theorem $\cD_{2}=\sigma(\cC)={\widetilde}{\cF}_{\infty}^{Y}$. This completes the proof.

Next, we will prove that the law of $Y$ under $\overline{\bP}^{i}$ is the same as that of $X$ under $\bP^{i}$. As usual, $\overline{\bE}^{i}(\cdot\,;s_{1},\ldots,s_{n})$ will denote the expectation associated with $\overline{\bP}^{i}(\cdot\,;s_{1},\ldots,s_{n})$, for $i\in\mathbf{E}$ and $0<s_{1}<\ldots<s_{n}$. In the sequel, if there is no ambiguity, we will omit the parameters $s_{1},\ldots,s_{n}$ in $\overline{\bP}^{i}$ and $\overline{\bE}^{i}$.

\[thm:YBarPiXPi\] For any $i\in\mathbf{E}$ and $0<s_{1}<\ldots<s_{n}$, under $\overline{\bP}^{i}$, $Y$ is a time-inhomogeneous Markov chain with generator $\mG=\{\mG_{t},t\geq 0\}$. In particular, $X$ and $Y$ have the same law under respective probability measures $\bP^{i}$ and $\overline{\bP}^{i}$.

Let $u_0,u_{1},\ldots,u_{m}$ be such that $$\begin{aligned}
0=u_{0}\leq u_{1}<\ldots<u_{k_{1}}\leq s_{1}<u_{k_{1}+1}<\ldots<u_{k_{2}}\leq s_{2}<\ldots\leq s_{n}<u_{k_{n}+1}<\ldots<u_{k_{n+1}}=u_{m}.\end{aligned}$$ By , for any $i_{1},\ldots,i_{m}\in\mathbf{E}$, $$\begin{aligned}
\overline{\bP}^{i}(Y_{u_{m}}=i_{m}{\;|\;}& Y_{u_{m-1}}=i_{m-1},\ldots,Y_{u_{1}}=i_{1})=\frac{\overline{\bP}^{i}\left(Y_{u_{1}}=i_{1},\ldots,Y_{u_{m}}=i_{m}\right)}
{\overline{\bP}^{i}\left(Y_{u_{1}}=i_{1},\ldots,Y_{u_{m-1}}=i_{m-1}\right)}\\
&=\frac{\prod_{\ell=0}^{n}\left(\prod_{p=k_{\ell}+1}^{k_{\ell+1}}\!e^{(u_{p}-u_{p-1})
\mG_{\ell}}(i_{p-1},i_{p})\right)}{\left[\prod_{\ell=0}^{n-1}\left(\prod_{p=k_{\ell}+1}^{k_{\ell+1}}\!e^{(u_{p}-u_{p-1})\mG_{\ell}}(i_{p-1},i_{p})\right)\right]\left(\prod_{p=k_{n}+1}^{k_{n+1}-1}\!e^{(u_{p}-u_{p-1})\mG_{\ell}}(i_{p-1},i_{p})\right)}\\
&=e^{(u_{m}-u_{m-1})\mG_{n}}(i_{m-1},i_{m}).\end{aligned}$$ On the other hand, by again, $$\begin{aligned}
\overline{\bP}^{i}  (Y_{u_{m}}&=i_{m}{\;|\;}Y_{u_{m-1}}=i_{m-1})=\frac{\overline{\bP}^{i}\left(Y_{u_{m}}=i_{m}\,Y_{u_{m-1}}=i_{m-1}\right)}{\overline{\bP}^{i}\left(Y_{u_{m-1}}=i_{m-1}\right)}\\
&=\frac{\sum_{i_{1},\ldots,i_{m-2}\in\mathbf{E}}\overline{\bP}^{i}\left(Y_{u_{1}}=i_{1},\ldots,Y_{u_{m}}=i_{m}\right)}{\sum_{i_{1},\ldots,i_{m-2}\in\mathbf{E}}\overline{\bP}^{i}\left(Y_{u_{1}}=i_{1},\ldots,Y_{u_{m-1}}=i_{m-1}\right)}\\
&=\frac{\sum_{i_{1},\ldots,i_{m-2}\in\mathbf{E}}\prod_{\ell=0}^{n}\left(\prod_{p=k_{\ell}+1}^{k_{\ell+1}}\!e^{(u_{p}-u_{p-1})\mG_{\ell}}(i_{p-1},i_{p})\right)}{\sum_{i_{1},\ldots,i_{m-2}\in\mathbf{E}}\left[\prod_{\ell=0}^{n-1}\left(\prod_{p=k_{\ell}+1}^{k_{\ell+1}}\!e^{(u_{p}-u_{p-1})\mG_{\ell}}(i_{p-1},i_{p})\right)\right]\left(\prod_{p=k_{n}+1}^{k_{n+1}-1}\!e^{(u_{p}-u_{p-1})\mG_{\ell}}(i_{p-1},i_{p})\right)}\\
&=e^{(u_{m}-u_{m-1})\mG_{n}}(i_{m-1},i_{m}).\end{aligned}$$ Analogous argument carries for any $u_0<u_{1}<\ldots<u_{m}$, which completes the proof.

In analogy to $\varphi_t$ and $\tau_t^+$ we now define an additive functional $\psi$ given as $\psi_{t}:=\int_{0}^{t}v(Y_{u})\operatorname{d \!}u,\ t\geq 0$, and we consider the following first passage time $\rho_{t}^{+}:=\inf\left\{\left.r\geq 0\,\right|\psi_{r}>t\right\}, \ t\geq 0$.

We end this part of this section with the following corollary to Theorem \[thm:YBarPiXPi\].

\[thm:XYWHRelation\] For any $(s_{1},\ldots,s_{n})$ in $\Delta_{n}$, $c>0$, and $t>0$, $$\begin{aligned}
\label{eq:rho0+}
\Pi_{c}^{+}(i,j;s_{1},\ldots,s_{n})&=\overline{\bE}^{i}\left(e^{-c\rho_{0}^{+}}{\mathbbm{1}}_{\{Y_{\rho_{0}^{+}}=j\}};s_{1},\ldots,s_{n}\right),\quad i\in\mathbf{E}^{-},\,j\in\mathbf{E}^{+},\\
\label{eq:rhot+} \Psi_{c}^{+}(t,i,j;s_{1},\ldots,s_{n})&=\overline{\bE}^{i}\left(e^{-c\rho_{t}^{+}}{\mathbbm{1}}_{\{Y_{\rho_{t}^{+}}=j\}};s_{1},\ldots,s_{n}\right),\quad i\in\mathbf{E}^{+},\,j\in\mathbf{E}^{+}.\end{aligned}$$ In 