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