Lack of right-continuity of the filtration adapted to Brownian motion

Question

Let us consider the standard Brownian motion and the natural filtration $(\mathcal{F}_t^B)$. It is known that $(\mathcal{F}_t^B)$ is not right-continuous at $t=0$. But what about $t>0$? Is it true that $(\mathcal{F}_t^B)$ is not right-continuous at $t>0$? If so, could you please explain why it is not right-continuous?

Answer 1

The answer is yes: $\{\mathcal{F}^B_t\}$ is not right continuous at every $t>0$.

To see this, let $(C[0,\infty),\mathcal{B}(C[0,\infty)),P_*)$ be our probability space, with $P_*$ denoting the Wiener measure, and under which, let $B_t(\omega):=\omega(t)$ be a standard Brownian motion.

For fixed $t>0$, the nonempty set

$$F:=\{\omega\in C[0,\infty): \omega \,\, \text{has a local maximum at $t$}\}$$

is in $\mathcal{F}^B_{t+}$, since for each $n\geq 0$,

$$F = \bigcup_{m=n}^\infty \,\, \bigcap_{r\in \mathbb{Q},|t-r|<1/m} \,\, \{\omega:\omega(t)\geq \omega(r)\} \,\, \in \mathcal{F}^B_{t+1/n}.$$

On the other hand, a typical set in $\mathcal{F}^B_t$ has the form $G=\{\omega\in C[0,\infty): (\omega(t_1),\omega(t_2),\cdots)\in \Gamma\}$, for $\{t_i\}_{i=1}^\infty\in [0,t]$ and $\Gamma\in\mathcal{B}(\mathbb{R}\times \mathbb{R}\times \cdots)$. We claim that $F\neq G$ for any $G\in\mathcal{F}^B_t$. Indeed, suppose that we have $F\cap G\neq \emptyset$ for some $G\in\mathcal{F}^B_t$. Then given any $\omega\in F\cap G$, the function

$$\tilde{\omega}(s):=\begin{cases} \omega(s), \,\, 0\leq s\leq t\\ \omega(t)+s-t, \,\, s\geq t\\ \end{cases}$$

is in $G$ but not in $F$, so $F$ cannot agree with any $G\in\mathcal{F}^B_t$. This concludes the proof.

An intuitive explanation is: Observing $B$ at time $t+$ is enough to determine whether the event $F$ happens, so we have $F\in\mathcal{F}^B_{t+}$. However, merely observing $B$ until time $t$ can not tell us whether $B$ has a local maximum at $t$, therefore $F\not\in\mathcal{F}_t^B$.

Moreover, we can show that $P_*(F)=0$ (see this post), so that $F$ is indeed a negligible set under $P_*$. This provides insights why the augmented filtration $\sigma(\mathcal{F}^B_t,\mathcal{N})$ becomes right continuous.