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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 $$\mathbb E\left[\left(\int_0^T X_t\,\mathrm dV_t\right) \left(\int_0^T Y_t\,\mathrm dW_t\right)\right] \overset{?}{=} \mathbb E\left[\int_0^T X_tY_t\,\mathrm d[V,W]_t\right]?$$

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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.

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  • $\begingroup$ Thank you for your reply. I can't follow though why the integrals become zero though once expectations are taken $\endgroup$ Oct 10, 2022 at 23:01
  • $\begingroup$ @SydAmerikaner, I have edited the answer to address your question. $\endgroup$ Oct 12, 2022 at 17:25
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Under the technical assumption that the two correlated Brownian motions are given as $W_t$ and $$ V_t=\rho W_t+\sqrt{1-\rho^2}\,W_t^\bot $$ where $W_t^\bot$ is a BM uncorrelated with $W_t$ the formula is correct and follows easily from linearity of the Ito integral in $dV_t$ and from $[V,W]_t=\rho\,t\,.$

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  • $\begingroup$ Thank you for your reply. I spent some time now thinking about your solution. If I get it correctly, the correlation matrix $R_t$ of the vector $(V_t,W_t)$ is decomposed by the Cholesky decomposition $R_t= L_tL_t^\top$, and the modified vector $L_t^{-1}(V_t,W_t)$ is analyzed instead. However, I don't think this result will help me as I would have to know $W_t^\top$ in order to implement the solution $\endgroup$ Oct 10, 2022 at 22:53
  • $\begingroup$ We know now that there is Ito isometry under the technical but mild assumption that $W^\bot$ exists. To implement this (you mean Monte Carlo simulate?) pick independent RVs and combine them with the assumption formula to correlate them. $\endgroup$
    – Kurt G.
    Oct 11, 2022 at 4:09

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