2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.