When $t\to X_t$ is an absolutely continuous process ($X_t= X_0+ \int_0^t Y_s dt$ for some measurable process $Y_t$) we have for all $t$ $$\sigma(Y_t) \subset \cap_{\epsilon >0}\sigma(X_{s}, s\in [t,t+\epsilon))$$ Is the converse inclusion true ? If yes can you give me a proof or a reference ? I was not able to find a counterexemple. Thank you.
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$\begingroup$ By "converse inclusion" do you mean is the RHS included in the LHS? It doesn't look like it. For example what if, with probability 1, $Y_s=0$ for all $s$? So $X_s=X_0$ for all $s$. Then the LHS is trivial but (unless $X_0$ is also constant) the RHS is not. But actually I don't even believe the original inclusion as you state it. Don't you need some extra assumption such as right-continuity of $Y$ at $t$? Otherwise how can the path of $X$ on $[t,\infty)$ tell you the value of $Y_t$? $\endgroup$– James MartinCommented Nov 29, 2014 at 17:28
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$\begingroup$ Later thought: if you replace "$X_s$" on the RHS by "$X_s-X_t$", then what you are asking for is slightly more reasonable. It's still false in general: for example, the event that $Y_{t+h}>Y_t$ for all sufficiently small $h>0$ is measurable w.r.t. the RHS and not the LHS. But for certain distributions - e.g.\ if $Y$ is a Brownian motion - I think you would have that the only events that are measurable with respect to the RHS and not the LHS would have probability 0. $\endgroup$– James MartinCommented Nov 29, 2014 at 23:54
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$\begingroup$ Hello, you are right we can assume that $$Y_t(\omega)= \lim_{\epsilon\to 0} \frac{X_{t+\epsilon}(\omega)- X_t(\omega)}{\epsilon}$$ for all $t$ and all $\omega$. Otherwise completion would be needed i guess. $\endgroup$– FiltraskCommented Nov 30, 2014 at 0:33
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