I am trying to evaluate the following expectation:
\begin{equation} \mathbb{E} \left( \int_{t_2}^{t_3} \sigma_\tau dW_\tau \right) \left( \int_{t_1}^{t_4} \sigma_\tau^2 d\tau \right) , \end{equation}
where $W_\tau$ is a Weiner process, $\sigma_\tau$ is a locally bounded, non-negative stochastic process, adapted to the filtration generated by $W_\tau$, and $t_1 \leq t_2 < t_3 \leq t_4$. Currently my best effort is as follows:
\begin{equation} \begin{split} & \mathbb{E} \left( \int_{t_2}^{t_3} \sigma_\tau dW_\tau \right) \left( \int_{t_1}^{t_4} \sigma_\tau^2 d\tau \right) \\ =& \mathbb{E} \int_{t_1}^{t_4} \left( \int_{t_1}^{t_4} \sigma_s^2 ds \right) \sigma_\tau \mathbf{1}_{t_2 < \tau \leq t_3} dW_\tau \\ =& \mathbb{E} \int_{t_1}^{t_4} \left( \int_{t_1}^{\tau} \sigma_s^2 ds \right) \sigma_\tau \mathbf{1}_{t_2 < \tau \leq t_3} dW_\tau + \mathbb{E} \int_{t_1}^{t_4} \left( \int_{\tau}^{t_4} \sigma_s^2 ds \right) \sigma_\tau \mathbf{1}_{t_2 < \tau \leq t_3} dW_\tau \\ =& \mathbb{E} \int_{t_1}^{t_4} \left( \int_{t_1}^{\tau} \sigma_s^2 ds \right) \sigma_\tau \mathbf{1}_{t_2 < \tau \leq t_3} dW_\tau + \mathbb{E} \int_{t_1}^{t_4} \left( \int_{t_1}^{\tau} \sigma_s \mathbf{1}_{t_2 < s \leq t_3} dW_s \right) \sigma_\tau^2 d\tau \\ =& \mathbb{E} \int_{t_1}^{t_4} \left( \int_{t_1}^{\tau} \sigma_s^2 ds \right) \sigma_\tau \mathbf{1}_{t_2 < \tau \leq t_3} \mathbb{E} dW_\tau | \mathcal{F}_\tau + \mathbb{E} \int_{t_1}^{t_4} \left( \int_{t_1}^{\tau} \sigma_s \mathbf{1}_{t_2 < s \leq t_3} \mathbb{E} dW_s | \mathcal{F}_\tau \right) \sigma_\tau^2 d\tau \\ =& 0 , \end{split} \end{equation} where I'm using Fubini/Tonelli in line 4, and the law of iterated expectations in line 5, conditioning on the filtration up to time $\tau$. Stochastic integrals is not a subject I am that familiar with, so I wanted to ask two questions:
1) Is my application of Fubini/Tonelli in line 4 valid?
2) Is there an easier way to evaluate this expectation? Perhaps a well-known theorem that turns it into a one-liner...
Assuming that $\sigma$ is good enough (stochastically differentiable), by the Clark--Ocone formula, $$ \sigma^2_u = \mathbb E[\sigma_u^2] + \int_0^u \mathbb E[D_s \sigma_u^2\mid \mathcal F_s] dW_s. $$ Hence, denoting $a = \int_{t_1}^{t_4} \mathbb E[\sigma_\tau^2] d\tau$ and using the stochastic Fubini theorem, $$ \int_{t_1}^{t_4} \sigma_u^2 du = a + \int_{t_1}^{t_4} \int_0^u \mathbb E[D_s \sigma_u^2\mid \mathcal F_s] dW_s\,du \\ = a + \int_0^{t_4} \int_{s\vee t_1}^{t_4} \mathbb E[D_s \sigma_u^2\mid \mathcal F_s]du\, dW_s. $$ Consequently, \begin{equation} \mathbb{E} \left( \int_{t_2}^{t_3} \sigma_s dW_s \right) \left( \int_{t_1}^{t_4} \sigma_s^2 ds \right) = \int_{t_2}^{t_3}\mathbb{E}\left[\sigma_s \int_s^{t_4} \mathbb E[D_s \sigma_u^2\mid \mathcal F_s] du\right]du\\ = \int_{t_2}^{t_3}\mathbb{E}\left[\sigma_s \int_s^{t_4} D_s \sigma_u^2 du\right]du, \end{equation} where the last equality is thanks to the $\mathcal F_s$-measurability of $\sigma_s$.
The only feasible assumption on $\sigma$ turning the last expectation to zero seems to be $D_s\sigma_u^2 = 0$ for $s<u\wedge t_4$, which implies that $\sigma_u$ is independent of $\{W_s,s\le t_4\}$. Under this assumption the equality of the expectation in question to zero is easy to see directly, without referring to stochastic derivatives.