property of brownian motion

Asked May 09 '17 at 12:09

Active May 09 '17 at 13:10

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I want to show this result:

$$\mathbb{P} \left[ \lim_{t \rightarrow \infty} \frac{B_t}{t} = 0 \right] = 1$$

For $B_t$ a standard brownian motion.

I have no idea how to do this. Naively I tried exchanging limit and probability but that doesn't work.

I just need a hint to get started.

EDIT: In general what is a good way to show almost sure convergence?

edited May 09 '17 at 13:10

asked May 09 '17 at 12:09

user405156

Is $t$ an integer or a real number? – Cm7F7Bb May 09 '17 at 12:35
t is a real number – user405156 May 09 '17 at 12:35
If $t$ was an integer, then we could simply use the SLLN since $B_t=B_t-B_0=B_1-B_0+B_2-B_1+\ldots+B_t-B_{t-1}$. – Cm7F7Bb May 09 '17 at 12:41
@tomak Do you mean Wiener process? If so, there exists a Law of Iterated Logarithm that will answer your question – Symmetrical May 09 '17 at 14:54
@Symmetrical No definitly no errors in what I put – user405156 May 09 '17 at 15:02
@Cm7F7Bb suppose $t$ where an integer, I still don't see. $lim\ B_n = lim\ n . \overline{X}_n$ but how do you control this? one term goes to infinity and the other one to 0. – user405156 May 09 '17 at 15:16
@tomak $B_i-B_{i-1}$ are iid random variables. By the SLLN, $t^{-1}\sum_{i=1}^t(B_i-B_{i-1})\to0$ almost surely as $t\to\infty$. – Cm7F7Bb May 09 '17 at 15:18
Yes ok I just forgot the $1/t$. Then maybe I can see $B_t = B_t - B_n + B_n$ the third term is known by what has been discussed and the sum of the two first has to be controlled in some way. – user405156 May 09 '17 at 15:22

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