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Execute a random walk from the origin on the integer lattice, but bias the four compass-direction probabilities from 14 each to prefer to step in a spiraling direction. Calling the four step vectors c0,c1,c2,c3, with ci=(cos(iπ/2),sin(iπ/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 to v. Then select step ci with probability 14(1+cin).


          SpiralVecs
          θ=60. 14(1+c2n)=(1+3/2)/40.47.
Unsurprisingly, the random walks spiral around the origin:
    3RandWalks
   
    5RandWalks
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.

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    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 d12 and 12 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. Commented Jun 23, 2017 at 14:49
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    I believe the answer is that this is recurrent. If you let Rn=|Xn|, this has the same kind of decaying drift to + as the standard 2D Brownian motion: Rn+1Rn+c/Rn. Commented Jun 23, 2017 at 20:37
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    The scaling limit of the norm seems to be a Bessel 3 process, so it should be transient... Commented Jun 23, 2017 at 20:47

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It seems to me that this random walk is recurrent. Denote Yn=, 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 ].

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Here is the theorem cited in Serguei Popov's answer:


Thm352
and here are the three assumptions:
L012


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  • \begingroup Just complementing: \bar{\mu}_1(t) and \underline{\mu}_2(t) are essentially the two displays in my answer above, with t=\|x\|. \endgroup Commented Jun 24, 2017 at 17:06
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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.

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  • \begingroup 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. \endgroup Commented Jun 23, 2017 at 22:22
  • \begingroup "the probability for increasing the distance from the origin is greater than 1/2": Yes, this is my intuition. But precise calculations are needed. \endgroup Commented Jun 23, 2017 at 23:04

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