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Let $(X_t)_{t\ge 0}$ and $(Y_t)_{t\ge 0}$ be a pair of stochastic processes taking values in $\mathbb{R}^n$ and in $\mathbb{R}^m$; defined on a filtered probability spaces $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge 0},\mathbb{P})$. Suppose that $$ \begin{aligned} X_t = & x + \int_0^t \, \mu(s,X_s)ds + \int_0^t\, \sigma(s,X_s)dW_s\\ Y_t = & y + \int_0^t \alpha(X_s,Y_s)ds + \int_0^t \, \beta(s,X_s,Y_s)dB_s \end{aligned} $$ where $(W_t)_{t\ge 0}$ and $(B_t)_{t\ge 0}$ are independant $\mathcal{F}$-adapted Brownian motions, and $\mu,\sigma,\alpha,\beta$ are uniformly (in time) Lipschitz smooth functions; such that $(X_t,Y_t)_{t\ge 0}$ has a unique strong solution.

Let $(\mathcal{F}^Y_t)_{t\ge 0}$ denote the right-continuous completed filtration generated by $(Y_t)_{t\ge 0}$ and let $\mathbb{P}(X_t|Y_t)$ denote the regular conditional distribution of $X_t$ given the $\sigma$-algebra generated by $Y_t$ (As Llya pointed out, this may be different from the $\sigma$-algebra generated by the process $(Y_s)_{s\ge 0}$ up to time $t$).
By the Markovianity of both processes, we have that
$$ \mathbb{P}(X_t|Y_t) = f(t,Y_t), $$ for some Borel function $f:[0,\infty)\times \mathbb{R}^m\rightarrow \mathcal{P}(\mathbb{R}^n)$. Under some integrability assumptions, we can even deduce that $f$'s image is in $\mathcal{P}_1(\mathbb{R}^n)$ equipped with the 1-Wasserstein metric $W_1$.

Under what conditions on the dynamics $\mu,\sigma,\alpha,\beta$ can we deduce that, for every $T>0$ the map $f|_{[0,T]\times \mathbb{R}^m}$ is Lipschitz?

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  • $\begingroup$ I am not an expert on continuous-time Markov processes, but I guess in discrete time if I have observations of $Y$ (being constructed dynamically using $Y$ and $X$), it is not enough for me to just know the latest value of $Y_t$ to get everything I can for the distribution of $X_t$, in most cases knowing the whole trajectory of $Y$ provides a strictly finer conditioning. I guess the same woule apply to continous-time case. Namely, $\Bbb P(X_t|\mathcal F_t^Y)$ is not a function of simply $Y_t$ and $t$. $\endgroup$
    – SBF
    Commented Dec 22, 2022 at 17:42
  • $\begingroup$ @Ilya This is a good point! I have modified the question to focus on the case I'm most interested in, which is conditioning on $\sigma(Y_t)$ not $\sigma(\{Y_s\}_{0\le s\le t})$. Thanks for pointing this out. $\endgroup$ Commented Dec 22, 2022 at 18:10

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