# What is meant by a filtration “contains the information” until time $t$?

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 $$X_s$$, $$s, it should contain all possible outcomes of $$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?

• Did you try to use the search? There are several questions on MSE which are closely related to yours.... – saz Mar 1 '19 at 10:58
• The only question similar to mine is: math.stackexchange.com/questions/1602445/… which has no answers or comments – Jonathan Lindgren Mar 1 '19 at 13:23
• My problem is that to me the filtration just looks like a collection of subsets of all possible outcomes. I don't see how it changes depending on what history the process has, ie what values it took, so I don't see how it defines the information available at time t – Jonathan Lindgren Mar 1 '19 at 13:25
• math.stackexchange.com/questions/3027236/standard-filtration the answer in this question for example, writes "Intuitively, Fs is all the "information" available to us up to time s.". Well, I don't see that intuition, not at all. – Jonathan Lindgren Mar 1 '19 at 13:26
• You might want to take a look this question which gives a characterization for a set to be in $F_t$. Moreover, it would be perhaps a good start to get first some intuition what the $\sigma$-algebra $\sigma(X)$ generated by a random variable $X$ is (i.e. "which information it contains"). Once you understand this, it's less difficult to understand the intuition behind filtrations. – saz Mar 1 '19 at 13:41

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.