Limiting distribution to Weibull

Question

I am trying to solve the following question: Let

$X_1,X_2,...$ i.i.d with $F(x) = P(X_i < x),~\\M_n := \max_{1\le{i}\le{n}}\ X_i,~\ $ $F(x_0) = 1$ and $F(x)<1$ for all x

Given; $\lim_{x\rightarrow x_0}\ (x_0-x)^{-\alpha}(1-F(x)) = b $. show that $$P((bn)^{\frac{1}{\alpha}}(M_n-x_0)<x) \rightarrow \mathbb{1}_{[x<\ 0\ ]}e^{-(-x)^\alpha}+\mathbb{1}_{[x\ge\ 0\ ]}.$$

My trial; $$P((bn)^{\frac{1}{\alpha}}(M_n-x_0)<x) = P(M_n < \frac{x}{(bn)^\frac{1}{\alpha}}+x_0) = (P(X_1 < \frac{x}{(bn)^\frac{1}{\alpha}}+x_0))^n \\ let \ x \rightarrow \ x_0 \ \ then\ (1-F(\frac{x}{(bn)^\frac{1}{\alpha}}+x_0))^n \rightarrow (b[x_0-(\frac{x}{(bn)^\frac{1}{\alpha}}+x_0)]^\alpha)\ = (\frac{(-x)^\alpha}{n})^n$$

I don't know how to proceed from here, help appreciated, thank you.

Answer 1

As the expression of the limit suggest, you have to distinguish the cases $x<0$ and $x\geqslant 0$. For $x\geqslant 0$, $F\left(\frac{x}{(bn)^{1/\alpha}}+x_0\right)=1$ for all $n$. For $x\lt 0$, let $x_n=\frac{x}{(bn)^{1/\alpha}}+x_0$. Then by assumption, we know that $$ (x_0-x_n)^{-\alpha}(1-F(x_n))\to b $$ or in other words, $$ (x_0-x_n)^{-\alpha}(1-F(x_n))= b+\epsilon_n, \epsilon_n\to 0. $$ Therefore, $(1-F(x_n))^n=(x-x_0)^{n\alpha}(b+\epsilon_n)^n$. Rearrange the expression and use the fact that for all $t$, $$ \left(1+\frac tn +\frac{\epsilon_n}n\right)^n\to e^t. $$

Answer 2

Thanks for anyone who tried, I found the answer and it is as follows; $$P((bn)^{\frac{1}{\alpha}}(M_n-x_0)<x) = P(M_n < \frac{x}{(bn)^\frac{1}{\alpha}}+x_0) = (P(X_1 < \frac{x}{(bn)^\frac{1}{\alpha}}+x_0))^n = [F(\frac{x}{(bn)^\frac{1}{\alpha}}+x_0)]^n \\= F^n(x_o)\mathbb{1}_{[x \ge 0]} + F^n(\frac{x}{(bn)^\frac{1}{\alpha}}+x_0) \mathbb{1}_{[x\le 0]} \\ as~~x \rightarrow x_0~~~ =\mathbb{1}_{[x \ge 0]} ~+~[ -b(x_0-\frac{x}{(bn)^\frac{1}{\alpha}}-x_o)+1]^n~ \mathbb{1}_{[x\le 0]} \\= \mathbb{1}_{[x \ge 0]}~ + ~ [1-\frac{(-x)^\alpha}{n}]^n~ \mathbb{1}_{[x\le 0]} \\ as~~ n \rightarrow \infty~~~ = \mathbb{1}_{[x \ge 0]}~ + ~\mathbb{1}_{[x\le 0]} e^{-(-x)^\alpha} $$