Probability Brownian motion is positive at two points

1.9k Views Asked by At

Let $0<s<t$ and $(B_r)_r$ is Brownian motion. Does anybody know what $P(B_s>0,B_t>0)$ is? I think I remember it was some $arctan$-law but I don't know the exact form. So I do not need a proof, stating the result would be completely sufficient, but I could not find it using google.

Thank you for your help.

1

There are 1 best solutions below

1
On BEST ANSWER

If $(X,Y)$ are jointly Gaussian with mean zero, then $P(X>0,Y>0)={\arccos(-\rho)\over 2\pi}$ where $\rho$ is the correlation of $X$ and $Y$.

Therefore, for Brownian motion and $0<s<t$ $$P(B_s>0, B_t>0)={\arccos(-\sqrt{s/t})\over 2\pi}.$$