Theorem: Excess Kurtosis of Weibull Distribution

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:
{{begin-eqn}}
{{eqn	| l = \gamma_2
	    | r = \frac {\paren {\beta^4 \map \Gamma {1 + \dfrac 4 \alpha} } - 4 \paren {\beta \, \map \Gamma {1 + \dfrac 1 \alpha} } \paren {\beta^3 \map \Gamma {1 + \dfrac 3 \alpha} }  + 6 \paren {\beta \, \map \Gamma {1 + \dfrac 1 \alpha} }^2 \paren {\beta^2 \map \Gamma {1 + \dfrac 2 \alpha} } - 3 \paren {\beta \, \map \Gamma {1 + \dfrac 1 \alpha} }^4} {\paren {\beta \, \paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^{\frac 1 2} }^4 } - 3
	    | c = 
}}
{{eqn	| r = \frac {\map \Gamma {1 + \dfrac 4 \alpha} - 4 \map \Gamma {1 + \dfrac 1 \alpha} \map \Gamma {1 + \dfrac 3 \alpha}  + 6 \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2 \map \Gamma {1 + \dfrac 2 \alpha} - 3 \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^4} {\paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^2 } - 3
	    | c = canceling $\beta^4$
}}
{{eqn	| r = \frac {\map \Gamma {1 + \dfrac 4 \alpha} - 4 \map \Gamma {1 + \dfrac 1 \alpha} \map \Gamma {1 + \dfrac 3 \alpha}  + 6 \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2 \map \Gamma {1 + \dfrac 2 \alpha} - 3 \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^4} {\paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^2 } - \frac {3 \paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^2} {\paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^2 }
	    | c = Multiply by $1$
}}
{{eqn	| r = \frac {\map \Gamma {1 + \dfrac 4 \alpha} - 4 \map \Gamma {1 + \dfrac 1 \alpha} \map \Gamma {1 + \dfrac 3 \alpha}  + 6 \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2 \map \Gamma {1 + \dfrac 2 \alpha} - 3 \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^4} {\paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^2 } - \frac {3 \paren {\paren {\map \Gamma {1 + \dfrac 2 \alpha} }^2 - 2 \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2 \map \Gamma {1 + \dfrac 2 \alpha} + \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^4} } {\paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^2 }
	    | c = Square of Sum
}}
{{eqn	| r = \frac {\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 }
	    | c = 
}}
{{end-eqn}}
{{qed}}
Category:Kurtosis
Category:Weibull Distribution
0ql49pv71ndyhipbrx63p5g9fkc17l5