I am still confused about the question about coupling. First of all, why do we need the definition of coupling? From my point of view, with a slight modification, we should be able to prove that the total variation of mu_n and nu_n is non-expansive. Then the convergence follows as fixed point. If we follow the coupling idea, one cumbersome problem occurs, what does X_n=Y_n means? Do you mean that if a sequence {Kn} converges to X in probability and X=Y, then {Kn} converges to Y in probability? If this is the definition, then grand coupling is not the unique coupling after X_n=Y_n, we can keep X_n and Y_n evolving independently. If you define X_n=Y_n as they have the same distribution (i.e. convergence in distribution), then, from my point of view, there is no need for defining coupling anymore. One can just focus on the non-expansiveness of the transition matrix to prove the theorem.
Hi Yingzhao,
X_n=Y_n means X_n=Y_n :) [the two random variables are the same (up to a set of probability 0, if you insist). Bottom line: it is saying much more than "X_n and Y_n have the same distribution", which would be a much weaker statement]
Please refer to my previous post on coupling: if you let X and Y evolve independently until the end of time, then one ineqauality will not be satisfied and you will not be able to prove the ergodic theorem.
All the best,
Olivier
Hi Prof. Olivier,
Thanks for your reply. If you define the "equal" as two identical random variables up to a set of probability 0, then the proof seems to be a bit shaky from my point of view. In the proof, we force X and Y to evolve in the same way once they hit each other. I know this is important for having the inequality, and once you have grand coupling then it converges. However, what if grand coupling is missing? In practice, there is no way to guarantee grand coupling will happen. In order to prove the theorem, we need to show that the convergence will happen what so ever, but the current proof only show that the convergence will happen if we are able to enforce grand coupling.
Furthermore, I do see the sophisticated flow behind this proof, but I don't agree that the "equal" is defined up to a set of probability 0. Let's think about the asymmetric cyclic random walk with odd nodes, it will converge to a stationary distribution, but the mass/measure is actually "rotating". Let us now consider this rotating phase, if X and Y both have the stationary distribution but with different phase. In terms of ergodic theorem, both X and Y have converged, but in terms of the equality definition, which is used in the proof, X and Y hasn't converge yet because they are not the same random variable. Therefore, I argue that the definition of equality in the proof is sometimes too strong and confusing.
Best,
Yingzhao
Let me respectfully disagree :)
The whole point of the proof is exactly to show that grand coupling (i.e. there exist a finite value of n for which X_n=Y_n) takes place with probability 1. The proof shows in detail that this is guaranteed thanks to the fact that both X and Y are irreducible, aperiodic and positive recurrent.
And according to the definition of the transition matrix of the coupled chain Z = X x Y, once X and Y meet, they live happily ever after... with probability 1...
Best,
Olivier
PS: An oral discussion would probably be more adapted here.
