Let $\{X_n: n \ge 1\}$ be a sequence of i.i.d random variables with $\mathbb{E}[|X_1|] < \infty$, and $\mathbb{E}[X_1] \neq 0$. Show that $$\frac{|X_n|}{n} \to 0 \quad \text{almost surely.}$$
Use this result to show $$\frac{\max_{1\le k \le n}|X_k|}{S_n} \to 0 \quad \text{almost surely,}$$ where $S_n := \sum_{j=1}^n X_j$.
To show the first part, I tried to prove the equivalent result that $$\forall \epsilon>0 \quad \mathbb{P}\left(\frac{|X_n|}{n} > \epsilon \quad \text{i.o.}\right) = 0,$$ which seemingly suggests the use of BC (1 or 2) lemma. The naive use of BC1 leads to nowhere since $$\sum_{n=1}^{\infty} \mathbb{P}\left(\frac{|X_n|}{n} > \epsilon\right)\le \sum_{n=1}^{\infty}\frac{\mathbb{E}[|X_n|]}{n\epsilon} = \frac{\mathbb{E}[|X_1|]}{\epsilon} \sum_{n=1}^{\infty} \frac{1}{n} = \infty,$$
(which needs $<\infty$ to work). I also tried to use BC2:
$$\sum_{n=1}^{\infty} \mathbb{P}\left(\frac{|X_n|}{n} \le \epsilon\right) = \sum_{n=1}^{\infty} \left(1- \mathbb{P}\left(\frac{|X_n|}{n} > \epsilon\right)\right)$$
$$= \lim_{M \to \infty} M – \sum_{n=1}^{M} \mathbb{P}\left(\frac{|X_n|}{n} > \epsilon\right)\ge \lim_{M \to \infty} M – \frac{\mathbb{E}[|X_1|]}{\epsilon} \sum_{n=1}^{M} \frac{1}{n} = \infty, $$
which results in $\mathbb{P}\left(\frac{|X_n|}{n} \le \epsilon \quad \text{i.o.}\right) = 1,$
again not helpful.
How can I crack the first part and also any idea for the second part is appreciated.
Best Answer
It is standard result that if $Y$ is a non-negative random variable then $\sum_n P\{Y>n\} <\infty$ iff $EY<\infty$. Taking $Y=\frac {|X_1|} {\epsilon}$ we get $\sum P\{|X_n| >n\epsilon \}=\sum P\{|X_1| >n\epsilon \}<\infty$. Apply Borel Cantelli Lemma and let $\epsilon \to 0$ through the sequence $1,\frac 1 2,\frac 1 3,\cdots$ to see that $\frac {|X_n|} n \to 0$ almost surely. If $a_n \geq 0$ and $\frac {a_n} n \to 0$ then $\frac {\max \{a_1,a_2,\cdots,a_n\}} n \to 0$. Now write $\frac {\max \{|X_1|,|X_2|,\cdots, |X_n|\}} {S_n}$ as $\frac {\max \{|X_1|,|X_2|,\cdots, |X_n|\}} n \times \frac n {S_n}$ and apply SLLN.