# Ito isometry for correlated Brownian motions

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

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.
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 $$[V,W]_t=\rho\,t\,.$$
• 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 Oct 10, 2022 at 22:53
• 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. Oct 11, 2022 at 4:09