# Regularity of law of conditional law of a Markov process equivalent to regularity of its paths

Let $$(X_t^x)_{t\in [0,\infty),\,x\in \mathbb{R}^n}$$ be a Markov process taking values in $$\mathbb{R}^m$$ and defined on some stochastic basis $$(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\in [0,\infty}), \mathbb{P})$$; where $$X_0^x=x$$ $$\mathbb{P}$$-a.s. Suppose also that $$\sup_{t\geq 0} \mathbb{E}[\|X_t^x\|]<\infty$$ for all $$x \in \mathbb{R}^n$$.

Let $$f$$ be the function sending $$(t,x)\in [0,\infty)\times \mathbb{R}^n$$ to the conditional law $$\mathbb{P}(X_t^x)$$. If $$X_t^x$$ has paths of finite $$p$$-variation a.s., for some fixed $$p\geq 1$$, then is $$f$$ Hölder-continuous? If so, how is the regularity of $$f$$ implies by the a.s. regulairty of its paths?

Edit: I quantify the distance between two laws by the Wasserstein distance on $$\mathcal{P}_1(\mathbb{R}^m)$$.

Edit: Under what (additional) conditions on $$X_t^x$$ do we have continuity?

• Is it a typo: values in $\mathbb{R}^m$ (and not in $\mathbb{R}^n$)? Mar 21, 2021 at 18:29
• Which metric do you want to consider on the set of conditional laws? Mar 21, 2021 at 18:35
• @JochenWengenroth The Wasserstein distance. Mar 21, 2021 at 18:39

No, this has no reason to be true. Take for example $$X_t^x = x+t$$ for $$x \ge 0$$ and $$x-t$$ for $$x < 0$$ ($$n=1$$). Paths are smooth, but $$f$$ is discontinuous at $$x=0$$.
• Is there an additional condition which you know of which would guarantee that the regularity of $X_t^x$ somehow translates to that of $f$? Mar 22, 2021 at 17:58