Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have an exam in the morning and there is still one question I cannot do.

$X_1, \ldots, X_n$ are iid random variables each having distribution with density $f_{X_i}(x;\theta)= 1/\theta$, for $x \in [0,\theta]$ where $\theta>0$ compute the CDF of the random variable $\max(X_1,\ldots X_n)$ and prove that $n(\theta-\max(X_1,\ldots,X_n)) \to W$ in distribution and state the CDF of $W$.

How can I do this? I have worked out that the CDF of $\max(X_1,\ldots,X_n)$ is $ (x/\theta)^n$ but that is all :(


share|cite|improve this question
The TeX was messed up (I think you'll learn some things from the new source; also, there's no need to put tildes between everything :)) so I tried to fix it. I know nothing about probability theory so you should check that I didn't change your meaning. Also consider making the title more descriptive. – Dylan Moreland Jun 12 '12 at 21:26
Good luck with your exam. I bet this question is not among the question in the exam tomorrow. – Gigili Jun 12 '12 at 21:28
Are you sure your CDF of max() is correct? I find a different answer starting from $P(max(X_1,\dots,X_n)>c) = \sum_{i=1}^n P(X_i >c, X_j\le c \;\;\forall j \neq i)$. – emrea Jun 12 '12 at 21:30
@EmreA Your derivation is incorrect: LHS > RHS for all $n>1$. – Erick Wong Jun 12 '12 at 22:02
@ErickWong, I don't see it, why? – emrea Jun 13 '12 at 3:30
up vote 1 down vote accepted

You've already worked out the CDF of $\max(X_1,...,X_n)$. Now just find the CDF of $n(\theta-\max(X_1,\ldots,X_n))$ (use the definition of CDF, it's not hard) and take the limit as $n\to\infty$ (use $\lim_{n\to\infty} (1+x/n)^n = e^x$).

share|cite|improve this answer
Thanks, so obvious now how to do it... but my calculations are going wrong, I worked out the CDF of the second one but i get $\ 1-(1-x/{n\theta})^n $ so then $\ 1-e^{-x/\theta}?$ – Rosie Jun 12 '12 at 22:36
Looks about right to me. Note that the your CDF for $\max(X_1,...,X_n)$ is valid for $0 \le x \le \theta$, which makes the later CDF valid for $0 \le x \le n\theta$. As $n \to \infty$ this range of validity extends to all of $[0,\infty)$. – Erick Wong Jun 12 '12 at 22:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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