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 ?

  • $\begingroup$ 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). $\endgroup$
    – Gono
    Jan 21, 2020 at 14:39
  • $\begingroup$ 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. $\endgroup$
    – Math-fun
    Jan 21, 2020 at 16:23

1 Answer 1


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 $\mathcal F_t$.


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