# Successful couplings and total variation convergence to equilibrium for time-homogeneous Markov processes

Let

• $$(\Omega,\mathcal A,\operatorname P)$$ be a probability space
• $$I=\mathbb N_0$$ or $$I=[0,\infty)$$
• $$(E,\mathcal E)$$ be a measurable space
• $$\mu$$ and $$\nu$$ be probability measures on $$(E,\mathcal E)$$
• $$X$$ be an $$(E,\mathcal E)$$-valued time-homogeneous Markov chain on $$(\Omega,\mathcal A,\operatorname P)$$ with transition semigroup $$(\kappa_t)_{t\in I}$$

Assume $$\mu$$ is invariant with respect to $$(\kappa_t)_{t\in I}$$ and $$\left|\mu-\nu\kappa_t\right|\xrightarrow{t\to\infty}0\tag1,$$ where the left-hand side denotes the total variation distance of $$\mu$$ and $$\nu\kappa_t$$ and $$\nu\kappa_t$$ denotes the composition of $$\nu$$ and $$\kappa_t$$.

I want to show that there is a version $$X^{(\eta)}$$ of $$X$$ with $$\operatorname P\left[X^{(\eta)}_0\in\;\cdot\;\right]=\eta$$ for $$\eta\in\left\{\mu,\nu\right\}$$ with $$\tau:=\inf\left\{t\in I:X^{(\mu)}_s=X^{(\nu)}_s\text{ for all }s\in I\text{ with }s\ge t\right\}<\infty\;\;\;\operatorname P\text{-almost surely}.\tag2$$

As pointed out by E-A, the basic idea is the following: For any $$t\ge0$$, there is (see Theorem 2.12) a probability measure $$\eta_t$$ (called a coupling of $$\mu$$ and $$\nu\kappa_t$$) on $$(E\times E,\mathcal E\otimes\mathcal E)$$ with $$\eta_t(B\times E)=\mu(B)\text{ and }\eta_t(E\times B)=(\nu\kappa_t)(B)\;\;\;\text{for all }B\in\mathcal E\tag3$$ and $$\left|\mu-\nu\kappa_t\right|=\eta_t(\Delta^c),\tag4$$ where $$\Delta:=\left\{(x,x):x\in E\right\}.$$

Clearly, if we would be able to show (are we?) that there is a $$(\mathcal A,\mathcal E\otimes\mathcal E)$$-measurable $$Y_t:\Omega\to E\times E$$ with distribution $$\eta_t$$ under $$\operatorname P$$, we would obtain $$\eta_t(\Delta^c)=\operatorname P\left[(Y_t)_1\ne(Y_t)_2\right]\tag5.$$ However, even when we can show this, why can we choose $$(Y_t)_{t\ge0}$$ such that $$X^{(\mu)}:=((Y_t)_1)_{t\ge0}$$ and $$X^{(\nu)}:=((Y_t)_2)_{t\ge0}$$ are versions of $$X$$?

Let me try this:

Coupling $$\rightarrow$$ TV goes to 0:

We can argue that $$P(\tau > t) \geq P(X_t \not = Y_t)$$, since if $$X_t$$ is not equal to $$Y_t$$, then $$\tau$$ has to be greater than t. We also know that $$P(X_t \not = Y_t)$$ is an upper bound on the TV (or rather, the TV is a lower bound on this quantity), since the $$P(X_n \not =Y_n) \geq | \mu(A) - \nu(A) |$$ for any $$A$$. (You can find the proof in a lot of sources; if I were to summarize it, you should write $$P(X_t \in A) = P(X_t = Y_t, X_t \in A) + P(X_t \not = Y_t, Y_t \in A)$$, and do the same for $$Y_t$$; the first terms will cancel. http://websites.math.leidenuniv.nl/probability/lecturenotes/CouplingLectures.pdf; you can also graphically observe what the total variation distance is by drawing densities together and you note that it is the half of the not-overlapping area, and that gives you a pictorial proof of this statement, since you know that you can't set $$X_t = Y_t$$ in any of those areas.)

TV goes to 0 $$\rightarrow$$ coupling:

Let us construct our coupling in the most natural sense: let $$X_t = Y_t$$ once $$X_{t_0} = Y_{t_0}$$ for some $$t_0$$. We need to start by arguing for $$P(X_t \not = Y_t | X_0, Y_0) \leq 1-\epsilon$$ for some $$t, \epsilon$$. Now, we know that we can construct a coupling such that $$P(X_t \not = Y_t) = TV(\mu, \nu_t)$$ for some fixed $$t$$. (Proof is in the pdf posted above; Theorem 2.12; you can also argue it from the picture again). Since the TV is going to 0, you can find a $$t, \epsilon$$ such that the claim is true. Now, you can repeat the same argument and invoke the time homogeneity of your Markov process to show that $$P(X_{2t} \not = Y_{2t} | X_t \not = Y_t) \leq 1 - \epsilon$$. Now, you can inductively show a geometric decay $$P(X_{2t} \not = Y_{2t}) \leq (1 - \epsilon)^k$$, which is enough to show that this time is almost surely finite. (Adapted this proof from https://www2.cs.duke.edu/courses/fall15/compsci590.4/slides/lec5.pdf; Section 4, Theorem 1, I might have messed up in certain places)

Feel free to point things out if there is anything wrong/unclear/unjustified. The first one is I think OK (it is a fairly standard proof); the second one involves constructing the coupling, so not sure if all of my steps were completely justified)

• Sorry for the late response. Can you tell me how $\hat X$ and $\hat X'$ in Theorem 2.12 of the first PDF are defined? – 0xbadf00d Jan 2 at 0:02
• The first direction is now clear to me. Thank you very much. However, I'm struggling with the other direction. I guess $X$ is supposed to be $X^{(\mu)}$ and your $Y$ is supposed to be $X^{(\eta)}$, right? The first question I have is why $\hat X$ and $\hat X'$ (from my former comment) even exist. The second one is: Why are $X^{(\mu)}$ and $X^{(\nu)}$ versions of $X$? (I've upvoted your answer and will accept it, once we've (hopefully) solved the remaining questions.) – 0xbadf00d Jan 3 at 0:12
• Still interested in an answer. Could you take a look at my comments? – 0xbadf00d Jan 11 at 18:02
• Yes; feel free to re-label them. Their existence is essentially proof of Theorem 2.12 in that PDF; in your notation $X$ and $\hat{X}$ are the same map from $\Sigma \rightarrow E$. It's just that one of them is defined with the measure $\mu$ and the other one is defined with $\nu$; the underlying space is otherwise the same. This also hopefully answers the second question too (they are fundamentally the same map). Feel free to ask a separate question about this though; I am not sure how technical of an answer I can give. – E-A Jan 12 at 21:46