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

Question

$\newcommand{\Ex}{\mathbb E}\newcommand{\diff}{\ \mathrm d}$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 $$ \mathbb P [X \in A, Y \in B | Z] = \mathbb P [X \in A | Z] \cdot \mathbb P [Y \in B | Z] \quad \text{a.s.} \quad \forall A,B \in \mathcal B (\mathbb R). $$

Answer 1

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.