TITLE: Nonmonotonicity of expected distance of a random walk
QUESTION [5 upvotes]: What's the simplest example of a reversible random walk $X_n$ on an infinite vertex-transitive graph such that the expected distance from the origin is not increasing, i.e. there exists $n$ such that:
$\mathbb{E} \,d(X_0, X_{n+1}) < \mathbb{E}\,d(X_0, X_n)$?
EDIT: Douglas Zare provided an example below, but the random walk is not simple (we are giving bigger weight to one of the outgoing edges) and the nonmonotonicity is essentially due to a parity issue, which is not that interesting.
So let me reformulate the question: does there exist a simple random walk on an infinite vertex-transitive graph (preferably: a Cayley graph) where nonmonotonicity would not arise due to some periodicity issue? The phenomenon I have in mind is that some graphs have "dead ends", i.e. vertices from which every step takes the random walker closer to the origin (equivalently: geodesics to such vertices cannot be extended). But I don't know whether this obstruction can be serious enough so that on average the distance will be nonmonotonous.

REPLY [3 votes]: Here are two examples with bipartite graphs.
Let the vertices be the integers. Take steps of $\pm(2n+1)$ with probability $2^{-n-2}$. This is bipartite and all vertices are connected to each vertex in the other part. At odd time steps, the distance from the origin is always $1$. At even time steps, the distance from the origin is usually $2$, with a small probability of $0$. For example, $\mathbb{E}d(x_0,x_2) = 1/6(0)+5/6(2) = 5/3 \gt \mathbb{E}d(x_0,x_3) = 1$ so the third step doesn't increase the average distance.
Let the vertices be $\lbrace n \rbrace_{n\in \mathbb Z} \cup \lbrace n' \rbrace_{n\in \mathbb Z}$. Let $n$ move to $n'$ with probability $p$ close to $1$, and to $(n-1)'$ or $(n+1)'$ with low probability. Symmetrically, let $n'$ move to $n$ with probability $p$ and to $(n-1)$ or $(n+1)$ with low probability. Taking two steps usually returns to the origin. $\mathbb{E}d(x_0,x_1) = 1$. $\mathbb{E}d(x_0,x_2) \le 2(1-p^2)$ which is less than $1$ for $p \gt 1/\sqrt{2}$ so the second step doesn't increase the average distance.