Given a Brownian Motion $B_t$ is it possible to reconstruct it from the knowledge of the local times $L^x_t$ ?

Using occupation time formula this would mean solving for some $f$ the following equation :

$$B_t=\int_{-\infty}^{+\infty}f(x)L^x_t.dx=\int_0^t f(B_s)ds$$

This seems achievable but I couldn't find out the solution or prove that there is none.

By the way, if someone can achieve this reconstruction of $B_t$ from $L^x_t$ using some other device than the occupation formula I would be equally interested.

Best regards

  • $\begingroup$ The closest $f$ I could find was $f(x)=x$ for which we get $\int_0^t B_s ds =\int_{-\infty}^{+\infty}x.L^x_tdx$. $\endgroup$ – The Bridge Nov 18 '11 at 17:03
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    $\begingroup$ Other than Yuri's answer, what's wrong with just writing $B_t=\frac{d}{dt}\int_{-\infty}^\infty xL^x_t\,dx$? $\endgroup$ – George Lowther Nov 19 '11 at 5:00
  • $\begingroup$ @George Lowther : What you say is true but I think it is hard to work with such an expression. Best regards $\endgroup$ – The Bridge Nov 19 '11 at 7:55

Knowing local times you can derive if the path $\gamma=\{(t,B_t): t\in[0,T]\}$ passes through any rectangle of the following form: $[k/2^n,(k+1)/2^n]\times[j/2^n,(j+1)/2^n]$. For fixed $n$, denote by $G_n$ the union of all these visited rectangles.

Since $B_t$ is uniformly continuous on $[0,T]$, we have $\gamma=\bigcap_n G_n$.

  • $\begingroup$ @yuri : Thank's for this very nice idea. But what if all I have is the realisation of $L^x_t$ (for any $x$ at fixed time) but no information on the Brownian path itself, do you think it is still possible to work out $B_t$'s value ? Best regards $\endgroup$ – The Bridge Nov 21 '11 at 8:41
  • $\begingroup$ @The Bridge: Ah, $t$ is fixed! Then no. Think about a trajectory that passes through a point 3 times before $t$ with 2 excursions between them. If you exchange the order of those two excursions you get a different trajectory that has same local times at terminal point $t$. This trajectory is "as probable as the original one". $\endgroup$ – Yuri Bakhtin Nov 21 '11 at 15:57
  • $\begingroup$ @Yuri : Sorry about the "fixed time" condition as this was not included as a condition in my original question (note that I accepted your answer). Regarding your comment, I am not completely convinced by the argument, because first I have all $L^x$ so if two paths coincide at a particular $x$ then at another level they should differ (at least intuitively), second the fact that many paths can lead to $B_t$ is not really a problem as long as the information included in all $L^x$ can lead to the right $B_t$. Best regards $\endgroup$ – The Bridge Nov 21 '11 at 17:25
  • $\begingroup$ @The Bridge: To be honest, I do not know how to explain my argument more precicely in a short message. It seems obvious that the occupation times do not change if you cut your trajectory in pieces, move them rigidly around (only along time axis) and reglue them again. $\endgroup$ – Yuri Bakhtin Nov 21 '11 at 21:24
  • $\begingroup$ @yuri : Hi I got your point about the "swap and glue" excursions path transform that is "invariant" for local times. The thing that I still don't clearly "see" is why does this process makes $B_t$ different after the swap and glue path transformation. Best regards $\endgroup$ – The Bridge Nov 22 '11 at 7:22

This paper seems to answer (something very close to) your question :

Warren, J. and Yor, M. (1998), The Brownian burglar: conditioning Brownian motion by its local time process. Seminaire de Probabilites XXXII, pages 328-342. (pdf link)

Abstract :

Imagine a Brownian crook who spent a month in a large metropolis. The number of nights he spent in hotels A,B,C...etc. is known; but not the order, nor his itinerary. So the only information the police has is total hotel bills.....

  • $\begingroup$ @pgassiat : Thank's for the reference but I wasn't able to link the mathematical results of the article with my question, even though, the "down to earth" explanation of Burglar's process seemed promising in this regard. But often Yor's articles stay way too far from my level of understandings. Best Regards $\endgroup$ – The Bridge Jan 23 '12 at 8:07

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