# Filtrations and right continuity

Here is a question that has been posted on another forum (in french) that I couldn't answer and about which I'm really curious.

So here it is, let's be given as a state space $\Omega=\{f\in C([0,1],R) s.t. f(0)=0\}$ associated with the borelian $\sigma$-field $\mathcal{F}=\mathcal{B}(O)$, where $O$ is the topology over $\Omega$ defined thanks to uniform convergence norm.

Let $X_t$ be the canonical process $X_s(f)=f(s)$ for $f\in \Omega$ and let $\mathcal{F}_t=\sigma(X_s;s\le t)$ be given as the natural filtration associated with the canonical process $X$.

Is it true that $\mathcal{F}_t$ is a right-continuous filtration?

PS: The usual way to construct continuous processes is by constructing first a process that has good distributional properties then to take a filtration that is both complete and right continuous and finally take a modification of the initial process that is continuous.

Regards

-
Byron's link is good. As a very simple example in this case, note that $\mathcal{F}_0 = \{\Omega, \emptyset\}$. However, it shouldn't be too hard to find some other events contained in $\mathcal{F}_{0+}$. – Nate Eldredge Jun 14 '11 at 18:01
@Byron : Thank's for the link, I only regret that you provide a counter example event without elaborating why it is in $\mathcal{F}_0^+$. Anyway George Lowther's comment gives a counter example which is almost trivial and fully answers the question. Best Regards – TheBridge Jun 14 '11 at 20:34
Hi I have a complementary quetion about Byron answer. He gives the event A=X differentiable at 0, as being in $\mathcal{F}_0^+$ but not in $\mathcal{F}_0$. By completing and taking the right continuous extension of the natural filtration of $X$, and by supposing that P(A)=1 (btw now 0-1 law applies so P(A) = 0 or 1), can we say that $X$ is not a markovian process ? Intuition behind this is that to know the transition density of $P_{0,t}(B)=P(X_t\in B)$ it is necessary to both now $X_0(=0)$ and $X'(0)$ so that $X(0)$ is not enough to caracterise the law of $X_t|\mathcal{F}_0$. Regards – TheBridge Jun 15 '11 at 7:16
In particular this would show that a differentiable process can't be markovian, which shows for example that Brownian Motion as a Markov process cannot be differentiable. So we get a proof of this without doing any calculus which would be nice. – TheBridge Jun 15 '11 at 7:18

## 1 Answer

I will try to answer on the question about non-differentiability of Markov processes. With regards to your question, consider equation $dx = dt$. This is the deterministic Markov process, however it's differentiable at any point.

I am curious what you definition of "Dirac measure", but I guess that PDPs (piecewise deterministic Markov processes) are stochastic a lot to satisfy you. Meanwhile, a lot of these processes are differentiable a.e.

Btw, rising that assumption on non-Dirac transitions, you don't deal with the fact that the process is Markov or not. I mean that you question was: if the process is Markov then it is nowhere differentiable. You can proceed rising more assumptions until we reach Markov diffusions or process with jumps at the every point. However, I am not sure that in that case the Markov property of the process will help you to find a new way to prove non-differentiability.

-
What you say is absolutely right but "moraly" a deterministic process defined by a differential equation is "more" than Markovian or if you prefer it is degenerate. In this case the transition operator has a law that corresponds to a "dirac" measure at the solution of the differential equation. If we want to exlude this type of behavior, I think that it is possible to define a criteria like the following : for every a.s. finite stopping time T, every $\epsilon>0$ then the transition kernel at $T+\epsilon$ is different from a dirac measure over one trajecory.Regards – TheBridge Jun 15 '11 at 13:47
@TheBridge, I hope I answered your question. – Ilya Jun 15 '11 at 13:56
Not really to be totally frank, but the question is not completely well designed in the first place I must admit. Still I beleive that it should be possible to do something about it probably using infinitesimal generator of the process. Regards. – TheBridge Jun 16 '11 at 22:03
@TheBridge: could you just ask a question which makes sense to you? Don't mix diffusions and Markov processes in you mind. There is no inclusion, just an intersection. Non-Markov diffusions are not differentiable and non-diffusion Markov processes are differentiable. – Ilya Jun 17 '11 at 9:29
I 'll try to keep that in mind, thank's for the advice. – TheBridge Jun 20 '11 at 7:57