equalities. In particular, ${\widetilde}{\bP}^{i}(N_{t+s}=n_{2}|N_{t}=n_{1},Y_{t}=j)$ does not depend on the choice of $j\in\mathbf{E}$.

As far as the first equality in , for any $t,s\geq 0$ and $0\leq n_{1}\leq n_{2}\leq n$, $$\begin{aligned}
{\widetilde}{\bP}^{i}\left(\left.N_{t+s}=n_{2}\,\right|N_{t}=n_{1}\right)&=\frac{{\widetilde}{\bP}^{i}\left(N_{t+s}=n_{2},N_{t}=n_{1}\right)}{{\widetilde}{\bP}^{i}\left(N_{t}=n_{1}\right)}=\frac{\sum_{\ell\in\mathbf{E}}{\widetilde}{\bP}^{i}\left(N_{t+s}=n_{2},N_{t}=n_{1},Y_{t}=j\right)}{\sum_{j\in\mathbf{E}}{\widetilde}{\bP}^{i}\left(N_{t}=n_{1},Y_{t}=j\right)}\\
&=\frac{\sum_{j\in\mathbf{E}}{\widetilde}{\bP}^{i}\left(\left.N_{t+s}=n_{2}\,\right|N_{t}=n_{1},Y_{t}=j\right){\widetilde}{\bP}^{i}\left(N_{t}=n_{1},Y_{t}=j\right)}{\sum_{j\in\mathbf{E}}{\widetilde}{\bP}^{i}\left(N_{t}=n_{1},Y_{t}=j\right)}\\
&=\frac{\sum_{j\in\mathbf{E}}{\widetilde}{\bP}^{i}\left(N_{t}=n_{1},Y_{t}=j\right)}{\sum_{j\in\mathbf{E}}{\widetilde}{\bP}^{i}\left(N_{t}=n_{1},Y_{t}=j\right)}\,e^{s{\widetilde}{\mG}_{N}}(n_{1},n_{2})=e^{s{\widetilde}{\mG}_{N}}(n_{1},n_{2}).\end{aligned}$$

We are ready to complete the proof of the proposition. Towards this end we observe that, for any $m\in\bN$, $0=t_{0}\leq t_{1}<\ldots<t_{m}$, and any $0\leq n_{1}\leq\ldots\leq n_{m}\leq n$, $$\begin{aligned}
{\widetilde}{\bP}^{i}(N_{t_{m}}=n_{m}\,{\;|\;}N_{t_{m-1}}  = & \, n_{m-1},\ldots,N_{t_{1}}=n_{1})=\frac{{\widetilde}{\bP}^{i}\left(N_{t_{1}}=n_{1},\ldots,N_{t_{m}}=n_{m}\right)}{{\widetilde}{\bP}^{i}\left(N_{t_{1}}=n_{1},\ldots,N_{t_{m-1}}=n_{m-1}\right)}\\
& =\frac{\sum_{j_{1},\ldots,j_{m}\in\mathbf{E}}{\widetilde}{\bP}^{i}\left(N_{t_{1}}=n_{1},Y_{t_{1}}=j_{1};\ldots;N_{t_{m}}=n_{m},Y_{t_{m}}=j_{m}\right)}{\sum_{j_{1},\ldots,j_{m-1}\in\mathbf{E}}{\widetilde}{\bP}^{i}\left(N_{t_{1}}=n_{1},Y_{t_{1}}=j_{1};\ldots;N_{t_{m-1}}=n_{m-1},Y_{t_{m}}=j_{m-1}\right)}\\
& =\frac{\sum_{j_{1},\ldots,j_{m}\in\mathbf{E}}\prod_{k=1}^{m}{\widetilde}{\bP}^{i}\left(\left.N_{t_{k}}=n_{k},Y_{t_{k}}=j_{k}\,\right|N_{t_{k-1}}=n_{k-1},Y_{t_{k-1}}=j_{k-1}\right)}{\sum_{j_{1},\ldots,j_{m-1}\in\mathbf{E}}\prod_{k=1}^{m-1}{\widetilde}{\bP}^{i}\left(\left.N_{t_{k}}=n_{k},Y_{t_{k}}=j_{k}\,\right|N_{t_{k-1}}=n_{k-1},Y_{t_{k-1}}=j_{k-1}\right)}\\
&=\sum_{j_{m}\in\mathbf{E}}{\widetilde}{\bP}^{i}\left(\left.N_{t_{m}}=n_{m},Y_{t_{m}}=j_{m}\,\right|N_{t_{m-1}}=n_{m-1},Y_{t_{m-1}}=j_{m-1}\right)\\
&={\widetilde}{\bP}^{i}\left(\left.N_{t_{m}}=n_{m}\,\right|N_{t_{m-1}}=n_{m-1},Y_{t_{m-1}}=j_{m-1}\right)\\
&={\widetilde}{\bP}^{i}\left(\left.N_{t_{m}}=n_{m}\,\right|N_{t_{m-1}}=n_{m-1}\right)=e^{(t_{m}-t_{m-1}){\widetilde}{\mG}_{N}}(n_{m-1},n_{m}),\end{aligned}$$ where we used the Markov property of $Z=(N,Y)$ under ${\widetilde}{\bP}^{i}$ in the third equality, and the result of Step 2 in the last two equalities. The proof is complete.

Let $\widetilde \bF^Y=(\widetilde {\cF}_{t}^{Y})_{t\geq 0}$ be the filtration generated by $Y$, and let ${\widetilde}{\cF}_{\infty}^{Y}=\sigma(\bigcup_{t\geq 0}{\widetilde}{\cF}_{t}^{Y})$. For each $i\in\mathbf{E}$, we will construct a probability measure $\overline{\bP}^{i}$ on $({\widetilde}{\Omega},{\widetilde}{\cF}_{\infty}^{Y})$ such that, the law of $Y$ under $\overline{\bP}^{i}$ is the same as the law of $X$ under $\bP^{i}$. Moreover, we will establish a connection between $\overline{\bP}^{i}$ and ${\widetilde}{\bP}^{i}$. For this purpose, we first let $$\begin{aligned}
S_{k}:=\inf\left\{t\geq 0\,|\,N_{t}=k\right\},\quad k=1,\ldots,n.\end{aligned}$$ We will now derive the joint density of $N$, and $(S_{1},\ldots,S_{n})$ under ${\widetilde}{\bP}^{i}$. For that, we set $$\begin{aligned}
\label{eq:StoT}
T_{1}:=S_{1},\quad T_{k}:=S_{k}-S_{k-1},\quad k=2,\ldots,n.\end{aligned}$$ It is shown in [@Syski1992 Section 1.1.4] that $T_{k}$’s are independent 