Markov Chain on an infinite binary tree

Question

Consider an infinite binary tree in which its vertices can be represented as finite strings of 0's and 1's. Let $\emptyset$ denote the top vertex aka the "root" of the tree. Let $0<p<\frac{1}{2}$ and define the transition probability as follows:

At each vertex, it has probability $p$ to move to one of its two children vertices, and probability $1-2p$ to move to its parent vertex. Since $\emptyset$ has no parent we let $p(\emptyset, \emptyset)=1-2p$.

For which values of $p$ is this chain recurrent, positive recurrent, or null recurrent?

EDIT: Some intuition now that I have given it more thought: For any node, consider only its "depth", aka $\emptyset$ has depth $0$, $0$ and $1$ has depth $1$, $10$, $11$,$01$, and $00$ has depth $2$, and so on. Then this tree simplifies down to a random walk on the number line $\mathbb{N}$. We know that such walk is transient if $p(x, x+1)>\frac{1}{2}$, null recurrent if $p(x, x+1)=\frac{1}{2}$, and positive recurrent if $p(x, x+1)<\frac{1}{2}$.

Translating this to the tree, it would be transient if $2p>\frac{1}{2}$, null recurrent if $2p=\frac{1}{2}$, and positive recurrent if $2p<\frac{1}{2}$. Is this the correct thought process? If so how can I formalize it?

Answer 1

Yes, this approach is correct (and is exactly what I was going to write as an answer before I read the “EDIT” part of the post).

To formalize it, you can form equivalence classes of states according to their depth in the tree and consider the “quotient chain” defined by

$$ \mathsf P([i]\to[j])=\sum_{k\in[j]}\mathsf P(i\to k)\;. $$

This is well-defined if the sum on the right-hand side is independent of the representative $i$ of the class [i], which it clearly is in the current case. This quotient chain is equivalent to a random walk on $\mathbb N$ as you describe.

Now, this by itself does not prove that the original chain has the same recurrence as the quotient chain. For instance we could define a chain on a two-dimensional square grid which is recurrent along one axis and transient along the other, and therefore transient as a whole; whereas if we quotient out the transient direction, the quotient is recurrent.

So we need one more piece in the argument. This is that $\emptyset$ is in a class by itself. That is, whenever the quotient chain is in state $[\emptyset]$, the original chain is in state $\emptyset$. Thus, state $\emptyset$ in the original chain has the same recurrence as state $[\emptyset]$ in the quotient chain. Since the chain is clearly irreducible (in fact this is already assumed in the question, as otherwise it wouldn’t make sense to ask about the recurrence of the entire chain, rather than of individual states or components), the recurrence of the entire chain is that of any state, and thus that of $\emptyset$.