Let $ X_n $ be a sequence of iid $ \exp(\lambda)- $ distributed rv´s with $ \lambda > 0 $. Show that
$ ( \max_{1\leq i\leq n} X_i ) - \lambda^{-1} \log(n) \overset{D}\rightarrow Z $
where $Z$ is a rv.
Compute the distribution function $F_2$ of $Z$
my idea:
Firt i set: $ \max_{1\leq i\leq n} X_i := M_n $.
$$ \Rightarrow P( M_n - \lambda^{-1} \log(n) \leq x ) = P( M_n \leq x + \lambda^{-1} \log(n) ) \overset { M_n iid }= (P( X_1 \leq x + \lambda^{-1} \log(n) ))^n \overset{ X_1 \mbox{ is }\exp(\lambda)\mbox{ distributed}} = (1-e^{-\lambda(x+\lambda^{-1}\log(n))})^n = \left( 1 - \frac{e^{-x\lambda}}{n} \right)^n \overset{n \rightarrow \infty} \rightarrow e^{-e^{-x\lambda}} $$
$ \Rightarrow Z $ is Weibull distributed.
Is that right? And how do i compute $ F_2 $ of $ Z $? do i have to choose $ x=2 $ and compute $ e^{-e^{-2}} $?
It seems that $F_2$ is only a notation for the distribution function of $Z$ (maybe $F_1$ and $F_3$ are reserved to the other extreme value distributions).
What you did works perfectly for all $x$ when you deduce that $$ \mathbb P\left(M_n - \lambda^{-1} \log(n) \leqslant x\right)=\left(\mathbb P\left( X_1 \leqslant x + \lambda^{-1} \log(n) \right)\right)^n. $$ But after, if we want to use the expression $1-e^{-t}$ for $\mathbb P\left( X_1 \leqslant t \right)$, we have to be sure that $t$ is not negative.
If would be fine when $x\geqslant 0$; when $x$ is negative we just have to restrict to $n$ large enough so that $x + \lambda^{-1} \log(n)\geqslant 0$.