A $2D$ random walk with step size $1$ starts on the edge of a disk of radius $r$. What is probability that the walk will return to the disk?

Question

Consider a two dimensional random walk with step size $1$ and each step in a random direction, with the angle $\theta$ uniformly distrbuted in $[0,2\pi)$. The walk starts on the perimeter of a disk of radius $r$.

What is probability that the walk will ever return to the disk, in terms of $r$?

By "the walk will ever return to the disk", I mean the walk will have a vertex on or within the perimeter of the disk, besides the initial one.

I came up with this question when thinking about the fact that a two dimensional lattice walk will return to the origin with probability $1$, as proved by Polya in 1921. I have not been able to find any reference that answers my question.

Context:

Answer 1

For $2$D random walks, Lord Rayleigh proved that for large number of steps, $N$, the PDF of the final position of each walk distributed around some initial point follows the Rayleigh distribution: $$f_R(r)=\frac{2r}{N}\exp\left(-\frac{r^2}{N}\right)$$

Since it is normalized only in the radius coordinate, we’ll have to normalize it for the angle too if we want to work in polar coordinates: $$\int_0^{2\pi}\int_0^\infty \frac{2r}{N}\exp\left(-\frac{r^2}{N}\right) r\text dr\text d\theta=\pi\sqrt{\pi N}$$ Therefore, we are looking at $$f_{R;\Theta}(r;\theta)= \frac{2r}{(\pi N)^{\frac{3}{2}}}\exp\left(-\frac{r^2}{N}\right). $$

Since the initial point is on the perimeter of the disk of radius $\rho$ (you said $r$, but $r$ and $R$ are already taken), without loss of generality and in view that the disk is polar-symmetric, we can center the coordinate system at, for convenience, $(x,y)=(-\rho,0)$. Thus, the distribution will be centered there and the region of return will be a disk to the right of the origin. In brief, the initial point would be at the origin and the region would be described like $r\in(0,2\rho\cos\theta)\ \wedge\ \theta\in(-\pi/2,\pi/2)$.

Finally, we calculate said probability of returning: $$ \begin{aligned} \mathsf{P}&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2\rho\cos\theta} \frac{2r^2}{(\pi N)^{\frac{3}{2}}}\exp\left(-\frac{r^2}{N}\right)\text dr\text d\theta\\ &\overset{r^2/N<\!<1}{=} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2\rho\cos\theta} \frac{2r^2}{(\pi N)^{\frac{3}{2}}}\sum_{k=0}^\infty\frac{1}{k!}\left(-\frac{r^2}{N}\right)^k\text dr\text d\theta\\ &= \frac{2}{(\pi N)^{\frac{3}{2}}}\sum_{k=0}^\infty\frac{(-1)^k}{k!N^k}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2\rho\cos\theta}r^{2k+2}\text dr\text d\theta\\ &= \frac{2}{(\pi N)^{\frac{3}{2}}}\sum_{k=0}^\infty\frac{(-1)^k2^{2k+3}\rho^{2k+3}}{k!N^k(2k+3)}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\cos\theta)^{2k+3}\text d\theta\\ &= \frac{2}{(\pi N)^{\frac{3}{2}}}\sum_{k=0}^\infty\frac{(-1)^k2^{2k+3}\rho^{2k+3}}{k!N^k(2k+3)}\int_{0}^{\pi}(\sin\theta)^{2k+3}\text d\theta \end{aligned} $$ Knowing that the integral satisfies $I(n)=\frac{n-1}{n}I(n-2)$ and subtituting $n=2k+3$, the integral is $$I(2k+3)=\frac{2k+2}{2k+3}I(2k+1)=\underbrace{2}_{I(1)}\frac{(2k+2)!!}{(2k+3)!!}$$.

At last we get $$\mathsf{P}(\rho)=\sum_{k=0}^\infty\frac{(-1)^k2^{2k+5}}{k!\pi^{\frac{3}{2}}N^{k+\frac{3}{2}}(2k+3)}\frac{(2k+2)!!}{(2k+3)!!}\rho^{2k+3}, \forall\rho $$

Answer 2

A circular sector centered at the starting point with some non-zero angle $\alpha$ and non-zero radius $\rho$ lies entirely within the disk of radius $r$. By symmetry and linearity of expectation, the expected number of returns to this sector is $\frac\alpha{2\pi}$ times the expected number of returns to the disk of radius $\rho$ about the starting point. Thus, if the expected number of returns to this disk of radius $\rho$ is infinite, so is the expected number of returns to the circular sector, and hence also the expected number of returns to the disk of radius $r$. This implies that the probability to return to that disk is $1$.

Since this only depends on the long-term behaviour, we can approximate the distance from the origin after $N$ steps by the Rayleigh distribution

$$ f(r)=\frac{2r}N\mathrm e^{-\frac{r^2}N} $$

given in Joan S. Guillamet F.’s answer. The probability to be within $\rho$ of the starting point after $N$ steps is thus approximated as

$$ \int_0^\rho\frac{2r}N\mathrm e^{-\frac{r^2}N}\mathrm dr=\left[-\mathrm e^{-\frac{r^2}N}\right]_0^\rho=1-\mathrm e^{-\frac{\rho^2}N}=\frac{\rho^2}N+O\left(N^{-2}\right)\;. $$

Since the harmonic series diverges, the expected number of returns is infinite.

Making this rigorous would require looking at the error in Rayleigh’s approximation, but that shouldn’t change the result.

This is essentially the approach that’s used to prove the recurrence of the simple random walk on a square lattice, e.g. in this answer. (Note the $3$-digit IDs of the answer and question, which are from the early days of the site in $2010$. :-)

Answer 3

The plane Brownian motion $B(t)$ visits any open subset of the plane with probability one. Consider the random times

$$T_1=\inf\{t>0, \|B(t)\|=1\},$$

$$ \ T_{n+1}=\inf\{t>T_n; \|B(t)-B(T_n)\|=1\}$$ The law of $(B(T_n))_{n=1}^{\infty}$ is the law of your random walk with discrete time. Take now a open disk $D(a,1)$. The Brownian motion will visit it an infinite number of times. Some reasoning shows that $N=\inf\{n; B(T_n)\in D(a,1)$ is finite.