Variance of Weibull Distribution

Theorem

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

Then the variance of $X$ is given by:

$\var X = \beta^2 \paren {\map \Gamma {1 + \dfrac 2 \alpha} - \paren {\map \Gamma {1 + \dfrac 1 \alpha} }^2}$

where $\Gamma$ is the Gamma function.

Proof

By Variance as Expectation of Square minus Square of Expectation, we have:

$\var X = \expect {X^2} - \paren {\expect X}^2$

By Expectation of Weibull Distribution, we have:

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

From Raw Moment of Weibull Distribution, we have:

The $n$th raw moment $\expect {X^n}$ of $X$ is given by:

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

Therefore, for $n = 2$ we have:

$\expect {X^2} = \beta^2 \map \Gamma {1 + \dfrac 2 \alpha}$

So:

\(\ds \var X\)

\(=\)

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

\(\ds \)

\(=\)

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

$\blacksquare$