Logarithmic Green-Function Divergence in Two Dimensions

Return Paths and the Time Index

For a random walk, recurrence is determined by how often the walk returns to its starting point. In two dimensions, the probability of a return at a particular even time becomes small, but not quickly enough for the sum of return probabilities to be finite. Counting the paths gives an exact formula; its asymptotic form then reveals the logarithmic divergence of the Green function.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((X_n)_{n\ge0}\) be the simple symmetric random walk on \(S=\mathbb{Z}^2\), started at \(X_0=(0,0)\). At each step it moves by one of \((1,0),(-1,0),(0,1),(0,-1)\), each with probability \(1/4\). Thus, from every state, the four transition probabilities are \(1/4\), and their sum is \(4(1/4)=1\). Let \(\mathcal{F}_n=\sigma(X_0,\ldots,X_n)\), and define \(p_m(0)=\mathbb{P}(X_m=(0,0))\).

A parity constraint comes first. Each step changes the parity of the sum of the two coordinates, so the walk can be at the origin only at an even time. To count even-time returns, transform a step \((\Delta x,\Delta y)\) into \((\Delta u,\Delta v)=(\Delta x+\Delta y,\Delta x-\Delta y)\). A step to the right, left, up, or down becomes, respectively, \((1,1),(-1,-1),(1,-1),(-1,1)\). These are the four equally likely pairs of signs. Consequently the two transformed coordinates have independent fair sign increments at each step.

Proof. Under the coordinate transformation, \(X_m=(0,0)\) exactly when both transformed coordinates are zero. Each transformed coordinate is the sum of \(m\) independent signs, each equally likely to be \(1\) or \(-1\). If \(m\) is odd, such a sum cannot be zero. If \(m\) is even, say \(m=2n\), a coordinate sum is zero exactly when \(n\) signs are \(1\) and \(n\) are \(-1\). There are \(\binom{2n}{n}\) such sign sequences among \(2^{2n}\) equally likely sequences, so the probability for one transformed coordinate is \(\binom{2n}{n}/2^{2n}\). The two transformed coordinate sequences are independent: at each step every pair of signs has probability \(1/4\), and the steps are independent. The probability that both sums are zero is therefore the square of the one-coordinate probability. This proves the formula. \(\square\)

The index in the formula matters. In terms of an even time \(m\), the binomial coefficient is \(\binom{m}{m/2}\). If the time is written as \(2n\), it is \(\binom{2n}{n}\). The alternative expression \(\left[\binom{n}{n/2}/2^n\right]^2\) describes \(p_n(0)\) when \(n\) is even, not \(p_{2n}(0)\).

Logarithmic Growth of the Green Function

The exact formula makes it possible to see not just that returns occur, but how their probabilities add up. The relevant asymptotic scale comes from the central binomial coefficient. We use Stirling’s formula in its standard form \(n!=\sqrt{2\pi n}(n/e)^n(1+O(1/n))\) as \(n\to\infty\); its proof is part of standard asymptotic analysis, not needed for the path-counting argument here.

Proof. Stirling’s formula applied to \((2n)!\) and \(n!\) gives $$ \frac{\binom{2n}{n}}{2^{2n}}=\frac{1}{\sqrt{\pi n}}\left(1+O\!\left(\frac1n\right)\right). $$ Squaring this expression and using the exact return formula yields \(p_{2n}(0)=\frac{1}{\pi n}(1+O(1/n))=\frac{1}{\pi n}+O(1/n^2)\). Since the odd-time return probabilities vanish, the partial Green sum through \(N\) is \(1+\sum_{n=1}^{\lfloor N/2\rfloor}p_{2n}(0)\). The sum of the error terms is bounded: \(\sum_{n\ge1}1/n^2\) converges. For \(M\ge1\), the harmonic sum \(H_M=\sum_{n=1}^M1/n\) satisfies \(\log(M+1)\le H_M\le1+\log M\), by comparing the sum with the integral of \(1/x\). Therefore \(H_M=\log M+O(1)\). Taking \(M=\lfloor N/2\rfloor\) gives \(\log M=\log N+O(1)\), so \(G_N(0,0)=\frac1\pi\log N+O(1)\). This tends to infinity, proving the claimed divergence. \(\square\)

