How can a Filtration be interpreted as Information / Filtration for repeated Coin toss

Question

Say we throw a coin three times. The sample space is then given by $\Omega=\{HHH, HHT, HTH, ..., THT, TTT\}$. The Filtration is given by:

$\mathbb{F}_0=\{\emptyset, \Omega\}$

$\mathbb{F}_1=\mathbb{F}_0 \cup \{\{HHH, HHT, HTH, HTT\},\{THH, THT, TTH, TTT\}\}$

$\mathbb{F}_2=\mathbb{F}_1 \cup \{\{HHH, HHT\}, \{HTH, HTT\},\{THH, THT\}, \{TTH, TTT\},\dots\}$

etc.

I understand the concept of a $\sigma$-algebra and also the formal definition of a Filtration. What I don't understand:

How exactly can this be interpreted to be information being released? As far as I can tell, the filtration is completely deterministic, how could it possibly contain information about past events? For example, whether the first toss is $H$ or $T$, the Filtration after the first throw is $\mathbb{F}_1$ regardless.

Answer 1

Sorry, I cannot post it as a comment. It seems to me that the core of your question is more how to interpret a sigma algebra as the information you can obtain from a random object.

I understand it in this way: $ \mathbb F_1$ is composed by the events you can distinguish after the first toss. You will be able to tell only which of those two events happened. As you go on in time you can divide your previous events into more events, which are more precise since are smaller. Note that you still have the information of how the first coin toss went, since $\mathbb F_1$ is in $\mathbb F_2$.

Another way to see it is The sigma algebra tells you the information you have in the sense that tells you which functions ( or observables) you are able to determine the value of: those measurable with respect to the sigma lagebra

Answer 2

The filtration is an information flow in the sense that you can spot the sets which we can assign a measure to. It doesn't mean you know exactly what happened after every coin flip but you have more and more events which you can assign a probability to.

In your case (if $ \mathcal{A}_i := \{ \text{i on first toss} \}, i \in \{H,T\} $)

1. If $\; n=0$ we have no information about the outcomes, hence we can only assign probabilities to the empty set and the whole sample space.

2. If $\; n=1$ we only know if the first coin flip was H or T (we have a bit more information than in 1.). Hence we can assign probabilities to more sets. in particular $\emptyset, \Omega, \mathcal{A}_T, \mathcal{A}_H$

3. If $\; n=2$ we know the first two outcomes, hence we can assign probabilities to even more events (which include the events in 2. as well), which intuitively means we have more information about the complete outcome.

4. If $\; n=3$ the information is complete. We can assign probabilities to all possible events of $\Omega$, i.e. $\mathbb{F}_3 = 2^{\Omega}$ which consists of $256$ sets.