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.

Answer 2

In probability theory, we can be interested in the outcome $\omega$ of a random experiment. One way to measure "information" is to identify and quantify the probability of an outcome or a set of outcomes (an event).

The set $\Omega$ of all outcomes has been called the outcome, sample, or possibility space and even termed as the certain event.

An element of a $\sigma$-algebra is called an event. A $\sigma$-algebra is a set of events. Then, a filtration $(\mathcal{F}_t)_{t \geq 0}$ is a collection of sets that contain a growing number of events ($\mathcal{F}_t \subset \mathcal{F}_{t+1}$).

Thus, the filtration $\mathcal{F}_t$ contains less events than $\mathcal{F}_{t+1}$ i.e. $\mathcal{F}_t$ contains less information than $\mathcal{F}_{t+1}$.

Example:

Suppose $X: \Omega \rightarrow \mathbb{R}_{+}$ is a random variable which measures the temperature in a box $\Omega = [0, 1]^2$, i.e. $X(\omega)$ gives us the temperature at the location $\omega$.

Goal

For $X(\omega^*) = c$, find the location of $\omega^*$. The probability measure $P$ is given.

Case 1:

Define the $\sigma$-algebra $\mathcal{F_0} = {\{\Omega, \emptyset\}}$. The probability space $(\Omega, \mathcal{F_0}, P)$ is obtained.

Under this probability space, the event $\{X = c\}$ describes a set of possible outcomes $X^{-1}(\{ c\}) = \{\omega \in \Omega, X(\omega) \in \{ c\}\}$.

However, recall that the definition of a random variable is to be measurable with respect to the probability space. Hence,

$X^{-1}(\{ c\}) \in \mathcal{F}_0 = {\{\Omega, \emptyset\}}$

This means, because of this $\sigma$-algebra $\mathcal{F_0}$, we have no clue on $\omega^*$. $X$ being $\mathcal{F_0}$-measurable only tells us that $X(\omega^*) \in \Omega$, i.e. the value of the random experiment originates from an element of the certain event $\Omega$. Hence, $\mathcal{F_0}$ carries almost none information. Moreover, $P(X^{-1}(\{ c\})) = 1$ or $0$ which is not very informative.

Case 2:

Choose now the $\sigma$-algebra $\mathcal{F_1} = \sigma(\{[0, \frac{1}{2}]^2, [\frac{1}{2}, 1]^2\})$ which is the smallest $\sigma$-algebra containing $\{[0, \frac{1}{2}]^2, [\frac{1}{2}, 1]^2\}$. This choice means the box is now divided into four regions.

By definition, if $X$ is a r.v. on $(\Omega, \mathcal{F_1}, P)$, it means that $X$ is $\mathcal{F_1}$-measurable which means that all values taken by $X$ can be associated to an event in $\mathcal{F_1}$. For instance, $\omega^* \in [\frac{1}{2}, 1]^2$, which means the temperature measure is located at the upper right sub-box of the domain $\Omega$. In that case, $\mathcal{F_1}$ is more informative.

Since $\mathcal{F_0} \subset \mathcal{F_1}$, more events can describe the value of $X$. Consequently, the choice of $\mathcal{F_1}$ increases the information on the outcome of the random experiment.

Thus, a filtration $(\mathcal{F}_t)_{t \geq 0}$ exhibits an increasing amount of information, in terms of events. An adapted stochastic process is a sequence of random quantities $(X_t)_{t \geq 0}$ where each random quantity $X_t$ carries an increasing amount of information (the number of events that can be extracted from the value taken by each random variable $X_t$ increases).

In particular, we can intuitively imagine that a filtration is a refinement of a mesh on $\Omega$. The finer the mesh, the more precisely the result $\omega \in \Omega$ can be ‘located’. In fact, the finer the $\sigma$-algebra, the more specific/complex the events.