Probability of 2D Brownian motion passing through a particular point.

Question

Let $B_t$ be a two-dimensional Brownian motion at time $t \in [0,\infty)$. Fix a point $p \in \mathbb{R}^2$. Is the probability that $B_t = p$ at some $t > 0$ equal to zero? If so, why?

Answer 1

Indeed, the probability is equal to $0$. One can prove this using conformal invariance of two-dimensional Brownian motion. Using you notation, this means that for $f\colon \mathbb{C}\to \mathbb{C}$ a non-constant analytic function, $f(B_t)$ is also a Brownian motion (possibly moving at a variable speed). Take $f(z) = -pe^z+p$, then using conformal invariance, $f(B_t)$ is a Brownian motion which doesn't hit point $p$.

The other approach would be the following. Take $r < |p| < R$ and let $A$ be the annulus around $p$ with radii $r$ and $R$. Let $\tau =\inf\{t\geq 0\colon B_t\in \partial A \}$ be the first time Brownian motion hits the boundary. Function $f(z)= \log |z-p|$ is harmonic on $\mathbb{C}\setminus\{p\}$, and it is well-known that for harmonic functions one has $\mathbb{E}_0[f(B_{\tau})] = f(0) = \log|p|$. Let $\pi$ be the probability that the Brownian motion would first hit the boundary of $A$ in smaller circle (the one of radius $r$). Then $$\mathbb{E}_0[f(B_{\tau})] = \pi\log r+(1-\pi)\log R,$$ so we have $\pi = (\log |p|-\log R) / (\log r-\log R)$ which tends to $0$ as $r\to 0$. The claim now follows easily from this.