I was reading this question Correlation between two stochastic processes and wonder if the coefficients can depend on the time and position. Namely, consider the following SDE: $$ X^{(i)}_t(\omega)=\xi^i+\sum_{j=1}^2\int_0^t \sigma_j^i(s,X_s(\omega)) dB_s^j+\int_0^t b^i(s,X_s(\omega))ds. $$ for $i=1,2$, where $(B^1,B^2)$ is the standard $2$-dimensional Brownian motion (i.e. $B^j$ are independent processes), $\sigma\colon[0,T]\times\mathbb{R}^2\to \mathbb{R}^2\times\mathbb{R}^2$ and $b\colon[0,T]\times\mathbb{R}^2\to \mathbb{R}^2$ are nice (globally Lipschitz, etc.), and $X_t=(X^{(1)}_t,X^{(2)}_t)$.
From Itô's lemma and since $X_t$ is the solution of the SDE above we have \begin{align} X^{(1)}_t \cdot X^{(2)}_t -X^{(1)}_0 X^{(2)}_0 & = \int_0^t X^{(2)}_s \, dX^{(1)}_s + \int_0^t X^{(1)}_s dX^{(2)}_s + \int_0^t 1 \, d\langle X^{(1)},X^{(2)}\rangle_s\\ &= \Big(\int_0^t \sum_{j=1}^2X^{(2)}_s\sigma_j^1(s,X_s)dB^j_s + \int_0^t \sum_{j=1}^2X^{(1)}_sb^2(s,X_s)ds \Big)\\ &\ \ +\Big(\int_0^t \sum_{j=1}^2X^{(1)}_s\sigma_j^2(s,X_s)dB^j_s + \int_0^t \sum_{j=1}^2X^{(2)}_sb^1(s,X_s)ds \Big) +\int_0^t d\langle X^{(1)},X^{(2)}\rangle_s \end{align} Following the proof in the question I cited above it seems that the approach can be generalized to the case. $$ X^{(i)}_t(\omega)=\xi^i+\sum_{j=1}^2\int_0^t \sigma_j^idB_s^j+\int_0^t b^iX_s(\omega)ds. $$ There, it was essential that in the non-martingale part in the right hand side we have $X^{(1)}_t X^{(2)}_t$. But when $\sigma$ and/or $b$ depends on time and/or position, are there known approach to know $E[X^{(1)}_tX^{(2)}_t]$? It seems when $\sigma$ is a constant matrix, if $\sigma$ is diagonal then $X^{(i)}_t$ are uncorrelated at each $t$. I wonder if the same holds when it depends on the position ($\sigma$ is a diagonal matrix but each element depends on $X_s$). In this case $X^{(1)}_t$ and $X^{(2)}_t$ are driven by independent Brownian motions but the integrands depends on $X$.