Brownian motion is a martingale, so it is known that it has the Markovian property. That is, is a Brownian motion independent of . But in fact we can show that this is independent of , 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:** is independent of .

*Proof:* Take a sequence of times and . 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 from above, as then we will definitely have independence from , and hope that we have enough machinery to carry through the statements through the limit down to s. So take , and then: 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 . Applying the previous argument in reserve gives that the limit of this is as desired.

In particular, this gives **Blumenthal’s 0-1 Law**, which states that 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 is independent of .

A consequence is the following. For a BM in one dimension, let . By the fact that BM is almost surely non-constant, for any sample path, at least one of these is 0, and by symmetry so these are greater than or equal to 1/2. But it is easy to see that the event , and so by the triviality of the sigma-field, this probability must be 1. With continuity, this means that every interval 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.

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