I'm trying to solve the problem below:

Let $X_1,...,X_n$ be independent with PDF $f(x)=e^{-x}$ if $x>0$ and zero otherwise and define $$X_{(n)} = \mathrm{max}\{X_1,..,X_n\}$$

  1. Find the CDF of $X_{(n)}$.

  2. Find a sequence of numbers $a_n$ so that the sequence of random variables $$X_{(n)}-a_n$$ is convergent in distribution and find the limiting distribution.


Since the CDF of a random variable $Y$ with exponential distribution and $\lambda = 1$ is $(1-e^{-x})$ we get that $$F_{X_{(n)}}=P(\mathrm{max}(X_1,...,X_n)<x)=P(X_1<x,X_2<x,...,X_n<X)$$ $$=P(X_1<x)\cdot P(X_2<x)\cdot \ldots \cdot P(X_n<x)=(1-e ^{-x})^n$$

so that is easy.

For the second question I'm not really sure how I can solve it since $F_{X_{(n)}}$ seems to satisfy the condition for CDF but when $n$ tends to infinity, it's limit is the null function for $x>0$. So I'm wondering how I can define those numbers $a_n$ to have convergence in distribution and to find the limiting distribution?


1 Answer 1


You will need a scaling parameter, too, so that the sequence you would need to look for would be more like $$b_n( X_{(n)} - a_n)$$ to prevent the distribution from escaping to infinity.

  • $\begingroup$ Well I have $$P(X_{(n)} \leq 1-\epsilon)=(1-e^{-1+\epsilon})^n$$ which tends to zero as n goes to infinity. I'm not finding any sensible way to define epsilon in terms of a variable t and get any well known cdf. Anybody with idea how I can plug some good value for $\epsilon$ and get that cdf? $\endgroup$
    – Raxel
    Commented Sep 20, 2013 at 0:37
  • $\begingroup$ Hint: use the historical definition of e as the limit of something. You should end up with a distribution that has a double exponent in it. $\endgroup$
    – StasK
    Commented Sep 20, 2013 at 18:00

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