= Solution
The angle is a bounded <local martingale>, hence a <uniformly integrable martingale>. The <Martingale convergence theorem> gives an almost sure and $L^1$ limit $\Theta_\infty\in[-\pi/2,\pi/2]$ with
$$
\mathbb E\Theta_\infty=\arctan x.
$$
It must lie at an endpoint. Indeed, the <Itô isometry> and boundedness give
$$
\mathbb E\int_0^\infty\cos^2\Theta_s\,ds
=\lim_{t\to\infty}\mathbb E[(\Theta_t-\Theta_0)^2]
\leq\pi^2<\infty.
$$
Thus the bracket integral is finite almost surely. If the angle converged to an interior point, its squared cosine would eventually be bounded below by a positive constant, forcing this integral to be infinite. Consequently
$$
\boxed{\Theta_\infty\in\{-\pi/2,\pi/2\}\quad\text{almost surely}.}
$$
This proves <endpoint convergence of a bounded angle diffusion>.
Let $p_x=\mathbb P(\Theta_\infty=\pi/2)$. Since $X_t=\tan\Theta_t$, this is exactly $\mathbb P(X_t\to+\infty)$, and the other endpoint means $X_t\to-\infty$. Taking expectations gives
$$
\arctan x=\frac\pi2p_x-\frac\pi2(1-p_x),
\qquad
\boxed{\mathbb P(X_t\to+\infty)=\frac12+\frac{\arctan x}{\pi}.}
$$
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