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I've been told the following result:

Prop. Let $X$ be a markov chain on $S$ (not necessarily finite) and transistion matrix $P=(p_{i,j})_{i,j\in S}$. Then if for some $i\in S$ we have that for all $j\in S$, $p^{(n)}_{i,j}\rightarrow \pi_j$ as $n\rightarrow \infty$, then $(\pi_j)_{j\in S}$ in an invariant measure (i.e. $\pi P=P$). Moreover if $S$ is finite then it's not only a measure but a probability distribution.

Well, if $S$ is finite then everything is easy to me, but i cannot understand the case in which $S$ is not finite. Indeed the argument that i came up with is the following:

$$\pi_j=\displaystyle \lim_{n\rightarrow \infty} p_{i,j}^{(n)}=\lim_{n\rightarrow \infty} \sum_{k\in S}p_{i,k}^{(n-1)}p_{k,j}=\sum_{k\in S} \lim_{n\rightarrow \infty} p_{i,k}^{(n-1)}p_{k,j}=\sum_{k\in S} \pi_{k}p_{k,j} $$

The problem here is the third equality; if $S$ in not finite then i'm exchanging the limit with a series and i don't know how to justify it. I also think that in this setting monotone/dominated convergence theorems cannot help beacause i'm not able to see that series as integral w.r.t some measure on which i can apply those theorems.

So is this result correct but require a different proof? Or is the result not correct for $S$ countable? Or is the limit exchange i pointed out correct and it's just me that i can't understand it?

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  • $\begingroup$ That sum is just an integral with respect to the counting measure on $S$. $\endgroup$ – Rhys Steele Jun 12 at 9:26
  • $\begingroup$ ok, i can see this but using the counting measure i cannot apply dominated convergence or monotone convergence (the sequence is not monotone and i can't find an integrable dominating function) $\endgroup$ – StabiloBoss Jun 12 at 10:09
  • $\begingroup$ All terms are nonnegative. So if you sum over only finitely many terms but then take a limit of the result you get $\pi\geq\pi P$. $\endgroup$ – Michael Jun 12 at 11:02
  • $\begingroup$ ok @Michael that's correct, but still not enough to have the equality that is needed. $\endgroup$ – StabiloBoss Jun 12 at 12:03
  • $\begingroup$ You can then prove equality: Imagine that one of the constraints is strict. But now sum over all $j$. Equivalently, multiply both sides on the right by the all-1 vector. You may want Fubini-Tonelli to justify switching sums of nonnegative terms. $\endgroup$ – Michael Jun 12 at 14:28
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The result is correct. Let $S = \{1, 2, 3, ...\}$ be countably infinite.

1) First prove that $$ \sum_{j \in S} \pi_j \leq 1$$

2) Next prove that $\pi \geq \pi P$ (via the hint in my comment above), equivalently:

$$ \pi_j \geq \sum_{k \in S} \pi_k P_{kj} \quad \forall j \in S \quad (Eq. 1)$$

3) Suppose (Eq. 1) holds with strict inequality for at least one $j\in S$. Summing over all $j \in S$ gives $$ \sum_{j \in S} \pi_j > \sum_{j \in S} \sum_{k \in S} \pi_k P_{kj}$$ where we have used the fact that the sums are finite to eliminate the $\infty \geq \infty$ case. Switch the sums of the nonnegative terms (Fubini-Tonelli) to reach a contradiction.

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  • $\begingroup$ PS: I suppose (Eq. 1) could be justified by Fatou's lemma but it is more basic to justify via $$ P_{ij}^{(n)} = \sum_{k=1}^{\infty} P_{ik}^{(n-1)}P_{kj} \geq \sum_{k=1}^m P_{ik}^{(n-1)}P_{kj}$$ $\endgroup$ – Michael Jun 12 at 14:54

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