Why does the Borel-Cantelli lemma finish the job? - Law of Large Numbers Brownian Motion

Question

The objective is to prove that \begin{align*} \text{$\lim_{t \to \infty} \frac{B_t}{t} =0 \qquad$ a.s.} \end{align*} By the strong Law of Large Numbers, we have that: \begin{align*} \text{$\lim_{t \to \infty} \frac{B_t}{t} = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n B_k - B_{k-1} = 0 \qquad$ a.s.} \end{align*} In doing so, the idea is to show that the trajectory of $B$ cannot deviate too much from $B_n$ on $[n, n+1]$. It can be shown that \begin{align*} \sum_{n=0}^\infty P\big(\big\{\sup_{t \in [n,n+1]} |B_t - B_n | \geq n^{\frac{2}{3}}\big\} \big) < \infty. \end{align*} Now, the question is, how does the Borel-Cantelli Lemma finish this prove?

So, the Borel-Cantelli Lemma is saying that \begin{align*} &P\big( \limsup_{n \to \infty} \big\{\sup_{t \in [n,n+1]} |B_t - B_n | \geq n^{\frac{2}{3}}\big\} \big) = 0, \qquad \text{so} \\ &P\big( \bigcup_{n=1}^\infty \bigcap_{k\geq n} \big\{\sup_{t \in [n,n+1]} |B_t - B_n | \geq n^{\frac{2}{3}}\big\} \big) = 0 \qquad \text{and} \\ &\lim_{n \to \infty} P\big( \bigcap_{k\geq n} \big\{\sup_{t \in [n,n+1]} |B_t - B_n | \geq n^{\frac{2}{3}}\big\} \big) = 0. \end{align*} How to find again the fraction $\frac{B_t}{t}$?

Answer 1

For $t>0$ denote by $n=n(t) \in \mathbb{N}_0$ the unique number such that $t \in (n,n+1]$. Then, by the triangle inequality,

$$\left| \frac{B_t}{t} \right| \leq \frac{|B_t-B_n|}{t} + \frac{|B_n|}{t} \leq \frac{1}{n} \sup_{s \in [n,n+1]} |B_s-B_n| + \frac{|B_n|}{n}. \tag{1}$$

We already know from the law of large numbers that the second term converges to $0$ as $t \to \infty$ (or, equivalently, $n=n(t) \to \infty$). For the first term, we note that, by the Borel-Cantelli lemma, there exists for almost all $\omega \in \Omega$ a constant $C>0$ such that

$$\sup_{s \in [n,n+1]} |B_s(\omega)-B_n(\omega)| \leq C n^{2/3}$$

for all $n \geq N(\omega)$ sufficiently large. Hence,

$$\frac{1}{n} \sup_{s \in [n,n+1]} |B_s(\omega)-B_n(\omega)| \leq C \frac{n^{2/3}}{n} \xrightarrow[]{n \to \infty} 0.$$

This proves that also the first term on the right-hand side of $(1)$ converges almost surely to $0$.