Prove that: $E[\int^{\tau}_{0} f(t)d\omega(t)]=0$ and $E\mid \int^{\tau}_{0} f(t)d\omega(t)\mid^2=E[\int^{\tau}_{0} f^2(t)dt]$.

Question

Suppose $f \in L^{2}_{\omega} [0, \infty]$, and $\tau$ is a stopping time such that $E[\int^{\tau}_{0} f^2(t)dt]<\infty$. Prove that:

$E[\int^{\tau}_{0} f(t)d\omega(t)]=0$ and $E\mid \int^{\tau}_{0} f(t)d\omega(t)\mid^2=E[\int^{\tau}_{0} f^2(t)dt]$.

where $\omega(t)$ is a Brownian motion and $L^{2}_{\omega} [0, \infty]$ is the class of all nonanticipative functions.

Progress: I can show the first equality but I'm stuck on the second one. Thank you!

Answer 1

$\int_0^\tau \! f(x) \, \mathrm{d}wt=\int_0^T \mathbb{1} \{ t \leq\tau \} f(x) \, \mathrm{d}wt$. You just need to show that $\mathbb{1} \{ t \leq\tau \} f(x) \in \mathbb{L}_w^2[0,T]$, which shouldn't be a problem, then you can apply Ito's isometry.