I'm studying order statistics and I came across this exercise.
I want to verify that:
$ F_{m,n} \in D(G_{1}) $
($F$ is in the max-domain of atraction of $G_{1}$).
By definition,
$G_{1}(x) = \begin{cases} e^{-x^{-\alpha}} && x>0, & \alpha>0 \\ 0 & & x \le 0 \end{cases} $
And by theorem I know that $ F \in D(G_{1}) \iff x^{*} = \infty $ and there exits an $x>0$ such that:
$$ \frac{1-F(tx)}{1-F(t)} \longrightarrow x^{-\alpha}$ \mbox{ as } t \longrightarrow \infty, \ \ \forall x >0 .$$
With $ x^{*}$ being the right end point of $F_{x}$, $\ x^{*}= \sup\{x:F_{x}(x)<1 \}$
My approach is trying to set $F(x)$ in that to the explicit form of $F_{m,n}(x)$ in the following way:
$\frac{m!}{(n-m)!(m-1)!}F(x)^{n-m}(1-F(x))^{m-1} $
But I'm not really sure how to get a function of $x$ from that. Is my approach viable? What can I do to continue?
Any hint would be really helpful, thank you very much!