I'm trying to prove that $X_{t} - X_{s}$ independent of $\mathcal{F}^{X}_{s}$ for $0 \leq s < t$, but I have a hard time getting started. My idea was to use expectations but I'm not quite convinced that I'm on the right path.
The problem is as follows. Assume the usual probability space.
- We have a random process $(X_{t})_{t\in \mathbb{R}_{+}}$ with independent increments.
- We have that the natural filtration is given by $\mathcal{F}_{t}^{X} = \sigma(\{X_{s}:0 \leq s < t\})$ for $t < \infty$ and $\mathcal{F}_{t}^{X} = \sigma(\{X_{s}:s \in \mathbb{R}_{+}\})$ for $t = \infty$.
I want now to prove that $X_{t} - X_{s}$ independent of $\mathcal{F}^{X}_{s}$ for $0 \leq s < t$.
Can someone explain me how to do so.
I think this holds only when $X_0$ is deterministic.
By definition, $(X_t)_{t\in\mathbb R}$ has independent increments iff for any $m\in\mathbb N$ and any choice $t_0<t_1<...<t_m$ the variables $$ (X_{t_1}-X_{t_0}),...,(X_{t_m}-X_{t_{m-1}}) $$ are stochastically independent.
Let $0\le s<t$ and take $t_m=t,t_{m-1}=s\,.$ Then $X_t-X_s$ is independent of $$ (X_{t_1}-X_{t_0})+...+(X_{t_{m-2}}-X_{t_{m-3}})=X_{t_{m-2}}-X_{t_0}\,. $$ Since $t_{m-2}<s$ is arbitrary and $t_0$ can be $0$ it follows from the assumption that $X_0$ is determinisic that $X_t-X_s$ is independent of any $X_{t_{m-2}}$ with $t_{m-2}<s\,.$ In other words, $X_t-X_s$ is independent of ${\cal F}^X_s\,.$