# Does a Brownian motion depend on a filtration or not?

I found many different definitions of Brownian motion. One is

Def 1: $$(B_t)$$ is a Brownian motion if it's a.s. continuous, $$B_0=0$$ a.s., has independents increments and $$B_t\sim N(0,t)$$.

An other one is

Def 2: $$(B_t)$$ is a Brownian motion if it's a.s. continuous,$$B_0=0$$ a.s., $$B_t-B_s$$ is independent of $$(B_u)_{0\leq u\leq s }$$ and $$B_t\sim N(0,t)$$.

I just saw now in a book that

Def 3: $$(\Omega ,\mathcal F,(\mathcal F_t),(B_t))$$ is a Brownian motion if it's a.s. continuous, $$B_0=0$$ a.s., $$B_t-B_s$$ is independent of $$\mathcal F_s$$ and $$B_t\sim N(0,t)$$.

So it looks that the last definition is a bit more general and for instance, in the last definition, there is a dependance with a filtration.

Question : Why do we rather consider definition 1 and 2 and not really definition 3 ? Could someone also explain in what definition 3 would be more helpful ?

• This is true if $$\mathcal{F}_s = \sigma(B_u, 0 \le u \le s)$$ And there are filtrations where both definitions differ (as long as it's really written without any further restrictions on the filtration).
– Gono
Jan 21, 2020 at 14:39
• It depends in which book you read the stuff and what is taught in the book ... The last definition is cleaner compared to your other definition. Jan 21, 2020 at 16:23

Definition 3 provides a little more flexibility. For example, if $$B_t=(B^{(1)}_t,B^{(2)}_t)$$ is a 2-dimensional Brownian motion (2-dim. version of your Def. 1 or 2) and if $$\mathcal F_t:=\sigma(B_s: 0\le s\le t)$$, then $$B^{(1)}$$ is a Brownian motion with respect to $$(\mathcal F_t)$$ in the sense of your Def. 3, but $$\sigma(B^{(1)}_t, 0\le s\le t)$$ is strictly continaed in $$\mathcal F_t$$.