Probability Brownian motion hitting ball in dimension 2 and 3.

Question

Can someone help me in solving this exercise ? I was thinking it is something dealing with the recurrent/transient property of the Brownian motion, but I am not sure.

Let $(B_t)_t \in [0,1)$ be a standard Brownian motion on $R^2$ and for $R ∈ (0,∞)$ let $B_R$ denote the ball of radius R centered at the origin. For every $t ∈ (0,∞)$, compute $P[B_t ∈ B_R]$, and thereby prove that (1) $P(B_t \notin B_{\sqrt{2\lambda t}}) = e^{-\lambda}$, (2) and that, for the Lebesgue measure $|B_R|$ of $B_R$, $\lim_{R \rightarrow 0} \frac{P[B_t ∈ B_R]}{|B_R|} = \frac{1}{2\pi t}$

What happens in dimension three?

Answer 1

First, note that $\|B_t/\sqrt{t}\|_2$ has the $\chi$ distribution with $2$ degrees of freedom. Therefore, $$ \mathsf{P}(B_t\in \mathsf{B}_R)=\mathsf{P}(\|B_t/\sqrt{t}\|_2\le R/\sqrt{t})=\int_0^{R^2/(2t)}e^{-x}\, dx=1-e^{-R^2/(2t)}, $$ and, consequently, $\mathsf{P}(B_t\notin \mathsf{B}_{\sqrt{2\lambda t}})=e^{-\lambda}$, and $$ \lim_{R\to 0}\frac{\mathsf{P}(B_t\in \mathsf{B}_R)}{|\mathsf{B}_R|}=\lim_{R\to 0}\frac{1-e^{-R^2/(2t)}}{\pi R^2}=\lim_{R\to 0}\frac{e^{-R^2/(2t)}}{2t\pi}=\frac{1}{2\pi t}. $$