Saturday, September 8, 2012

Probability 6 - Predictable process

Sixth post in the series.

This post is natural sequel to the last post on relation of information and filtration.

Predictable Process

\(H_n\) is called a predictable process on filtration \(\{\mathcal{F}_n\}\) if for every \(n\), \(H_n \in \mathcal{F}_{n-1} \), in another words \(H_n\) is measurable on \(\mathcal{F}_{n-1}\)

Why is it justified to call \(H_n\) predictable ?

First, it is never said that the filtration \(\{\mathcal{F}_n\}\) is natural filtration of the sequence \(\{H_n\}\) itself. So lets think of it as filtration induced by some other random process \(X_n\). Predictability of \(H_n\) tell us that if we know the result of this \(X_{n-1}\) we will be able to tell what the next \(H_n\) is. If we know \(X_{n-1}\)  we can say what is the preimage set (level set) of it in \(\mathcal{F}_{n-1}\), since \(H_n\) is measurable on \(\mathcal{F}_{n-1}\), we will be able also to deduce what \(H_n\) is. Because, in a sense, measurability means all the following: Values are constant on “minimal” sets in \(\mathcal{F}_{n-1}\), values respect the “minimal” sets of \(\mathcal{F}_{n-1}\) or, one may say, \(X_{n-1}\) and \(H_n\) agree on their level sets. Once you know the outcome of  \(X_{n-1}\), you know what the set of \(\omega\)-s \(\in \Omega\) that came out, but due to just discussed properties of \(H_n\), \(H_n\) is constant on all those \(\omega\)-s so you know what it will be as well.

\(X_{n-1}\) result is know before step \(n\), and it turns out that \(H_n\) is also know before step \(n\) so it is possible to predict on step \(n-1\) what \(H_n\) will be.

It fills that using same claims it is possible to show that \(H_n\) is constant, isn’t it?

No, it is not the case, if the filtration we talked about was natural filtration of \(H_n\) it self, then it would be true since in this case once we have \(H_0\) we can tell \(H_1\), but once we have \(H_1\) we can tell \(H_2\), every next result is predictable from the previous result. But it is when the filtration is natural filtration of \(H_n\) it self. However, in most of the cases the filtration is of some other sequence, just like I told in the beginning of the previous paragraph, in this case we acquire \(X_{n-1}\) and can tell the \(H_{n}\), then we acquire \(X_{n}\) and can tell \(H_{n+1}\). But since the \(\{X_n\}\) are not predictable, we cant deduce all the \(\{H_n\}\) right away we only have \(H_n\) one step ahead.

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