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