Skewness of Weibull Distribution

Theorem

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

Then the skewness $\gamma_1$ of $X$ is given by:

$\gamma_1 = \dfrac {\map \Gamma {1 + \dfrac 3 \alpha} - 3 \map \Gamma {1 + \dfrac 1 \alpha} \map \Gamma {1 + \dfrac 2 \alpha} + 2 \map \Gamma {1 + \dfrac 1 \alpha}^3} {\paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^{\frac 3 2 } }$

where $\Gamma$ is the Gamma function.

Proof

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

$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$

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:

$\expect {X^3} = \beta^3 \map \Gamma {1 + \dfrac 3 \alpha}$

So:

\(\ds \gamma_1\)

\(=\)

\(\ds \frac {\beta^3 \map \Gamma {1 + \dfrac 3 \alpha} - 3 \beta \, \map \Gamma {1 + \dfrac 1 \alpha} \beta^2 \paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2} - \paren {\beta \, \map \Gamma {1 + \dfrac 1 \alpha} }^3} {\paren {\beta \, \paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^{\frac 1 2 } }^3 }\)

\(\ds \)

\(=\)

\(\ds \frac {\map \Gamma {1 + \dfrac 3 \alpha} - 3 \map \Gamma {1 + \dfrac 1 \alpha} \paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^3} {\paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^{\frac 3 2 } }\)

canceling $\beta^3$

\(\ds \)

\(=\)

\(\ds \frac {\map \Gamma {1 + \dfrac 3 \alpha} - 3 \map \Gamma {1 + \dfrac 1 \alpha} \map \Gamma {1 + \dfrac 2 \alpha} + 2 \map \Gamma {1 + \dfrac 1 \alpha}^3} {\paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}^{\frac 3 2 } }\)

$\blacksquare$