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
$$
X\_t(\omega) = Y\_{T-t}(\omega)
$$
holds $\mathbb{P}\otimes m$-a.e?



Time-Reversal of BSDE = SDE
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.


