A question on independence of increments How could I prove the following?
Let $X=(X_t)_{t \in[0,1]}$ be a real-valued stochastic process on a probability space
$(\Omega,F,P)$ with $X_0=0$ a.s Show that the following statement as are equivalent
1 $X$ has independent increments ie. $\forall n \in \mathbb{N}$ and for every choice of 
$0 \leq t_0 < t_1...< t_n \leq 1$ we have that $X_{t_i}-X_{t_{i-1}} $ where $i=1,2,..n$ are independent
2 For every $0\leq t<u$,  $X_u-X_t$ is independent of $F_t^{X}=\sigma(X_s,s\leq t)$
How could I go about proving this,can someone give me a hint? I have been trying to use the definition of independence of Rv's ie X, Y are independent if the $\sigma$-algebra generated by the are independent but I am stuck. 
 A: Hints:


*

*(2) $\Rightarrow$ (1): Recall that (real-valued) random variabeles $Z_1,\ldots,Z_n$ are independent if, and only if, $$\mathbb{E} \exp \left( \imath \, \sum_{j=1}^n \xi_j Z_j \right) = \prod_{j=1}^n \mathbb{E}\exp(\imath \, \xi_j Z_j)$$ for all $\xi_1,\ldots,\xi_n \in \mathbb{R}$. Now fix $0 \leq t_0 < \ldots \leq t_n$. By assumption, $X_{t_n}-X_{t_{n-1}}$ is independent from $$\sum_{j=1}^{n-1} \xi_j (X_{t_j}-X_{t_{j-1}}).$$ Consequently, $$\begin{align*} \mathbb{E}\exp \left( \imath \, \sum_{j=1}^n \xi_j (X_{t_j}-X_{t_{j-1}}) \right) &= \mathbb{E} \left[ \exp \left( \imath \, \sum_{j=1}^{n-1} \xi_j (X_{t_j}-X_{t_{j-1}}) \right) \exp(\imath \, \xi_n (X_{t_n}-X_{t_{n-1}}) \right] \\ &= \mathbb{E} \exp \left( \imath \, \sum_{j=1}^{n-1} \xi_j (X_{t_j}-X_{t_{j-1}}) \right) \cdot \mathbb{E}\exp(\imath \, \xi_n (X_{t_n}-X_{t_{n-1}})). \end{align*}$$ Can you take it from here?

*(1) $\Rightarrow$ (2): Note that $$\begin{align*} \mathcal{F}_t^X &= \sigma \left( \bigcup_{n \in \mathbb{N}} \bigcup_{u_1<\ldots<u_n \leq t} \sigma(X_{u_1},\ldots,X_{u_n}) \right) \\ &= \sigma \left( \bigcup_{n \in \mathbb{N}} \bigcup_{u_1<\ldots<u_n \leq t} \sigma(X_{u_1},\ldots,X_{u_n}-X_{u_{n-1}}) \right), \end{align*}$$ i.e. $$\mathcal{D} := \bigcup_{n \in \mathbb{N}} \bigcup_{u_1<\ldots<u_n \leq t} \sigma(X_{u_1},\ldots,X_{u_n}-X_{u_{n-1}})$$ is a $\cap$-stable generator of $\mathcal{F}_t^X$. Conclude that the independence of $(X_{u_1},\ldots,X_{u_n}-X_{u_{n-1}})$ and $X_u-X_t$ for any sequence $0 \leq u_1 < \ldots < u_n \leq t$ already implies the claim.

