# Spiral lattice random walk

Execute a random walk from the origin on the integer lattice, but bias the four compass-direction probabilities from $\frac{1}{4}$ each to prefer to step in a spiraling direction. Calling the four step vectors $c_0,c_1,c_2,c_3$, with $c_i=(\cos (i \, \pi/2), \sin (i \, \pi/2))$, adjust the probabilities as follows. Let $v$ be the vector from the origin to the last point on the path, and $n$ the unit normal to $v$, counterclockwise $90^\circ$ to $v$. Then select step $c_i$ with probability $\frac{1}{4} (1+ c_i \cdot n)$. $\theta=60^\circ$. $\frac{1}{4}(1+c_2 \cdot n) = (1+\sqrt{3}/2)/4 \approx 0.47$.
Unsurprisingly, the random walks spiral around the origin:  $2000$-step random walks. Origin: green. Last point: red. Three examples, followed by five examples at reduced scale.

Q. Does Pólya's recurrence theorem hold for these walks? Do the walks return to the origin with probability $1$?

All but one of the above examples (the penultimate) returned to the origin, but usually rather quickly.

• I'm not immediately sure what the answer is, but I think the spiral thing is a bit of a red herring. What you should look at is does this random walk give the same parameter for the corresponding Bessel process. For d dimensional BM the parameter is $\frac{d-1}{2}$ and $\frac12$ is the critical parameter for recurrence, corresponding to $d=2$. I don't have time right now to calculate the parameter for the spiral RW. – Ori Gurel-Gurevich Jun 23 '17 at 14:49
• I believe the answer is that this is recurrent. If you let $R_n=|X_n|$, this has the same kind of decaying drift to $+\infty$ as the standard 2D Brownian motion: $R_{n+1}\approx R_n + c/\sqrt{R_n}$. – Anthony Quas Jun 23 '17 at 20:37
• The scaling limit of the norm seems to be a Bessel 3 process, so it should be transient... – Martin Hairer Jun 23 '17 at 20:47

It seems to me that this random walk is recurrent. Denote $Y_n=\|X_n\|$, where $(X_n, n\geq 0)$ is your "spiral" walk. Then, as $x\to \infty$, my calculations imply that $$\mathbb{E}(Y_{n+1}-Y_n\mid Y_n=x) = \frac{1}{4\|x\|} + O(\|x\|^{-2}),$$ and $$\mathbb{E}((Y_{n+1}-Y_n)^2\mid Y_n=x) = \frac{1}{2} + O(\|x\|^{-1}).$$ Then, (null) recurrence follows from Theorem 3.5.2 of [Menshikov, Popov, Wade, "Non-homogeneous random walks", C.U.P.-2017, http://www.ime.unicamp.br/~popov/book_lyapunov.pdf ].

Here is the theorem cited in Serguei Popov's answer: and here are the three assumptions: • Just complementing: $\bar{\mu}_1(t)$ and $\underline{\mu}_2(t)$ are essentially the two displays in my answer above, with $t=\|x\|$. – Serguei Popov Jun 24 '17 at 17:06

Forgive this simple-minded approach, but it may be instructive when made rigorous by someone who knows what they are doing.

I think Anthony Quas's suggestion of looking at radius good, but I do not understand his use of the word 'recurrent'. The probability of going east-west is 1/2, as is going north-south. However, most of the time, the probability for increasing the distance from the origin is greater than 1/2. This is because one of (say) east west is biased toward increasing distance, while the other (say) is half increasing and half decreasing. So for near the origin, there is a chance of returning, but (as Martin Hairer suggests) once you get far enough away, the dynamic seems transient.

I turn the calculation of "far enough away" over to those with more experience. I suspect the probability of returning to within S distance of the origin given one is at distance R in this dynamic is exponential in S-R (I assume S less than R).

Gerhard "Wonders About Inward Spiral Dynamics" Paseman, 2017.06.23.

• I am purposely ignoring the expected change of distance from origin which might lead to a different interpretation. Again, those who are more experienced should explain the probability angle correctly. Gerhard "Dynamic Model For Thought Processes?" Paseman, 2017.06.23. – Gerhard Paseman Jun 23 '17 at 22:22
• "the probability for increasing the distance from the origin is greater than 1/2": Yes, this is my intuition. But precise calculations are needed. – Joseph O'Rourke Jun 23 '17 at 23:04