Why does a filtration represent information over time?

Question

The book I am reading says that Ft contains all events whose occurrence or not is fixed by time t. Why is this the case?

Let say A is an event that has not been fixed yet, B is an event that has been fixed. Then A union B has to be in Ft, but A union B is not fixed, is it? How should I think about this statement?

Another example of this: if E is an event that contains all elements of $\Omega$ that have been fixed. Then E is in Ft and E complement contains no event that has been fixed. E complement is still in Ft. But E U $E^c$ is an event that is neither fixed, nor unfixed, but it is still in Ft? Sth is off.

Answer 1

First, let me say that if $\{\mathcal{F}_{t}\}_{t\geq0}$ is a filtration, with $\mathcal{F}_{t}\subseteq\mathcal{F}$ for all $t\geq0,$ then if $A\in\mathcal{F},$ $B\in\mathcal{F}_{t},$ then $A\cup B$ need not be in $\mathcal{F}_{t}.$ For example, if $\mathcal{F}_{t}=\{\varnothing,\Omega\}$, $B=\varnothing,$ then $A\cup B=A$ could be anything in $\mathcal{F},$ which might be much richer than $\mathcal{F}_{t}.$

I believe the goal of this line "$\mathcal{F}_{t}$ contains all events whose occurrence or not is fixed by time $t$" is to give some intuition to the rather technical and general idea of a filtration. A common context where this makes a lot of sense is when the filtration is that generated by some stochastic process: Suppose $\{X_{t}\}_{t\geq0}$ is a collection of $\mathcal{F}$-measurable random variables, and for $t\geq 0,$ let $\mathcal{F}_{t}$ be the smallest sub-$\sigma$-field of $\mathcal{F}$ for which all of $\{X_{s}\}_{0\leq s\leq t}$ are measurable. Then in other words, "anything" you can say about $X$ up to time $t,$ for example, $X_{t_{1}}>4,$ $X_{t_{2}}<X_{t_{3}},$ or $\max_{0\leq s\leq t}X_{s}\leq0$ (where $0\leq t_{1},t_{2},t_{3}\leq t$), all correspond to subsets of $\Omega$ which belong to $\mathcal{F}_{t}.$ Conversely, if there is some $A\in\mathcal{F}_{t}$ such that $A\not\in\mathcal{F}_{s},$ for some $s<t,$ this is saying that $A$ cannot be decided just from knowing $\{X_{r}\}_{0\leq r\leq s}.$ Since it can be decided from knowing $\{X_{r}\}_{0\leq r\leq t},$ deciding $A$ must somehow rely on information about $X_{r}$ for some $s<r\leq t,$ fitting with the intuition of the statement.

Answer 2

I think it's good to see an example. Let's consider the trivial example of a symmetric random walk starting at $0$ with two steps.

Our probability space is $\Omega =\{HH,HT,TT,TH\}$, $\mathcal F=P(\Omega)$, $\mu(\omega)=1/4$, $\forall \omega\in \Omega$.

Our time interval is $\{0,1,2\}$.

Our stochastic process is:

$X_0(\omega)=0$, $\forall \omega \in \Omega$.

$X_1(HH)=X_1(HT)=1$, $X_1(TH)=X_1(TT)=-1$.

$X_2(HH)=2$, $X_2(HT)=0=X_2(TH)$, $X_2(TT)=-2$.

So you flip a coin twice and go up or down. Now what is the natural filtration (smallest filtration that $X_t$ is measurable)?

For $t=0$ we don't know what path we're taking. Our process is a.s. constant. The smallest sigma algebra that a constant random variable is $\mathcal F_0\{\emptyset,\Omega\}$. Equivalently we can only distinguish between the process and something else. We know if we're looking at the process or not. If we look and we see it starting at 3, we know that we aren't looking at the process. Equivalently that's the $\emptyset$.

For $t=1$ the smallest sigma algebra is $\mathcal F_1=\{\emptyset, \Omega, \{HH,HT\},\{TT,TH\}\}$. Equivalently, we're able to distinguish $\{HH,HT\}$ and $\{TT,TH\}$ but not between $\{HH\}$ and $\{HT\}$ say (that would require looking at what happens at time $2$). At time $1$ all we know is what we knew before (are we looking at the process or not) and if we got a $H$ or $T$.

For $t=2$ we have that $\mathcal F_2=P(\Omega)$. That is by time $2$, we are able to distinguish all paths.