# Can we re-write the natural filtration of a Brownian Motion?

I'd be astounded if this isn't a duplicate, but I've yet to find anything equivalent. Whenever I've seen the natural filtration of a Brownian Motion stated explicitly, it's been in the form of $$\mathcal{F}_t=\sigma\{W_s:s\leq t\}$$ for any $$t\geq0$$. In other words, the natural filtration of a Brownian Motion up to some point $$t$$ is the sigma-field generated by all of the Brownian Motions from every time not after $$t$$.

My question is, why is that last bit necessary? Why do we not just have $$\mathcal{F}_t=\sigma\{W_t\}$$ for any $$t\geq0$$? By definition $$W_t$$ is $$\sigma\{W_t\}$$-measurable and $$W_t$$ can clearly take all of the values that any $$W_s$$ (with $$s\leq t$$) can, so I can't see what is gained by adding the fields generated by the "earlier" random variables. Even in cases where we want to talk about some point $$b\leq t$$, I can't see why we would use $$\mathcal{F}_b=\sigma\{W_a:a\leq b\}$$ when we could use $$\mathcal{F}_b=\sigma\{W_b\}$$. Indeed, I can't see why these two fields would not be identical. Is there an obvious element that I'm missing?

• Yes, that's a filtration, but it loses information compared to the "natural" one. Keep in mind that events are evaluated with respect to a sigma-algebra, so losing the past means all you know is where the process is right now. – nomen Feb 25 at 17:06
• @nomen Isn't that a contradiction? You've just said that my proposed filtration is a filtration that is smaller than the natural one. – J. Mini Feb 25 at 17:08
• There are lots of filtrations... the constantly empty filtration is the smallest. The smallest one that captures the "past" is called the natural filtration. – nomen Feb 25 at 17:11
• @nomen Ahh, so you're saying that the process is not adapted to the filtration that I've proposed? – J. Mini Feb 25 at 17:13

The proposed filtration is not a filtration at all. If it was, then we would have $$\mathcal{F}_s\subseteq\mathcal{F}_t$$ for all $$s\leq t$$. However, this is clearly not the case. Consider $$\{W_0=0\}\subseteq\mathcal{F}_0$$ where $$\mathcal{F}_0=\sigma\{W_0\}$$. Clearly this set is in $$\mathcal{F}_1$$ if we take $$\mathcal{F}_1=\sigma\{W_s:s\leq 1\}$$ (and therefore we have $$\mathcal{F}_0\subseteq\mathcal{F}_1$$), but this is not the case if we take $$\mathcal{F}_1=\sigma\{W_1\}$$ (and therefore we have $$\mathcal{F}_0\not\subseteq\mathcal{F}_1$$), meaning that the proposed filtration is not a filtration.