# Almost sure convergence of $\frac1{\sqrt n}\max_{1\le i\le n}|X_i|$

Let $$X_1,X_2,\ldots$$ be i.i.d random variables. Assuming $$EX_1^2 < \infty$$, show that $$\frac{\max_{1 \leq i \leq n} |X_i|}{\sqrt n}\stackrel{\text{a.s.}}\longrightarrow 0.$$

Let $$M_n = \max_{1 \leq i \leq n} |x_i|$$.

I tried different ways, but none seemed to work.

I can show that $$\frac{\max_{1 \leq i \leq n} |x_i|}{\sqrt n}$$ converges in probability to 0. And tried to say since it is a monotone sequence, it also converges almost surely. But I get stuck in showing the monotonicity (only $$M_n \leq M_{n+1}$$, not really $$\frac{M_n}{\sqrt n} \leq \frac{M_{n+1}}{\sqrt n}$$).

Then, I tried to use the Borel-Cantelli lemma to show

i). $$\sum_{i = 1}^\infty P\left(\frac{M_n}{\sqrt n}>\epsilon\right) < \infty$$

or ii). $$\sum_{i = 1}^\infty P\left(\frac{M_n}{\sqrt n} \leq \epsilon\right) = \infty$$

I have $$P\left(\frac{M_n}{\sqrt n} > \epsilon\right) \leq nP(X_i > \epsilon \sqrt n) \leq n \frac{E(X^2)}{n \epsilon^2} = \frac{E(X^2)}{\epsilon^2}$$ But the sum of this series then does seem to converge.

Then, I tried to see $$P\left(\frac{M_n}{\sqrt n} \leq \epsilon\right) = 1 - P\left(\frac{M_n}{\sqrt n} \geq \epsilon\right) = 1 - \left(P\left(\frac{X_i}{\sqrt n} \geq \epsilon\right)\right)^n \geq 1 - \left(\frac{E(X^2)}{n \epsilon^2}\right)^n$$

$$\sum_{n = 1}^{\infty } P\left(\frac{M_n}{\sqrt n} \leq \epsilon\right) \geq \sum_{n = 1}^{\infty } 1- \left(\frac{E(X^2)}{n \epsilon^2}\right)^n = \infty, n \rightarrow \infty$$

Is this last approach enough to conclude that $$\frac{\max_{1 \leq i \leq n} |x_i|}{\sqrt n}$$ converges almost surely to 0??? I feel it seems I am missing something.

If I understand correctly, you first used the fact that you have a maximum, but $$P(M_n < x)= P(X and not $$P(M_n > x) \ne P(X > x)^n$$. Afterwards, you applied the Chebyshev-inequality, and lastly the Borel-Cantelli lemma. However the Borell-Cantelli lemma states that if the sum of the probabilities of a series of events is finite, then the limsup is zero, which implies that in this case our series converges almost surely. But from the fact that sum of probabilities is infinite, you can't say much.
For example consider $$(X_n)_{n=1}^{\infty}$$ s.t. $$P(X_n=1)=\frac 1n$$ and $$P(X_n=0)=1 - \frac 1n$$. In this case $$\sum_{n=1}^{\infty} P(X_n = 1) = \sum \frac 1n = \infty$$, but you absolutely can't say that $$X_n$$ converges to 1 a.s.  Now back to the solution.
I believe it's easier to deal with this problem if you square the whole thing first. Let $$Y_i=X_i^2$$, then $$E{Y_i} < \infty$$ $$\max_i{\frac{|X_i|}{\sqrt{n}}}\to 0 \Longleftrightarrow \max_i \frac{Y_i}{n}\to 0$$
Then you can use the following events: $$A_k = \{\frac{Y_k}{k} \ge \epsilon \}$$ Now if we show the following, then we are done. $$\sum_k P(A_k) \le \frac{EY_1}{\epsilon} < \infty$$, because by the B.C. lemma this means only finitely many $$A_k$$ occurs for every positive $$\epsilon$$ which also means $$\frac{Y_k}{k} \to 0$$ a.s.
We can show the above relationship as follows for every $$Z>0$$. $$\sum_{k=1}^{n} P(Z \ge k) = E\bigg(\sum_{k=1}^{n} I(Z \ge k)\bigg) = E\big(\lfloor Z \rfloor\big) \le E(Z)$$
You substitute $$Z=\frac{Y_i}{\epsilon}$$ and there you have it $$:)$$