# Kalman filter distribution of observation process

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. 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?

• Of course it is Gaussian.... Linear transformation of Gaussian process is Gaussian. – ofer zeitouni Jun 16 at 10:41
• @oferzeitouni Yes, I have noticed this after posting (infact it's in any standard control-theoretic derivation of the Kalman-Bucy filter) but I cant find a clear expression of its mean and covariance.. – Zorn's Lama Jun 16 at 10:45
• 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 closed-form expression. – 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 $$\mathbb{E}(X_tX_t^T )=\int_0^t e^{A(t-s)}C_sC_s^Te^{A^T(t-s)}ds.$$ 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 $$U_t$$.