## Continuous Time Filtration Pathologies

Let $(\Omega, \mathcal{F}, P)$ be a probability space. A stochastic process is a (Borel) measurable map $X: \Omega\times [0,\infty) \rightarrow \mathbb{R}^d$. The elements $\omega \in \Omega$ can be thought of as experiments, which each yield as output some path or function in $\mathbb{R}^d$, while $t \in [0,\infty)$ parametrizes time. A process $X$ is said to be continuous if for each $\omega$, the path $X(\omega): [0,\infty) \rightarrow \mathbb{R}^d$ is continuous in the normal function sense. Also, for each $t \in [0, \infty)$, the map $X_t : \Omega \rightarrow \mathbb{R}^d$ will be a normal random variable.

Basically, a filtration is an ascending chain of $\sigma$-algebras indexed by time. So, formally, it is a collection of $\sigma$-algebras $\{\mathcal{F}\}_{t \in [0,\infty)}$, where if $a < b$, then $\mathcal{F}_a \subset \mathcal{F}_b$. Every guy in the filtration is supposed to be contained in the “universal” $\sigma$-algebra, which is $\mathcal{F}$.

After creating a filtration in continuous time, it is possible to create two auxiliary filtrations. After looking at the construction, note that both of these filtrations would be trivial in the discrete time case. For each $t$, define $\mathcal{F}_{t+} = \cap_{\epsilon > 0} \mathcal{F}_{t + \epsilon}$. If you philosophically consider $\mathcal{F}_t$ to be the information accrued up to time $t$, then $\mathcal{F}_{t+}$ represents the information you can obtain if you are allowed to peek infinitesimally into the future at time $t$. A filtration is said to be right continuous if $\mathcal{F}_t = \mathcal{F}_{t+}$ for all $t$. You can also define a $\{\mathcal{F}_{t-}\}$ filtration, which will represent the information that you have just before time $t$.

When you are considering a stochastic process, if you want it to model anything, it has to be true that by time $t$, you know what the value of $X$ is at time $t$. Technically, you require that for all $t \in [0, \infty)$, $X_t$ should be measurable with respect to $\mathcal{F}_t$. If this condition holds, people say that the process $X$ is adapted (to the filtration $\mathcal{F}_t$).

If you’re given a stochastic process $X$, there is a canonical way to make a filtration so that $X$ is adapted with respect to it. This process is basically the same as defining the $\sigma$-algebra generated by a random variable. What you do is define $\{ \mathcal{F}^X_t\} = \sigma(X_s : 0 \leq s \leq t)$. You might be initially tempted to say that this is overkill, and that one should define $\mathcal{F}^X_t = \sigma(X_t)$. This doesn’t work though, because to have a filtration, you need to make sure that the containment property holds. Keeping up the information theme, filtrations have this rule because as time progresses, you should never lose information about what your process has done up to that point.

So the point of all this is that it is not necessarily true that a continuous stochastic process generates a continuous filtration. This fact fails for a pretty simple but interesting reason. Let $\Omega = C[0,\infty)$, the space of continuous functions into $\mathbb{R}^d$. There is a natural way to put a Borel-type measure on this space, but its not necessary for this problem, so I’m not going to talk about it. The process we will consider on $\Omega$ is pretty awesome in its simplicity, and its very important. It is called the coordinate-mapping process, and it is defined by $X(\omega, t) = \omega(t)$. Rembmer, that here $\omega$ represents a continuous function.

Now consider this set. Let $F = \{\omega : X(\omega) \text{ has a local maximum at } t \}$. We will show that for any $t$, $F \in \mathcal{F}_{t+}$, but that it is not in $\mathcal{F}_t$. First let me give an intuitive reason why this is true, and then a formal justification. Basically, suppose that you are traveling along a function, and you happen to get to this point, which is a local maximum. Until you move forward some, any small amount, you have no idea that its a local maximum, because you have no way of knowing whether or not the function will start to decrease or keep on getting bigger. Since you don’t know at time $t$, $F$ shouldn’t be in $\mathcal{F}$. Since you will know if you’re in $F$ by stepping infinitesimally far into the future, $F$ should be in $\mathcal{F}_{t+}$.

Formal proofs from Karatzas/Shreve. Showing that $F$ is in $\mathcal{F}_{t+}$: Write, for any $n\geq 0,$

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

This equation is just saying that you are a local maximum if and only if in some small enough neighborhood, all the rational points around are further down. We can get away with the rational points chicanery because our process is continuous. This is necessary because set-theoretic business has to be countable.

To show that $F \not \in \mathcal{F}_t$, we do sabotage. Let $G \in \mathcal{F}_t$, and suppose that $F \cap G \neq \emptyset$. What we’ll do is take something in their intersection and tweak it so that it is not in $F$ but still in $G$. For such an $\omega$, define $\tilde{\omega}(s)$ to be equal to $\omega(s)$ for $0 \leq s \leq t$, and $\tilde{\omega}(s) = \omega(t) + s -t$ for $s \geq t$. What we’ve done is destroy the local maximum without changing the function before time $t$. Since $\tilde{\omega}(s)$ (check) doesn’t have a local maximum at $t$, it is not a member of $F$. To be slightly informal, $G$ only cares about what happens up to time $t$. Whether or not a function belongs to $G$ depends only on what it does by time $t$. Since we didn’t do anything to $\tilde{\omega(s)}$ up to that time, then it must still be a member of $G$. Hence, $F \neq G$, and so $F$ can’t be an element of $\mathcal{F}_t$