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Assume that for each $n\in\mathbb{N}$, there's a stochastic function $f_n$ of type $\mathbb{R}^{m}\to\Delta\mathbb{R}^{m}$, and for each $x\in\mathbb{R}^{m}$, the distributions $\frac{f_n(x)-x}{\frac{1}{n}}$ will weakly converge as $n$ limits to $\infty$, s.t. the n'th distribution is about $O(\frac{1}{n^2})$ away from the limiting distribution w.r.t. the usual earthmover distance. Also, the function which maps a given starting point $x$ to its (rescaled) limiting distribution $\lim_{n\to\infty}\frac{f_n(x)-x}{\frac{1}{n}}$ is known to be Lipschitz.

The particular thing I'm trying to prove is that, for any finite interval of time $[0,I]$, $\epsilon,\delta$, then in the $n\to\infty$ limit, all but $\delta$ of the trajectories you get from applying $f_n$ $In$ times to a starting point are $\epsilon$-close to the trajectory you get from the differential equation $\frac{dx}{dt}=\lim_{n\to\infty}\mathbb{E}\left[\frac{f_n(x)-x}{\frac{1}{n}}\right]$ (ie, in a tiny time interval, the point just moves deterministically to the average of its next positions, instead of moving randomly)

Now, there's a strong informal intuition for why it would work out this way. Because the function which maps a given starting point $x$ to its (rescaled) limiting distribution $\lim_{n\to\infty}\frac{f_n(x)-x}{\frac{1}{n}}$ is Lipschitz, that means that for large $n$, it should be be possible to iterate the function a whole bunch of times in a row without the distribution on next-positions changing much. Splitting the application of the function into a deterministic "drift" term that goes to the average next position, and a random term with a mean of 0, the resulting distribution should look something like "sum up all the drift terms, and by Martingale Central Limit Theorem shenanigans, the random terms will mostly cancel themselves out and look like a Gaussian with the same covariance matrix."

But since the "spread" of the Gaussian is about $\sqrt{\text{number of iterations}}\cdot\text{step-size}$, and the step size scales as $\frac{1}{\text{total number of iterations}}$, the "spread" after $n$ steps would be about $\sqrt{n}\cdot\frac{1}{n}=\frac{1}{\sqrt{n}}$ and so in the $n\to\infty$ limit, the Gaussian sharpens up to a single point, a dirac-delta distribution. The variance is decreasing too fast as $n$ ramps up, and so, in the limit, the differential equation is deterministic, not stochastic, and only the "drift" terms survive from the repeated application of $f_n$ for large $n$.

The issue is, although the informal argument is quite clear, I don't know what theorems would be involved in rigorously proving this conjecture. The stochastic differential equation theorems I've looked at seem to primarily be about how to analyze stochastic differential equations once you already have them, not about rigorously showing that a given sequence of discrete stochastic processes have a certain SDE as their limit. A response like "if you prove this bound and that bound and this other bound on your stochastic function we can directly apply Theorem 3.24 from BlahBlah to show your desired result" would be ideal.

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  • $\begingroup$ What is $\Delta \mathbb R^m$? $\endgroup$
    – user479223
    Nov 26, 2022 at 23:39
  • $\begingroup$ The space of probability distributions over m-dimensional euclidean space. $\endgroup$
    – Alex Appel
    Nov 27, 2022 at 0:03
  • $\begingroup$ The rest of your question doesn't make sense, then. First of all the expression $f_n(x)-x$ is not defined as $f_n(x)$ is a measure and $x$ is a vector $\endgroup$
    – user479223
    Nov 27, 2022 at 0:30
  • $\begingroup$ That's the probability distribution over points that is produced by sampling a point from the probability distribution $f_n(x)$ and subtracting the point $x$ from the resulting point. $\endgroup$
    – Alex Appel
    Nov 27, 2022 at 1:24
  • $\begingroup$ In that case, $f_n(x)$ is an $\mathbb R^m$ valued random variable. It doesn't take values in $\Delta \mathbb R^m$ $\endgroup$
    – user479223
    Nov 27, 2022 at 1:27

1 Answer 1

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I am guessing in "The particular thing I'm trying to prove is that,..." you are talking about the convergence of discrete generator to continuous one. The natural topology for these questions is Skorokhod. See here for some ideas and references/keywords: https://math.stackexchange.com/questions/4225750/continuous-limit-of-a-discrete-stochastic-process

We denote by $P_n$ the transition kernel of $X^n$ given by $$P_nf(x) = \mathbb{E}\left[f\left(X_{t_1}^n\right)\left| X_0 = x\right.\right], $$ for functions $f$ of a suitable class (we will take that class to be $C^2\left(\left[0,1\right]\right)$). We will show that the discrete generators of $X^n$, $$\mathcal{L}_n = \frac{P - I}{\Delta t^n}$$ where $I$ is the identity operator (i.e. $If = f$ for any function), converges to $\mathcal{L}$.

As mentioned there some good references are:

  • Billingsley's Convergence of Probability Measures.
  • Ethier and Kurtz's Markov Processes.

If you are dealing with SPDEs, you can also check out the literature in Wong-Zakai type theorems.

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  • $\begingroup$ Thanks, this was extremely helpful! Tightness+"there can only be one limit point because the generators converge" sufficed quite nicely for my purposes, and I've got a few helpful references to chew on now. All I'm missing at this point is how to go back from the infinitesimal generator of the limiting stochastic process (known) to the actual differential equation for the limiting stochastic process, to show the limiting differential equation is what I conjectured it to be. $\endgroup$
    – Alex Appel
    Nov 27, 2022 at 2:54
  • $\begingroup$ Theorem 6.5 in Ethier and Kurtz was exactly what I was looking for. Problem resolved. $\endgroup$
    – Alex Appel
    Nov 28, 2022 at 7:25

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