Can we construct close martingales if their terminal marginal laws are close?

Question

Let $M=(M_t)_{0\le t\le 1}$ be a real-valued continuous martingale. Let $\mu := {\rm Law}(M_1)$ and $\varepsilon \in (0,1)$. For any $\nu$ satisfying $W_2(\mu,\nu)\le \varepsilon$, can we construct two continuous martingales $X, Y$ (on suitable probability space) such that

$${\rm Law}(X)={\rm Law}(M),\quad {\rm Law}(Y_1)=\nu,\quad \mathbb E\big[\sup_{0\le t\le 1}|X_t-Y_t|^2\big]\le 8\varepsilon^2?$$

PS : We assume $\mu,\nu$ both admit finite second order moment, and $W_2$ denotes the Wasserstein distance of order $2$.

Some thoughts : As mentioned below, the difficult step is to construct a "continuous" martingale by taking conditional expectation. I did some attempts as follows :

Step 1. Take first a discrete martingale, denoted by $M^n:=(M^n_{t_k})_{0\le k\le n}$ such that ${\rm Law}(M)\approx {\rm Law}(M^n)$;

Step 2. Using the result Can every discrete martingale be embedded in a continuous martingale? where we notice that we approximate further $M^n$ by a continuous martingale $N^n$ whose natural filtration can be quite "Brownian";

Step 3. Approximate $N^n_1$ by $Y_1$ (using Skorokhod representation theorem) and then take conditional expectation of $Y_1$ with respect to the filtration generated by $N^n$.

Can we make the above arguments rigorous?

Answer 1

A partial answer:

By the $L^2$ bound on $\mu$ and Doob's inequality, $M_t$ is $L^2$ bounded, and a fortiori uniformly integrable. Hence it is closable - that is, $M_t = \mathbb E[M_1 | \mathcal M_t]$, where $\mathcal M_t$ is the natural filtration of $M$.

Now let $X$ be such that $\text{Law}(X) = \text{Law}(M)$ and $X_1, Y_1$ be an optimal coupling of $\mu$ and $\nu$. By definition we then have $\mathbb E[|X_1 - Y_1|^2] \leq \varepsilon^2$.

Define the martingale $Y$ by $Y_t := \mathbb E[Y_1| \mathcal X_t]$, where $\mathcal X_t$ is the natural filtration of $X$. Then $X - Y$ is an $\mathcal X_t$ martingale, and we have by Doob's inequality

$$\mathbb E[\sup_{0 \leq t \leq 1} |X_t - Y_t|^2] \leq 4 \mathbb E[|X_1 - Y_1|^2] \leq 4\varepsilon^2,$$

as desired.

It is left to show that $Y$ is continuous. Unfortunately, I could only get that for every $T \in [0, 1]$, $Y$ is continuous at $T$ almost surely - but the null sets can depend on $T$, thus it cannot be trivially upgraded to almost sure continuity at all times. I outline the argument for fixed $T$ below.

To get left continuity, apply the almost sure martingale convergence theorem to the uniformly integrable martingale $Y$ restricted to $[0, T]$. The theorem says that $Y_t \to Y_T$ almost surely as $t \to T^-$, hence the left continuity.

Meanwhile, applying the backward martingale convergence theorem to the $L^2$ bounded process $Z_t := Y_{1-t}$ between $0$ and $1-T$, we get that $Y_t \to Y_T$ almost surely as $t \to T^+$, hence the right continuity.