Setup:
Let $\gamma \in(0,1)$, ${\bf F},{\bf Q} \in \mathbb R^{n\times n}$, ${\bf H}\in \mathbb R^{n\times r}$, and ${\bf R}\in \mathbb R^{r\times r}$ be given and suppose that ${\bf P}$,${\bf W}$,${\bf X}\in \mathbb R^{n\times n}$, and ${\bf K}$,${\bf L}\in \mathbb R^{n\times r}$ satisfy
\begin{align} {\bf P} &={\bf F}({\bf I}_{n}-{\bf K} {\bf H}^\top){\bf P}{\bf F}^\top+{\bf Q}, \;\;\;\;\;\;\;\;\text{where}\;\;\;\;{\bf K}\equiv {\bf P} {\bf H}\left({\bf H}^\top{\bf P} {\bf H}+ \frac{1}{\gamma}{\bf R} \right)^{-1} \tag1\\[4ex] {\bf W} &={\bf F}({\bf I}_{n}-{\bf L} {\bf H}^\top){\bf W}{\bf F}^\top+{\bf Q}, \;\;\;\;\;\;\;\;\text{where}\;\;\;\;{\bf L}\equiv {\bf W} {\bf H}({\bf H}^\top {\bf W} {\bf H}+ {\bf R})^{-1} \tag2\\[4ex] {\bf X} &={\bf K}{\bf H}^\top {\bf W}+({\bf I}_n-{\bf K}{\bf H}^\top){\bf F}{\bf X}({\bf I}_n-{\bf H} {\bf L}^\top){\bf F}^\top \tag3\\ {\color{white}X} \end{align}
Moreover, assume that ${\bf P}$, ${\bf W}$, ${\bf R}$, and ${\bf Q}$ are symmetric, and ${\bf P}$, ${\bf W}$ and ${\bf X}$ are invertible.
Want to prove:
$${\bf K}=(\gamma{\bf I}_{n}+(1-\gamma) {\bf X} {\bf W}^{-1}){\bf L} \tag4$$
Some ideas and comments:
In the scalar case (with $n=r=1$) what works is to subtract $(2)$ from $(1)$ which eliminates ${\bf Q}$, then use that to solve for ${\bf F}^2$, i.e. (suppressing the $^\top$ notation) $$ {\bf F}^2 = \frac{(\gamma{\bf H}^2{\bf P}+{\bf R})({\bf H}^2{\bf W}+{\bf R})({\bf P}-{\bf W})}{{\bf R}^2({\bf P}-{\bf W})+(1-\gamma){\bf H}^2{\bf R}{\bf P}{\bf W})},$$ substitute this into $(3)$ and solve for ${\bf X}$ which yields $${\bf X} = \frac{(\gamma {\bf R}({\bf P}-{\bf W})+\gamma(1-\gamma){\bf H}^2{\bf P}{\bf W}}{(1-\gamma)(\gamma {\bf H}^2{\bf P}+{\bf R}))}. $$ Finally, solving for ${\bf X}$ using $(4)$ yields the same thing.
Notice that the equations $(1)$ and $(2)$ are Riccati equations on ${\bf P}$ and ${\bf W}$ respectively, which means that there are known methods (described here) to solve for ${\bf P}$ and ${\bf W}$. Though I was not able to make use of those solutions.
Equation $(3)$ is similar to a Lyapunov equation on ${\bf X}$, which means that a vectorization method (described here) allows one to obtain an equation for ${\bf X}$.
I think the proof will come from a procedure similar to the one that works for the scalar case. That is, subtracting $(2)$ from $(1)$, using this equation to eliminate ${\bf F}$ from $(3)$, then showing that the resulting equation implies $(4)$.
If you can prove it under additional assumptions that could also be helpful.
Here is a simple Matlab code that allows you to test the result:
gamma = 0.5;
F = [2, 2, 3; 4, 5, 6; 7, 8, 9]; % n x n
H = [1, 2; 3, 1; 2, 1]; % n x r
R = [3 , 1; 1, 3]; % r x r, symmetric
Q = [2, 0, 0; 0, 5, 0; 0, 0, 9]; % n x n, symmetric
n = length(F);
r = length(R);
I = eye(n);
P = I;
W = I;
X = I;
for i=1:1000
K = (P*H)/(H'*P*H+(R/gamma));
P = F*(P-K*H'*P)*F'+Q;
L = (W*H)/(H'*W*H+R);
W = F*(W-L*H'*W)*F'+Q;
X = K*H'*W+(I-K*H')*F*X*(I-H*L')*F';
end
disp(['K - (gamma*I+(1-gamma)*X*W^(-1))*L = ',...
num2str(sum(sum(abs(K-(gamma*I+(1-gamma)*X*(W^(-1)))*L))))])
This question is a boiled down version of another question: Forecast equivalence between two steady-state Kalman filters (with ${\bf L} \equiv {\bf L}_1$, ${\bf W} \equiv {\bf W}_{11}$, and ${\bf X} \equiv {\bf W}_{12}$).
