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Let $W$ be a Brownian motion and $\alpha$, $\beta$ be two progressively measurable processes taking values in $\mathbb R_+$ s.t. $\alpha_t\le \beta_t$ for all $t\ge 0$. Define respectively $X$, $Y$ by

$$X_t:=\int_0^t \alpha_s dW_s,\quad Y_t:=\int_0^t \beta_s dW_s,\quad \forall t\ge 0.$$

Can we find a probability space on which one has a Brownian motion $B$ and two stochastic processes $\tau, \sigma$ s.t. $0\le\tau_t\le \sigma_t$ for all $t\ge 0$ and

$$\DeclareMathOperator\Law{Law}\Law(X)=\Law\bigl(B_{\tau}:=(B_{\tau_t})_{t\ge 0}\bigr),\quad\Law(Y)=\Law\bigl(B_{\sigma}:=(B_{\sigma_t})_{t\ge 0}\bigr)?$$

PS : Here we may assume any integrability conditions for $\alpha$, $\beta$ if needed.

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    $\begingroup$ @IosifPinelis Yes. I think "taking values in $\mathbb R_+$" means that $\alpha_t\ge 0$ (if my English is not that bad :)) $\endgroup$
    – Fawen90
    Feb 17, 2023 at 16:24
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    $\begingroup$ In the DS theorem, the two Brownian motions constructed are adapted to different filtrations. $\endgroup$ Feb 17, 2023 at 23:07
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    $\begingroup$ If there was a common Brownian motion, it would be need to be adaptable to two different filtrations and thus, for example, contradict the Markov property. $\endgroup$ Feb 17, 2023 at 23:08
  • $\begingroup$ You can do this up to any finite time T, but it is a cheat. Just imbed the first process, the wait for the brown Ian motion to return to zero, and imbed the second, even using a different Brownian motion if you like. Of course, you could them in either order. $\endgroup$
    – mike
    Feb 18, 2023 at 7:32
  • $\begingroup$ @mike Thank you mike for the answer, while it is not completely clear to me. Do you mind detailing this embedding procedure by providing an answer? I do appreciate $\endgroup$
    – Fawen90
    Feb 18, 2023 at 12:14

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