Excess Kurtosis of Weibull Distribution

Theorem
Let $X$ be a continuous random variable with the Weibull distribution with $\alpha, \beta \in \R_{> 0}$.

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


 * $\gamma_2 = \dfrac {\map \Gamma {1 + \dfrac 4 \alpha} - 4 \map \Gamma {1 + \dfrac 1 \alpha} \map \Gamma {1 + \dfrac 3 \alpha} + 12 \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2 \map \Gamma {1 + \dfrac 2 \alpha} - 6 \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^4 - 3 \paren {\map \Gamma {1 + \dfrac 2 \alpha} }^2 } {\paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^2 }$

where $\Gamma$ is the Gamma function.

Proof
From Kurtosis in terms of Non-Central Moments, we have:


 * $\gamma_2 = \dfrac {\expect {X^4} - 4 \mu \expect {X^3} + 6 \mu^2 \expect {X^2} - 3 \mu^4} {\sigma^4} - 3$

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

By Expectation of Weibull Distribution we have:


 * $\mu = \beta \, \map \Gamma {1 + \dfrac 1 \alpha}$

By Variance of Weibull Distribution we have:


 * $\sigma = \beta \, \paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^{\frac 1 2}$

From Raw Moment of Weibull Distribution, we have:


 * $\expect {X^n} = \beta^n \map \Gamma {1 + \dfrac n \alpha}$

Hence: