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Consider a Frechet distribution with the following cumulative distribution function:

$$Pr(X \leq x) = e^{-x^{-\alpha}} ~ \quad\mbox{if}\quad~ x>0$$

The expected value is $E(X) = \Gamma(1 - \frac{1}{\alpha})$ with the gamma function: $\Gamma(a) = \int_0^{\infty} x^{a-1}e^{-x} dx$.

I am struggling to prove the above expected value.

$$E(X) = \int_0^{\infty} x \alpha x^{-\alpha -1} e^{-x^{-\alpha}} dx$$

Using a change of variable $y=x^{-\alpha}$ so $dy = - \alpha x^{-\alpha-1} dx$ and $x = y^{\frac{-1}{\alpha}}$.

Then, I would be tempted to write $E(X) = \int_0^{\infty} \textbf{-} y^{\frac{-1}{\alpha}} e^{-y} dy$ instead of $E(X) = \int_0^{\infty} y^{\frac{-1}{\alpha}} e^{-y} dy$, because $dy = \textbf{-}$ something.

Should be trivial but why did the sign minus disappear?

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    $\begingroup$ When you do the transformation, what happens to your limits? $\endgroup$ – Glen_b Sep 27 '15 at 9:43
  • $\begingroup$ Thanks @Glen_b but what do you mean by "what happens to your limits" in this specific case? $\endgroup$ – emeryville Sep 27 '15 at 16:40
  • $\begingroup$ When you do a change of variable with a definite integral, what happens to the limits on the integral? $\endgroup$ – Glen_b Sep 27 '15 at 17:00
  • $\begingroup$ @Glen_b Do you want to me to understand that I can simply write: $-\int_0^\infty y^{\frac{-1}{\alpha}} e^{-y} dy = \int_{-\infty}^0 y^{\frac{-1}{\alpha}} e^{-y} dy = \int_0^\infty y^{\frac{-1}{\alpha}} e^{-y} dy$ ? $\endgroup$ – emeryville Sep 27 '15 at 18:47
  • $\begingroup$ @emeryville $y=\frac{1}{x^a}$. As $x \longrightarrow 0$, $y \longrightarrow \infty$ and as $x \longrightarrow \infty$, $y \longrightarrow ...$ $\endgroup$ – rightskewed Sep 27 '15 at 19:30
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$$E(X) = \int_0^{\infty} x \alpha x^{-\alpha -1} e^{-x^{-\alpha}} dx$$

Let $y=x^{-\alpha}$ so $dy = - \alpha x^{-\alpha-1} dx$ and $x = y^{\frac{-1}{\alpha}}$. For $x \in (0, \infty)$, $y \in (\infty, 0)$

Thus,

$$ \begin{align*} E(X) &= \int_0^{\infty} x e^{-x^{-\alpha}} (\alpha x^{-\alpha -1} dx)\\ &= \int_\infty^{0} y^{\frac{-1}{\alpha}} e^{-y} (-dy)\\ &= \int_{0}^{\infty} y^{\frac{-1}{\alpha}} e^{-y} dy\ \text{by interchanging limits}\\ &= \int_{0}^{\infty} y^{\big(1-\frac{1}{\alpha}\big)-1} e^{-y} dy\\ &= \Gamma(1-\frac{1}{\alpha}) \end{align*} $$

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