# If $B=(B_t,t\ge 0)$ is a Brownian motion and $(\mathcal{F}_t,t\ge 0$ is its generated filtration, then $X_t-X_s$ are independent of $\mathcal{A}_s$

A real-valued stochastic process $B=(B_t,t\ge 0)$ is called Brownian motion $:\Leftrightarrow$

1. $B_0=0$
2. $B$ has independent and stationary increments, i.e. $$\left(X_{t_i}-X_{t_{i-1}}\right)_{i=1,\ldots,n}\;\;\;\text{are independent}$$ for all $n\in\mathbb{N}$ and $0\le t_0<\ldots<t_n$ and $$\mathcal{L}\left[X_{s+t}-X_s\right]=\mathcal{L}\left[X_t-X_0\right]$$ for all $s,t\ge 0$ (where $\mathcal{L}[X]$ denotes the distribution of $X$)
3. $B_t\text{ ~ }\mathcal{N}_{0,t}$, i.e. $B_t$ is normally-distributed with expectation $0$ and variance $t$, for all $t>0$
4. The paths $t\mapsto B_t$ are almost surely continuous

Now, I've read that $B$ is a martingal. My first question was: "Martingal with respect to which filtration?". I assume we're considering the generated filtration $$\mathbb{F}=(\mathcal{F}_t,t\ge 0)\;\;\;\text{with }\mathcal{F}_t=\sigma(B_0,\ldots,B_t)$$ However, the proofs that I've seen so far generally start with: "Since the increments $X_t-X_s$ are independent of $\mathcal{F}_s$ for all $t>s$, $\ldots$"

But why is this true? I absolutely don't get it.

• generally if not explicitely stated, one considers the filtration generated by the process: $F_t = \sigma(X_s, x \le t)$. Apr 28, 2015 at 18:31
• @mookid As I've written: That's what I thought so. Apr 28, 2015 at 18:34
• in that case, this is a consequence of the definition of the sigma-algebra generated by a family of random variables. Apr 28, 2015 at 18:35
• @mookid Could you please create an answer. I absolutely don't see how this follows from the definition. Apr 28, 2015 at 18:36
• let me write it. Apr 28, 2015 at 18:37

The sigma algebra $F_t$ is generated by the variables $X_s,x≤t$. As a sigma algebra, you can consider it generated by the elements $$\sum_{i=1}^p a_i X_{t_i}, p\in\Bbb N_{>0}, a_i\in\Bbb R, t_i\le t$$
Hence to prove the equation $$\forall G\in F_t \ \ \forall f\ \ \ \ E(G\times f(X_{t+h} - X_t)) = EG\times E f(X_{t+h} - X_t))$$
you can consider only the elements having the form $$G= \sum_{i=1}^p a_i X_{t_i}, p\in\Bbb N_{>0}, a_i\in\Bbb R, t_i\le t$$ (which is trivial) and extend it using the monotone class theorem.