Mean and Variance of the Weibull Distribution
The density of the Weibull Distribution is given by:
$$f(x) = \alpha x^{\alpha-1}e^{-x^{\alpha}}$$
The Gamma function is defined as: $$\Gamma(\alpha)=\int_{0}^{\infty}x^{\alpha-1}e^{-x} \,dx$$
Show that $E(X)=\Gamma(\frac{1}{\alpha}+1)$ and $Var(X)=\Gamma(\frac{2}{\alpha}+1)-\Gamma^2(\frac{1}{\alpha} + 1)$
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$\begingroup$For constant $k$, we have the following $$E(X^k)=\int\alpha x^{\alpha+k-1}e^{-x^\alpha}dx$$ Using substitution $u=x^\alpha\Leftrightarrow x=u^{1/\alpha}$ results in $$du=\alpha x^{\alpha-1}dx \Rightarrow dx=\frac{du}{\alpha x^{\alpha-1}}=\frac{du}{\alpha u^{\frac{\alpha-1}{\alpha}}}=\left(u^{\frac{1}{\alpha}-1}\right)\frac{du}{\alpha}$$ leading to $$\begin{align}E(X^k)&=\int\alpha \color{blue}{x^{\alpha+k-1}}\color{red}{e^{-x^\alpha}}\color{green}{dx}\\&=\int\alpha \color{blue}{u^{(\frac{k}{\alpha}+1)-\frac{1}{\alpha}}}\color{red}{e^{-u}}\color{green}{\left(u^{\frac{1}{\alpha}-1}\right)\frac{du}{\alpha}}\\&=\int u^{\left(\frac{k}{\alpha}+1\right)-1}e^{-u}\ du\\&=\Gamma\left(\frac{k}{\alpha}+1\right)\end{align}$$ Thus the mean is $$E(X)=\Gamma\left(\frac{1}{\alpha}+1\right)$$ and the variance is $$Var(X)=E(X^2)-(E(X))^2=\Gamma\left(\frac{2}{\alpha}+1\right)-\Gamma^2\left(\frac{1}{\alpha}+1\right)$$
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