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Suppose $P(x,dy)$ and $Q(x,dy)$ are two Markov transition kernels on a topological space $E$ equipped with Borel $\sigma$-algebra $\mathcal B(E)$. Suppose for every $x \in E$, $P(x,\cdot)$ and $Q(x, \cdot)$ are equivalent, i.e. for any $A \in \mathcal B(E)$ and $x \in E$, $P(x,A) > 0 \Leftrightarrow Q(x,A) > 0$.

Suppose there exists a unique measure which is invariant for $Q$ (up to multiplicative constant). I explicitly do not wish to exclude the case that this invariant measure has infinite mass.

Is it true that, up to a multiplicative constant, there exists at most one invariant measure for $P$?

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  • $\begingroup$ Can you be define "invariant measure for Q"? Is it what's usually called a stationary measure? $\endgroup$
    – Jairo Bochi
    Commented Jun 16, 2016 at 18:29
  • $\begingroup$ Yes, invariant measure = stationary measure. $\endgroup$
    – Joris Bierkens
    Commented Jun 16, 2016 at 19:18

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Since you don't want to exclude the case of infinite stationary measures, it is very easy to produce a counterexample. The simple random walk on $\mathbb Z$ has a unique stationary measure (which coincides with the counting one), whereas for $p\neq q$ the random walk with the transition probabilities $p(n,n+1)=p$ and $p(n,n-1)=q$ has two minimal stationary measures.

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  • $\begingroup$ Of course, nice example, the stationary measures being the constant distribution as well as the one for which $\mu(n+1) = \alpha \mu(n)$ for some $\alpha \neq 1$. Thanks! $\endgroup$
    – Joris Bierkens
    Commented Jun 16, 2016 at 19:28

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