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Recall that in practice, to simulate a Brownian motion on $[0,1]$, we usually use the interpolated process $X^n=(X^n_t)_{t\in[0,1]}$ between the jumps of a random walk $(S_k)_{k=1,...,n}$ with $n$-steps, where $n$ is pretty big ($X_{k/n} = S_k$). Donsker's invariant principle then states, the bigger the $n$, the closer the law of the interpolated process is towards the law of $B$.

In this post, I may ask a dumb question but perhaps it makes sense: Is there a type of approximation of Brownian by Markov processes taking values in finite state spaces?

To make it more precise, let $B = (B_t)_{t\in [0,1]}$ is a Brownian motion on $[0,1]$. I would like to use a process $X^n = (X^n_t)_{t \in [0,1]}$ to "approximate" $B$, where $\forall t \in [0,1]$, $X^n_t \in E^n$, where $E^n$ is a finite set with cardinality $n$. When parameter $n$ increases, the "closer" $X^n$ is towards $B$. Here, I am looking for a new type of "closeness" (or "approximation" in some new sense) from $X^n$ to $B$. Any references, insights are highly appreciated.

(Note that interpolated process in the approximation of $B$ is not finite-valued)

My motivation comes from finance, where the stock prices are modelled as diffusions and we would like to approximate them by Markov processes taking values in a finite state spaces. The reason is that working with finite state spaces is better for simulations, predictive analysis and algorithm designs.

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  • $\begingroup$ Since brownian motion is the limit of a random walk, $log(P_t) = log P_{t-1} + \epsilon_t$ approximates brownian motion for a small enough step size. See this thread for a possibly better explanation. Hamilton takes this RW limiting approach argument in his text so maybe check that out also. quant.stackexchange.com/questions/43195/… $\endgroup$
    – mark leeds
    Commented Jul 8 at 12:36
  • $\begingroup$ To all the people who downvoted, why didn't you give any explanation? Why downvote a genuine question? $\endgroup$ Commented Jul 8 at 17:15
  • $\begingroup$ @markleeds That's not what I ask. I ask for continuous time Markov chains... $\endgroup$ Commented Jul 8 at 17:16
  • $\begingroup$ Hi jeffrey: oops. my mistake and apologies for noise. $\endgroup$
    – mark leeds
    Commented Jul 9 at 2:35

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