Independent increments $\iff$ increments independent of natural filtration? Let $X=(X_t)_{t\in T}$ be a stochastic process with $X_0=0$, and $(\mathcal{F}^X_t)_{t\in T}$ the natural filtration, that is, $\mathcal{F}^X_t=\sigma(X_r : r\le t)$. Is it true that the following are equivalent?

*

*$X$ has independent increments, i.e. for $t_1<\ldots<t_n$ in $T$ we have $(X_{t_n}-X_{t_{n-1}}),\ldots (X_{t_2}-X_{t_1})$ are independent

*for every $s<t$ in $T$ we have $X_t-X_s$ is independent of $\mathcal{F}^X_s$.

For $1)\implies 2)$ I first thought I could just say that if $1)$ is true then $X_t-X_s$ independent of $X_r-0$ for every $r\le s$ and then conclude that $\sigma(X_t-X_s)$ is independent of $\sigma(X_r : r\le s)$ but it seems I cannot conclude this right away as I saw in this example.
For $2)\implies 1)$ I was thinking that if $t_1<\ldots<t_n$ then $2)$ implies that $X_{t_n}-X_{t_{n-1}}$ is independent of $\sigma(X_r : r\le t_{n-1})$. But then in particular $X_{t_n}-X_{t_{n-1}}$ is independent of $\sigma(X_{t_1},\ldots,X_{t_{n-1}})$. But considering the measurable map $f(x_1,\ldots,x_{n-1})=(x_{n-1}-x_{n-2},\ldots,x_2-x_1)$ we get that $X_{t_n}-X_{t_{n-1}}$ is independent of $(X_{t_{n-1}}-X_{t_{n-2}},\ldots,X_{t_2}-X_{t_1})$ but I don't think think this allows me to conclude that all the increments are independent.
 A: $\newcommand{\indep}{\perp\!\!\perp}$
In light of recent things I learned, I will try to answer my own question.
The more difficult part is to prove
1)$\implies$ 2) :
Consider $s_1<\ldots<s_n<s<t$. If the increments $X_t-X_s, X_s-X_{s_n},\ldots, X_{s_2}-X_{s_1},X_{s_1}-\underbrace{X_0}_{=0}$ are independent then in particular $\sigma(X_t-X_s)$ is independent of $\sigma(X_{s_n}-X_{s_{n-1}},\ldots,X_{s_2}-X_{s_1},X_{s_1})$. However, by definition of independence if a sigma-field $\mathcal{F}$ is independent of sigma-fields $\mathcal{G}_i$  for $i\in I$ then $\mathcal{F}$ is independent of the class $\bigcup\limits_{i\in I}\mathcal{G}_i$. We thus have $$\sigma(X_t-X_s)\indep\bigcup\limits_{n\in\mathbb{N}}\bigcup\limits_{0\le s_1<\ldots<s_n\le s}\sigma(X_{s_n}-X_{s_{n-1}},\ldots,X_{s_2}-X_{s_1},X_{s_1}).$$ However, we also have $$\sigma(X_{s_n}-X_{s_{n-1}},\ldots,X_{s_2}-X_{s_1},X_{s_1})=\sigma(X_{s_1},\ldots,X_{s_n}).$$ $"\subset"$ : $X_{s_{i+1}}-X_{s_i}$ are $\sigma(X_{s_1},\ldots,X_{s_n})$-measurable since the sum of measurable functions are measurable.
$"\supset"$ :$X_{s_i}=X_{s_1}+X_{s_2}-X_{s_1}+\ldots+ X_{s_i}-X_{s_{i-1}}$ is $\sigma(X_{s_n}-X_{s_{n-1}},\ldots,X_{s_2}-X_{s_1},X_{s_1})$-measurable again because the sum of measurable maps are measurable. So until here we have $$\sigma(X_t-X_s)\indep\bigcup\limits_{n\in\mathbb{N}}\bigcup\limits_{0\le s_1<\ldots<s_n\le s}\sigma(X_{s_1},\ldots,X_{s_n}).$$
Moreover, $\bigcup\limits_{n\in\mathbb{N}}\bigcup\limits_{0\le s_1<\ldots<s_n\le s}\sigma(X_{s_1},X_{s_n})$ is an $\cap$-stable system. Indeed take $A, B$ in this set. Then by the definition of union, $A\in \sigma(X_{a_1},\ldots,X_{a_k})$ and $B\in\sigma(X_{b_1},\ldots,X_{b_{\ell}})$ but since both these sigma fields are contained in $\sigma(X_{a_1},\ldots,X_{a_k},X_{b_1},\ldots,X_{b_{\ell}})$ in particular $A,B\in\sigma(X_{a_1},\ldots,X_{a_k},X_{b_1},\ldots,X_{b_{\ell}})$ and so $A\cap B\in \sigma(X_{a_1},\ldots,X_{a_k},X_{b_1},\ldots,X_{b_{\ell}})$ since sigma-fields are closed unter intersection.
Then there is a theorem that states that if  a $\sigma$-field $\mathcal{F}$ and a class $C$ are independent and $C$ is a $\cap$-stable system then $\mathcal{F}$ is independent of $\sigma(C)$ (the proof of this theorem is based on the $\pi-\lambda$-theorem). It then holds that $$\sigma(X_t-X_s)\indep\sigma\left(\bigcup\limits_{n\in\mathbb{N}}\bigcup\limits_{0\le s_1<\ldots<s_n\le s}\sigma(X_{s_1},\ldots,X_{s_n})\right).$$ To conclude we note that $\mathcal{F}^X_s=\sigma\left(\bigcup\limits_{n\in\mathbb{N}}\bigcup\limits_{0\le s_1<\ldots<s_n\le s}\sigma(X_{s_1},\ldots,X_{s_n})\right)$
$"\subset"$ : Taking $n=1$ in the union, we get  $\mathcal{F}^X_s:=\sigma\left(\bigcup\limits_{u\le s}\sigma(X_u)\right)\subset \sigma\left(\bigcup\limits_{n\in\mathbb{N}}\bigcup\limits_{0\le s_1<\ldots<s_n\le s}\sigma(X_{s_1},\ldots,X_{s_n})\right)$
$"\supset"$ : For every $0\le s_1<\ldots<s_n\le s, X_{s_1},\ldots,X_{s_n}$ are $\mathcal{F}^X_s$-measurable, so $\sigma(X_{s_1},\ldots,X_{s_n})\subset \mathcal{F}^X_s$ and also the union is included in $\mathcal{F}^X_s$
