Let $X$ be a right-continuous process with values in $(E,\mathcal{E})$, defined on $(\Omega, \mathcal{F}_t,P)$. Suppose that $X$ has stationary, independent increments.
I now want to show the following with knowledge that $X$ is in fact a Markov process:
Let $\tau$ be a finite $(\mathcal{F}_t)_t$-stopping time. Then the process $X(\tau) = (X_{\tau + t} - X_\tau)_{t \geq 0}$ is independent of $\mathcal{F}_\tau$ and it is a Markov process, adapted to the filtration $(\mathcal{F}_{\tau+t})_t$. The distribution $P_\tau$ of $X(\tau)$ is the same as the distribution of $X- X_0$ under $P_0$.
My attempt of a proof:
First how could I prove independence? However when moving on we can put $Y_t = X_{\tau+t} - X_\tau$, $t \geq 0$. For $t_1<\ldots < t_n$ and functions $f_1,\ldots, f_n \in b \mathcal{E}$ (bounded functions) we have \begin{align*} \mathbb{E}_\nu \left( \prod_k^n f_k (Y_{t_k}) \mid \mathcal{F}_\tau \right) &= \mathbb{E}_\nu \left( \prod_k^n f_k (X_{\tau+t_k} - X_\tau) \mid \mathcal{F}_\tau \right) \\ &= \mathbb{E}_{X_\tau} \left( \prod_k^n f_k (X_{t_k} - X_0)\right) \end{align*} $P_\nu$-a.s., by the strong Markov property. As consequence the proof is complete once we have shown that for an arbitrary $x\in E$ $$ \mathbb{E}_x \left( \prod_{k=1}^n f_k(X_{t_k} - X_0)\right) = P_{t_1}f_1 \cdots P_{t_n-t_{n-1}} f_n(0),$$ which is the characterisation of a Markov process. Induction is needed but I can't seem to figure out the details. Any help is greatly appreciated thanks.