# Natural filtration of the Brownian motion is not right-continuous

Let $$B$$ be a Brownian motion on a filtered probability space. Let $$\mathcal{F}_t^B$$ be the natural filtration associated to $$B$$. The fact that the natural filtration of the Brownian motion is not right-continuous, i.e. $$\mathcal{F}_t^B\neq\mathcal{F}^B_{t^{+}}$$, is for sure widely discussed, even on this forum. Nevertheless, I did not find a convincing argument. Among many, let me quote a couple of examples that I have found. The first is
$$A_{t,n}\doteq\left\{\omega\in\Omega\left|B_{t+1/n}(\omega)>B_t(\omega)\right.\right\}\in\mathcal{F}^B_{t+1/n}\Rightarrow A_{\infty}\doteq\bigcap_{n\in\mathbb{N}}A_{t,n}\in\mathcal{F}_{t^{+}}^{B}$$ and it is claimed that $$A_{\infty}\notin\mathcal{F}_t^{B}$$. However, how can I be sure that $$A_{\infty}\neq\emptyset$$ or $$A_{\infty}\neq\Omega$$ ? Because, in both cases, I would have $$A_{\infty}\in\mathcal{F}_t^{B}$$, so the example is not valid.

Other type of examples are $$A_t\doteq\left\{\omega\in\Omega|\exists \delta >0:B_{s}(\omega)\leq B_{t}(\omega)\forall s\in(t-\delta,t+\delta) \right\},$$ i.e. $$A_t$$ corresponds to the event of the Brownian motion having a local maximum in $$t$$. Again, for sure, $$A_t\in\mathcal{F}_{t^{+}}$$. But how can I be sure that $$A_t\neq\emptyset$$ and $$A_t\neq\Omega$$?

## 2 Answers

The issue is that you haven't defined $$\Omega$$. The example you gave have $$\mathbb{P}(A_\infty) = 0$$, so you could define the Brownian motion on the new probability space $$\tilde \Omega := \Omega \setminus A_\infty$$ and do the same process on $$\tilde \Omega$$ (i.e. define $$\tilde A_{t,n} = \{ \omega \in \tilde \Omega : B_{t + 1/n}(\omega) > B_t(\omega)\}$$, $$\tilde A_\infty = \bigcap \tilde A_{t,n}$$) and end up with $$\tilde A_\infty = \emptyset$$. The point is that, without saying what $$\Omega$$ is, we cannot say for sure that $$A_\infty \ne \emptyset$$.

The typical way to define $$\Omega$$ is $$\Omega = C([0,T])$$, the set of continuous functions on $$[0,T]$$. Now it is clear that $$A_\infty \ne \emptyset$$ because $$A_\infty$$ contains, for example, all the functions that are increasing on $$[t,t+1]$$.

• Thanks, but I miss something. Why, in the example I gave, it holds that $\mathbb{P}(A_{\infty})=0$ ? Finally, can you give me some references that help to understand why $A_{\infty}$ contains all of the functions that are increasing on $[t,t+1]$ if $\Omega=C([0,T])$ ? Jan 31, 2023 at 17:48
• To clarify what is meant by $\Omega = C([0,T])$, we are considering $\omega: [0,T] \rightarrow \mathbb{R}$ to be a continuous function and $B_t(\omega) = \omega(t)$. Therefore, in $\omega$ is an increasing function on $[t,t+1]$, we have $\omega(t + 1/n) > \omega(t)$ for all $n \in \mathbb{N}$, and correspondingly $B_{t + 1/n}(\omega) > B_t(\omega)$. The fact that $\mathbb{P}(A_\infty) = 0$ is because it is easy to show $\mathbb{P}(A_\infty) \le \frac 12 < 1$, and Blumenthal's 0-1 law ensures it is either $0$ or $1$. Jan 31, 2023 at 18:37

You can tell that these sets aren't trivial by estimating their probabilities. Note that if $$\mathbb P[A]\in (0,1)$$ then $$A\neq\emptyset$$ and $$A\neq \Omega$$.

For the first example, note that $$A_\infty \subseteq A_{t,n}$$. Now by definition of Brownian Motion we have $$\mathbb P[A_{t,n}]=\mathbb P[B_{t+1/n}>B_t]=\mathbb P[B_{1/n}>0]=\Phi_{0,1/n}(0)<1$$ Hence $$\mathbb P[A_\infty]<1$$ and $$A_\infty\neq\Omega$$.

By similar methods you can show that the other set $$A_t$$ you defined is not equal to $$\Omega$$.

There are probably simpler ways to show these results, this was just my first thought.

• Sorry, but $\mathbb{P}(B_s > 0, \forall s \in (0,1]) = 0$ and similarly $\mathbb{P}(A_\infty) = 0$. The stopping time $\tau$ you defined is $0$ a.s. Jan 31, 2023 at 16:24
• @user6247850 You are correct, i must have misremembered. I edited my answer accordingly. Jan 31, 2023 at 16:40