Let $(Y,Z)$ be a solution the the BSDE on a stochastic base $(\Omega,\mathcal{F},(\mathcal{F}_t)_t,\mathbb{P})$: $$ Y_t = \int_t^T f(s,Y_s,Z_s)ds + Z_t dW_t \qquad Y_T = \xi \in \mathcal{F}_T^W; $$ adapted to the filtration $(\mathcal{F}_t^W)_{t\in [0,T]}$. Is it possible to write down an SDE $$ X_t = \xi + \int_0^t F(t,X_t)dt + \int_0^T \Sigma(t,X_t)dV_t $$ whose solution process $X_t$ is adapted to some other filtration $(\mathcal{G}_t)_t$ for which $\mathcal{G}_0=\mathcal{F}_T^W$, $V$ is another Brownian motion, and $$ X_t(\omega) = Y_{Tt}(\omega) $$ holds $\mathbb{P}\otimes m$a.e?

$\begingroup$ Do you mean $X_t = \xi + \int_0^t F(s,X_s)ds + \int_0^t \Sigma(s,X_s)dV_s$ instead of $X_t = \xi + \int_0^t F(t,X_t)dt + \int_0^T \Sigma(t,X_t)dV_t$ $\endgroup$– mikeDec 10, 2020 at 16:10
1 Answer
Probably not. It seems to me that $Y_t \in \mathcal{G}_0$ for all t and therefore so is $X_t$ and $\xi + \int_0^t F(s,X_s)ds$. But in this case $$X_t  \xi + \int_0^t F(s,X_s)ds = \int_0^t \Sigma(s,X_s)dV_s$$ is a $\mathcal{G}_0$ adapted martingale and so constant. That leaves $X_t$ and therefore $Y_t$ being a process of bounded variation.