# Remarkable fact about Brownian Motion #2: Blumenthal’s 0-1 Law and its Consequences

Brownian motion is a martingale, so it is known that it has the Markovian property. That is, $(B_{t+s}-B_s,t\geq 0)$ is a Brownian motion independent of $\mathcal{F}_s$. But in fact we can show that this is independent of $\mathcal{F}_s^+=\cap_{t>s}\mathcal{F}_t$, the sigma algebra that (informally) deals with events which are determined by the process up to time s and in the infinitesimal period after time s.

Is this surprising? Perhaps. This larger sigma algebra contains, for example, the existence and value of the right-derivative of the process at time s. But the events determined by the local behaviour after time s are clearly also determined by the whole process after time s, so the independence property looks likely to give a 0-1 type law, like in the proof of Kolmogorov’s 0-1 law.

Theorem: $(B_{t+s}-B_s,t\geq 0)$ is independent of $\mathcal{F}_s^+$.

Proof: Take a sequence of times $s and $A\in\mathcal{F}_s^+$. It will suffice to show that the joint law of the new process at these times is independent of event A. So take F a bounded continuous function. The plan is to approximate $B_s$ from above, as then we will definitely have independence from $\mathcal{F}_s$, and hope that we have enough machinery to carry through the statements through the limit down to s. So take $s_n\downarrow s$, and then: $\mathbb{E}[F(B_{t_1+s}-B_s,\ldots,B_{t_k+s}-B_s)1_A]=\lim_n \mathbb{E}[F(B_{t_1+s_n}-B_{s_n},\ldots,B_{t_k+s_n}-B_{s_n})1_A]$ as continuity gives a.s. pointwise convergence, and we can lift to expectations by Dominated Convergence. The function in the limit on the right separates by independence as $\mathbb{E}[F(_{t_1+s_n}-B_{s_n},\ldots,B_{t_k+s_n}-B_{s_n})]\mathbb{P}(A)$. Applying the previous argument in reserve gives that the limit of this is $\mathbb{E}[F(B_{t_1+s}-B_s,\ldots,B_{t_k+s}-B_s)]\mathbb{P}(A)$ as desired.

In particular, this gives Blumenthal’s 0-1 Law, which states that $\mathcal{F}_0^+$ is trivial. This is apparent by setting s=0 in the above result, because then the process under discussion is the original process, and so $\mathcal{F}_0^+$ is independent of $\mathcal{F}\supset \mathcal{F}_0^+$.

A consequence is the following. For a BM in one dimension, let $\tau=\inf\{t>0:B_t>0\}, \sigma=\inf\{t>0:B_t<0\}$. By the fact that BM is almost surely non-constant, for any sample path, at least one of these is 0, and by symmetry $\mathbb{P}(\tau=0)=\mathbb{P}(\sigma=0)$ so these are greater than or equal to 1/2. But it is easy to see that the event $\{\tau=0\}\in\mathcal{F}_0^+$, and so by the triviality of the sigma-field, this probability must be 1. With continuity, this means that every interval $(0,\epsilon)$ contains a zero of the Brownian motion almost surely. Patching together on rational intervals (so we can use countable additivity) gives that BM is almost surely monotonic on no interval. A similar argument can be used to show that BM is almost surely not differentiable at t=0. For example, the existence and (conditional on existence) value of the derivative at t=0 is a trivial event, so by symmetry, either the derivative is a.s. =0, or a.s. doesn’t exist. Ruling out the former option can be done in a few ways. Predictably, in fact it is almost surely differentiable nowhere, but that is probably something to save for another post.