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Consider a random walk on a infinitely countable connected graph. We assume that each vertex has finitely many neighbors and that we have a uniform bound of the number of neighbors at each vertex. The probability to move from x to a neighbor y of x is equal to the inverse of the number of neighbors of x.

Can we prove the existence of a invariant probability measure for the corresponding Markov chain? Under which assumptions?


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Are you saying the number of neighbors (i.e. the degree) of each vertex is the same, or just uniformly bounded? If not the same, is it random? – gt6989b Feb 21 '13 at 21:02
The degree of each vertex is uniformly bounded but not random. – Link Feb 21 '13 at 21:06
Thanks for your answer Ilya. If I take T the first hitting time of a vertex x, is the expectancy of T, with respect to the law of the markov chain started according to the invariant measure finite? Is it uniformly bounded over all vertices x? – Link Feb 21 '13 at 21:14
This law may not be a probabilistic law, is it ok? – Ilya Feb 21 '13 at 21:24
Already for the integer lattice, there is no invariant probability distribution as the markov chain is irreducible and null recurrent. The same is true for the SRW on all integer lattices. Whether this is the case for any sufficiently connected graph seems like a difficult problem. – A Blumenthal Feb 21 '13 at 21:38

With the given transition law, no invariant probability measure can exist on any connected, undirected graph with finite degrees and a countably infinite set of vertices.

Suppose an invariant probability measure did exist and assigned vertex $x$ probability $\pi_x$. Set $q_x:=\pi_x/\deg x$. Then $$\sum_x q_x=\sum_x \frac{\pi_x}{\deg x}\le \sum_x \pi_x=1.$$ Therefore, $\sum_x q_x$ must be finite, and since some $\pi_x$ must be positive, $\sum_x q_x$ is also positive.

Let $V$ be the set of vertices. According to the Markov law, since the measure is invariant, $$\sum_{y\in V} \pi_y p_{yx} =\pi_x, \qquad \text{for all } x\in V,\tag1$$ where $p_{yx}$ is the probability to make a transition to $x$, starting at $y$.

By assumption, $p_{yx}$ is $1/\deg y$ if $x$ and $y$ are neighbors and $0$ otherwise, so from $(1)$, $$ \sum_{y \text{ neighbors } x} q_y = \pi_x = q_x \deg x, \qquad \text{for all } x\in V. \tag2 $$

Suppose that there are neighboring vertices $x_0$ and $x_1$ with $q_{x_0}\ne q_{x_1}$. Without loss of generality, let $q_{x_1}>q_{x_0}$. Then from $(2)$, $q_{x_1}$ is the average of $q_y$ over the neighbors $y$ of $x_1$, so there must be some neighbor $x_2$ of $x_1$ with $q_{x_2}>q_{x_1}$. Applying this reasoning repeatedly, we get a sequence $x_0$, $x_1$, $x_2$, $\dots$ of vertices, necessarily distinct, with $q_{x_{i+1}}>q_{x_i}$ for all $i$. This is impossible as then $\sum_i q_{x_i}=\infty$.

So, any pair of neighboring vertices $x$ and $y$ must have $q_x=q_y$. Since the graph is connected, this means that $q_x$ is constant at some value, say $Q$. This is also impossible as then either $Q=0$ and so $\sum_x q_x=0$ or $Q>0$ and so $\sum_x q_x=\infty$. Therefore there is no invariant probability measure.

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Thanks for the answer David! – Link Feb 21 '13 at 21:59
The very end is a little odd. One could go back to the invariant measure $\pi$, noting that $\pi_x=Q\mathrm{deg}\, x$ for every $x$, with $Q\ne0$ since $\pi\ne0$, hence the total mass of $\pi$ is $Q\sum\limits_x\mathrm{deg}\,x$, which is infinite for every $Q\ne0$ (the fact that $\sum\limits_xq_x$ cannot be nonzero and finite is not a contradiction in itself). – Did Feb 21 '13 at 22:03
Yes, I made a mistake there. I think that the proof is now correct. – David Moews Feb 21 '13 at 22:26
@DavidMoews Nice answer. I hope you don't mind, I "tagged" your equation numbers. I think it looks a bit better. – Byron Schmuland Feb 22 '13 at 3:03

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