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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 :(


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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

1 Answer 1

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 $\lim_{n\to\infty} (1+x/n)^n = e^x$).

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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

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