The coefficient \(1/\pi\) describes the leading growth when time is counted through \(N\), including the zero contributions from odd times. The sum grows more slowly than a positive power of \(N\), but it is unbounded: each even-time return probability is of order \(1/n\), and the harmonic series diverges.

Why Divergence Means Recurrence

For this walk, define the first return time \(T_0^+=\inf\{n\ge1:X_n=(0,0)\}\), with \(\inf\varnothing=\infty\), and let \(f=\mathbb{P}_0(T_0^+<\infty)\). The origin is recurrent when \(f=1\): starting there, the walk returns with probability one. The following argument connects this definition directly to the Green function, including the case \(f=1\).

Proof. Let \(R_0=0\), and for \(k\ge1\) let \(R_k\) be the time of the \(k\)-th return to the origin, setting \(R_k=\infty\) if fewer than \(k\) returns occur. The event \(\{R_k<\infty\}\) means that there are at least \(k\) returns. At each finite return time, the discrete-time Markov property implies that the future walk has the law of a fresh walk started at the origin. To see why it applies at a return time \(T\), split any event concerning the history and future according to \(T=r\); on each event \(\{T=r\}\), the ordinary Markov property at deterministic time \(r\) gives the transition law from \(X_r=(0,0)\). Thus the probability of another eventual return, conditional on the history up to a finite return, is \(f\). Induction gives \(\mathbb{P}_0(R_k<\infty)=f^k\), including \(k=0\), for which the probability is \(1\). Let \(V=\sum_{m\ge0}\mathbf{1}_{\{X_m=(0,0)\}}\) be the total number of visits, including time \(0\). Since all terms are nonnegative, summing expectations gives \(\mathbb{E}_0[V]=\sum_{m\ge0}p_m(0)=G(0,0)\). Also \(V=\sum_{k\ge0}\mathbf{1}_{\{R_k<\infty\}}\): both sides count the initial visit and each subsequent return. Taking expectations again for nonnegative terms yields \(G(0,0)=\sum_{k\ge0}f^k\). If \(f<1\), this geometric series is \(1/(1-f)<\infty\); if \(f=1\), every term is \(1\) and the series diverges. Therefore \(G(0,0)=\infty\) exactly when \(f=1\). \(\square\)

Combining this criterion with the logarithmic divergence proves that the two-dimensional simple symmetric random walk returns to the origin with probability one. This is a statement about eventual return, not a claim that the walk stays near the origin or that it returns at every time. In fact, its probability of being at the origin at any particular odd time is zero.

A Common Indexing Pitfall

A common error is to substitute \(n\) for the time in the even-time formula without changing the binomial coefficient. For example, the incorrect expression \(p_{2n}(0)=[\binom{n}{n/2}/2^n]^2\), used at \(n=2\), gives \(p_4(0)=[\binom21/2^2]^2=(2/4)^2=1/4\). But the correct count at four steps is \(p_4(0)=[\binom42/2^4]^2=(6/16)^2=9/64\). The discrepancy comes from confusing the time \(m\) with the half-time index \(n\). A reliable check is to count the sign sequences for one transformed coordinate: at time \(4\), there are \(\binom42=6\) returning sequences out of \(2^4=16\), not \(2\) out of \(4\).

The proof uses only the return probabilities and their sum. It does not require a formula for the distribution of the first return time. The distinction matters: \(p_m(0)\) is the probability of being at the origin at time \(m\), while \(T_0^+\) records the first positive time of return. The Green-function argument relates these quantities by counting all visits, rather than by identifying them.