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 nonrandom matrixvalued functions $A,C,H,K$ of appropriate dimension satisfying the usual conditions of the KalmanBucy filter. It's clear that $X_t$ follows a (multidimensional) OrnsteinUhlembeck process so is distributed according to this wiki post. 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 covariance?

$\begingroup$ Of course it is Gaussian.... Linear transformation of Gaussian process is Gaussian. $\endgroup$ – ofer zeitouni Jun 16 at 10:41

$\begingroup$ @oferzeitouni Yes, I have noticed this after posting (infact it's in any standard controltheoretic derivation of the KalmanBucy filter) but I cant find a clear expression of its mean and covariance.. $\endgroup$ – Zorn's Lama Jun 16 at 10:45

$\begingroup$ Note that the drift part and the martingale part of $Y_t$ are independent. Use Ito isometry to find the variance of the martingale part. The variance of the drift part can be turned into a double integral involving the covariance function of the multivariate OU process, which if I recall does not have a closedform expression. $\endgroup$ – S.Surace Jun 16 at 11:02
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 $$\mathbb{E}(X_tX_t^T )=\int_0^t e^{A(ts)}C_sC_s^Te^{A^T(ts)}ds.$$ Moreover for $u>t$ $$\mathbb{E}(X_tX_u^T )=\int_0^t e^{A(ts)}C_sC_s^Te^{A^T(us)}ds.$$ For $Y$ we have $$\mathbb{E}(Y_tY_t^T)= \int_0^t\int_0^t H_s\mathbb{E}(X_sX_u^T)H_u^Tdsdu +\int_0^t K_s K_s^Tds \\ = 2\int_0^t\int_0^t\int_0^t H_se^{A(sv)}C_vC_v^Te^{A^T(uv)}H_u^T 1_{v\leq s \leq u}dsdudv +\int_0^t K_s K_s^Tds $$
$$\quad$$
In the case $A$ non constant we have to replace the $e^{(ts)A}$ 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 $U_t$.