Proving that $\lim\limits_{t \rightarrow \infty}B_t/t=0$ a.s. without using the law of iterated logarighm.

Question

I want to prove that $\lim\limits_{t \rightarrow \infty}B_t/t=0$ a.s directly, without using the law of iterated logarithm.

I have read a hint that one can use the law of large numbers. If I look at the sequence $B_n/n, n\in \mathbb{N}$ I can write it as $$\sum\limits_{i=1}^n\frac{B_i-B_{i-1}}{n},$$ this converges to zero by the strong law of large numbers.

So if I write

$$\frac{B_t}{t}=\frac{B_{\lfloor t\rfloor}}{\lfloor t \rfloor}\cdot\frac{\lfloor t \rfloor}{t}+\frac{B_t-B_{\lfloor t \rfloor }}{t},$$

I have that the first term converges a.s. to zero. So I only need to show that

$$\frac{B_t-B_{\lfloor t \rfloor }}{t}$$ converges a.s. to zero. Do you see how to prove this directly?

Answer 1

We want to show that $\frac{\sup_{s \in [0,1)} \lvert B_{n+s}-B_n\rvert}n$ tends to zero almost surely as $n$ goes to $\infty$.

We have that $$ \mathbb P\Big(\sup_{s \in [0,1)} \lvert B_{n+s}-B_n\rvert \geq n^{2/3}\Big) = \mathbb P\Big(\sup_{s \in [0,1)} \lvert B_s\rvert \geq n^{2/3}\Big) \leq 2\mathbb P\Big(\sup_{s \in [0,1)} B_s\geq n^{2/3}\Big) = 2\mathbb P\Big(\lvert B_1 \rvert \geq n^{2/3}\Big) \leq \frac{2 \mathbb E\big[B_1^2\big]}{n^{4/3}}.$$

Borel-Cantelli implies that almost surely there is $\bar n = \bar n(\omega)$ such that for any $n\geq \bar n$ it holds that $\sup_{s \in [0,1)} \lvert B_{n+s}-B_n\rvert \leq n^{2/3}$, and thus $$\frac{\sup_{s \in [0,1)} \lvert B_{n+s}-B_n\rvert}n \leq n^{-1/3}.$$

Answer 2

Essentially what you need is the following (general) result: suppose you have a continuous-in-time stochastic processes $(X_t)_{t\geq 0}$ such that $\mathbb{E}(X_t) = 0$ and $\lim_{t\to \infty} \mathrm{Var}(X_t) = 0$, then $X_t \to 0 $ almost surely.