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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

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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.)

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    $\begingroup$ Am I missing something, or is this probability simply given by the unique bounded harmonic function that is zero on the negative x axis and 1 one the positive x axis, which is $1-\pi^{-1}\arctan(b/a)$? $\endgroup$ – Kostya_I Nov 14 '17 at 17:32
  • $\begingroup$ @Kostya_I : That's right. $\endgroup$ – Iosif Pinelis Nov 14 '17 at 17:58
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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

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