derive formulas which include the Bayes estimates discussed in Section \[sec:Conditional-Expectations\] as well as the prognose of the information process $\beta$ at some time $u$, the latter being related with the Markov property which has been proven in Section \[sec:Definition-and-First\] (see Theorem \[TEO:The-information-process\]). First we will state a lemma that will be used in the proof of the main results of this section.

\[lem:AUXlemma\]Let $0\leq t<u$ and $g$ be a measurable function on $(0,+\infty)\times\mathbb{R}$ such that $g(\tau,\beta_{u})$ is integrable. Then it holds $\mathbf{P}$-a.s. $$\begin{aligned}
\mathbf{E}\left[g\left(\tau,\beta_{u}\right)\mathbb{I}_{\left\{ t<\tau\right\} }|\mathcal{F}_{t}^{P}\right] & =\mathbf{E}\left[g\left(\tau,\beta_{u}\right)|\beta_{t}\right]\mathbb{I}_{\left\{ t<\tau\right\} }\\
 & =\mathbf{E}\left[\mathbf{E}\left[g\left(\tau,\beta_{u}\right)|\sigma\left(\tau\right)\vee\sigma\left(\beta_{t}\right)\right]|\beta_{t}\right]\mathbb{I}_{\left\{ t<\tau\right\} }\\
 & =\mathbf{E}\left[\left(\mathbf{E}\left[g\left(r,\beta_{u}^{r}\right)|\beta_{t}^{r}\right]\right)_{r=\tau}|\beta_{t}\right]\mathbb{I}_{\left\{ t<\tau\right\} }.\end{aligned}$$

It is clear that the first equality holds true due to the fact that $g(\tau,\beta_{u})\mathbb{I}_{\{ t<\tau\} }$ is $\mathcal{F}_{t,\infty}$-measurable and, hence, Lemma \[lem:basic MKV\] can be applied. The second equality is obvious. Let $h$ be an arbitrary bounded Borel function. Using Lemma \[lem:If–is\] we have that $$\begin{aligned}
\mathbf{E}\left[g\left(\tau,\beta_{u}\right)h\left(\beta_{t}\right)\mathbb{I}_{\left\{ t<\tau\right\} }\right] & =\int_{\left(t,+\infty\right)}\mathbf{E}\left[g\left(r,\beta_{u}^{r}\right)h\left(\beta_{t}^{r}\right)\right]dF\left(r\right)\\
 & =\int_{\left(t,+\infty\right)}\mathbf{E}\left[\mathbf{E}\left[g\left(r,\beta_{u}^{r}\right)h\left(\beta_{t}^{r}\right)|\beta_{t}^{r}\right]\right]dF\left(r\right)\\
 & =\int_{\left(t,+\infty\right)}\mathbf{E}\left[\mathbf{E}\left[g\left(r,\beta_{u}^{r}\right)|\beta_{t}^{r}\right]h\left(\beta_{t}^{r}\right)\right]dF\left(r\right)\\
 & =\mathbf{E}\left[\mathbf{E}\left[g\left(r,\beta_{u}^{r}\right)|\beta_{t}^{r}\right]_{r=\tau}h\left(\beta_{t}\right)\mathbb{I}_{\left\{ t<\tau\right\} }\right]\,,\end{aligned}$$ that is, $$\mathbf{E}\left[\left(g\left(\tau,\beta_{u}\right)-\mathbf{E}\left[g\left(r,\beta_{u}^{r}\right)|\beta_{t}^{r}\right]_{r=\tau}\right)h\left(\beta_{t}\right)\mathbb{I}_{\left\{ t<\tau\right\} }\right]=0.$$ But $h$ is arbitrary and thus $$\begin{aligned}
\mathbf{E}\left[g\left(\tau,\beta_{u}\right)|\beta_{t}\right] & =\mathbf{E}\left[\mathbf{E}\left[g\left(r,\beta_{u}^{r}\right)|\beta_{t}^{r}\right]_{r=\tau}|\beta_{t}\right],\end{aligned}$$ $\mathbf{P}$-a.s. on $\{ t<\tau\} $, for $t<u$.

The next theorem is prepared by the following result.

\[cor:Let–be-BIS\] Let $t\geq0$ and $g$ be a measurable function such 