## 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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