Krdytkyu

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 $(kappa_t)_{tin I}$
  

Assume $mu$ is invariant with respect to $(kappa_t)_{tin I}$ and $$left|mu-nukappa_tright|xrightarrow{ttoinfty}0tag1,$$ where the left-hand side denotes the total variation distance of $mu$ and $nukappa_t$ and $nukappa_t$ denotes the composition of $nu$ and $kappa_t$.

I want to show that there is a version $X^{(eta)}$ of $X$ with $operatorname Pleft[X^{(eta)}_0in;cdot;right]=eta$ for $etainleft{mu,nuright}$ with $$tau:=infleft{tin I:X^{(mu)}_s=X^{(nu)}_stext{ for all }sin Itext{ with }sge tright}<infty;;;operatorname Ptext{-almost surely}.tag2$$

As pointed out by E-A, the basic idea is the following: For any $tge0$, there is (see Theorem 2.12) a probability measure $eta_t$ (called a coupling of $mu$ and $nukappa_t$) on $(Etimes E,mathcal Eotimesmathcal E)$ with $$eta_t(Btimes E)=mu(B)text{ and }eta_t(Etimes B)=(nukappa_t)(B);;;text{for all }Binmathcal Etag3$$ and $$left|mu-nukappa_tright|=eta_t(Delta^c),tag4$$ where $$Delta:=left{(x,x):xin Eright}.$$

Clearly, if we would be able to show (are we?) that there is a $(mathcal A,mathcal Eotimesmathcal E)$-measurable $Y_t:Omegato Etimes E$ with distribution $eta_t$ under $operatorname P$, we would obtain $$eta_t(Delta^c)=operatorname Pleft[(Y_t)_1ne(Y_t)_2right]tag5.$$ However, even when we can show this, why can we choose $(Y_t)_{tge0}$ such that $X^{(mu)}:=((Y_t)_1)_{tge0}$ and $X^{(nu)}:=((Y_t)_2)_{tge0}$ 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)