$$\begin{aligned}
\mathbf{E}\left[H\left(\beta_{t+h},\beta_{t}\right)\mathbb{I}_{\left\{ \tau>t\right\}}L\left(\eta_{1},...,\eta_{n}\right)\right] &=\mathbf{E}\left[H\left(\beta_{t+h},\beta_{t}\right)\mathbb{I}_{\left\{ \tau>t\right\} }\right]\mathbf{E}\left[L\left(\eta_{1},...,\eta_{n}\right)\right]\end{aligned}$$

Using the formula of total probability and Lemma \[lem:If–is\] we get $$\begin{aligned}
\lefteqn{\mathbf{E}\left[H\left(\beta_{t+h},\beta_{t}\right)\mathbb{I}_{\left\{ \tau>t\right\} }L\left(\eta_{1},...,\eta_{n}\right)\right]}\\
&=\int_{\left(t,+\infty\right)}\mathbf{E}\left[H\left(\beta_{t+h},\beta_{t}\right)\mathbb{I}_{\left\{ \tau>t\right\} }L\left(\eta_{1},...,\eta_{n}\right)|\tau=r\right]dF\left(r\right)\\
&=\int_{\left(t,+\infty\right)}\mathbf{E}\left[H\left(\beta_{t+h}^{r},\beta_{t}^{r}\right)L\left(\eta_{1},...,\eta_{n}\right)\right]dF\left(r\right)\\
&=\int_{\left(t,+\infty\right)}\mathbf{E}\left[H\left(\beta_{t+h}^{r},\beta_{t}^{r}\right)\right]dF\left(r\right)\mathbf{E}\left[L\left(\eta_{1},...,\eta_{n}\right)\right]\\
&=\mathbf{E}\left[H\left(\beta_{t+h},\beta_{t}\right)\mathbb{I}_{\left\{ \tau>t\right\} }\right]\mathbf{E}\left[L\left(\eta_{1},...,\eta_{n}\right)\right].\end{aligned}$$ The proof of the lemma is finished.

We now prove (\[eq:equivalentMK1\]) for our special choice of $A$. From Lemma \[lem:The-random-variables\] above we have $$\begin{aligned}
\int_{A\cap\left\{ \tau>t\right\} }f\left(\beta_{t+h}\right)d\mathbf{P} & =\mathbf{E}\left[H\left(\beta_{t+h},\beta_{t}\right)\mathbb{I}_{\left\{ \tau>t\right\} }L\left(\xi_{1},...,\xi_{n}\right)\right]\\
 & =\mathbf{E}\left[H\left(\beta_{t+h},\beta_{t}\right)\mathbb{I}_{\left\{ \tau>t\right\} }L\left(\eta_{1},...,\eta_{n}\right)\right]\\
 &  =\mathbf{E}\left[f\left(\beta_{t+h}\right)g\left(\beta_{t}\right)\mathbb{I}_{\left\{ \tau>t\right\} }\right]\mathbf{E}\left[L\left(\eta_{1},...,\eta_{n}\right)\right]\\
 & =\mathbf{E}\left[\mathbf{E}\left[f\left(\beta_{t+h}\right)|\beta_{t}\right]g\left(\beta_{t}\right)\mathbb{I}_{\left\{ \tau>t\right\} }\right]\mathbf{E}\left[L\left(\eta_{1},...,\eta_{n}\right)\right]\\
 & =\mathbf{E}\left[\mathbf{E}\left[f\left(\beta_{t+h}\right)|\beta_{t}\right]g\left(\beta_{t}\right)\mathbb{I}_{\left\{ \tau>t\right\} }L\left(\eta_{1},...,\eta_{n}\right)\right]\\
 & =\mathbf{E}\left[\mathbf{E}\left[f\left(\beta_{t+h}\right)|\beta_{t}\right]g\left(\beta_{t}\right)\mathbb{I}_{\left\{ \tau>t\right\} }L\left(\xi_{1},...,\xi_{n}\right)\right]\\
 & =\int_{A\cap\left\{ \tau>t\right\} }\mathbf{E}\left[f\left(\beta_{t+h}\right)|\beta_{t}\right]d\mathbf{P},\end{aligned}$$ which proves that (\[eq:equivalentMK1\]) is true and this ends the proof.

Note that the Markov property is trivially extended to the completed filtration $\mathbb{F}^{P}$.

\[sec:Conditional-Expectations\] Bayes Estimates of the Default Time $\tau$
===========================================================================

The basic aim of the present section is to provide estimates of the a priori unknown default time $\tau$ based on the observation of the information process $\beta$ up to time $t$. For fixed $t\geq 0$, the observation is represented by the $\sigma$-algebra $\mathcal{F}^P_t$ and, because of the Markov property, the observation of $\beta_t$ would be sufficient. To this end it is natural to exploit the Bayesian approach.

The idea is to use the knowledge gained from the observation of the flow $(\beta_t, 