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

  • $\begingroup$ Maybe I'm missing something, but I don't see why you should need or expect $W$ to be adapted to a filtration of the sort you describe. I think you're already done. $\endgroup$ – Steve Huntsman May 9 '10 at 3:20
  • $\begingroup$ First, I would like to be sure that I can chose the Brownian motion W independently from the initial data U_0. Second, I need to prove that the couple (x^N_0, W^N) does converge, which is not clear to me. I need that because we can write U^N=I(x^N_0,W^N) where I is a continuous function: in order to apply the continuous mapping theorem, I need to show that the sequence (x^N_0,W^N) does converge in laws. $\endgroup$ – Alekk May 9 '10 at 11:04

Check out the book by Ethier & Kurtz on Markov processes. As far as I remember, the chapter on diffusion approximations should have everything you might need.

  • $\begingroup$ Ethier and Kurtz would be perfect if I was working in R^d, but I am actually working with Hilbert space valued Markov chains. The whole story should not change much, but I do not know the proof of E&K enough (based on generators of diffusion etc..) to see if this is a straightforward generalization or not. By the way, if there is a good reference for SPDE approximations of Markov chains, that would be very helpful. Thank you. $\endgroup$ – Alekk May 10 '10 at 20:11

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