2 dimensional brownian motion hitting time If we have two independent brownian motion in $x$ and $y$ direction. At time zero we sit at $(a,b)$ with $a>0, b>0$.
What is the probability that we will hit positive $x$ axis before hitting the negative $x$ axis?
I tried to look at some posts but no clue yet...
2d-brownian-motion-hitting-a-point
2d Brownian motion first passage time
 A: The probability in question is $1-p$, where $p$ is the probability that we will hit the negative  $x$-semiaxis before hitting the positive $x$-semiaxis. Next, $p$ is the probability  that (we will hit the positive $y$-semiaxis before hitting the positive $x$-semiaxis, and then we will hit the negative $x$-semiaxis before hitting the positive $x$-semiaxis). So, by the strong Markov property and the symmetry, 
\begin{equation*}
 p=q\,\tfrac12,
\end{equation*}
where $q$ is the probability  that we will hit the positive $y$-semiaxis before hitting the positive $x$-semiaxis. In turn,
\begin{equation}
 q=P(\tau_a<\tau_b)=\int_0^\infty P(\tau_a<t)\,dP(\tau_b<t), \tag{1}
\end{equation}
where $\tau_a$ and $\tau_b$ are independent random variables (r.v.'s) such that for any real $c>0$ the distribution of the r.v. $\tau_c$ is that of the time for a standard Brownian motion (starting at $0$) to first reach point $c$: 
\begin{equation*}
 P(\tau_c<t)=2(1-\Phi(c/\sqrt t))
\end{equation*}
for $t>0$, by the reflection principle, where $\Phi$ is the standard normal distribution function. Thus, by (1), 
the probability in question is 
\begin{equation}
 1-q/2=
 1-\int_0^\infty (1-\Phi(a/\sqrt t))\varphi(b/\sqrt t)\frac{b\,dt}{t^{3/2}}=1-\frac1\pi\,\arctan\frac ba,   
\end{equation}
where $\varphi:=\Phi'$. 
(One way to evaluate the latter integral is as follows: differentiate it in $a$, then use the substitution $t=x^{-2}$ to see that the derivative of the integral is $-\frac b{\pi(a^2+b^2)}$, and finally integrate this derivative back in $a$. Another way is to use the substitution $t=x^{-2}$ right away, to rewrite the integral as 
\begin{equation*}
 \int_0^\infty P(Z_1>ax)P(|Z_2|\in b\,dx)=P(Z_1>a|Z_2|/b)=P(C<b/a)=\frac1\pi\,\arctan\frac ba,
\end{equation*}
where $Z_1$ and $Z_2$ are independent standard normal r.v.'s, so that $C:=|Z_2|/Z_1$ is a Cauchy r.v. 
Yet another way is suggested by Kostya.)
A: Let $Z_t$ be the two dimensional Brownian motion.  This is easily solved once you notice that $arg(Z_t)$ (the angle between $\vec{OZ_t}$ and the positive x-axis) is a martingale.  This is intuitively correct because $Z_t$ is isotropic.  Let $\tau$ be the stopping time when $Z_\tau$ first hit the x-axis, then apply the Optional Stopping Theorem,
$\text{E}[arg(Z_\tau)] = \text{P}\{arg(Z_\tau)=0\}\cdot 0 + \text{P}\{arg(Z_\tau)=\pi\}\cdot \pi = \text{E}[arg(Z_0)]=\text{arc}\tan\frac{b}{a}$
Therefore,
$\text{P}\{arg(Z_\tau)=0\} = 1 - \text{P}\{arg(Z_\tau)=\pi\} = 1 - \frac{1}{\pi}\text{arc}\tan\frac{b}{a}.$  QED
