$\newcommand{\Ex}{\mathbb E}\newcommand{\diff}{\ \mathrm d}$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](https://www.tandfonline.com/doi/abs/10.1080/07362994.2019.1622426) 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](https://math.stackexchange.com/questions/26042/understanding-conditional-independence-of-two-random-variables-given-a-third-one#comment56059_26055) is



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



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


