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Pólya's recurrence theorem

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The drunkard wandering along a sidewalk almost surely returns home. The drunkard wandering across a city almost surely returns home. The bird flying through three-dimensional space, taking unit-length steps in random directions, may never return: there is a positive probability of escaping to infinity, never to come back.

This is Pólya’s recurrence theorem (Pólya, 1921): on the integer lattice Zd\mathbb{Z}^d, a simple random walk (each step picks one of the 2d2d unit-vector directions uniformly) returns to the origin with probability 11 if and only if d2d \le 2.

The phase transition is precisely at d=2d = 2. In d=3d = 3 the walk returns with probability about 0.34050.3405, known as Pólya’s constant. The probability shrinks further in higher dimensions.

Three independent walks running in 1D, 2D, and 3D below. The current walk in each panel restarts when it hits the origin (a return) or after 30003000 steps (give up). The counter shows the empirical fraction of walks that have returned.

1D
walks: 0 · returns: 0 · rate:
2D
walks: 0 · returns: 0 · rate:
3D
walks: 0 · returns: 0 · rate:

Watch the rates after a couple of minutes. The 1D walk returns almost every time: a classical identity says the probability of not having returned within the first nn steps is exactly (nn/2)/2n2/(πn)\binom{n}{n/2}/2^{n} \sim \sqrt{2/(\pi n)}, which is only about 1.5%1.5\% at n=3000n = 3000, so the 1D rate sits well above 90%90\%. The 2D walk also returns with probability 11 eventually, but far more slowly: the no-return probability after nn steps decays only like π/lnn\pi/\ln n, a logarithmic crawl, so even with the 30003000-step cap a real fraction of 2D walks have not yet come back and the rate settles clearly below the 1D one. The 3D walk is the qualitatively different case: it is transient, and its return probability converges to Pólya’s constant 0.3405\approx 0.3405, landing just below that value because the cutoff drops the rare walks that would have returned later. The split between the dimensions is the result.

The argument

Let SnS_n denote the position of the walk after nn steps. The walk is recurrent (returns to the origin infinitely often almost surely, i.e. with probability 11) iff the expected number of returns is infinite:

E[#{n:Sn=0}]  =  n=0Pr[Sn=0].\mathbb{E}\bigl[\#\{n : S_n = 0\}\bigr] \;=\; \sum_{n=0}^\infty \Pr[S_n = 0].

So the dichotomy reduces to convergence vs. divergence of nPr[Sn=0]\sum_n \Pr[S_n = 0].

For the simple symmetric random walk on Zd\mathbb{Z}^d, a local version of the central limit theorem (which estimates the chance of landing on one specific lattice point, not just the spread of the position) gives

Pr[S2n=0]    Cdnd/2as n,\Pr[S_{2n} = 0] \;\sim\; \frac{C_d}{n^{d/2}} \quad \text{as } n \to \infty,

with an explicit constant Cd>0C_d > 0. Plugging in,

nPr[S2n=0]    Cdn1nd/2.\sum_n \Pr[S_{2n} = 0] \;\sim\; C_d \sum_n \frac{1}{n^{d/2}}.

The series 1/nd/2\sum 1/n^{d/2} converges iff d/2>1d/2 > 1, i.e., d3d \ge 3.

So:

The precise return probability in d=3d = 3 has the closed form

Pr[return]  =  11u(0,0,0),\Pr[\text{return}] \;=\; 1 - \frac{1}{u(0,0,0)},

where u(0,0,0)u(0,0,0) is the value of the lattice Green’s function at the origin:

u(0,0,0)  =  1(2π)3[π,π]3dxdydz113(cosx+cosy+cosz).u(0,0,0) \;=\; \frac{1}{(2\pi)^3} \int_{[-\pi,\pi]^3} \frac{dx\,dy\,dz}{1 - \tfrac{1}{3}(\cos x + \cos y + \cos z)}.

Watson (1939) evaluated this triple integral in terms of elliptic integrals, getting u(0,0,0)1.516u(0,0,0) \approx 1.516 and so the return probability 0.3405\approx 0.3405.

Where this matters

References



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