$X_1, X_2,\ldots, X_n$ 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?
Best Answer
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.