Anisotropic Nearest-Neighbor Walks on Z^d

Directional Weights and the Fourier Symbol

A nearest-neighbor walk on \(\mathbb{Z}^d\) need not choose every coordinate direction equally often. It can favor some axes while remaining symmetric under reversal of each step. The key question is whether that anisotropy changes recurrence: does a walk return to its starting point with probability one, or can it eventually escape? We will see that positive directional weights change the Fourier symbol and the constants in the return estimates, but not the dimension threshold.

Fix an integer \(d\ge1\) and positive numbers \(\alpha_1,\ldots,\alpha_d\) with \(\sum_{r=1}^d\alpha_r=1\). On a probability space \((\Omega,\mathcal{F},\mathbb{P})\), let \((X_n)_{n\ge0}\) be the discrete-time Markov chain with state space \(S=\mathbb{Z}^d\), started at \(X_0=0\), whose one-step increment is \(+e_r\) with probability \(\alpha_r/2\) and \(-e_r\) with probability \(\alpha_r/2\). Here \(e_r\) is the \(r\)th coordinate unit vector. Thus each coordinate axis is chosen with probability \(\alpha_r\), and either sign is then chosen with equal probability. The probability of a step in either direction along axis \(r\) is positive, and the probabilities of all \(2d\) possible steps sum to \(\sum_r 2(\alpha_r/2)=1\).

Write \(p_{xy}=\mathbb{P}(X_{n+1}=y\mid X_n=x)\), which does not depend on \(n\). For each \(x\), its nonzero transition probabilities are \(p_{x,x+e_r}=p_{x,x-e_r}=\alpha_r/2\); their sum is \(1\). The natural filtration is \(\mathcal{F}_n=\sigma(X_0,\ldots,X_n)\). Since each coordinate step in either direction has positive probability, any lattice point can be reached from any other by a finite sequence of allowed steps. The chain is irreducible.

For \(\theta=(\theta_1,\ldots,\theta_d)\in[-\pi,\pi]^d\), define the Fourier symbol to be the characteristic function of one increment, evaluated at \(\theta\). Symmetry makes it real:

The weights appear inside the sum. In particular, the symbol is not generally the unweighted average of the coordinate cosines. The next result connects this symbol to recurrence.

We prove the theorem using the expected total number of visits to the origin. Define \(p_{00}^{(n)}=\mathbb{P}_0(X_n=0)\) and \(G=\sum_{n\ge0}p_{00}^{(n)}\). This sum includes the visit at time zero. Fourier inversion gives

Indeed, if the increments are \(Y_1,\ldots,Y_n\), then \(X_n=\sum_{\ell=1}^nY_\ell\). The characteristic function of this sum is \(\phi(\theta)^n\), by independence of the increments. Integrating \(e^{i\theta\cdot X_n}\) over the cube with the normalized measure gives \(1\) when \(X_n=0\) and \(0\) for every other integer vector, since the integral of \(e^{ik\theta_r}\) over \([-\pi,\pi]\) is zero for a nonzero integer \(k\). Taking expectations yields the formula.

Proof. The symbol formula follows directly from the increment probabilities: the contributions of \(+e_r\) and \(-e_r\) are \((\alpha_r/2)e^{i\theta_r}\) and \((\alpha_r/2)e^{-i\theta_r}\), whose sum is \(\alpha_r\cos\theta_r\). We now prove the recurrence classification.

For \(0\le s<1\), define \(G(s)=\sum_{n\ge0}s^n p_{00}^{(n)}\). The Fourier formula and the geometric series give

For fixed \(s<1\), the geometric series converges uniformly because \(|s\phi(\theta)|\le s<1\), so it can be integrated term by term. The integral is well-defined: \(1-s\phi(\theta)\ge1-s>0\).

