Fifth post in the series.

__Conditional Expectation as best guess of the next result__

Think of some random Variable \(X_n\) and its \(\sigma\)-algebra \(\mathcal{F}_n\) (algebra of its level sets)

look at the figure above, each patch represents a “minimal” set in \(\mathcal{F}_n\). \(X_n\) gives each of those sets constant value. Namely if \(A\) is one of the patches then \(\forall \omega \in A \in \mathcal{F}_n\), \(X_n(\omega) = \alpha_{A}\).

Now what is \(\mathbb{E}[X_n|\mathcal{F}_{n-1}]\)?. Note that \(\mathcal{F}_{n-1}\subseteq \mathcal{F}_{n}\) so lets think of it as -

Look how the new \(\sigma\)-algebra is more coarse, some sets that are minimal here were finer divided in \(\mathcal{F}_{n}\). \(X_n\) is not measurable on \(\mathcal{F}_{n-1}\), But \(\mathbb{E}[X_n|\mathcal{F}_{n-1}]\) is, and \(\mathbb{E}[X_n|\mathcal{F}_{n-1}]\) gives us the best guess we can make about the out come of \(X_n\). For example If on step \(n-1\), \(X_n\) happened to fall on the sets that are minimal on both \(\sigma\)-algebras then we know it cant change on step \(n\) but if it hadn’t then we will be able to narrow down possible outcomes of step \(n\) but still there will be some uncertainty, our best guess will be the ecpected value of \(X_n\) on the possible outcomes in \(n\), which is exactly the conditional expectation. Step \(n-1\) does not provide all the information but it give us the probabilities of our best guesses we will be able to do on step \(n-1\).

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## 1 comment:

thanks for the great intuition!

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