Consider the following process on $\mathbb{C}$:


* Start at the point 1.
* At each step, move by adding $e^{i\theta}$, where $\theta$ is uniformly drawn from $\mathbb{S}^1$.
* Stop at the first positive time $T$ where you move inside the *open* unit disc.


What is the distribution of the stopping times?


*Some notes*:


* $T = 1$ happens 1/3 of the time, by simple geometry.
* $T = 2$ happens 1/9 of the time, by evaluating an explicit integral. (Let $g(T,x)$ be the probability that starting at $x$ you will hit the unit disc with exactly stopping time $T$, you can write an integral relating $g(T+1,x)$ to $g(T,x+z)$ with $z\in \mathbb{S}^1$. When $T = 1$ the function is explicitly known by simple geometry. When $T = 2$ and $|x| = 1$ the value can be obtained by direct integration, but for $|x| > 1$ I don't know how to evaluate the integral involved. And hence I also cannot get the exact value for $T = 3$ at $x = 1$. [Later this weekend I may try to evaluate these chain of integrals numerically and provide some more numeric data.])
* Numerical simulations of the random walk seems to suggest that $P(T)$ follows a power rule $T^\alpha$ where $\alpha \approx \log\_2(0.4)$. But I only checked up to around $T = 1000$, as the numerical simulation is not very efficient.



Distribution of stopping time for a 2D random walk
For the continuous counterpart, if a 2-D Brownian particle is started at $x$ with $|x| > 1$, the density function of the hitting time of the unit disk decays as
$$ \frac{1}{t (\log t)^2} $$
as $t \to \infty$. More precisely, it is comparable to
$$ \frac{|x|-1}{|x|} e^{-(|x|-1)^2/(2t)} \frac{(|x|+t)^{1/2}}{t^{3/2}} \frac{1 + \log |x|}{(1 + \log(1 + \tfrac{t}{|x|})) (|x| + \log t)} \, . $$
with absolute constants in both bounds. This incredibly precise estimate, as well as similar results in higher dimensions, were proved in:


* T. Byczkowski, J. Małecki, M. Ryznar, *Hitting Times of Bessel Processes*. Potential Anal 38, 753–786 (2013). <https://doi.org/10.1007/s11118-012-9296-7>


I would expect the large-time decay of $P(T)$ is thus similar to $T^{-1} (\log T)^{-2}$.


