Let $y$ and $z$ be given by the equations \begin{equation} \mathrm{d}y_t = A(y_t) \mathrm{d}t + B(y_t) \mathrm{d}W^1_t, \end{equation} and \begin{equation} \mathrm{d}z_t = C(z_t) \mathrm{d}t + D(z_t) \mathrm{d}W^2_t, \end{equation} where $(W^1)$ and $(W^2)$ are Brownian motions, $A$, $C$ are vector fields and $B$, $D$ take values in the space of matrices of appropriate dimensions.
What is the equation for the tensor product $y\otimes z$? My guess is that \begin{equation} \begin{split} \mathrm{d} (y_t \otimes z_t) =& [ (A\otimes I) (y_t \otimes z_t) + (I\otimes C) (y_t \otimes z_t) ]\mathrm{d}t\\ &+ (B\otimes I)(y_t \otimes z_t) \mathrm{d}W^1_t + (I\otimes D) (y_t \otimes z_t) \mathrm{d}W^2_t\\ &+ (B\otimes D) (y_t \otimes z_t) \mathrm{d}\langle W^1, W^2\rangle_t \end{split} \end{equation} where the quadratic covariation in the last line is taken component-wise, i.e. \begin{equation} \langle W^1,W^2 \rangle_t^{ij} = \langle W^{1,i},W^{2,j} \rangle_t. \end{equation} Is that correct? I couldn't find anything about this in the literature. I would be happy about any hint, where to find such things (even for the deterministic anologue). Also, how does one generalize this result to infinite dimensions?
Any help appreciated!
Best, Luke