Considering the following SDE with multiplicative as well as additive stochastic uncertainty, $$ \dot{x}=A x+\sum_{\ell=1}^m \sigma_{\ell} B_{\ell} x \xi_{\ell}+H \eta $$ where $x \in \mathbb{R}^n, H \in \mathbb{R}^n$; for $\ell=1, \cdots, m$, $\xi_{\ell}=\frac{d \Delta_{\ell}}{\iota}, \eta=\frac{d \zeta}{u}$ with $\Delta_1, \cdots , \Delta_m$ and $\zeta$ being the standard independent Wiener process.
How to prove the following Lemma:
Let the covariance matrix, $\bar{Q}(t)=E\left[x(t) x(t)^{\top} \mid \psi\right]:=\int_{\mathbb{R}^n} x x^{\top} \psi(x, t) d x$, and $\bar{Q}(0):=\bar{Q}_0<$ $\infty$, then $\bar{Q}(t)$ satisfies the following matrix differential equation (MDE)
$$ \dot{\bar{Q}}=\bar{Q} A^{\top}+A \bar{Q}+\sum_{\ell=1}^m \sigma_{\ell}^2 B_{\ell} \bar{Q} B_{\ell}^{\top}+H H^{\top} $$
Here is the detailed proof:
Taking the time derivative on both sides and after simplification, we obtain [1, Theorem 11.9.1] $$ \frac{d E[V \mid \psi]}{d t}=\int_{\mathbb{R}^n}\left\{\frac{1}{2} \sum_{i, j=1}^n\left[\sum_{\ell=1}^m \sigma_{\ell}^2\left(b_{\ell}^i x\right)\left(b_{\ell}^j x\right)+h_i h_j\right] \frac{\partial^2 V}{\partial x_i \partial x_j}+\sum_{i=1}^n\left(a_i x\right) \frac{\partial V}{\partial x_i}\right\} \psi(x, t) d x=E[\mathcal{L} V \mid \psi] . $$
where the term $\mathcal{L} V$ is given by $$ \mathcal{L} V=x^{\top}\left(A^{\top} P+P A+\sum_{\ell=1}^m \sigma_{\ell}^2 B_{\ell}^{\top} P B_{\ell}\right) x+H^{\top} P H . $$
The time derivative of $E[V \mid \psi]$ is obtained by doing integration by parts where we make use of Remark 4 . In particular, we make use of the fact that the products, $\psi V, \frac{\partial \psi}{\partial x_i} V, \psi \frac{\partial V}{\partial x_i}$ vanish exponentially as $|x| \rightarrow \infty$ and hence, the higher order moments become zero. By substituting $\mathcal{L} V$ in the above euqation, and using the linearity of trace, expectation and commutativity inside trace, we obtain, $$ d\left(\operatorname{tr}\left(E\left[x x^{\top} \mid \psi\right] P\right)\right) / d t=\operatorname{tr}\left(\left(A^{\top} P+P A+\sum_{\ell=1}^m \sigma_{\ell}^2 B_{\ell}^{\top} P B_{\ell}\right) E\left[x x^{\top} \mid \psi\right]+H H^{\top} P\right) . $$
By definition of expectation, $E\left[x x^{\top} \mid \psi\right]=\bar{Q}$, we have, $$ \operatorname{tr}(\dot{\bar{Q}} P)=\operatorname{tr}\left(\left(\bar{Q} A^{\top}+A \bar{Q}+\sum_{\ell=1}^m \sigma_{\ell}^2 B_{\ell} \bar{Q} B_{\ell}^{\top}+H H^{\top}\right) P\right) . $$
This can be rewritten in terms of an inner product as $$ \left\langle\dot{\bar{Q}}-\left(\bar{Q} A^{\top}+A \bar{Q}+\sum_{\ell=1}^m \sigma_{\ell}^2 B_{\ell} \bar{Q} B_{\ell}^{\top}+H H^{\top}\right), P\right\rangle=0 $$
since $P>0$, $\dot{\bar{Q}}=\bar{Q} A^{\top}+A \bar{Q}+\sum_{\ell=1}^m \sigma_{\ell}^2 B_{\ell} \bar{Q} B_{\ell}^{\top}+H H^{\top}$
[1] A. Lasota and M. C. Mackey, Chaos, Fractals, and Noise: Stochastic Aspects of Dynamics. New York: Springer-Verlag, 1994
Since I have no access to the context of ref [1], I'm just confused of the above derivation of $\frac{d E[V \mid \psi]}{d t}$, could someone offer me a more detailed calculation, thanks.