Calculate the probability with respect to Brownian motion. Given two positive constants $a,b>0$. Suppose both $B_1$ and $B_2$ are 2D standard complex Brownian motions which start from $0$. Let $T_a=\inf\{t>0:|B_1|=a\}$,$T_b=\inf\{t>0:|B_2|=b\}$. Define $T_a<T_b$ as the even $E_{a,b}$. In other words. the even $E_{a,b}$ means $|B_1|>a$ before $|B_2|>b$.
How to calculate the probability $\mathbb{P}(E_{a,b})$ which puzzles me for a long time. 
If you also don't know how to solve this problem. Recommend some relevant documents or books to me will be OK. Anyway, any help would be appreciated. Thank you very much.
 A: I'll assume $B_1$, $B_2$ are independent. To avoid ambiguity, write $T^1_a$ and $T^2_b$ for the respective exit times. 


*

*Observe that  $T^1_a$ is equal in distribution to $a^2 \tau$, where $\tau$ is the exit time of the radial part of 2D BM, starting from $0$,  from the interval $[0,1)$. The radial part of 2D BM is a Bessel process of dimension $2$. 
Similarly, $T^1_b$ is equal in distribution to $b^2 \tau$. 

*Therefore what you're looking for is a formula for 
$$ P (a^2 \tau' < b^2 \tau),$$
where $\tau'$ is an independent copy of $\tau$. 

*Let $\rho=b^2/a^2$, and let $F$ be the PDF of $\tau$. Then by conditioning on $\tau'$, we're looking at 
$$ E [ F(\rho \tau)]$$ 
Sanity check:


*

*Recall that if $X$ is a continuous RV and $F$ is its PDF, then $F(X)$ is uniform $[0,1]$. In particular, if $\rho=1$, the answer is $\frac 12$, as expected; also by dominated convergence: 

*If $\rho \to \infty$, then the limit is $1$.

*If $\rho \to 0$, then the limit is $0$.



Hope this helps. 
