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?

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

1 Answer 1


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 $x=0$.

  • $\begingroup$ Is there an additional condition which you know of which would guarantee that the regularity of $X_t^x$ somehow translates to that of $f$? $\endgroup$ Mar 22, 2021 at 17:58
  • $\begingroup$ I am not sure I understand why you would expect there to be a natural class of Markov processes for which these are linked. Of course, for most diffusions both are true, so it is easy to come up with natural classes for which the implication holds simply because its premise and its conclusion are both true, but that's presumably not what you want. $\endgroup$ Mar 23, 2021 at 8:51
  • $\begingroup$ I guess my intuition came from the Gaussian case; specifically from fBM...but maybe something else is at play there... $\endgroup$ Mar 23, 2021 at 9:37

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