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I have a question regarding that topic: Are Ito Integrals adapted to the Brownian Motion Filtration.

Given a probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_t,P)$, we could define a 1-dim Brownian motion $W_t$ adapted to $(\mathcal{F}_t)_t$ with its own filtration $\mathcal{F}_t^W$. $X$ an $(\mathcal{F}_t)_t$-adapted process. For the new process defined by the ito integral $Y_t=\int_0^t X_s dW_s$, is it enough for $Y_t$ to be $\mathcal{F}_t^W$-adapted that $X_s\neq 0$, $s\in [0,t]$ a.s.? Or is there also a counterexample?

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  • $\begingroup$ Why would you expect that $Y$ is $\mathcal{F}^W$-adapted if the integrand is not adapted to this filtration...? $\endgroup$ – saz Sep 22 '18 at 8:04
  • $\begingroup$ I thought, that the filtration of $X$ is somehow simple if there is $X^{-1}$.. $\endgroup$ – L. Wang Sep 22 '18 at 9:11
  • $\begingroup$ No... consider for instance an independent random variable $Z$ and set $X_t := Z$, $$\mathcal{F}_t := \sigma(W_s; s \leq t, Z).$$ Then $Y_t = Z W_t$ is not adapted to $\mathcal{F}^W$. $\endgroup$ – saz Sep 22 '18 at 16:08
  • $\begingroup$ Oh, I understand... but at the following point I am still unsure: If $X$ is $\mathcal{F}^W$-adapted, then $Y$ is also $\mathcal{F}^W$-adapted, i.e. $\mathcal{F}_t^Y \subseteq \mathcal{F}_t^W$. Assume again $X_s\neq 0$, $s\geq 0$ a.s. Is it right that $\mathcal{F}_t^Y= \mathcal{F}_t^W$ holds for every $t \geq 0$? We have that $dY_t=X_t dW_t$ and so with our assumption we get $dW_t=X_t^{-1} dY_t$. Is that enough for showing $\mathcal{F}_t^W \subseteq \mathcal{F}_t^Y$? Thank you in advance! $\endgroup$ – L. Wang Sep 22 '18 at 18:44
  • $\begingroup$ Do you or does anybody else have any feeling or idea whether this is true? Sorry for annoying:/ $\endgroup$ – L. Wang Sep 24 '18 at 16:38

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