This question Ito isometry with two independent Brownian motions asks for an Itô isometry for two independent Brownian motions $V_t,W_t:[0,T]\times\Omega\rightarrow\mathbb R$. It turns out that the independence of the Brownian motions renders the integrals independent and the expectation of the product becomes the product of expectations.
In the case where $W_t = V_t$, it follows from the classic Itô isometry that $$E\left[\left(\int_0^T X_t\,\mathrm dW_t\right)\left(\int_0^T Y_t\,\mathrm dW_t\right)\right] = \mathbb E\left[\int_0^T X_tY_t\,\mathrm d[W,W]_t\right],$$ where $[\cdot,\cdot]_t$ denotes the covariation.
In my case, however, I have $V_t\neq W_t$ and I can't assume that the Brownian motions are independent. In fact, I know that the correlation between $V_t$ and $W_t$ is equal to some constant $\rho$. Is there a result that gives me something like $$\mathbb E\left[\left(\int_0^T X_t\,\mathrm dV_t\right) \left(\int_0^T Y_t\,\mathrm dW_t\right)\right] \overset{?}{=} \mathbb E\left[\int_0^T X_tY_t\,\mathrm d[V,W]_t\right]?$$
Best Answer
If $M_t=\int_0^t X_s\,dV_s$ and $N_t=\int_0^t Y_s\,dW_s$, then by Problem 3.3.12 in Karatzas and Shreve, $$ M_t N_t = \int_0^t M_s\,dN_s + \int_0^t N_s\,dM_s + [M, N]_t, $$ giving \begin{multline} \left(\int_0^t X_s\,dV_s\right)\left(\int_0^t Y_s\,dW_s\right)\\ = \int_0^t M_s Y_s\,dW_s + \int_0^t N_s X_s\,dV_s + \int_0^t X_s Y_s\,d[V, W]_s. \end{multline} At Ito integral of the form $\int_0^t\theta_s\,dB_s$, where $B$ is a Brownian motion, is something called a local martingale. If, additionally, $E\int_0^t|\theta_s|^2\,ds<\infty$ for all $t$, then the Ito integral is a martingale. Martingales have the property that their expectation is constant over time. Since the expectation is zero at time $t=0$, it is always zero.
Thus, if $E\int_0^t|M_sY_s|^2\,ds<\infty$ and $E\int_0^t|N_sX_s|^2\,ds<\infty$, then the Ito integrals on the right-hand side of the above display are martingales. Taking expectations gives the result you are looking for.