How Filtrations Represent Information from Sequential Coin Tosses

# The meaning of filteration ( coin toss example )


# The meaning of filteration ( coin toss example )












Reference book is 'Steven Shreve: Stochastic Calculus and Finance'

What I don't understand is $F_3$ below picture

I understand that 'filteration' have accumulative information. So when we tossed the coin one time($F_1$), if the outcome is head, then last possible future two toss is {HH,HT,TH,TT}. Therefore $F_1$ has {H'HH',H'HT',H'TH',H'TT'}

However I cannot understand the $F_3$. So it seems I totally misunderstand the meaning of filteration, even more $σ-field$.

I think when we tossed the coin 3 times, then possible outcome is total 8. {HHH,HHT,HTH,...,TTT} So I mean : $F_3 = 8$

However the reference says '$F_3 = 2^Ω = 2^8 = 256$ It means $F_3$ has a lot of combination with HHH, HHT, HTH, ..., TTT So it means some combination, for example '{HHH,HTH,TTH}' is possible.

But how this can be interpreted?

Think about it with real case, we just toss the coin 3 times. And In $F_3$, We already have seen all the information(outcome of 3 times toss).

In the case $F_2$, We have already seen 2 tosses. So we need to have a room for '1' random toss. Think about the Head & Head case. We have a room for the last Head or Tail({HH'H'}, {HH'T'})

But in the case $F_3$, we never have additional room for more random toss. Think about when we got the outcome 3 Head in a row. Then We cannot make additional toss. therefore just {HHH} is done. So I think $F_3$'s number should be 8.

Please help me understand this.

## Answer by Bob Jansen (score 0, accepted)

https://quant.stackexchange.com/a/70033

I don't have example with me but I believe $\Omega$ is the set of all possible outcomes of three coin tosses, correct me if I'm wrong.

I don't fully understand your question but hopefully this explanation how Shreve came to 256 helps.

Disclaimer aside. Indeed, three coin tosses can have $2^3 = 8$ results and each of these outcomes is an element of $\Omega$. There are $2^8 = 256$ distinct ways to pick 8 elements (the element is either in or out). The set of all possible subsets is the set $F_3$.

I agree this can be confusing with once it's very clear due to the word set appearing everywhere with different meanings.

## Answer by Peter Lind (score 2)

https://quant.stackexchange.com/a/70039

The filtration is intuitively an information flow, so you have more information after each coin flip. After each coin flip, you represent your information with a $\sigma$-algebra to pinpoint the sets, which we can assign a measure to.

In this example, you know your experiment consists of three coin flips, so the sample space $\Omega$ is the power set with $2^{2^3}=256$ sets. The power set and $\mathcal{F}_3$ coincide in this example, so we have $256$ measurable sets after observing the three coin flips. To see the information flow:

- $\mathcal{F}_0=\{\emptyset, \Omega\}$, we do have not any coin flips, hence we do not have any information about the outcomes. Therefore we can only assign the probability of getting the empty set or the whole sample space

- $\mathcal{F}_1=\{\emptyset, \Omega, A_H, A_T \}$, we have one coin flip. We can observe that the first flip is either an H or T. The rest of the sequence is unknown, hence we can only assign the probability of getting the empty set, the whole sample space, $A_H$, and $A_T$.

- $\mathcal{F}_2=\{\emptyset, \Omega, A_{HH}, A_{TT}, A_{TH}, A_{HT},$ and set that can be built from these with unions $\}$. We now have two coin flips. We can observe that the two flips are either HH, TT, HT, or TH. The rest of the sequence is unknown, hence we can only assign the probability of getting the empty set, the whole sample space, $A_{HH}, A_{TT}, A_{TH}, A_{HT}$, and set that can be built from these with unions.

- $\mathcal{F}_3$ same recipe.

Best, Peter Lind