It was suggested* to me that if we have a stochastic process with independent increments, and $T > t$, then $$ E(X_{T-t}| \mathcal{F}_t) = X_{T-t} $$ where $\mathcal{F}_t$ is the filtration at time $t$. This relation allegedly holds because $X_{T-t}$ is $\mathcal{F}_t$-measurable.
I am unable to reconcile this relation. I have had some experience with filtrations in discrete models, so as far as I know, the only processes $X_T$ that are $\mathcal{F}_t$-measurable when $T>t$, are the predictable processes, for example a deterministic interest rate process.
Could someone help me elucidate this relation?
To elucidate relations, let's take a brownian motion $B_t$ as an example and look carefully when we use which property. When looking up the definition of brownian motion here (http://www.randomservices.org/random/brown/Standard.html) we get the definitions of stationarity increments and independant increments at the same time.
From the independancy of increments we can deduce the relation \begin{equation} \mathbb{E}[B_{T}-B_{t}|\mathcal{F}_t]=B_{T}-B_{t}. \end{equation} From the stationarity of the increments we further know \begin{equation} (B_{T}-B_{t}) \tilde{} B_{T-t} \end{equation} and could deduce \begin{equation} \mathbb{E}[B_{T}-B_{t}]=\mathbb{E}[B_{T-t}]. \end{equation} Brownian motion also has other properties, like e.g. the Martingale property, i.e. \begin{equation} \mathbb{E}[B_{t_2}|\mathcal{F}_{t_1}]=B_{t_1}. \end{equation} for t_1<=t_2. But your suggestion is wrong and the brownian motion serves as counter example: Assume your equation would hold for brownian motion. Take $T=3$ and $t=1$: \begin{equation} \mathbb{E}[B_{2}|\mathcal{F}_1]=B_{2}. \end{equation} That contradicts the martingale property, since $B_1\neq B_2$ with probability $>0$ (Exercise!).