What is the difference between "filtration for a Brownian motion" and "filtration generated by a Brownian motion"?

Solution 1:

A filtration generated by Brownian motion simply means the smallest filtration with respect to which Brownian motion is adapted i.e. $$\mathcal{F}^B_t = \sigma \{B_s, \: s\leq t\}$$

On the other hand, a filtration for the Brownian motion, $\mathcal{F}_t$ is one to which the Brownian motion is adapted AND we have that for all $s<t$, $B_t - B_s$ is independent of $\mathcal{F}_s$. So notice that the filtration generated by the Brownian motion also satisfies these.

Now for your example. Let $X$ be a random variable independent of your Brownian motion. Then if we enlarge the Brownian filtration by adding in information about $X$: $$\mathcal{F}_t := \sigma \{X, B_s \: s\leq t\}$$ Then this is still a filtration for Brownian motion.