# Interacting particle system: how are the particles independent conditionally to the knowledge of their initial positions?

$$\newcommand{\Ex}{\mathbb E}\newcommand{\diff}{\ \mathrm d}$$Let

• $$(\Omega, \mathcal F, \mathbb P)$$ be a probability space.
• $$B=(B^1, \ldots, B^N)$$ independent one-dimensional Brownian motions.
• $$X=(X_0^1, \ldots, X_0^N)$$ independent real-valued random variables.
• $$X$$ independent of $$B$$
• $$(P_t, t\ge0)$$ a Markov semi-group.
• $$V, F:\mathbb R \to \mathbb R$$ smooth functions with compact supports.
• $$*$$ the convolution operation.

We consider a particle system $$X_t^i = X_0^i + \sigma B_t^i - \int_0^t V (X_s^i) \diff s - \int_0^t F * \eta_s (X_s^i) \diff s,$$ where $$\eta_s := \bigg( \frac{1}{N} \sum_{j=1}^N \delta_{X_0^j} \bigg ) P_s.$$

It is menitoned at page $$14$$ of this paper that

The particles $$X^r, 1 \leq r \leq N$$, are not independent but they are independent conditionally to the knowledge of the initial random variables $$X_0^1, \ldots, X_0^i, \ldots, X_0^N$$.

This statement is very intuitive to me because the dependence of $$X^r, 1 \leq r \leq N$$ comes from the random measure $$\eta_s$$. After conditioning, this measure becomes "non-random". However, I could not see how to establish the above statement rigorously.

Could you elaborate on how to obtain above claim?

My definition of conditional independence is

$$X,Y$$ are conditionally independent given $$Z$$ if and only if $$\mathbb P [X \in A, Y \in B | Z] = \mathbb P [X \in A | Z] \cdot \mathbb P [Y \in B | Z] \quad \text{a.s.} \quad \forall A,B \in \mathcal B (\mathbb R).$$

Let $$\{\nu_x\}_{x \in \mathbb R^n}$$ be the regular conditional probability measures on $$\Omega$$ associated with $$X$$, and $$\mu_X$$ the law of $$X$$ on $$\mathbb R^n$$.

Denote by $$E$$ the event $$\left \{ \nu_X ( \bigcap_i \, \{X^i \in A_i\} ) = \prod_i \nu_X (X^i \in A_i) \, , \, \forall A_i \in \mathcal B(C[0, T])\right \}.$$

By definition of conditional independence, we need to show that $$\mathbb P(E) = 1.$$

But for $$\mu_X$$-a.e. $$x$$, the $$X^i_0$$ are deterministic under $$\nu_x$$, and hence also the process $$\eta_s$$. As such, for $$\mu_X$$ a.e. $$x$$, under $$\nu_x$$ each $$X^i$$ is a standard diffusion SDE driven by $$B^i$$ with non-random coefficients and deterministic initial condition, for which it is known there is a strong solution.

Here independence of $$B$$ from $$X$$ guarantees that $$B$$ is still an independent collection of Brownian motions under each $$\nu_x$$.

Thus there exist deterministic maps $$\Phi_{i, x}$$ such that $$X^i = \Phi_{i, x} (B_i)$$ for all $$i$$ almost surely under $$\nu_x$$ for $$\mu_X$$-a.e. $$x$$. Independence of the $$X^i$$ under $$\nu_x$$ for $$\mu_X$$-a.e $$x$$ thus follows from that of the $$B_i$$.

In other words, denoting by $$S$$ the set

$$\{ x \in \mathbb R^n \, | \, \nu_x ( \bigcap_i \, \{X^i \in A_i\} ) = \prod_i \nu_x (X^i \in A_i) \, , \, \forall A_i \in \mathcal B(C[0, T]) \}$$

we have $$\mu_X (S) = 1$$, and so

$$\mathbb P (E) = \int_{\mathbb R^n} \mathbf 1_S (X(\omega)) \, d\mathbb P (\omega) = \int_{\mathbb R^n} \mathbf 1_S (x) \, d\mu_X (x) = 1.$$

Thus we conclude conditional independence of the processes $$X^i$$ as desired.

• In this question, I can only show that $B$ is still a collection of random variables conditionally independent given $X$ (i.e., under the random measure $\nu_X$). Could you explain how you obtain such independence under every $\nu_x$? Commented Mar 23, 2023 at 9:39
• Oh, to be more precise, I mean under $\mu_X$ almost every $\nu_x$. This follows directly from your answer here, since (in your notation) $\nu(Z, X^{-1}(A) \cap Y^{-1}(B)) = \nu(Z, X^{-1} (A)) \nu(Z, Y^{-1}(B))$ a.s. if and only if $\nu(z, X^{-1} (A) \cap Y^{-1} (B)) = \nu(z, X^{-1}(A)) \, \nu(z, Y^{-1}(B))$ for $\mu_Z$ a.e. $z$. Commented Mar 23, 2023 at 12:20
• Ah I got the independence under $\nu_x$ for $\mu_X$-a.e. $x \in \mathbb R^n$. Could you explain how $B$ is still a Brownian motion under $\nu_x$ (for $\mu_X$-a.e. $x \in \mathbb R^n$)? Commented Mar 23, 2023 at 13:23
• By definition of conditional probability, we have $\mathbb P(\{B \in A\} \cap \{X \in C\}) = \int_C \nu_x (B \in A) \, d\mu_X$. On the other hand, by independence, $\mathbb P(\{B \in A\} \cap \{X \in C\}) = \mathbb P(B \in A) \mathbb P(X \in C)$. We can write the latter as $\int_C \mathbb P(B \in A) \, d\mu_X$. Since this holds for all Borel $C$, we have $\nu_x (B \in A) = \mathbb P(B \in A)$ for $\mu_X$ a.e. $x$. Ranging $A$ across a countable set of generators for the Borel sigma algebra yields that the law of $B$ under $\nu_x$ is the same as that under $\mathbb P$, for $\mu_X$ a.e. $x$. Commented Mar 23, 2023 at 13:40
• You’re most welcome! :D Commented Mar 23, 2023 at 14:34