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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}} $?

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    $\begingroup$ If $a_n \leq b_n$ you cannot say that $a_n$ and $b_n$ have the same limit. You have not obtained the right limit. $\endgroup$ – Kavi Rama Murthy Feb 25 at 9:36
  • $\begingroup$ i think i found my fault, now it has to be right, or not? $\endgroup$ – Kaya Feb 25 at 10:04
  • $\begingroup$ Your computation seems to be right. I do not know what by $F_2$ meant is. $\endgroup$ – Davide Giraudo Feb 25 at 10:07
  • $\begingroup$ thank you! me neither, there is no definition in the task for $F_2$. $\endgroup$ – Kaya Feb 25 at 10:18
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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 $x + \lambda^{-1} \log(n)\geqslant 0$.

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