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?

  • 1
    $\begingroup$ 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

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

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


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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.