Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space on which we define Brownian motion $B$ and let us denote by $\mathcal{F}_t$ its natural filtration. Assume we have Ito process $dX_t = \alpha(t,X_t, \omega)dt + \beta(t, X_t, \omega)dB_t.$

My question is if $Y_t = \mathbb{E}[X_t \mid \mathcal{F}_t]$ can be expressed $\omega$-wise as a continuous map $\phi$ of the path of Brownian motion $B$ up to time $t$.

By martingale representation theorem or by applying conditional expectation to the differential of $X$, one can get a differential expression of $Y$. Still, I'm pretty sure continuity can't be guaranteed by the tools at disposal using Ito calculus since we have the convergence of integrals only in probability.

Can the continuity of map $\phi$ be justified by the tools from rough path theory? I know that by Lyons' Universal Limit Theorem, one gets the continuity of the Ito map corresponding to the controlled rough SDE $dy_t = f(y_t)dx_t$ for fairly mild conditions on $f$. However, I'm not certain this result can be applied to my case due to the loss of "probabilistic" information one faces when choosing the "deterministic" rough path approach. Any help would be greatly appreciated.


1 Answer 1


The answer to the question as stated is no. Take the SDE coefficients $\alpha$ and $\beta$ to be deterministic, then $X_t$ is $\mathcal F_t$ measurable, so that $\mathbb E[X_t | \mathcal F_t] = X_t$ for all $t$.

The question then asks if you can write the solution to an Ito SDE as a continuous function of the sample paths in sup norm, to which the answer is “no” in the literal sense, unless you add additional data as in rough paths theory.

  • 1
    $\begingroup$ I thought so. But can one then at least write conditional expectation as a continuous map of the rough path (the original path and the lift)? So, can we make this map continuous by adding additional information.? $\endgroup$
    – Bombadil
    Commented May 8 at 13:15
  • $\begingroup$ @Bombadil That is a very good question. I’m not sure myself. $\endgroup$
    – Nate River
    Commented May 8 at 13:17

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