\notag \\
&=& \mathbb{E}_{\Sigma} [ \{ I \otimes (Y(t) - Y(s) )  \notag \\
&& - \int_s^t \pi_r( I \otimes (L+L^\ast) + \sigma_+ \otimes S \xi(r) +
\sigma_- \otimes S^\ast \xi^\ast(r) ) dr \} I \otimes K ]  \notag \\
&=& \mathbb{E}_{\Sigma} [ \{ I \otimes ( Y(t) - Y(s) )  \notag \\
&& - \int_s^t ( I \otimes (L(r)+L^\ast(r) ) + \sigma_+ \otimes S(r) \xi(r) +
\sigma_- \otimes S^\ast(r) \xi^\ast(r) ) dr \} I \otimes K ]  \notag \\
&=& \mathbb{E}_{\Sigma} [ \{ I \otimes (V(t) - V (s)) - \int_s^t ( \sigma_+
\otimes S(r) \nu \xi(r) + \sigma_- \otimes S^\ast(r) \nu^\ast \xi^\ast(r) )
dr \} I \otimes K ] =0.  \notag\end{aligned}$$ To see that this last expression is zero, we make use of Lemma \[lemma:Z-compensated-mtg\], the multiplicative factorization of the vacuum state, the fact that $V(t)$ has zero expectation in the vacuum state, to find that $$\begin{aligned}
\mathbb{E}_{\Sigma} [ I \otimes (V(t) - V(s)) I \otimes  K ] &=& w_{11} \mathbb{E}_{11}[
(V(t) - V(s)) K ] + w_{00} \mathbb{E}_{00}[ (V(t) - V(s)) K ]  \notag \\
&=& w_{11} ( \int_s^t \mathbb{E}_{10}[ S(r)K] \xi(r)dr + \int_s^t \mathbb{E}_{01}[ S^\ast(r) K] \xi^\ast(r)dr ) ,  \notag\end{aligned}$$ and $$\begin{aligned}
&& \mathbb{E}_{\Sigma} [ \int_s^t 