Centered independent increments process is a martingale

Question

Let $(X_n)$ be an centered integrable process with independent increments (which as far as I understand means that $(X_{n+1}-X_n)_{n\in \mathbb N}$ is independent). While showing that $(X_n)$ is a martingale w.r.t. $(\mathcal F_n)$ where $\mathcal F_n=\sigma(X_1, \ldots, X_n)$ we use the following step: $$\mathbb E (X_{n+1}-X_n \ | \ \mathcal F_n)=\mathbb E (X_{n+1}-X_n )$$ which seems to be implied by the fact that $\sigma(X_{n+1}-X_n)$ and $\sigma(X_1, \ldots, X_n)$ are independent. But why are they independent? Well, we need to use the independence of increments somewhere, so if we'd manage to show that $$\sigma(X_1, \ldots, X_n)=\sigma(X_n-X_{n-1}, \ldots, X_1-X_0)$$ (taking $X_0 =0$) then we'd be done. I've tried showing it with induction yet didn't succeed.

Answer 1

We use the fact that if there exists measurable function $f$ such that $Y = f(X_1, \cdots, X_n)$, then $\sigma(Y)\subset \sigma(X_1, \cdots, X_n)$.

Thus obviously $\sigma(X_1 - X_0, \cdots, X_n - X_{n-1}) \subset \sigma(X_1, \cdots,X_n)$, since $X_0 = 0$ is deterministic.

On the other hand, $X_k = \sum_{i=1}^k(X_{i} - X_{i-1})$, thus $\sigma(X_1, \cdots,X_n) \subset \sigma(X_1 - X_0, \cdots, X_n - X_{n-1}) $