Prove that lim sup $|B(n)| = \infty$ almost surely, where $B(n)$ is standard Brownian motion.

273 Views Asked by At

Okay, I have cleared up some of my proof. Sorry for confusion (it was first time to be in this StackExchange, to make an excuse).

HINT(given): Use the independent increments property of brownian motion and continuity property.

What I want to do is to show that probability of having each random moment of Brownian motion smaller than $\epsilon$ is zero.

$P(\sup (B(t_2) - B(t_1) < \epsilon), sup (B(t_3) - B(t_2)) < \epsilon, ..., ) = \prod P (sup (B(t_i ) - B(t_i-1)) < \epsilon), $

by independent increment property.

I think I can use the continuity property to show that the product of such is equal to zero, and then probability of having each Brownian smaller than $\epsilon$ is zero. Thus, $\limsup \left| B(n) \right| = \infty.$

I want to know:

i) Am I on the right track?

ii) How can I move to the rest of the proof?

Thanks in advance.