An Ito diffusion process can be written as $$dX_t = \mu(X_t, t)dt + \sigma(X_t, t)dB_t$$ where $\mu$ is the drift term, $\sigma$ is the diffusion term, and $B_t$ is a standard Brownian motion.
Suppose $X_t$ is two-dimensional, $X_t = (X^1_{t}, X^2_{t})$. I know the drift term is simply the componentwise mean, $\mu(X_t,t) = (\mu^1_{t}, \mu^2_{t})$. What about the diffusion coefficient?
Edit: I see that $\sigma$ is the square root of the covariance matrix. I'm trying to find the covariance matrix, and did this calculation, denoting $\mu_x = \mu(X_t, t)$ and $\mu_y = \mu(Y_t,t)$): \begin{align*} \mathbb{E}[(dX_t - \mu(X_t))(dY_t - \mu(Y_t))] &= \mathbb{E}[dX_t dY_t] - \mu_y \mathbb{E}[dX_t] + \mu_x \mathbb{E}[dY_t] - \mu_x \mu_y\\ &= \mathbb{E}[dX_t dY_t] - \mu_x\mu_y \end{align*}
(The second equality is due to $\mathbb{E}[dX_t] = \mu_xdt$, $\mathbb{E}[dY_t] = \mu_ydt$ so that the middle two terms cancel)
From here, though, how does one compute $\mathbb{E}[ dX_y dY_t]$?