Let $(X\_t,Y\_t)$ be a pair of stochastic processes such that
$$
\begin{aligned}
dX\_t =& A\_t X\_t dt + C\_t dW\_t,\\
dY\_t = & H\_t X\_t dt + K\_tdB\_t
\end{aligned}
$$
for some non-random matrix-valued functions $A,C,H,K$ of appropriate dimension satisfying the usual conditions of the Kalman-Bucy filter. It's clear that $X\_t$ follows a (multidimensional) Ornstein-Uhlembeck process so is distributed according to this [wiki post](https://en.wikipedia.org/wiki/Ornstein%E2%80%93Uhlenbeck_process). However, what is the distribution of $Y\_t$? Obviously, it's Gaussian (see standard proofs on Kalman filtering) so the meat of the question is...what is its mean and co-variance?



Kalman filter distribution of observation process
If we assume $A$ constant $$\frac{d}{dt}\mathbb{E}(X\_t )=A \mathbb{E}(X\_t ) $$so $\mathbb{E}(X\_t)=e^{tA}X\_0$ and $\mathbb{E}Y\_t = Y\_0 + \int\_0^t
H\_s e^{sA}X\_0ds$.


For the variance, we can assume $X\_0=0$ and $Y\_0=0$. And we have $$\frac{d}{dt}\mathbb{E}(X\_tX\_t^T )=A\mathbb{E}(X\_tX\_t^T )+\mathbb{E}(X\_tX\_t^T )A^T+C\_tC\_t^T $$so $$\mathbb{E}(X\_tX\_t^T )=\int\_0^t e^{A(t-s)}C\_sC\_s^Te^{A^T(t-s)}ds.$$ Moreover for $u>t$ $$\mathbb{E}(X\_tX\_u^T )=\int\_0^t e^{A(t-s)}C\_sC\_s^Te^{A^T(u-s)}ds.$$
For $Y$ we have $$\mathbb{E}(Y\_tY\_t^T)= \int\_0^t\int\_0^t H\_s\mathbb{E}(X\_sX\_u^T)H\_u^Tdsdu +\int\_0^t K\_s K\_s^Tds \\ = 2\int\_0^t\int\_0^t\int\_0^t H\_se^{A(s-v)}C\_vC\_v^Te^{A^T(u-v)}H\_u^T 1\_{v\leq s \leq u}dsdudv +\int\_0^t K\_s K\_s^Tds $$


$$\quad$$


In the case $A$ non constant we have to replace the $e^{(t-s)A}$ by $U\_tU\_s^{-1}$ the solution of $\frac{d}{dt}U\_t=A\_tU\_t$. Unfortunatly if the familly of $A\_t$ don't commute then there is no simple formula for $U\_t$.


