\otimes XS ) \nu \xi(t) + \tilde \pi_t( \sigma_-A \otimes
S^\ast X) \nu^\ast \xi^\ast(t) \} ]dt  \notag\end{aligned}$$ and $$\begin{aligned}
&& \mathbb{E}_\Sigma [ A \otimes (d\tilde \pi_t(A\otimes X) c_f(t)) ] 
\\
 &=& \mathbb{E}_\Sigma [ (I \otimes c_f(t) \{ \mathcal{F}_t(A\otimes X) + \mathcal{H}_t(A\otimes X) \tilde \pi_t( I \otimes
(L+L^\ast))\nonumber  \\
&& +\mathcal{H}_t(A\otimes X) \pi_t( ( \sigma_+ \otimes S) \nu \xi(t) + (\sigma_-
\otimes S^\ast ) \nu^\ast \xi^\ast(t) ) \}  \notag \\
&& + (I \otimes f(t) c_f(t) ) \{ \tilde \pi_t( A \otimes X) \tilde \pi_t( I \otimes
(L+L^\ast) ) + \mathcal{H}_t  \notag \\
&& + \tilde \pi_t( A \otimes X)\tilde  \pi_t( \sigma_+ \otimes S \nu \xi(t) + \sigma_-
\otimes S^\ast \nu^\ast \xi^\ast(t) ) \} ] dt.  \notag\end{aligned}$$ Now equating coefficients of $c_f(t)$ and $f(t) c_f(t)$ we solve for $\mathcal{F}_t(A\otimes X)$ and $\mathcal{H}_t(A\otimes X)$ to obtain the filter equation.

We now prove the martingale property $\mathbb{E}_{\Sigma} [ I \otimes ( W(t) -
W(s) ) \, \vert \, I \otimes \mathscr{Y}_s ] = 0$, that is, $
\mathbb{E}_{\Sigma} [ I \otimes ( W(t) - W(s) ) (I \otimes K) ] =0 
$ for all $K \in \mathscr{Y}_t$. Now $$\begin{aligned}
&& \mathbb{E}_{\Sigma} [ I \otimes ( W(t) - W(s) ) (I \otimes K) ]  