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Is the random walk on $\mathbf{Z}^2$ started at $\left\lbrace \left(0,0\right) \right\rbrace $, in which the probability of advance to right is $p$, the probability of retreat to left is $p$, the probability of advance to up is $q$ and the probability of retreat to down is $q$, with $p \not = q$, recurrent?

Let $\left( S_n\right) $ the random walk, then the random walk is recurrent if and only if $\sum_{n=1}^{\infty}P(S_{2n}=\left(0,0 \right) )=+\infty$

$P(S_{2n}=\left(0,0 \right) )=\sum_{j=0}^{n}\frac{\left( 2n\right)! }{j!j!\left(n-j \right)! \left(n-j \right)! }p^{2j}q^{2n-2j}=q^{2n} {2n \choose n}\sum_{j=0}^{n}{n\choose j}^2 \left(\frac{p^{2}}{q^{2}} \right)^{j} $

is

$\sum_{n=1}^{\infty}{q^{2n} {2n \choose n}\sum_{j=0}^{n}{n\choose j}^2 \left(\frac{p^{2}}{q^{2}} \right)^{j}}=+\infty$ ?

Did
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Carlos Andres Henao Acevedo
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  • Do you mean that with probability $p$ you move left, with probability $p$ you move right and with probability $1-2p$ you don’t move? Otherwise what happens if left and right are both decided? – adfriedman Aug 30 '18 at 03:18
  • if we only see the movements to the right and left, it is clearly recurrent, but this is on the line $x = 0$. – Carlos Andres Henao Acevedo Sep 01 '18 at 03:37

1 Answers1

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In $\mathbb Z$ and $\mathbb Z^2$, a finitely supported probability measure $\mu$ yields a recurrent random walk if and only if it is centered, meaning that $$\sum_{x\in \mathbb Z^d}x\mu(x)=0.$$

This is usually proved using Fourier transforms. Indeed, you can prove that the $n$th power convolution of $\mu$ satisfies the local following local limit theorem: $$\mu^{*n}(e)\sim CR^{-n}n^{d/2}$$ and $R=1$ if and only if $\mu$ is centered.

Actually, $R$ is the inverse of the minimum of the function $$\phi(u)=\sum_{x\in \mathbb Z^d}\mu(x)\mathrm{e}^{u\cdot x}.$$ This function is strictly convex and it reaches its minimum where its gradient vanishes. You can check that the gradient is given by $$\nabla \phi(u)=\sum_{x\in \mathbb Z^d}x\mu(x)\mathrm{e}^{u\cdot x}.$$ Thus, the minimum is reached at 0 if and only if $$\sum_{x\in \mathbb Z^d}x\mu(x)=0$$ that is $\mu$ is centered. In other words, the minimum is equal to $\phi(0)=1$ if and only if $\mu$ is centered.

Now, if $d\geq 3$, you see that $\mu^{*n}(e)$ is always summable, whether $R=1$ or not, but if $d\leq 2$, then $\mu^{*n}(e)$ is summable if and only if $R>1$, if and only if $\mu$ is non centered.

Finally, summability of $\mu^{*n}$ is by definition equivalent to finiteness of the Green function, which is equivalent to the random walk being transient.

For all the details I have not proved, see for example Woess's book random walks on infinite graphs and groups, in particular Chapter III.13 for local limit theorems on $\mathbb Z^d$.

M. Dus
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