What is meant by a filtration "contains the information" until time $t$?

Question

I have problems understanding the concept of a filtration in stochastic calculus. I understand that for example the natural filtration $F_t$ contains only outcomes up to time $t$, but since it is a sigma algebra it contains all possible events. For instance for $X_s$, $s<t$, it should contain all possible outcomes of $X_s$, and even all subsets of possible outcomes of $X_s$, right? How can it then contain information about which events occurred and which did not, when it contains all possible events up to time $t$? What really is meant when people write that "the filtration contains the information of outcomes up to time $t$" and uses $|F_t$ to indicate conditional expectation values? Or have I completely misunderstood what is meant when one says that a filtration is a sigma algebra?

Answer 1

In order to understand the intuition behind filtrations, it's a good idea to start with a very particular case: the $\sigma$-algebra generated by a single random variable $X:\Omega \to \mathbb{R}$, i.e.

$$\sigma(X) = \{ \{X \in B\}; B \in \mathcal{B}(\mathbb{R})\} \tag{1}$$

which is the smallest $\sigma$-algebra $\mathcal{F}$ on $\Omega$ such that $X: (\Omega,\mathcal{F}) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ is measurable. First of all, you should notice two facts:

• The $\sigma$-algebra $\sigma(X)$ is a deterministic object, i.e. it does not depend on $\omega \in \Omega$.
• The $\sigma$-algebra $\sigma(X)$ does not uniquely characterize a $\sigma$-algebra $X$. In particular, it does not allow to "reconstruct" the outcomes of a random variable $X$.

Example: For $A \subseteq \Omega$ consider the random variables $$X := 2 \cdot 1_A \quad \text{and} \quad Y := 5 \cdot 1_{A^c}.$$ It follows from the very definition $(1)$ that $$\sigma(X) = \sigma(Y) = \{\emptyset, \Omega,A,A^c\},$$ i.e. the two random variables generate the same $\sigma$-algebra although the random variables are quite different. In particular, we cannot expect to use $\sigma(X)$ to reconstruct the random variable $X$.

This leads to the natural question in which sense we can understand $\sigma(X)$ as "information" about a random variable $X$. There is the following characterization of $\sigma(X)$:

An event $A \subseteq \Omega$ is an element of $\sigma(X)$ if, and only if, after observing the outcome $X(\omega)$ of our random variable we can tell whether the event $A$ happened or not, i.e. whether $\omega \in A$ or $\omega \notin A$.

Example: Let $U,V$ be two independent random variables taking the values $0$ and $1$ with probability $1$ and set $R:= U+V$. Then $\{U=1\}$ is not contained in $\sigma(R)$. Why? Once we have observed $R(\omega)$, we cannot tell whether $\omega \in \{U=1\}$, for instance if $R(\omega)=1$ we do not know whether $U(\omega)=1$ or $V(\omega)=1$.

The so-called factorization lemma states that a random variable $Y$ is $\sigma(X)$-measurable if, and only if, there exists a measurable function $h$ such that $$Y=h(X).$$ Intuitively this means that a random variable $Y$ is $\sigma(X)$-measurable if and only if after oberserving our random variable $X(\omega)$ we have all the necessary information to determine $Y(\omega)$. The $\sigma$-algebra $\sigma(X)$ hence stores the information which additional "knowledge" we can get once we have observed the outcome $X(\omega)$ of the random variable. We can consequently read the conditional expectation $$\mathbb{E}(Y \mid \sigma(X))$$ as the expectation of $Y$ given that we have observed $X$. For instance if $Y$ is $\sigma(X)$-measurable, then $$\mathbb{E}(Y \mid \sigma(X)) = Y$$ because - according to our intuition - $Y(\omega)$ is fully determined by $X(\omega)$ (which we already observed).

For the canonical filtration $$\mathcal{F}_t := \sigma(X_s; s \leq t)$$ the situation is not that much different. Similar to the characterization for $\sigma(X)$ we have the following result (see here for a rigorous statement)

A set $A$ is in $\mathcal{F}_t$ if, and only if, after observing $X_s(\omega)$, $s \leq t$, we can decide whether $\omega \in A$ or $\omega \notin A$.

Example: Let $U$ be an exponentially distributed random variable and define $$X_t(\omega) := 1_{(U(\omega),\infty)}(t) = \begin{cases} 0, & \text{if $t \leq U(\omega)$} \\ 1, & \text{if $t > U(\omega)$}. \end{cases}$$ Then $$\tau := \sup\{t \geq 0; X_t = 0\}$$ is not a stopping time with respect to the canonical filtration $(\mathcal{F}_t)_{t \geq 0}$, i.e. $\{\tau \leq t\} \notin \mathcal{F}_t$. Why? Say, we observed our process for some time $t$ and it equals zero up to time $t$,i.e. $X_s(\omega)=0$ for all $s \leq t$. Can we decide whether $\omega \in \{\tau \leq t\}$ or not? No, because we do not know whether $X$ is going to jump to $1$ directly after our final observation (i.e. $X_s(\omega)$ for all $s>t$) or whether it will stay zero for another period of time.

According to the above characterization, we can understand the conditional expectation

$$\mathbb{E}(Y \mid \mathcal{F}_t)$$

as the expectation of $Y$ given that we have already observed $X_s$ for $s \leq t$. The "exreme" cases are clearly that

• $Y$ is independent of $\mathcal{F}_t$; in this case the conditional expectation equals $\mathbb{E}(Y)$ because our observations do not give us any additional knowledge about $Y$,
• $Y$ is $\mathcal{F}_t$-measurable; in this case the conditional expectation equals $Y$ because $Y$ is fully determined by our observations.