Simple random walk on the 3-1 tree is recurrent Hello guys,
There is an infinite tree structure named "the 3-1 tree", denoted by $T_{3-1}$. The tree is constructed as follows:
The origin vertex (which can be referred to as the zeroth level) has two sons. In each level $n$ we have $2^n$ vertices. In order to construct the next level, we split the $2^n$ vertices into two groups (the left half which consists of $2^{n-1}$ vertices and the right half which consists of the other $2^{n-1}$ vertices). From each left vertex we have three edges leaving it (therefore it has three sons) and for each right vertex we have only one edge leaving (therefore it has only one son). This way, in every level there are indeed $2^n$ vertices and a total of $2^{n+1}$ edges leaving that level. This is the structure.
It is a known fact that simple weighted random walk (all edges have weight 1, i.e. $c(x,y)=1$ $\forall$ $x,y\in T_{3-1}$) on $T_{3-1}$ is recurrent. I would like to prove this using simple tools such as electrical networks, martingales and standard probability tools.
Thank you very much!
 A: For any vertex $v$ which is not in the all-left ray, there is some generation $n$ so that all descendants of $v$ in the $n$th generation are in the right half. (If the fraction of vertices to the left of $v$ in its level is $\alpha$, this happens by $\lceil \log_{3/2} 1/(2\alpha) \rceil$ generations below $v$, since the fraction of vertices to the left expands by a factor of $3/2$ until the generation when it reaches $1/2$.) This means the number of descendants of $v$ in each generation is eventually constant, so the resistance in the downward direction is infinite. Therefore, with probability $1$ a random walk starting at $v$ will reach the parent of $v$. 
So, a random walk on the $3-1$ tree almost surely retracts to a random walk on the all-left ray. Since this random walk is recurrent, the random walk on the $3-1$ tree is recurrent.
A: You can find this example discussed also in my book with Russ Lyons, "Probability on Trees and networks", http://mypage.iu.edu/~rdlyons/prbtree/prbtree.html
As Douglas points out, For any vertex $v$
 which is not in the all-left ray, there is some generation $n$ so that all descendants of $v$ in the $n$th  generation are in the right half. Thus a flow of finite energy from the root cannot assign any positive mass to an edge off the left ray, since some of this mass has to eventually go down an infinite ray with no branching which yields infinite energy. But a flow confined to the leftmost ray certainly has infinite energy.  
A: I was trying to solve this problem by showing that every flow from $o$ to $\infty$ 
on this graph must have infinite energy. 
Therefore, by Lyons criterion the graph must be recurrent.
But I didn't succeed finishing the proof.
Does any of you guys have suggestions?
Thanks
