# Standard result on existence and uniqueness of BSDE

Below is the standard result on existence and uniqueness of BSDE of the form

$\displaystyle Y_{t} = Y_{T} + \int_{t} ^{T} g(s, Y_{s}, Z_{s}) ds - \int_{t}^{T} Z_{s} d W_{s}, \ Y_{T} = \xi. \ \ \ \ \ (1)$

Notations follow the paper [Pen10].

• Let ${(\Omega, \mathcal F, \mathbb P, \{\mathcal F_{t}\}, W)}$ be a filtered probability space with ${\mathbb R^{d}}$-valued Brownian motion ${W}$.
• Let ${L^{p}_{\mathbb P} (\mathcal F_{t}, \mathbb R^{m})}$ be the collection of random variables ${X\in \mathcal F_{t}/ \mathcal B(\mathbb R^{m})}$ satisfying ${\mathbb E [ |X|^{p} ] < \infty}$.
• ${M_{\mathbb P}^{p}(0,T, \mathbb R^{m})}$ is the collection of the ${\mathcal F_{t}}$-adapted processes ${X: \Omega \times [0,T] \mapsto \mathbb R^{m}}$ satisfying

$\displaystyle \mathbb E [\int_{0}^{T} |X_{t}|^{p} dt] <\infty$

• ${S_{\mathbb P}^{p}(0,T, \mathbb R^{m})}$ is the collection of the ${\mathcal F_{t}}$-adapted processes ${X: \Omega \times [0,T] \mapsto \mathbb R^{m}}$ satisfying

$\displaystyle \mathbb E [\sup_{[0,T]} |X_{t}|^{p}] <\infty$

Theorem 1 ([Pen10]) Suppose ${g(\cdot, y, z) \in M_{\mathbb P}^{2}(0,T, \mathbb R^{m})}$ for all ${T, y, z}$ and ${g\in Lip_{y,z}}$. Then, for any ${\xi \in L_{\mathbb P}^{2}(\mathcal F_{T}, \mathbb R^{m})}$, there exists unique ${(Y,Z) \in M_{\mathbb P}^{2}(0,T, \mathbb R^{m}\times \mathbb R^{m\times d})}$.

Under the same setting, there is a similar result in

[YZ99] J. Yong and X. Y. Zhou. Stochastic controls: Hamiltonian systems and HJB equations, vol 43. Springer, 1999.

Theorem 2 (Theorem 7.3.2 of [YZ99]) Suppose ${g(\cdot, y, z) \in M_{\mathbb P}^{2}(0,T, \mathbb R^{m})}$ for all ${T, y, z}$ and ${g\in Lip_{y,z}}$. Then, for any ${\xi \in L_{\mathbb P}^{2}(\mathcal F_{T}, \mathbb R^{m})}$, there exists unique ${(Y,Z) \in S_{\mathbb P}^{2}(0,T, \mathbb R^{m}) \times M_{\mathbb P}^{2}(0,T, \mathbb R^{m\times d})}$.

It is obvious that ${S_{\mathbb P}^{2}(0,T, \mathbb R^{m})}$ is a subset of ${M_{\mathbb P}^{2}(0,T, \mathbb R^{m})}$. Can we say they are equal?

NO. Consider following counter-example. Let ${\omega_{i} \in C[0,1]}$ be given by

$\displaystyle \omega_{i} (t) = 0, \forall 0\le t < 1 - 2^{-i}, \quad \omega_{i}(t) = 2^{2i} (t - 1 + 2^{-i}), \forall 1-2^{-i} \le t \le 1.$

Let ${\Omega = \{\omega_{i} : i = 1, 2, \ldots\}}$, and ${\mathbb P\{ \omega_{i} \} = 2^{-i}}$. Set ${X(\omega, t) = \omega(t)}$ for all ${\omega \in \Omega}$, and ${\mathcal F_{t}}$ is generated by ${X}$. Then, one can check

$\displaystyle X\in M_{\mathbb P}^{2}(0,1, \mathbb R) \setminus S_{\mathbb P}^{2}(0,1, \mathbb R).$

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