Problem: Let $X_1,X_2,\dots$ be i.i.d. random variables with distribution function $F(x)$. Denote the maximum of the first $n$ elements by $M_n$. Show that if
$$\lim_{x\to\infty}x^\alpha\left[1-F(x)\right]=b$$
with fixed positive constants $\alpha,b$ then $n^{-1/\alpha}M_n$ converges in distribution and identify the limiting distribution.
What I have so far: Put $Y_n=n^{-1/\alpha}M_n$. Then we have
\begin{align*}
F_{Y_n}(x)
&=P\left(\max_{1\leq k\leq n}X_k\leq n^{1/\alpha}x\right)\\
&=[F(xn^{1/\alpha})]^n\\
&=[1-P(X_1>xn^{1/\alpha})]^n\\
&=\left[1-\frac{x^\alpha n}{x^\alpha n}P(X_1>xn^{1/\alpha})\right]^n.
\end{align*}
Now I need to evaluate the limit above. It seems to me that this limit evaluates to $e^{-b/x^\alpha}$, but upon graphing this function, I note that it is not a valid CDF. Therefore, this cannot be the right answer.
Could anyone help with a hint on how to evaluate the limit above rigorously?
Thank you for your time and appreciate any feedback.
Best Answer
To be honest, you have finished all the necessary work( except a typo in the end).
And $F(x)=e^{-b/x^{\alpha}}$ is obviously an acceptable CDF( for $\alpha>0$).
Note that: $F(x)=0$ when $x \le 0$