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$X_1, X_2,\ldots, X_n$ are independent with uniform distribution on $[0,a]$.

Prove that $\max(X_1, X_2,\ldots,X_n) \rightarrow a$ almost surely.

Ar first I look for the probability distribution i.e. $$F_\max(t)=P(\max(X_1, X_2,\ldots,X_n)<t)=P(X_1<t)^n=(F_{X_1}(t))^n=\left(\frac{t}{a}\right)^n.$$

So if $t<a$ it converges to $0$ and if $t\geq a$ it converges to $1$.

How can I say that $\max(X_1, X_2,\ldots,X_n) \rightarrow a$ almost surely. Is there a better way to do it?

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  • $\begingroup$ you need almost sure convergence, i.e. $$ \mathbb{P}\left[ \lim_{n \to \infty} \max(X_1, \ldots, X_n) = a \right] = 1 $$ $\endgroup$ – gt6989b Jun 24 '15 at 16:27
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    $\begingroup$ You can use this sufficient condition (for any $Y_n$ random variables): If for all $\epsilon>0$ we have $\sum_{n=1}^{\infty} Pr[|Y_n-a| > \epsilon] < \infty$, then $\lim_{n\rightarrow\infty} Y_n=a$ with probability 1. $\endgroup$ – Michael Jun 24 '15 at 16:29
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    $\begingroup$ This is a standard sufficient condition, and follows since $$Pr[\cup_{n=m}^{\infty} \{|Y_n-a|>\epsilon\}] \leq \sum_{n=m}^{\infty} Pr[|Y_n-a|>\epsilon]$$ Taking limits of both sides as $m\rightarrow\infty$ means the left goes to zero whenever the right does, and the right goes to zero if it is the tail of a summable sequence. $\endgroup$ – Michael Jun 24 '15 at 16:35
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The solution was mostly covered by the comments, but here is a more complete answer:

Your solution does not show a.s. convergence, as convergence in distribution does not imply a.s. convergence. What needs to be shown is that (denote $\max(X_1, \dots X_n)$ by $M_n$) $$ P(\lim_{n \rightarrow \infty} M_n = a) = 1 $$ This happens to be the equivalent of (this is not directly evident but can be shown) $$ P(\mid M_n -a \mid > \epsilon \text{ infinitely often as } n\rightarrow \infty) = 0 $$ By Borel Cantelli Lemma, this holds if $$ \sum^{\infty}_{n=1}P(\mid M_n -a \mid > \epsilon) <\infty $$ As you showed yourself, $P(\mid M_n -a \mid > \epsilon) = \left(\frac{a-\epsilon}{a} \right)^n$ so you have $$ \sum^{\infty}_{n=1}P(\mid M_n -a \mid > \epsilon) = \frac{a}{\epsilon}<\infty $$ and we are done.

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  • $\begingroup$ Can you please explain as to how the RHS of the second last step was arrived? $\endgroup$ – Harry Oct 29 '19 at 10:00
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Another argument is to note that $P(|M_n - a| < \epsilon) \to 1$ as $n \to \infty$ for $\epsilon > 0$ and so $M_n \to a$ in probability. But since the events $\{ |M_n - a| < \epsilon \}$ are monotone increasing this implies $M_n \to a$ with probability one.

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