Given disjoint nonempty subsets $X_1, X_2$ of the state space of a finite irreducible Markov chain, there are unique measures $\mu_1$ on $X_1$ and $\mu_2$ on $X_2$ such that (a) starting from a $\mu_1$-random point $x$, if one walks until one first hits $X_2$ at some point $y$, the point $y$ is distributed according to $\mu_2$, and (b) vice versa (with the roles of $X_1$ and $X_2$ reversed). E.g., if the Markov chain is (unbiased) random walk on $Z/5Z$ and $X_1$ is {0,1} and $X_2$ is {3,4}, then $\mu_1(0)=\mu_2(4)=2/3$ and $\mu_1(1)=\mu_2(3)=1/3$. Is there a way to think about these measures in terms of standard constructions in discrete probability theory?
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asked Feb 17, 2011 at 22:26
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$\begingroup$ I take it you are looking at finite state space? And irreducible? In which case you can look at the sequence of hitting times $T_1,T_2,\ldots$ of, alternately, $X_1$ and $X_2$. Then, if $M_t$ is your markov chain, $M_{T_k}$ is an irreducible chain on $X_1\cup X_2$ with unique stationary distribution $(\mu_1+\mu_2)/2$. $\endgroup$– George LowtherCommented Feb 17, 2011 at 22:59
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$\begingroup$ I added the words "finite" and "irreducible". Thanks, George. George's comments made me realize that my problem has a simple answer: Just create a new Markov chain with one or two states corresponding to each state of the original chain, according to whether the walk's most recent visit to $X_1 \cup X_2$ visited $X_1$ or $X_2$. Then $\mu_1$ and $\mu_2$ are stationary measures for return-maps of this chain. Sorry not to have thought harder before posting... $\endgroup$– James ProppCommented Feb 18, 2011 at 1:40
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Just create a new Markov chain with one or two states corresponding to each state of the original chain, according to whether the walk's most recent visit to $X_1 \cup X_2$ visited $X_1$ or $X_2$. $\mu_1$ and $\mu_2$ are stationary measures for return-maps of this chain.
answered Feb 18, 2011 at 1:42