Let $\mathbb{S}$ be countable and $X_n:\mathbb{S}^\mathbb{N} \rightarrow \mathbb{S}$ be the coordinate map, i.e., $X_n(\omega) = \omega_n$. Let $p:\mathbb{S}\times\mathbb{S} \rightarrow [0,1]$ be an irreducible transitional probability so that $X_n$ is a Markov chain and let $\pi\in (0,\infty)$ be a reversible measure with respect to $p$. Let $T^x = \inf \{n \ge 1: X_n = x\}$.


In this case, is there a relation between $P^x(T^y<\infty)$ and $P^y (T^x<\infty)$?

Some thought process:

I initially thought that it would satisfy $\pi(x) P^x(T^y<\infty) = \pi(y) P^y(T^x<\infty)$, since if I took a path $x=x_0,x_1,...,x_n = y$ for $x_1,...,x_{n-1} \ne y$ and computed $$ \pi(x) \prod_{i=1}^n p(x_{i-1},x_i) = \pi(y) \prod_{i=1}^n p(x_i,x_{i-1}) $$ But then I realized that $x_i = x$ is possible for some $1 \le i <n$ so that $x=x_0,x_1,...,x_n=y$ is a path from $x$ to the first time hitting $y$, while the reverse path $y=x_n,x_{n-1},...,x_0=x$ is not a path from $y$ to the first time hitting $x$.


A comment that I write here for lack of reputation: Note that the event $(T^x < \infty)$ means " do we ever hit x" and does not really break your argument. The only thing that changes is that you now have a bijection of paths with different hitting times but since you sum over all times anyway it does not matter.

  • $\begingroup$ I don't quite see why the statement is true. Consider a birth-death chain on $\{0,1,2\}$ with $p(0,1)=1,p(1,0)=p(1,2)=1/2$ and $p(2,1)=1$. Then $\pi(0)=1,\pi(1)=2,\pi(2)=1$ is a reversible measure and $P^1(T^2<\infty) = P^2(T^1<\infty) = 1$. $\endgroup$ May 19 '18 at 3:42
  • $\begingroup$ Sry I was not clear. The counter you raise is not valid. If the equation you gave were true then summing over all such paths works since hitting time is finite in both paths. $\endgroup$ May 19 '18 at 7:57
  • $\begingroup$ It seems that you are right. Looking at individual paths is useless since every single one has probability 0 and there are uncountably many of them. Instead we have to look at sets of paths with a certain start. The most natural partitioning of the sets would then be by first hitting time "x to y". However simply reversing the beginning sequence does not lead to disjoint sets precisely because of the point you made, thus leading to overcounting. The relation between $P^x(T^y < \infty )$ and $P^y(T^x < \infty )$ is simply that they are both equal to 1 by irreducibility. $\endgroup$ May 19 '18 at 19:12
  • $\begingroup$ The probability of a single (finite) path is usually nonzero, and there are only countable of them (in countable $\mathbb{S}$. For example, the path $0\rightarrow 1 \rightarrow 2$ has probability $1/2$. $\endgroup$ May 20 '18 at 7:07
  • $\begingroup$ Also, $P^x(T^y<\infty)$ is not necessarily 1 for merely irreducible $p$. A sufficient condition for that to be true would be something like $p$ is recurrent, but I'm not assuming that. $\endgroup$ May 20 '18 at 7:09

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