Exchangeability of an interacting particle SDE system

Question

Let $(W^1,\ldots,W^N)$ be $N$ independent Brownian motions, $\sigma > 0$ and $k$ a Lipschitz function. Consider the following system of SDE's \begin{equation} \mathrm{d}X_t^{i} = -\frac{1}{N} \sum\limits_{j=1}^N k(X_t^{i}-X_t^{j}) \mathrm{d}t + \sigma \mathrm{d}W_t^{i}, \quad \quad i=1,\ldots, N, \end{equation} which is subject to i.i.d. initial conditions.

The system is well-defined and has a strong solution by the classic theory.

Now, in every paper on interacting particle systems the author's claim that this system is exchangable. This seems reasonable. Unfortunately, I have never seen a complete proof of this statement and I am struggling with the proof myself.

If you have an references or could provide a proof of the exchangability I would appreciate it! I think you need to use the pathwise uniqueness + Yamada-Watanabe Theorem, which provides a $\mathbb{R}^N$-valued function of the form $h(X_0,W_{.})$,($X_0,W$ are the vectors of the whole system).

Answer 1

This follows quite quickly, as mentioned here too on page 5:

the entire system is symmetric, in the sense that if we permute the “names” $i = 1, . . . , n$, we end up with the same particle system. In other words, $(X_1, . . . , X_n)$ is exchangeable.

The SDE he studies in 3.1 is more complicated where the symmetry is not quite immediate. But in your system, everything is symmetric already.

A non-exchangeable system is $$dX_{1}=(X_{2}+X_{1})dt+dW,dX_{2}=2X_{1}dt+dW$$ because if I swap the indices, the $X_{2}$ will satisfy a different equation. Whereas in the above OP system, if you swap $X_{1},X_{2}$, they satisfy the exact same equation with the same-in-law initial data.

If you are wondering, how to do the exercise for 3.1, he simply uses uniqueness in law and that the initial data is iid. So if he swaps $X_{1},X_{2}$, then since they satisfy the same equation and have same initial data (in law), by existence and uniqueness, they give out the same path in law.

But yes one can also use Yamada-Watanabe as stated in Shreve-Karatzas 3.23 corollary in 5.4

Here we have

$$(X_{1},...,X_{n})=h(\xi_{1},...,\xi_{n}, W_{1},...,W_{n})$$

and so if we swap say $X_{1},X_{2}$, then $X_{1}$ will correspond to $\xi_{2},W_{2}$, which is equal in law to the pair $\xi_{1},W_{1}$ and so

$$(X_{2},X_{1},...,X_{n})=h(\xi_{1},...,\xi_{n}, W_{1},...,W_{n})$$

$$\stackrel{d}{=}h(\xi_{2},\xi_{1},...,\xi_{n},W_{2},W_{1},...,W_{n})$$

$$=h(\tilde{\xi}_{1},\tilde{\xi}_{2},\xi_{3}...,\xi_{n},\tilde{W}_{1},\tilde{W}_{2},W_{3}...,W_{n})$$

$$=(X_{1},X_{2},...,X_{n}),$$

where throughout we used the iid nature of $\xi,W$ and that $h$ is symmetric in its components because the original equations are.

Answer 2

Inspired by the answer of Thomas Kojar, I want to try to answer my question with the following proof.

Let us have a general interacting particle system

\begin{equation} \mathrm{d} X_t^{i} = b(t,X_t^{i}, \mu_t^{N}) \mathrm{d}t + \sigma (t,X_t^{i}) \mathrm{d} W_t^{i} \end{equation} for $i=1,\ldots,N$ and initial data $(\xi^1,\ldots,\xi^N)$. Here $\mu_t^{N}$ is the empirical measure given by \begin{equation} \mu_t^{N} = \frac{1}{N} \sum\limits_{i=1}^N \delta_{X_t^{i}} \end{equation}

Let now $(\mathbf{X_t}, t \ge 0)$ be the whole system, $B$ and $\Sigma$ be given by \begin{equation} \Sigma(t,\mathbf{x}) := diag(\sigma(t,x_1),\ldots, \sigma(t,x_N)) \end{equation} and \begin{equation} B(t,\mathbf{x})):= (b(t,x_1,L_n(\mathbf{x})),\ldots,b(t,x_N,,L_n(\mathbf{x}))^T \end{equation} with $L_n(\mathbf{x}):=\frac{1}{N} \sum\limits_{i=1}^N \delta_{x_{i}}$ and $\mathbf{x}=(x_1,\ldots,x_N)^T$. We assume that the above system has strong solution. Let $h$ be the function provided by Shreve-Karatzas 3.23 Corollary 5.4 such that $$\mathbf{X_t}= h(\xi^{1},\xi^{2},...,\xi^{n},W^{1},W^{2},...,W^{n}).$$ Next, let $g$ be the same function but acossiated to the particle system $ (X^2,X^1,\ldots,X^N) $, i.e. we changed the positions of the first two particles. The question we need to ask is whether the functions $g$ and $h$ are connected. Indeed, we ask ourselfs, which system does the following function $h(\xi^{2},\xi^{1},...,\xi^{n},W^{2},W^{1},...,W^{n})$ solve? By Shreve-Karatzas 3.23 Corollary 5.4 we know it is a strong solution to the SDE with coeffecients $B,\Sigma$ but the intial data and Brownian motion are changed. The observation now is that the first component is still "the same" as the second component, i.e. $$ L_n(\mathbf{x}):=\frac{1}{N} \sum\limits_{i=1}^N \delta_{x_{i}} = \frac{1}{N} \bigg( \delta_{x_{2}} + \delta_{x_{1}} + \sum\limits_{i=3}^N \delta_{x_{i}} \bigg). $$ It seems trivial but that is the "symmetry" in this system, which is induced by the empirical measure. Hence the first component of the solution $h(\xi^{2},\xi^{1},...,\xi^{n},W^{2},W^{1},...,W^{n})$ must be $X^2$ by the strong uniqueness property of the whole system. Similar, we obtain the same statement for the second component. Hence, $$(X^2,X^1,\ldots,X^N) = h(\xi^{2},\xi^{1},...,\xi^{n},W^{2},W^{1},...,W^{n}).$$ Now, the toy version of exchangeability with only two indices follows by the exchangeability of the initial data and the Bronian motion. More precise, \begin{align} (X^{2},X^{1},...,X^{n}) &=h(\xi^{2},\xi^{1},...,\xi^{N}, W^{2},W^{1},...,W^{n}) \\ & \stackrel{d}{=}h(\xi^{1},\xi^{2},...,\xi^{n},W^{1},W^{2},...,W^{n}) \\ &= (X^{1},X^{2},...,X^{n}). \end{align} The proof should also work for general permutations but takes more notation.

As Thomas Kojar pointed out, this should not work for a "non-symmetric" system.