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I apologize if this is to specific but i've already talked to two of my professors without much success and I really need to understand this subject. The following theorem is stated in Durrett page 254

Let x be a recurrent state, and let $T=\inf \{n\geq 1: X_n =x\}$ then $\mu _x (y) = \sum_{n=0} ^\infty P_x (X_n =y, T > n)$ defines a stationary measure.

He proves it and then writes a "technical note" saying that "To show we are not cheating, we should prove that $\mu _x (y)<\infty$". Why do we need that? One of my professors (who is usually right about things) talked about it being impossible for a stationary measure to give infinite mass to a point (I'm not quite sure what mass means in that context), but I really can't find any reason (in the book anyway) for that to be true.

I could really need to ask a few short questions to a person familiar with Durrett and his chapter about Markov chains.

p.s. I have the relevant pages as a pdf which I can supply if anyone would like.

Thanks in advance,

Henrik

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    @did That was a pedagogical question, I know the answer ... Anyway, I was thinking that the fact a stationnary measure doesn't give infinite mass to a singleton was part of the definition, but apparently this is not explicitely said in Durrett's book.2012-09-06

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Let $\nu$ denote a stationary measure. If $\nu(y)$ is infinite for some state $y$, then, for every state $z$ such that $p(y,z)\ne0$, the inequality $\nu(z)\geqslant\nu(y) p(y,z)$ shows that $\nu(z)$ is infinite as well. If the Markov chain is irreducible, this proves that $\nu(z)$ is infinite for every state $z$. This measure $\nu$ is stationary but not very interesting. What Durrett says is that, to prove that the fancy formula he proposes for $\mu_x$ does not lead to an empty statement, one should check that $\mu_x(y)$ is finite for some state $y$ (or, equivalently when the chain is irreducible, for every state $y$).

(The mass of a measure $\nu$ at a point $y$ is $\nu(\{y\})$, often denoted by $\nu(y)$ when $\nu$ is discrete, by an abuse of notation.)

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    Pretty much: measures instead of vectors, kernels instead of matrices, integration instead of matrix-multiplication, *et voilà !*2012-09-06
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Since x is a recurrent state it occurs infinitely often. Does Durrett say whether or not x is positive recurrent? If it is positive recurrent the average return time is finite. If it is null recurrent the average return time is infinite even though the state is recurrent. In the positive recurrent case, T is guaranteed to be finite. So there will only be some finite number of values for n that will be less than T. For each nn) is the probability of transitioning from y to x in K$_n$ steps. This is a finite probability.

for each n P$_x$(X$_n$=y, for T>n)=∑P$_x$(X$_n$=y, k>n|T=k) P(T=k) where the sum is taken over k>=n+1. At this point I think positive recurrence is necessary and should imply that P(T=k) goes to 0 quickly enough as k approaches infinity so that the sum on the right hand side of the equation is finite and approaches 0 as n approaches infinity.

You need P$_x$(X$_n$=y, for T>n) to go to 0 sufficiently fast so that the sum that produces μ$_x$(y) is finite.

This is sketchy I know but I hope that someone else can pick it up and fill in the details for you. I think that presenting this at least gives you an idea as to why the measure is finite and also that if certain quantities do not go to 0 sufficiently fast (as I think would happen in the null recurrent case) the measure could be infinite.

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    I don't think I can help with that.2012-09-06