4 introduced martingale tag
3 rephrasing + post scriptum

I am trying to prove that a certain sequence of Markov chains $x^N_k$ converges towards a diffusion process. The invariant measure of $x^N$ is $\pi^N$ and the Markov chain $x^N$ is started in stationarity $x^N_0 \sim \pi^N$. It is also assumed that $\pi^N$ converges (in laws) to a limiting probability distribution $\pi$.

The idea of the diffusion approximation is then pretty simple, and runs roughly as follows: we can write $$x^N_{k+1}-x^N_k = \mu(x^N_k) \Delta_N + m^N_k \sqrt{\Delta_N}$$ where $\Delta_N$ is a time step that goes to zero as $N \to \infty$ and $\mu(\cdot)$ is a deterministic function and $m^N_k$ is a martingale difference with respect to the natural filtration generated by $x^N$. It happens that a martingale central limit theorem applies in this case and that it can be shown that $$W^N(t) = \sqrt{\Delta_N} \sum_{k < [\frac{t}{\Delta_N}]} m^N_k$$ converges in laws to a Brownian motion $W$. This is why the sequence of rescaled Markov chain $U^N(t) = x^N(t/\Delta_N)$ should converge in laws to the diffusion $$dU_t = \mu(U_t) \ dt + dW_t, \quad U_0 \sim \pi,$$ where $W$ is independent from the initial position $U_0$.

If we can show that the couple $(x^N_0, W^N)$ converges in laws to $(U_0,W)$ where $U_0$ is independent from $W$, a continuous mapping argument gives the conclusion.

Question: in short, it is assumed that $x^N_0 \sim \pi^N$ converges in law to $U_0 \sim \pi$ and it can be shown that $W^N \to W$. This seems always true that $W$ is independent from $U_0$, is it ? How can one ensure that $W$ is adapted to a filtration $\mathcal{F}_t$ with $U_0 \in \mathcal{F}_0$ ?

Rephrasing: consider a sequence of martingales $M^N$, not starting from $0$, adapted to the filtrations $\mathcal{F}^N$. Let us also assume that the distribution $\pi^N$ of $M^N_0$ converges in law to a distribution $\pi$. If a martingale invariance principle applies to the sequence of martingales $X^N_k = M^N_k - M^N_0$, is it true that the original sequence $M^N$ also converges in laws to a modified "Brownian motion" $W$ with $W_0 \sim \pi$ ?

Ps: the whole point of this pedestrian approach is to avoid the usual generator-based theorems that ensure that a diffusion approximation is valid.re

2 added 457 characters in body

I am trying to prove that a certain sequence of Markov chains $x^N_k$ converges towards a diffusion process. The invariant measure of $x^N$ is $\pi^N$ and the Markov chain $x^N$ is started in stationarity $x^N_0 \sim \pi^N$. It is also assumed that $\pi^N$ converges (in laws) to a limiting probability distribution $\pi$.

The idea of the diffusion approximation is then pretty simple, and runs roughly as follows: we can write $$x^N_{k+1}-x^N_k = \mu(x^N_k) \Delta_N + m^N_k \sqrt{\Delta_N}$$ where $\Delta_N$ is a time step that goes to zero as $N \to \infty$ and $\mu(\cdot)$ is a deterministic function and $m^N_k$ is a martingale difference with respect to the natural filtration generated by $x^N$. It happens that a martingale central limit theorem applies in this case and that it can be shown that $$W^N(t) = \sqrt{\Delta_N} \sum_{k < [\frac{t}{\Delta_N}]} m^N_k$$ converges in laws to a Brownian motion $W$. This is why the sequence of rescaled Markov chain $U^N(t) = x^N(t/\Delta_N)$ should converge in laws to the diffusion $$dU_t = \mu(U_t) \ dt + dW_t, \quad U_0 \sim \pi,$$ where $W$ is independent from the initial position $U_0$.

If we can show that the couple $(x^N_0, W^N)$ converges in laws to $(U_0,W)$ where $U_0$ is independent from $W$, a continuous mapping argument gives the conclusion.

Question: in short, it is assumed that $x^N_0 \sim \pi^N$ converges in law to $U_0 \sim \pi$ and it can be shown that $W^N \to W$. This seems always true that $W$ is independent from $U_0$, is it ? How can one ensure that $W$ is adapted to a filtration $\mathcal{F}_t$ with $U_0 \in \mathcal{F}_0$ ?

Rephrasing: consider a sequence of martingales $M^N$, not starting from $0$, adapted to the filtrations $\mathcal{F}^N$. Let us also assume that the distribution $\pi^N$ of $M^N_0$ converges in law to a distribution $\pi$. If a martingale invariance principle applies to the sequence of martingales $X^N_k = M^N_k - M^N_0$, is it true that the original sequence $M^N$ also converges in laws to a modified "Brownian motion" $W$ with $W_0 \sim \pi$ ?

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