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.


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