Translation of every coordinate by \(\pi\), interpreted modulo \(2\pi\), preserves integration over the torus and changes \(\phi\) to \(-\phi\). Averaging the integral with its translated version gives

To verify the averaging step, the average of the two integrands is \(\tfrac12\{(1-s\phi)^{-1}+(1+s\phi)^{-1}\}=(1-s^2\phi^2)^{-1}\). As \(s\) increases to \(1\), the last integrand increases pointwise to \((1-\phi^2)^{-1}\), with value \(+\infty\) where \(|\phi|=1\). Also, \(\sum_n s^n p_{00}^{(n)}\) increases to \(G\). The monotone convergence theorem therefore gives

where both sides may be infinite. The identity \[ \frac{1}{1-\phi^2} =\frac12\left(\frac{1}{1-\phi}+\frac{1}{1+\phi}\right) \] holds wherever the denominators are nonzero. The two terms on the right have equal integrals by the same coordinate translation. Hence \(G\) is finite exactly when the integral of \((1-\phi)^{-1}\) is finite.

It remains to determine the integrability of that last expression. On \([-\pi,\pi]\), the elementary inequalities \[ \frac{2x^2}{\pi^2}\le1-\cos x\le\frac{x^2}{2} \] show that, for \(\theta\) near the origin,

where \(\alpha_{\min}=\min_r\alpha_r>0\), \(\alpha_{\max}=\max_r\alpha_r\), and \(|\theta|^2=\sum_r\theta_r^2\). Thus the integrand is bounded above and below by positive constant multiples of \(|\theta|^{-2}\) near zero. Away from zero on the torus, \(1-\phi\) is positive: a weighted average of cosines with positive weights equals \(1\) only if every cosine equals \(1\). The integrand is therefore bounded on every closed region separated from the origin.

In polar coordinates near zero, the integral of \(|\theta|^{-2}\) has the same convergence behavior as \(\int_0^\varepsilon r^{d-3}\,dr\), where \(\varepsilon>0\) is small. For \(d=1\) the exponent is \(-2\) and the integral diverges; for \(d=2\) it is \(-1\) and diverges; for \(d\ge3\) it is at least \(0\) and converges. Consequently \(G=\infty\) exactly when \(d\le2\).

To finish, let \(f=\mathbb{P}_0(T_0^+<\infty)\), where \(T_0^+=\inf\{n\ge1:X_n=0\}\). Let \(V\) be the total number of visits to the origin, including time zero. Repeated conditioning at successive return times gives \(\mathbb{P}_0(V\ge k)=f^{k-1}\) for \(k\ge1\): after each return the chain starts again at the origin with the same probability \(f\) of another return. Therefore \(\mathbb{E}_0[V]=\sum_{k\ge1}f^{k-1}\), which is finite if \(f<1\) and infinite if \(f=1\). On the other hand, writing \(V=\sum_{n\ge0}\mathbf{1}_{\{X_n=0\}}\) and taking expectations gives \(\mathbb{E}_0[V]=G\). Thus recurrence, the condition \(f=1\), is equivalent to \(G=\infty\), proving the classification. Translation invariance gives the same return classification from every starting state. \(\square\)

Worked Examples

What Anisotropy Changes—and What It Does Not

The weights affect the exact return probabilities and the shape of the local quadratic form. They do not change the number of dimensions in which the Green integral is finite, provided every \(\alpha_r\) is strictly positive. The positivity assumption matters: it supplies a positive lower bound for the quadratic form in every coordinate direction. If a coordinate weight were zero, the walk would stay in a lower-dimensional sublattice, and the dimension of that accessible subspace—not the ambient dimension alone—would determine the relevant threshold.

Common mistake. Replacing the weighted symbol by the unweighted average \(\tfrac12(\cos\theta_1+\cos\theta_2)\) in the two-dimensional example gives the wrong transition model. A direct check at two steps exposes the error. To return to the origin in two steps, the walk must take opposite steps along the same axis. Axis \(1\) contributes \(2(\tfrac16)^2=\tfrac1{18}\), and axis \(2\) contributes \(2(\tfrac13)^2=\tfrac29\). Thus \(p_{00}^{(2)}=\tfrac1{18}+\tfrac29=\tfrac5{18}\). The unweighted model would instead give \(2(\tfrac14)^2+2(\tfrac14)^2=\tfrac14\). The correct Fourier symbol retains each \(\alpha_r\), even though the recurrence threshold ultimately does not depend on their particular positive values.