Assume the standard probability setting and $B_t \in \mathbb{R}^d$ be the Brownian motion. Let $\xi$ be some random variable and $(X_{\xi}(t),Y_{\xi}(t),Z_{\xi}(t))$ be the solution to the following mean field FBSDE
\begin{equation} \left\{ \begin{aligned} X_{\xi}(t) &= \xi + \int^t_0 b(X_{\xi}(s),Y_{\xi}(s))ds+\sigma W_t;\\ Y_{\xi}(t) &= g(X_{\xi}(T)) +\int_t^T f(X_{\xi}(s),\mathcal{L}(X_{\xi}(s)),Y_{\xi}(s))\;d s -\int_t^TZ_{\xi}(s) dW_s. \end{aligned} \right. \end{equation} for some smooth enough functions $b$, $g$, $f$ and $\mathcal{L}(X_{\xi}(s))$ is the law of $X_{\xi}(s)$. Suppose I can show that the solution is strongly unique, that is $(X_{\xi_1}(t),Y_{\xi_1}(t),Z_{\xi_1}(t))=(X_{\xi_2}(t),Y_{\xi_2}(t),Z_{\xi_2}(t))$ a.s. if $\xi_1=\xi_2$ a.s.. How can we know that $\mathcal{L}(X_{\xi_1}(s))=\mathcal{L}(X_{\xi_2}(s))$ if $\xi_1$ and $\xi_2$ have the same law?
I think it is something related to the Yamada–Watanabe theorem. And I found a statement in the proof of lemma 5.6. of the paper "Forward–backward stochastic differential equations and controlled McKean–Vlasov dynamics". It said that "We now claim that the law of $(\xi, Y_{\xi}(t))$ only depends upon the law of $\xi$. This directly follows from the version of the Yamada–Watanabe theorem for FBSDEs [11]. Since uniqueness holds pathwise, it also holds in law, so that given two initial conditions with the same law, the solutions also have the same laws."
However, the Yamada–Watanabe theorem for FBSDEs in [11] concerns only FBSDE but not mean field FBSDE. So, how can we prove the result from the Yamada–Watanabe theorem for FBSDEs in [11] to the mean field one?
[11] DELARUE, F. (2002). On the existence and uniqueness of solutions to FBSDEs in a nondegenerate case. Stochastic Process. Appl. 99 209–286.
