Let $\{W_t, t\geq 0\}$ be a Brownian motion, and has a.s. continuous sample paths.

Let $\{\mathcal{F}^W_t, t\geq 0\}$ be the canonical filtration, i.e. $\mathcal{F}^W_t=\sigma(W_s, 0\leq s\leq t)$.

So why is $\{\mathcal{F}^W_t, t\geq 0\}$ left-continuous? i.e. $\displaystyle{\bigcup_{s<t}}\mathcal{F}^W_s= \mathcal{F}^W_t$.

Thank you so much!


1 Answer 1


If you want to keep using a.s. continuous sample paths, you need to augment the filtration.

If you don't want to augment the filtration, suppose $W_t$ is always continuous.

Then apply $\lim_{s\uparrow t}W_s = W_t$, almost surely(if you complete the filtration) or always(if you suppose $W_t$ is always continuous)

  • $\begingroup$ Thank you! Could you please give more detail if the process is always continuous? Merci beau coup! $\endgroup$
    – Tony.Liu
    Oct 14, 2014 at 22:40
  • $\begingroup$ @Tony.Liu We only need to confirm that the information contained in $\cup_{s<t}\mathcal{F}_s^W$ can make the value of $W_t$ available, which is true because the above limit. $\endgroup$ Oct 14, 2014 at 22:44
  • $\begingroup$ Thanks for your hint which helps me to get the point! Good night! $\endgroup$
    – Tony.Liu
    Oct 14, 2014 at 23:04
  • 1
    $\begingroup$ @Tony.Liu You are welcome! Bonne nuit! $\endgroup$ Oct 14, 2014 at 23:05

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