Time-Reversal of BSDE = SDE

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?

• 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 $X_t = \xi + \int_0^t F(t,X_t)dt + \int_0^T \Sigma(t,X_t)dV_t$
– mike
Dec 10, 2020 at 16:10

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.