# Is $\mathcal G_t = \sigma(\int_{0}^{s} W_s \,du:s\leq t)$ a right-continuous filtration?

As the question states:

Is $$\mathcal G_t = \sigma\left(\int_{0}^{s} W_s \,du:s\leq t\right)$$

where $$W_s$$ is the Brownian motion, a right-continuous filtration?

I think the answer is no, as $$\mathcal G_t \neq \bigcap_{n=1}^{\infty}\mathcal G_{t+\frac{1}{n}}$$; it is possible for more information to be gained infinitesimally far into the future.

Is my understanding correct?

Thanks!

$$\mathcal{G}_t=\sigma\left(\int_{0}^{s} W_u \,du:s\leq t\right)=\sigma\left(Y_s:s\leq t\right)$$ is right-continuous unless mistaken under the usual conditions for $$W$$.

First observe that $$\mathcal{F}_t = \sigma\left(W_s: s\leq t\right)$$ is right-continuous by ... hypothesis and $$W$$ is also supposed to be continuous.

So if we prove that $$\forall t>0, \mathcal{G}_t=\mathcal{F}_t$$ then $$\mathcal{G}_{t^+}= \mathcal{F}_{t^+}$$, and we are done as as $$\mathcal{F}_{t^+}= \mathcal{F}_t$$ from usual conditions.

To see this observe that for every $$\omega$$ :

$$W_t(\omega)=Lim_{n\to \infty}\frac{1}{n}\int_{t-1/n}^{t}W_s(\omega)ds$$ by (Lebesgue differentiation theorem).

But as every term in this sequence is $$\mathcal{G}_{t}$$-measurable, so is the converging limit and we have:

$$\mathcal{F}_t \subset \mathcal{G}_{t}$$

The other way around is also true as for every $$\omega$$, $$Y_t$$ is in this case the limit of a Riemann sum as Brownian path are continuous:

$$Y_t(\omega)=Lim_{n\to \infty}\sum_{i=1}^n \frac{1}{n}.W_{i.t/n}(\omega)$$

And as every term of the sum are in $$\mathcal{F}_t$$, so is the sum and its converging limit and we have:

$$\mathcal{G}_t \subset \mathcal{F}_{t}$$

Q.E.D.

Comment:

thanks for the answer. However, why is $$\mathcal F_t = \sigma(W_s : s > \leq t)$$ right-continuous? I was thinking of a counterexample as such: the event A = $$\{ \omega : W_t(\omega)\mathrm{\, has \, a \, local \, > maximum \, at \, time \, t} \}$$. However,$$A \in \sigma(W_s : s \leq t > + \epsilon)$$. Thus $$\mathcal F_t = \sigma(W_s : s \leq t)$$ is not right-continuous.

My answer (too long to fit the space in a comment so I add it here with the comment above reproduced):

I haven't checked your example but it is known that the "pure" Brownian motion (i.e constructed without augmentation of the filtration that makes it satisfy usual conditions) has its "natural" filtration that is not right continuous, this is a big problem because for martingales to have a right-continuous version it is necessary for the usual conditions to hold and so we need the filtration to be right-continuous (see for example Karatzas and Shreve "Brownian Motion and Stochastic Calculus" - theorem 3.13 page 16 - 2nd edition).

Also note that completeness is in fact enough for usual conditions to hold for the Brownian Motion, as right continuity is then entailed by completeness for strong Markov processes (and the BM is strong Markov) see same reference proposition 7.7 page 90 (Protter's book has also a proof with Markov property only (i.e. not strong Markov) but they need to be Lévy processes though if I remember well).

So this is why it is SO IMPORTANT to have usual conditions because none want to work with martingales that have no right-continuous versions, I even wonder if it would possible to construct stochasitc integral without this property. Maybe George Lowther might achieve such a feat ;-)

• thanks for the answer. However, why is $\mathcal F_t = \sigma(W_s : s \leq t)$ right-continuous? I was thinking of a counterexample as such: the event A = $\{ \omega : W_t(\omega)\mathrm{\, has \, a \, local \, maximum \, at \, time \, t} \}$ $\notin \sigma(W_s : s \leq t)$. However, $A \in \sigma(W_s : s \leq t + \epsilon)$. Thus $\mathcal F_t = \sigma(W_s : s \leq t)$ is not right-continous – finmathstudent Jul 22 '19 at 16:42
• $\sigma(W_s: s\le t)$ is not right continuous; it becomes so if you complete it in the appropriate way. On the other hand, as TheBridge shows, you do have $\sigma(W_s: s\le t)=\mathcal G_t$. So the latter is not right continuous either. Your local-maximum example illustrates this. – John Dawkins Jul 23 '19 at 19:22