Variance of Weibull Distribution
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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:
\(\ds \var X\) | \(=\) | \(\ds \beta^2 \map \Gamma {1 + \frac 2 \alpha} - \paren {\beta \, \map \Gamma {1 + \frac 1 \alpha} }^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \beta^2 \paren {\map \Gamma {1 + \frac 2 \alpha} - \paren {\map \Gamma {1 + \frac 1 \alpha} }^2}\) |
$\blacksquare$