5
$\begingroup$

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?

Improve this question
$\endgroup$
3
  • $\begingroup$ Is it a typo: values in $\mathbb{R}^m$ (and not in $\mathbb{R}^n$)? $\endgroup$
    – Dieter Kadelka
    Commented Mar 21, 2021 at 18:29
  • $\begingroup$ Which metric do you want to consider on the set of conditional laws? $\endgroup$
    – Jochen Wengenroth
    Commented Mar 21, 2021 at 18:35
  • $\begingroup$ @JochenWengenroth The Wasserstein distance. $\endgroup$
    – Bernard_Karkanidis
    Commented Mar 21, 2021 at 18:39

1 Answer 1

Reset to default
1
+100
$\begingroup$

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 $x=0$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Is there an additional condition which you know of which would guarantee that the regularity of $X_t^x$ somehow translates to that of $f$? $\endgroup$
    – Bernard_Karkanidis
    Commented 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. $\endgroup$
    – Martin Hairer
    Commented 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... $\endgroup$
    – Bernard_Karkanidis
    Commented Mar 23, 2021 at 9:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .