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I've come across this problem which has completely stumped me. It goes as follows:

Let $(X_n)$ be a sequence of independent and identically distributed exponential random variables with parameter $\lambda$. Let $M_n$ denote max{$X_1,...,X_n $}. Show there exists a random variable $Z$ such that $M_n - \frac{1}{\lambda} \log (n) $ converges in distribution to $Z$.

Now this problem seems really hard, so I tried proving it for convergence in probability to see if that would work and imply convergence in distribution, but I couldn't get very far. Moreover, calculating the expectations also seems non-trivial due to the fact that it would be hard to integrate $M_n = \frac{1}{\lambda} \log n$.

Any suggestions and ways to approach this problem would be greatly appreciated as I've been stuck on this for a long time. Should I try using the Skorohod equivalent statement?

Thanks.

Debreu
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1 Answers1

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Unfortunately we do not have convergence in probability, so that approach is not going to succeed. The basic trick is to note that $M_n\le a$ if and only if $X_i\le a$ for each $i=1,\ldots,n$ and then use independence: $$P(M_n-\tfrac1\lambda\log n\le x)=P(M_n\le\tfrac1\lambda\log n+x)=\prod_{i=1}^nP(X_i\le\tfrac1\lambda\log n+x).$$ Now remembering that $P(X_i\le a)=1-e^{-\lambda a}$, we find $$P(M_n-\tfrac1\lambda\log n\le x)=\left(1-e^{-\lambda(\frac1\lambda\log n+x)}\right)^n=\left(1-\frac{e^{-\lambda x}}n\right)^n\to e^{-e^{-\lambda x}}$$ as $n\to\infty$. Hence $M_n\to Z$ in distribution where $P(Z\le x)=e^{-e^{-\lambda x}}$.

Jason
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  • Thanks for this, it was quite an elegant response. I think I lacked a proper understanding of convergence in distribution. You clarified it a lot for me I think. I didn't realise that I just had to show convergence of the cdf. – Debreu May 02 '17 at 14:51
  • That makes sense, and in that case I think trying a few more exercises like this is a good idea! There are other ways to check convergence in distribution, too: you can use convergence of the characteristic function, or look into something called the Portmanteau theorem. – Jason May 02 '17 at 15:07
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    By the way, the distribution of $Z$ is the Gumbel distribution with parameters $\mu=0$ and $\beta=1/\lambda$. – Cm7F7Bb Feb 08 '18 at 14:31
  • Would anyone be kind to explain more about the statement "Unfortunately we do not have convergence in probability"? – MathArt Feb 21 '22 at 10:19
  • @MathArt https://en.m.wikipedia.org/wiki/Convergence_of_random_variables – Jason Feb 22 '22 at 01:55