# Sequence of i.i.d $\exp(\lambda)$ distributed rv´s. Show that it converges against a rv Z

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 $$e^{-e^{-2}}$$?

• 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. – Kavi Rama Murthy Feb 25 at 9:36
• i think i found my fault, now it has to be right, or not? – Kaya Feb 25 at 10:04
• Your computation seems to be right. I do not know what by $F_2$ meant is. – Davide Giraudo Feb 25 at 10:07
• thank you! me neither, there is no definition in the task for $F_2$. – Kaya Feb 25 at 10:18

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 $$x + \lambda^{-1} \log(n)\geqslant 0$$.