Intuition about Blumenthal's 0-1 law

Question

I'm studying Brownian motion from Durrett. I'm trying to understand what Blumenthal's 0-1 law really says about what Durrett calls the germ field, $\mathcal{F}_0^+$.

Let $\mathcal{F}_t^+ = \cap_{s > t} \mathcal{F}_s^0$ and $\mathcal{F}_t^0 = \sigma(B_s,s\le t)$ where $B_t$ is Brownian motion. Also let $P_x$ denote the usual measure on $\mathcal{C}[0,\infty)$ making $B_t(\omega) = \omega(t)$ a Brownian motion starting at $x$.

Blumenthal's 0-1 law: If $A \in \mathcal{F}_0^+$ then for all $x \in \mathbb{R}$, $P_x(A) \in \{0,1\}$.

I have no trouble understanding this proof, but I'm not sure I understand why this is surprising/significant.

Obviously, $\mathcal{F}_0^0$ is trivial. Is there any intuitive way to explain why we might not expect $\mathcal{F}_0^+$ to be trivial as well? Even though $\mathcal{F}_0^+$ seems to involve a tiny bit of information into the future past $t=0$, since Brownian motion is continuous it would seem that we ought to expect no difference between the two fields.

Answer 1

To help hone your intuition, it might help to think about the following simple example, which shows that the Blumenthal law is using much more than just the continuity of Brownian motion.

Let $Z$ be a random variable with $P(Z=1) = P(Z=-1) = 1/2$, i.e. a coin flip. Set $X_t = tZ$, so that the process $X_t$ just moves with constant speed $\pm 1$ depending on the value of $Z$. Note that $X_t$ is continuous in $t$. Let $\mathcal{F}_t^0 = \sigma(X_s : 0 \le s \le t)$ be the natural filtration.

Now $X_0 = 0$ so $\mathcal{F}_0^0$ is trivial. On the other hand, for every $t > 0$ we have $\mathcal{F}_t^0 = \sigma(Z)$. So $\mathcal{F}_0^+$ is not trivial; it really contains more information than $\mathcal{F}_0^0$. For instance, it contains the nontrivial events $\{Z=1\}$ and $\{Z = -1\}$; it contains enough information to see how the coin flip came out.

Intuitively, by observing the process at time 0 you don't learn anything. But if you get to observe it at some time after time 0, no matter how short, you can see whether it is above or below 0, and thus deduce the value of $Z$.