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As no answer or comment to Can we construct close martingales if their terminal marginal laws are close? we consider a simplified version (discrete-time) as below:

Let $M=(M_k)_{0\le k\le n}$ be a real-valued martingale. Let $\mu := {\rm Law}(M_n)$ and $\varepsilon \in (0,1)$. For any $\nu$ satisfying $W_2(\mu,\nu)\le \varepsilon$, can we construct two discrete-time martingales $X, Y$ (on suitable probability space) such that

$${\rm Law}(X)={\rm Law}(M),\quad {\rm Law}(Y_n)=\nu,\quad \mathbb E\big[\sup_{0\le k\le n}|X_k-Y_k|^2\big]\le 8\varepsilon^2?$$

Here $\mu,\nu$ are assumed to have finite second order moment, and $W_2$ denotes the Wasserstein distance of order $2$.

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    $\begingroup$ If continuity is no longer an issue since we are discrete time, what goes wrong with the method in the answer to your original post? $\endgroup$
    – Nate River
    Commented Jan 3 at 16:54
  • $\begingroup$ @NateRiver Do you mind specifying the details? $\endgroup$
    – Fawen90
    Commented Jan 3 at 16:57
  • $\begingroup$ I mean, the details are given in the answer, just replace the time index with the non negative integers instead of $\mathbb R_+$. $\endgroup$
    – Nate River
    Commented Jan 3 at 16:58
  • $\begingroup$ @NateRiver I see the idea. What I wish to make rigorous is that, if your construct such $X,Y$, you need to find a common filtration for which both $X,Y$ are martingales. Of course from your construction, we need additional randomness that is not generated by the natural filtration of $X$... $\endgroup$
    – Fawen90
    Commented Jan 3 at 17:14
  • $\begingroup$ Hm, it will suffice then to construct a $\mathcal F_1$-measurable random variable $Y_1$, then conditioning $Y_1$ on the natural filtration of $X$ gives us a martingale. But the optimal coupling forces extra randomness, which is a problem. $\endgroup$
    – Nate River
    Commented Jan 3 at 17:19

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