Let $X_1,\ldots,X_n \overset{i.i.d}{\sim} \ Uniform(0,1)$ and write $M_n = \max(X_1,\ldots,X_n)$. If we have proved that $M_n \overset{a.s.}{\rightarrow} 1$, then we know that $(M_n)$ converges in law, but how to find this limiting distribution, knowing that $n(1-M_n) \overset{d}{\rightarrow} Exponential(1)$ ?

Thank you.

  • $\begingroup$ Yes but you do not answer my question.. Is it possible to find the distribution of $M_n$ ? Maybe the best thing we can say is: if $x \leq 0$, then the limit is $0$ and if $x \geq 1$, then the limit is $1$ and for $x \in (0,1)$ and $n$ "large", $P(M_n \leq x) \approx \exp(-n(1-x))$. Is it correct ? $\endgroup$ – MCrassus Nov 19 '14 at 13:40
  • $\begingroup$ So you mean that if we know the distribution of $M_n$ for a fixed $n$ then it suffices for studying the distribution of $M_n$ (also if $n$ is very large), right ? So I do not understand why is it so important to know the limiting distribution of $n(1-M_n)$.. We want to show $M_n$ converges in distribution. Why is it possible to find the limiting distribution of $n(1-M_n)$ and not $M_n$ precisely ? Is it because of the definition of convergence in distribution (continuity points of the limiting distribution) ? $\endgroup$ – MCrassus Nov 19 '14 at 14:15
  • 2
    $\begingroup$ Oh, I just realize I misread what you wrote. I'm deleting my comments, and sorry about the noise. The limiting distribution of $M_n$ is of course the Dirac distribution at 1. $\endgroup$ – Stefan Hansen Nov 19 '14 at 14:32
  • $\begingroup$ Question to be closed? $\endgroup$ – Did Dec 13 '14 at 7:48

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