Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
3  
    
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
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.