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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 $Var(X)=\Gamma(\frac{2}{\alpha}+1)-\Gamma^2(\frac{1}{\alpha} + 1)$

user137481
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    Have you tried anything? – angryavian Aug 19 '14 at 02:51
  • I've actually solved it. I posted it for anyone interested in solving it. Proof should be about less than 10 lines. – user137481 Aug 19 '14 at 02:53
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    Just a heads up, that's the density for a Weibull with scale parameter ($\beta$) equal to 1. A more general p.d.f would be $\frac{\alpha}{\beta} \left(\frac{x}{\beta}\right)^{\alpha - 1} \exp\left{-\left(\frac{x}{\beta}\right)^\alpha\right}$, $x \geq 0$. – Waldir Leoncio Mar 20 '16 at 09:35

1 Answers1

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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 $$Var(X)=E(X^2)-(E(X))^2=\Gamma\left(\frac{2}{\alpha}+1\right)-\Gamma^2\left(\frac{1}{\alpha}+1\right)$$

Alijah Ahmed
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  • can you explain your substitution? I am actually working on this problem and this is where I am struggling –  Jun 30 '15 at 02:35
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    @AlyssaWallace Please see edits to my answer. By way of explanation, I have added a few more lines and some colouring. – Alijah Ahmed Jun 30 '15 at 20:13