4
$\begingroup$

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.

Solution

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?

|cite|improve this question
$\endgroup$
2
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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 Sep 20 '13 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 Sep 20 '13 at 18:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.