# Probability of a brownian motion leaving some area

Let $B_t$, $t\geq0$ be a standard $n$-dimensional Brownian motion, that is $B_t(\omega)\in\mathbb{R}^n$ and let $\Lambda\subset\mathbb{R}^n$ be some ball such that the Brownian motion starts within $\Lambda$, i.e. $B_0\in\Lambda$. Now I'm supposed to prove $$P(B_s\in\Lambda^c\text{ for some }0\leq s\leq t)\leq 2P(B_t\in \Lambda^c).$$ I have no idea how to start. It's of course easy to estimate $P(B_s\in\Lambda^c)$ for any fixed $s$ but I don't see how that helps.

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Are you familiar with Désiré André's reflection principle? –  Did Jan 17 '13 at 23:47
Does it start from the center of the ball? If so you can just look at the norm $||B_t||$ which should be a Bessel process. You could compare this to a 1-D Brownian motion. –  Mihai Nica Jan 18 '13 at 5:00
I know the $1$-dimensional reflection principle. The Brownian Motion can start anywhere within the ball. Actually I want to prove that the above probability tends to $0$ as the ball fills $\mathbb{R}^n$. The inequality is just a hint. –  Julian Jan 18 '13 at 6:50
Well, this last fact is quite easier to prove: use dominated Lebesgue convergence theorem for $X_n=\mathbf 1_{A_n}$ with $A_n=[\exists s\leqslant t, B_s\notin\Lambda_n]$ and $\Lambda_n\to\mathbb R^d$. This applies to every process with continuous paths. –  Did Jan 18 '13 at 8:24
I think the same argument as for the relection principle works: conditioned qn hitting the boundary of the sphere at x you are equally likely to end up on either side of the tangent plane at x. –  mike Jan 18 '13 at 14:30