Simplifying the expectation of the product of two stochastic integrals

Question

Let $f(t, \omega), g(t, \omega)$ be functions that are independent of the increments of the Brownian motion $w(t, \omega)$ in the future. That is, $f(t, \omega), g(t, \omega)$ are independent of $w(t + s, \omega) - w(t, \omega)$ for all $s > 0$. I want to show that for any $0 \leq \tau \leq t \leq T$,

$$E\left(\int_\tau^t f(s,\omega) dw(s,\omega) \int_{\tau}^t g(s,\omega)dw(s,\omega)\right) = \int_{\tau}^{t} E[f(s,\omega)g(s,\omega)] ds$$

This is somewhat like a linearity property in some sense. I'm not really sure how to show it. If it helps, I know how to show the following:

$$E\left(\int_\tau^t f(s,\omega) dw(s,\omega)\right) = 0,$$

but I don't quite see how to apply that here (it doesn't seem related). I was thinking this might have to do with Fubini's theorem (since the expectation is really just an integral), but I'm not so sure.

Any help is appreciated. Maybe there is some clever way to apply Ito's Lemma here.

Answer 1

Providing only a sketch here, using Ito integral definition (and commuting limit, summations and expectation), the result boils down to studying the expectation term:

$$ E\left[ f_{t_{i-1}} (W_{t_i}-W_{t_{i-1}}) \cdot g_{t_{j-1}} (W_{t_j} - W_{t_{j-1}}) \right]. $$

If the intervals don't overlap, $i\not= j$, and say $t_i \leq t_{j-1}$, then $f_{t_{i-1}} g_{t_{j-1}}(W_{t_i}-W_{t_{i-1}})$ is independent of $W_{t_j}-W_{t_{j-1}}$, so the expectation term above is equal to

$$ E\left[ f_{t_{i-1}} g_{t_{j-1}} (W_{t_i}-W_{t_{i-1}})\right] \cdot E\left[ (W_{t_j} - W_{t_{j-1}}) \right] = 0.$$

If the intervals overlap, $i=j$, then $f_{t_{i-1}} g_{t_{i-1}}$ is independent of $W_{t_i}-W_{t_{i-1}}$, so the expectation term above is equal to

$$ E\left[ f_{t_{i-1}} g_{t_{i-1}} \right] E\left[ (W_{t_i} - W_{t_{i-1}})^2 \right] = E\left[ f_{t_{i-1}} g_{t_{j-1}} (t_{i} - t_{i-1})\right]. $$

Have also a look at Ito isometry (and its textbook proofs).