Excess Kurtosis of Weibull Distribution
Jump to navigation
Jump to search
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:
\(\ds \gamma_2\) | \(=\) | \(\ds \frac {\paren {\beta^4 \map \Gamma {1 + \dfrac 4 \alpha} } - 4 \paren {\beta \, \map \Gamma {1 + \dfrac 1 \alpha} } \paren {\beta^3 \map \Gamma {1 + \dfrac 3 \alpha} } + 6 \paren {\beta \, \map \Gamma {1 + \dfrac 1 \alpha} }^2 \paren {\beta^2 \map \Gamma {1 + \dfrac 2 \alpha} } - 3 \paren {\beta \, \map \Gamma {1 + \dfrac 1 \alpha} }^4} {\paren {\beta \, \paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^{\frac 1 2} }^4 } - 3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\map \Gamma {1 + \dfrac 4 \alpha} - 4 \map \Gamma {1 + \dfrac 1 \alpha} \map \Gamma {1 + \dfrac 3 \alpha} + 6 \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2 \map \Gamma {1 + \dfrac 2 \alpha} - 3 \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^4} {\paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^2 } - 3\) | canceling $\beta^4$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\map \Gamma {1 + \dfrac 4 \alpha} - 4 \map \Gamma {1 + \dfrac 1 \alpha} \map \Gamma {1 + \dfrac 3 \alpha} + 6 \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2 \map \Gamma {1 + \dfrac 2 \alpha} - 3 \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^4} {\paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^2 } - \frac {3 \paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^2} {\paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^2 }\) | Multiply by $1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\map \Gamma {1 + \dfrac 4 \alpha} - 4 \map \Gamma {1 + \dfrac 1 \alpha} \map \Gamma {1 + \dfrac 3 \alpha} + 6 \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2 \map \Gamma {1 + \dfrac 2 \alpha} - 3 \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^4} {\paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^2 } - \frac {3 \paren {\paren {\map \Gamma {1 + \dfrac 2 \alpha} }^2 - 2 \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2 \map \Gamma {1 + \dfrac 2 \alpha} + \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^4} } {\paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^2 }\) | Square of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\map \Gamma {1 + \dfrac 4 \alpha} - 4 \map \Gamma {1 + \dfrac 1 \alpha} \map \Gamma {1 + \dfrac 3 \alpha} + 12 \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2 \map \Gamma {1 + \dfrac 2 \alpha} - 6 \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^4 - 3 \paren {\map \Gamma {1 + \dfrac 2 \alpha} }^2 } {\paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^2 }\) |
$\blacksquare$