2
$\begingroup$

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

  • 0
    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.2011-06-15

1 Answers 1

1

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.

  • 0
    I 'll tr$y$ to keep that in mind, thank's for the advice.2011-06-20