2
$\begingroup$

We say that a filtration $\{\mathcal F_t\}$ is right-continuous if $\mathcal F_{t^+}=\mathcal F_t$ where $$\mathcal F_{t^+}=\bigcap_{s>t}\mathcal F_s.$$

Let $(X_t)$ a stochastic process and suppose that $\mathcal F_t=\sigma \{X_s\mid s\leq t\}$. Do we have that $(X_t)$ is continuous $\iff$ $\{\mathcal F_t\}$ is continuous ?

If not, could someone explain me in what right-continuous filtration are important ? I'm not sure to really understand why we consider them.

$\endgroup$

1 Answer 1

3
$\begingroup$

The answer, sadly, is no. There is a marvelous answer to that part here: $\mathcal F_t=\mathcal F_{t^+}$ $\iff $ $(X_t)$ is right continuous

Now, as for why it is interesting to consider the right-continuous augmentation of a filtration: Consider a continuous process $Y_t$ and the event that "$Y$ closes a loop for the first time". If $Y$ is one-dimensional this is really the event that "$Y$ turns around". But at time exactly $t$, I don't actually know whether I'm turning around (say $Y$ is differentiable, we can just see that $Y'=0$), but at time $t+\varepsilon$, I do know, no matter what $\varepsilon$ is (because I can now check whether $Y'$has flipped sign in the differentiable case).

So the first loop-closing time is only a stopping time, if your filtration is right continuous. So the right continuity of a filtration is more like a technical assumption that allows the proper measurability properties of certain events.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .