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](https://math.stackexchange.com/questions/88555/2d-brownian-motion-hitting-a-point)


[2d Brownian motion first passage time](https://mathoverflow.net/questions/49208/two-dimensional-brownian-motion-first-passage-time)



2 dimensional brownian motion hitting time
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.)


