# Mutual dependencies of BSDE solutions with markovian drivers with different starting points

Let $$(\Omega,\mathcal F, P)$$ be a complete probability space with a Brownian motion $$(W_t)_{0\le t\le T}$$ and the Brownian standard filtration $$(\mathcal F_t)_t$$ with $$\mathcal F_T = \mathcal F$$.

Let $$f\colon [0;T]\times \mathbb R \times \mathbb R \to \mathbb R$$ be a Lipschitz continuous and bounded function. The theorem of Pardoux-Peng implies that, for each $$(t,x) \in [0;T]\times\mathbb R$$, there is a unique continuous, predictable and square-integrable process $$(Y^{t,x}_s)_{s\in[0;T-t]}$$ satisfying the BSDE

$$Y^{t,x}_s = 1 + \mathbb E\bigg(\int_s^{T-t} f(t+u,x+W_u,Y^{t,x}_u) \mathrm du \Big\vert \mathcal F_s \bigg)$$

Now I would like to prove that, for all $$\omega \in \Omega$$, $$s,t\in[0;T]$$ with $$s+t\le T$$, it is

$$Y^{t,x}_s(\omega) = \mathbb E\big(Y^{t+s,x+W_s(\omega)}_0\big).$$

The intuition is that

1. $$Y^{0,0}$$ is kind of a solution to the "full" problem on $$[0;T]$$, whereas $$Y^{t,x}$$ is a partly solution for the same BSDE only on $$[t;T]$$ under the condition that the Brownian motion at time $$t$$ takes the value $$x$$,
2. any $$Y^{t,x}_s$$ should not depend on the past of the Brownian motion, but only on the Brownian motion at time $$s$$ because the driver is also markovian.

Unfortunately, I have no idea how to formally prove it. I know that it is common to use the notation $$(\bar Y^{t,x}_s)_{t\le s\le T}$$ instead of my $$(Y^{t,x}_s)_{0\le s\le T-t}$$. However, I don't quite get the formal definition of those because we would have to use not only different Brownian motions, but also different filtrations, which is pretty confusing.