# Is the Levy Transform Ergodic? [closed]

First, let me start by saying, before you spend too much time on this, that it is still an open question, so I am just trying to gather some ideas and views on the problem.

Let $(\Omega ,\mathcal{A},\mathbb{P)}$ be a probability space and $(B_{t})_{t\geq 0}$ a Brownian motion defined on this space and starting at $0$. Let us introduce the local time at $0$, denoted by $(L_{t})_{t\geq 0}$. A simple consequence of Tanaka formula and Levy characterisation of Brownian motion is that : $$\tilde{B}_{t}=\left| B_{t}\right| -L_{t}=\int_{0}^{t}\mathrm{sgn}(B_{s})\mathrm{d}B_{s}$$ is a Brownian motion. In other words, the Levy transform : \begin{align*} T:\mathbb{W}& \longrightarrow \mathbb{W} \\ B& \longmapsto \tilde{B} \end{align*} defined almost everywhere on the Wiener space $\mathbb{W}=\mathcal{C} \mathbb{(R}_{+},\mathbb{R}\mathbb{)}$ leaves invariant the Wiener measure.

Let $\mathcal{F}_{t}=\sigma (B_{s},s\leq t)$ denote the natural filtration of the Brownian motion $B$, completed by the negligible $\sigma$-field $\mathcal{N}$, $\mathcal{F}_{t}^{(1)}=\sigma (\tilde{B}_{s},s\leq t)$ and similarly $\mathcal{F}_{t}^{(n)}$ the natural filtration of $T^{n}(B)$. Then one can wonder if the asymptotic filtration is trivial, i.e. if : \begin{equation*} \forall t\geq 0,\quad\bigcap_{n\geq 0}\mathcal{F}_{t}^{(n)}=\mathcal{N}. \end{equation*} In an ergodic theory context this would mean that the transformation $T$ is exact and would imply that $T$ is ergodic. There is an indication that it is as the discrete case is. But the many attempts in the continuous case have been unsuccessful, although we are making steps in the right direction, for example the fact that orbits are dense is known.

See the book by Revuz and Yor, and articles by Dubins, Smorodinsky, Malriq, and others.

Thanks!

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What is known about the spectrum of this operator? Can you please put your formulas into LaTeX and provide some references on this problem? –  Raskolnikov Dec 2 '10 at 11:29
I don't know that math.SE is a reasonable place to ask open questions. Shouldn't you be asking on MathOverflow, if at all? –  Qiaochu Yuan Dec 2 '10 at 13:15
It is mentioned here also: books.google.com/… –  user1119 Dec 2 '10 at 22:58
yes, Marc Yor was my teacher and he introduced me to the question. –  solojazz Dec 9 '10 at 13:01
What is your question? As you know very well, the answer to the question you ask in your title is We do not know, yet. And I see no other question in your post. –  Did Mar 20 '11 at 16:13