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


1 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.

  • $\begingroup$ 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 $\nu_x$? $\endgroup$
    – Akira
    Commented Mar 23, 2023 at 9:39
  • $\begingroup$ 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. $z$. $\endgroup$
    – Nate River
    Commented Mar 23, 2023 at 12:20
  • $\begingroup$ 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. $x \in \mathbb R^n$)? $\endgroup$
    – Akira
    Commented Mar 23, 2023 at 13:23
  • 1
    $\begingroup$ 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. $x$. $\endgroup$
    – Nate River
    Commented Mar 23, 2023 at 13:40
  • 1
    $\begingroup$ You’re most welcome! :D $\endgroup$
    – Nate River
    Commented Mar 23, 2023 at 14:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.