Forth post in the series.

How, filtration \(\{\mathcal{F}\}\) of \(\sigma\)-algebras is related to the claim “that filtration corresponds to information one has at time \(n\)”

__Let me build the stage__

Think of you self as been at time zero and you intend perform simple experiment where you toss coin five times in a row at times \(n=1, n=2, n=3, n=4, n=5\) and record the results. Before you toss the coin for the first time there is total uncertainty about the overall out come of the experiment, after you toss the coin for the first time you will be able to tell the result of the first toss while the rest will still remain uncertain, when you toss the coin next time you now able to tell two out comes and so on… until you completely know the result.

This process can be related to information being gaining as time passes or equivalently the uncertainty being decreasing with each toss.

So lets now turn this little bit “around” and speak of probability of you answering to the question of “what the overall result of you tossing will turnout”. This, rather weird angle of looking at the problem, will turn out to be very natural for us.

Before you toss there is probability \(1\) for you answering that every outcome is equally possible, after the first toss you will know how the coin happened to fall, so there is probability \(0.5\) of you answering that it has to be the head first and the rest is unknown and similarly with probability \(0.5\) for the tail being first while the rest is unknown. Next time you will tell with \(25\%\) chance that it is twice heads (the rest is unknown) \(25\%\) head and tail (the rest is unknown) and so on…

Ok, now, lets go back to the usual probability space of coin tossing. What this space looks like? It is reasonable to assume that it is generated by the following sets:

\(\{\emptyset, \Omega, (00000), (00001), (00010), (00100), (01000), (10000), (00011), (00101),\dots,(11111))\}\)

Now what model we can adopt for the “space of probabilities of your answers”?

Check this out:

At time \(n=0\) we have \[\{\emptyset, \Omega, \{(00000), (00001), (00010), (00100),\\ (01000), (10000), (00011), (00101),\dots,(11111)\} \}\], pay attention to the addition of the curly brackets “\(\{\)” and “\(\}\)” we regard all the outcomes between them as single element! So actually we can write it as \(\{\emptyset, \Omega\}\) since \(\{(00000), (00001), (00010), (00100), (01000), (10000), (00011), (00101),\dots,(11111))\} \) just equals to the whole of \(\Omega\) so no need to write it twice. Lets designate \(\sigma-algebra\) generated by this set as \(\mathcal{F}_0\).

You probably know see where its all going…

\(\mathcal{F}_1\) will be generated by \[\{\emptyset, \Omega, \{(00000), (00001), (00010), (00100), (01000), (00011), (00101),\dots,(01111))\},\\\{(10000), (10001), (10010), (10100), (11000), (10011), (10101),\dots,(11111))\} \}\] and so on.

Note: that \(\mathcal{F}_0 \subseteq \mathcal{F}_1 \subseteq \mathcal{F}_2 \subseteq \dots \subseteq \mathcal{F}_5 \subseteq \mathcal{F}\).

Now I will let the fog to dissipate:

__The answer __

Algebras (set of sets) \(\mathcal{F}_n\) encode the possible questions that can be answered at time \(n\) (indeed at time \(n=1\) we can only ask what was the outcome of the first toss so the elements in \(\mathcal{F}_1\) are build by alternating first toss) while the measure defined on this algebra gives the probability for various answers. In this regard filtration encodes also that as time will go by the number of possible answers will increase decreasing the “uncertainty”.

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