TITLE: Null-recurrence of a random walk
QUESTION [6 upvotes]: In a random walk on $\mathbb{Z}$ starting at $0$, with probability 1/3 we go +2, with probability 2/3 we go -1. Please prove that all states in this Markov Chain are null-recurrent.
Thoughts: it is clear all states are inter-communicating, all with periodicity 3, therefore proving state 0 is null-recurrent is enough.

null-recurrence


One lengthy solution for simple random walk and null-recurrence





Where do I get stuck with 2/3, 1/3 unsymmetric random walk?
Cannot find a series expansion that simplifies the binomial form of $P_{ij}(s)$

REPLY [2 votes]: One shall know that statement (not so obvious btw) :

for an irreducible chain $E \colon $ there exists a stationnary
distribution $\iff \exists i \in E$ such that $i$ is positive
recurrent $\iff  \forall i \in E$, $i$ is positive recurrent.

Here, I want to prove that the chain you mentionned can't have a stationary distribution. Of course, here I'm proving that the chain is not positive recurrent, but a simple argument proves that the chain is not transient.
It would be by using the difference equations. Your case is certainly not straight forward (as the length of the other answers suggest). So for the sake of understanding the method I'll simply consider the normal random walk ( $ P_{i,i+1} = \frac 1 2 = P_{i,i-1}$ ).
Let's assume that for the classic random walk, there exists a stationary distribution. Then it satisfies $$\forall j \in E :  \pi_j = \sum_{i \in E} \pi_i P_{i,j} = \frac 1 2 \pi_{j-1} + \frac 1 2  \pi_{j+1}$$
and we know how to solve such equations (difference equations). The answer is given by $$ \pi_j =  a + bj $$ with constants $a,b$. However, because of the "boundary" conditions :
$$\pi_j > 0, \forall j \implies b = 0, a > 0 $$
$$ \#E = \infty \implies \forall j, \pi_j \neq a > 0  $$
thus yielding that $ \pi $ does not exist.