Currently, I am reading a paper about planar Brownian Motion. However, I am just getting to know Brownian Motion so I am not familiar with all well-known properties of BM or stochastic processes.
I am looking at a Brownian Motion that is somewhere in a neighborhood of several circles $\partial B_{r_i}(z_0), z_0\in\mathbb R^2, \ i=1,\dots,n,$ such that $r_1<\dots<r_n.$ The author says that when $BM$ starts on a circle $\partial B_{r_i},$ the probability BM hits $\partial B_{r_{i+1}}$ before $\partial B_{r_{i-1}}$ is $\frac 1 2$.
Although this seems like quite a simple statement, I don't really why this is. My intuition says that if the distances i.e. the $r_i$ were the same, then this statement is clear. However, the $r_i$ don't necessarily have to be equal.
If you could point out a few arguments for this, I'd be really glad. Any help is appreciated.
Edit: Upon request; the values are given as $r_i = R (\frac{\epsilon}{R})^\frac{i}{K}$ where $R\in(0, \frac 1 2)$, $K \geq 1$ and $i = 0,1,\dots,K.$
Let $0<y<|z|<w<\infty$ with $z\in\mathbb{R}^2$, and $z_0=0$. Using the fact that $u(x)=\ln|x|$ is harmonic on $\mathbb{R}^2\setminus \{0\}$ and Ito's lemma, one can show that $$ N_t:=u\!\left(B_{t\wedge T_y}\right)-u(z)=\sum_{i=1}^2\int_0^t\frac{\partial}{\partial x_i} u(B_{s\wedge T_y})\,d B_{s\wedge T_y}^i, $$ where $T_r:=\inf\{t\ge 0:|B_t|=r\}$, is a local martingale null at 0. Stopping $N_t$ at $T_w$, one gets a uniformly integrable martingale null at 0 which converges in $L_1$. Therefore, $$ u(z)=\mathsf{E}^z[u(B_{T_y\wedge T_w})]=\mathsf{P}^z(T_y<T_w)\ln y+\mathsf{P}^z(T_y\ge T_w)\ln w, $$
and so, when $B_t$ starts at $z$, $$ \mathsf{P}^z(T_y<T_w)=\frac{\ln w-\ln|z|}{\ln w-\ln y}. $$
In your case, $r_i=R(\varepsilon/R)^{i/K}$ for $R\in(0,1/2)$ and some $K\in\mathbb{N}$ (see Equation (5) on page 2). So $$ \frac{\ln r_{i+1}-\ln r_i}{\ln r_{i+1}-\ln r_{i-1}}=\frac{1}{2}, $$ as required.