## Friday, September 7, 2012

### Probability 5 - Conditional expectation the best guess

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:

Degang WU said...

thanks for the great intuition!