Suppose a two dimensional Brownian motion begins with (1,1) and it will stop moving when it hits y-axis. What's the probability of stopping at positive part of y-axis?
I only know the reflection principle of first passage times of one dimensional Brownian motion, but have no idea to deal with such problem.
Update:
We change the center of coordinate system to (1,1), namely the old X-axis and Y-axis become both -1. Denote $\tau^y_m,$ the first passage time of $m$ in Y-direction; $W^x_t,$ the Brownian motion in X-direction. $\tau^y_m$ and $W^x_t$ are independent. Then our problem is to solve $$P(W^x_{\tau^y_{-1}}>-1).$$ The PDF of $W^x_{\tau^y_{-1}}$ can be calculated as $$f(a)=f(W^x_{\tau^y_{-1}}=a) = \int_0^{\infty}f(W^x_t=a|\tau^y_{-1}=t)f(\tau^y_{-1}=t)dt=\int_0^{\infty}f(W^x_t=a)f(\tau^y_{-1}=t)dt,$$ here we simply use $f$ to generally represent the PDF of its random variable. I am not very confident on my deduction and does that mean we have to calculate the PDF of $\tau^y_m?$ It is an interview question, I think it is impossible to be such complicate.