# Martingale representation theorem up to a stopping time

Let $$W$$ be a standard one dimensional Brownian motion, and let $$\mathcal F_t$$ be its completed natural filtration.

Let $$\tau$$ be an $$\mathcal F_t$$ stopping time with $$\tau < T$$ almost surely for some $$T > 0$$. Suppose $$\xi$$ is an $$\mathcal F_\tau$$ measurable $$L^2$$ random variable.

Question: Does there exist some $$\mathcal F_t$$ predictable process $$H$$ such that

$$\xi = \mathbb E[\xi] + \int_0^\tau H_s \, dW_s$$

almost surely?

Idea:

I tried to proceed as follows - since $$\xi$$ is $$\mathcal F_\tau$$ measurable and $$\tau < T$$ a.s., $$\xi$$ is clearly $$\mathcal F_T$$ measurable. Now the standard martingale representation theorem gives some predictable $$H$$ such that

$$\xi = \mathbb E[\xi] + \int_0^T H_s \, dW_s$$

almost surely. But since $$\xi$$ is $$\mathcal F_\tau$$ measurable, it should follow that $$H_s = 0$$, $$d\mathbb P \times d\mu$$ a.e. whenever $$s > \tau$$, whence

$$\xi = \mathbb E[\xi] + \int_0^\tau H_s \, dW_s$$

as desired.

However, I am not fully sure how to rigorously show the claim $$H_s = 0$$ whenever $$s > \tau$$.

If $$\xi\in \mathbb D^{1,2}$$ is in the Sobolev-Watanabe space then we can apply Clark-Ocone formula to get that

$$\xi=E[\xi]+\int_0^T E[D_s\xi|\mathcal F_s]dW_s$$

where $$D_s$$ is the Malliavin derivative. For $$s\in [0,T]$$ we may write $$\xi=\xi 1_{\{\tau > s\}}+\xi 1_{\{\tau \leq s\}}$$. Then

\begin{align*} \xi&=E[\xi]+\int_0^T E[D_s(\xi 1_{\{\tau > s\}}+\xi 1_{\{\tau \leq s\}})|\mathcal F_s]dW_s\\ &=E[\xi]+\int_0^T E[D_s(\xi 1_{\{\tau > s\}})|\mathcal F_s]dW_s+\int_0^T E[D_s(\xi 1_{\{\tau \leq s\}})|\mathcal F_s]dW_s \end{align*}

$$\xi 1_{\{\tau \leq s\}}$$ is $$\mathcal F_s$$-measurable so $$D_s (\xi 1_{\{\tau \leq s\}})=0$$. Also for $$s>\tau$$ we have that $$\xi 1_{\{\tau > s\}}=0$$ so $$D_s (\xi 1_{\{\tau > s\}})=0$$ and for $$s<\tau$$ we have that $$\xi 1_{\{\tau > s\}}=\xi$$. So

$$\xi=E[\xi]+\int_0^\tau E[D_s\xi|\mathcal F_s]dW_s.$$

• Wonderful solution. Dec 13, 2022 at 0:47

Even if $$\xi$$ is just integrable (and $$\mathcal F_\tau$$-measurable) you can consider (a continuous version of) the martingale $$X_t:=\Bbb E[\xi\mid\mathcal F_t],\qquad 0\le t\le T.$$ By the martingale representaion theorem there is a predictable $$H$$ with $$\int_0^T H_s^2\,ds<\infty$$ a.s. such that $$X_t=\Bbb E[\xi]+\int_0^t H_s\,dW_s,\qquad\forall t\in[0,T],$$ almost surely. In particular, because $$\xi$$ is $$\mathcal F_\tau$$ measurable, $$\xi=\Bbb E[\xi\mid \mathcal F_\tau]=X_\tau=\Bbb E[\xi]+\int_0^\tau H_s\,ds.$$ You can modify the definition of $$H$$ if necessary — set $$H_s(\omega)=0$$ if $$s>\tau(\omega)$$ — without affecting the stochastic integral.