2
$\begingroup$

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.
Improve this question
$\endgroup$
7
  • $\begingroup$ whats ur question? $\lim_{T \to \infty} \log P(T)/\log T$? $\endgroup$
    – mathworker21
    Commented Sep 17, 2021 at 13:39
  • 1
    $\begingroup$ Weird. My guess would be that the decay of $P(T)$ is the same as for the hitting time of a disk by the 2-D Brownian motion, that is, $T^{-1} (\log T)^{-2}$, see Byczkowski–Małecki–Ryznar 2013, DOI:10.1007/s11118-012-9296-7. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Sep 17, 2021 at 14:22
  • 1
    $\begingroup$ On a second thought, these two functions are rather hard to distinguish... :-) $\endgroup$
    – Mateusz Kwaśnicki
    Commented Sep 17, 2021 at 14:25
  • 1
    $\begingroup$ Not exactly the same question, but close: math.stackexchange.com/q/612392/241 $\endgroup$
    – Steve Huntsman
    Commented Sep 17, 2021 at 14:26
  • $\begingroup$ @mathworker21: any description of the asymptotic of $P(T)$ would be great, if $P(T)$ is explicitly known even better. $\endgroup$
    – Willie Wong
    Commented Sep 17, 2021 at 15:24

1 Answer 1

Reset to default
2
$\begingroup$

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:

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

Improve this answer
$\endgroup$
5
  • $\begingroup$ Thanks! Apparently the $T$ asymptotics was already known due to Hunt jstor.org/stable/1992918?seq=1#metadata_info_tab_contents and the main sharpening in the paper you mentioned is of the spatial dependence. $\endgroup$
    – Willie Wong
    Commented Sep 17, 2021 at 16:27
  • $\begingroup$ Yes, the problem is of course very old — although I did not know about Hunt's paper, thanks for sharing it. TThe paper I mentioned was simply the first one that came to my mind. (The authors are my colleagues and I remember their seminar talk.) $\endgroup$
    – Mateusz Kwaśnicki
    Commented Sep 17, 2021 at 16:50
  • $\begingroup$ I only saw Hunt's paper as a reference when I started reading the paper that you linked to. The older work happens to be a little easier for me to digest, this being a bit outside my normal interests. $\endgroup$
    – Willie Wong
    Commented Sep 17, 2021 at 17:03
  • $\begingroup$ I also should thank you for the observation of the similarity between the functions $1/(t \ln(t)^2)$ and $t^{-1.3}$. I was mentally prepared for $t^{-1}$ + corrections, and it just never occurred to me (not being a numerics person) that 1000 is still really damn far from infinity. $\endgroup$
    – Willie Wong
    Commented Sep 17, 2021 at 17:08
  • $\begingroup$ Actually, I would have never expected myself that these two functions could be so similar in the range $[10, 1000]$. That's right, logarithms make asymptotic analysis much more complicated. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Sep 17, 2021 at 17:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .