gamma Distribution, we have:


 * $\mu = \dfrac \alpha \beta$

By Variance of Gamma Distribution, we have:


 * $\sigma = \dfrac {\sqrt \alpha} \beta$

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 Gamma Distribution, this is given by:


 * $\map {M_X} t = \paren {1 - \dfrac t \beta}^{-\alpha} = \beta^\alpha \paren {\beta - t}^{-\alpha} $

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: