Let $E$ be an Abelian group. Let $X$ be a right-continuous process with values in $(E,\mathcal{E})$ (where $\mathcal{E}$ denotes the $\sigma$-algebra on $E$), defined on $\Omega, \mathcal{F}_t,P)$. Suppose that $X$ has stationary, independent increments.
I now want to show the following :
i) $X$ is Markov process with initial distribution $P(X_0 \in \cdot)$.
ii) 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 $\mathcal{F}_\tau$. 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 $P_0$.
Yet I don't know how to start proving i) for ii) however 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 f_k (Y_{t_k}) \mid \mathcal{F}_\tau \right) &= \mathbb{E}_\nu \left( \prod_k f_k (X_{\tau+t_k} - X_\tau) \mid \mathcal{F}_\tau \right) \\ &= \mathbb{E}_{X_\tau} \left( \prod_k 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. I think I need to use induction to prove this, yet don't know the details.
Could anyone help me with i) and help me with details of ii). Many thanks!
For part (i), we don't even need right-continuity or stationary increments from $X$. Notice that if $t>s$, then for any (bounded-measurable) function $f: E \to \mathbb{R}$, we can write $$E[f(X_t)|\mathcal{F}_s] = E[f(X_t-X_s+X_s)|\mathcal{F}_s] = E[g(X_t-X_s,X_s)|\mathcal{F}_s]$$ where $g(x,y) = f(x+y)$. Now, we know that $X_s$ is $\mathcal{F}_s$-measurable and $X_t-X_s$ is independent of $\mathcal{F}_s$. Therefore, using this question (Conditional Expectation of Functions of Random Variables satisfying certain Properties; see the comment below the question), we know that $$E[g(X_t-X_s,X_s)|\mathcal{F}_s] = E[g(X_t-X_s,X_s)|X_s] = E[f(X_t)|X_s]$$It follows that $(X_t)_t$ is a Markov Process. The initial distribution is clearly $P^{X_0}$.
For part (ii), I'll come back later, assuming someone else hasn't already answered...