Does a Brownian motion remain in any given open set for a given interval of time with positive probability?

Question

Let $B$ be a standard $d$-dimensional Brownian motion. Given $b>a>0$ and an open ball $U$ in $\mathbb{R}^d$, I want to be able to comment on the probability that $B$ remains in $U$ during the time interval $(a, b)$. I am aware that we can derive a complicated-looking density for the exit time corresponding to that ball(I think I have seen that in the $d = 1$ case, the distribution of the time taken by the Brownian motion to exit an interval around its starting point), and then possibly condition on the value of $B$ at time $a$(and using the density of $B_a$) and use Markov property to get a formula for the above probability. But is it possible to somehow directly prove that this probability is always non-zero(because I anyway need only this conclusion at the moment)? I feel that it should follow from a simpler argument(without much calculations) as we shouldn't really need the exact density of the exit time to determine whether that probability is non-zero.

Answer 1

Assume without loss of generality that $U$ contains the cube $\prod\limits_{i=1}^d(x_i-2r,x_i+2r)$ and consider the smaller cube $L=\prod\limits_{i=1}^d(x_i-r,x_i+r)$, then the event you are interested in contains the event $$[B_a\in L]\cap\bigcap_{i=1}^d[\forall t\leqslant b-a,|B^i_t-B^i_a|\lt r]. $$ These are independent and $P(B_a\in L)\ne0$ since $L$ has positive Lebesgue measure. Each of the remaining events (in the intersection) has the same probability hence it remains to show that, if $W$ is a standard one-dimensional Brownian motion starting from $0$, then, for every positive $T$ and $r$, $$ [\forall t\leqslant T,|W_t|\lt r] $$ has positive probability. This is $[\tau_r\gt T]$, where $\tau_r$ denotes the first hitting time of $r$ by $|W|$ thus it remains to show that, for every $T$, $$P(\tau_r\gt T)\ne0.$$ Depending on the properties of Brownian motion that one can use, the proof of this property may vary. An approach uses the fact that, for every $u$, $$M^u_t=\cos(uW_t)\,\mathrm e^{u^2t/2}$$ defines a martingale $M^u$ starting from $M^u_0=1$. If $\tau_r$ is almost surely bounded, $M^u$ is bounded before $\tau_r$ hence the stopping time theorem yields $E(M^u_{\tau_r})=1$. By symmetry $W_{\tau_r}$ is uniformly distributed on $\{-r,r\}$ and independent on $\tau_r$ hence $$\cos(ur)\,E(\mathrm e^{u^2\tau_r/2})=1.$$ By hypothesis, $E(\mathrm e^{u^2\tau_r/2})\leqslant E(\mathrm e^{u^2T/2})$ is finite for every $u$ hence, if $\cos(ur)=0$, then we have a problem, Houston... QED.

Answer 2

You can easily prove the result for $d>1$ from the result for $d=1$ because

$$ P(a<t<b\implies B(t)\in U) \\\ge P(a<t<b\implies B(t)\in B(x_0, r)) \\\ge P(a<t<b\implies \forall i\le d\ \ B^i(t)\in (x_0^j\pm r/A)) \\= P(a<t<b\implies \forall \ B^1(t)\in (x_0^j\pm r/A)) ^d $$ if $A$ is big enough.

Now for $d=1$:

just get the joint probability of the min and the max of $B$ on $(a,b)$, I do not see anything else.