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