# Forecast equivalence between two steady-state Kalman filters

I have two related steady-state Kalman filter problems that I want to prove satisfy a condition associated with their respective Kalman gains.

I am not really looking for a complete proof since this is likely require quite a bit of work. However, any ideas about where to start or useful results that I could use would be much appreciated.

Problem 1:

$${\bf P} ={\bf F}({\bf I}_{n}-{\bf K} {\bf H}){\bf P}{\bf F}^\top+{\bf Q}, \;\;\;\;\;\;\;\;\text{where}\;\;\;\;{\bf K}\equiv {\bf P} {\bf H^\top}\left({\bf H}{\bf P} {\bf H}^\top+ \frac{1}{\gamma}{\bf R} \right)^{-1}$$

where ${\bf P}$, ${\bf F}$, and ${\bf Q}$ are $n\times n$, ${\bf H}$ is $r\times n$, ${\bf K}$ is $n\times r$, ${\bf R}$ is $r\times r$ and symmetric, ${\bf Q}$ is diagonal.

Problem 2:

$${\bf W} ={\bf A}({\bf I}_{2n}-{\bf L} {\bf B}){\bf W}{\bf A}^\top+{\bf C}, \;\;\;\;\;\;\;\;\text{where}\;\;\;\;{\bf L}\equiv {\bf W} {\bf B}^\top\left({\bf B}{\bf W} {\bf B}^\top+ {\bf R} \right)^{-1}$$

where ${\bf W}$, ${\bf A}$ and ${\bf C}$ are $2n\times 2n$, and ${\bf B}$ is $r\times 2n$, and ${\bf L}$ is $2n\times r$.

Relationship between the two problems:

$${\bf A} \equiv \left[ \begin{matrix} {\bf F} & {\bf0} \\ {\bf K} {\bf H} {\bf F} & ({\bf I}_{n}-{\bf K} {\bf H}){\bf F} \end{matrix} \right], \;\;\;\; {\bf B}^\top \equiv \left[ \begin{matrix} {\bf H}^\top & {\bf 0} \end{matrix} \right], \;\;\;\; {\bf C} \equiv \left[ \begin{matrix} {\bf Q} & {\bf Q}{\bf H}^\top{\bf K}^\top \\ {\bf K}{\bf H}{\bf Q} & {\bf K}{\bf H} {\bf Q}{\bf H}^\top{\bf K}^\top \end{matrix} \right]$$

also let $${\bf L} \equiv \left[ \begin{matrix} {\bf L}_{1} \\ {\bf L}_{2} \end{matrix} \right], \;\;\;\; {\bf W} \equiv \left[ \begin{matrix} {\bf W}_{11} & {\bf W}_{21}' \\ {\bf W}_{21} & {\bf W}_{22} \end{matrix} \right]$$

Want to prove:

$${\bf K} = \gamma {\bf L}_{1} + (1-\gamma) {\bf L}_{2}$$

All matrices are real.

Doing a bit of algebra, it is easy to show that

\begin{align} {\bf L}_{1} &={\bf W}_{11}{\bf H}^\top({\bf H}{\bf W}_{11}{\bf H}^\top+{\bf R})^{-1} \\[1.5ex] {\bf L}_{2} &={\bf W}_{21}{\bf H}^\top({\bf H}{\bf W}_{11}{\bf H}^\top+{\bf R})^{-1} \end{align}

and that

\begin{align} {\bf W}_{11} &= {\bf F}({\bf I}_n-{\bf L}_{1}{\bf H}){\bf W}_{11}{\bf F}^\top+{\bf Q} \\ {\bf W}_{21} &={\bf K}{\bf H}{\bf W}_{11}+({\bf I}_n-{\bf K}{\bf H}){\bf F}{\bf W}_{21}({\bf I}_n-{\bf H}^\top {\bf L}_{1}^\top){\bf F}^\top \end{align}