] ,  \label{eq:Kt-10} \\
\mathbb{E}_{01}[ K(t) ] &=& \mathbb{E}_{01}[ \int_0^t M_0(s) ds ] + \mathbb{E}_{00}[ \int_0^t M_- (s) \xi(s) ds ] ,  \label{eq:Kt-01} \\
\mathbb{E}_{00}[ K(t) ] &=& \mathbb{E}_{00}[ \int_0^t M_0(s) ds ] .
\label{eq:Kt-00}\end{aligned}$$

Using (\[eq:xi-phi\]), the expressions $B^-_t(\xi) = \int_0^t \xi^\ast(s)
dB(s)$, $B^{-\ast}_t(\xi) = \int_0^t \xi(s) dB^\ast(s)$, and the Ito rule we have $$\begin{aligned}
\mathbb{E}_{11}[ d K(t) ] &=& \mathbb{E}_{00} [ d ( B^-_t (\xi) K(t)
B^{-\ast}_t (\xi) + r(t) K(t) ) ]  \notag \\
&=& \mathbb{E}_{00} [ B^-_t (\xi) M_0(t) B^{-\ast}_t (\xi) + r(t) M_0(t)
\notag \\
&& + M_+(t) B^{-\ast}_t(\xi) \xi^\ast(t) + B^-_t (\xi) M_-(t) \xi(t) +M_1(t) |\xi(t)|^2] dt
\notag \\
&=& \mathbb{E}_{11} [ M_0(t) ] dt + \mathbb{E}_{00} [ M_+(t) B^{\ast} (\xi)
\xi^\ast(t) + B(\xi) M_-(t) \xi(t) + M_1 |\xi(t)|^2] dt .\end{aligned}$$ This last line is justified since $M_\pm$ are adapted and $\mathbb{E}_{00}[
B^+_t(\xi) ] = 0$. That is, $$\mathbb{E}_{11}[ dK(t) ] = \mathbb{E}_{11} [ M_0(t) ] dt + \mathbb{E}_{01}[
M_+(t) ] \xi^\ast(t) dt + \mathbb{E}_{10} [ M_-(t) ] \xi(t) dt +  \mathbb{E}_{00}[
\int_0^t M_1 (s) \vert \xi(s) \vert^2 ds ] .$$ This proves (\[eq:Kt-11\]). The remaining expressions are proven in a similar manner. [$\Box$]{}

We will first express the master equation in Heisenberg form 