# Why isn't conditional expectation called restricted random variable?

Definition of Condition Expectation( of random variable) can be seen here: http://en.wikipedia.org/wiki/Conditional_expectation

My silly question is: Intuitively, conditional expectation seems to be a restriction of a measurable function on the sub-$\sigma$-algebra. Why don't people call it a restriction instead of such a confusing name. After all, one expects 'expectation' to be a number instead of a function.

Thanks

Modified: Sorry for the vagueness. Here I'm talking about conditional expectation w.r.t a sub-$\sigma$-algebra. And what I mean by 'restriction' is the following:

A random variable is a measurable function defined on the whole sample space. Once defined, it becomes definite.

However if we don't know how this function is defined, there is another way to collect information of it by observing how it works on the events. More precisely, we observe the integral of this function on each set in the corresponding $\sigma$ algebra. After we complete the observation on all sets, we have a good idea on this function(random variable).---But I don't know if we know everything about it?

With the above point of view, the conditional expectation w.r.t a sub-$\sigma$-algebra seems to finish a observation on some but not all sets in the original $\sigma$-algebra. In that sense, it is a restriction.

I want to know if this intuition is correct.

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Conditional expectation is NOT a restriction of a random variable. The CE generalizes the absolute expectation, which corresponds to the trivial sigma-algebra containing the empty set and its complement. The conditional with respect to an event with positive probability IS a number, but the CE with respect to a sigma-algebra is meant to encode many such numbers. –  Michael Greinecker Sep 20 '11 at 12:47
Conditional expectation is a number, more precisely, a family of numbers. An inclusion of sigma-algebras can be thought of geometrically as a fibration of measurable spaces and a conditional expectation takes a function on the total space, integrates it fiberwise and produces a function on the base space of the fibration. More information can be found in this answer: mathoverflow.net/questions/20740/… –  Dmitri Pavlov Sep 20 '11 at 12:48

Here's an instructive example. Let $\sigma_1 = \{ \emptyset ,[0,1/2),[1/2,1),[0,1)\}$, and $\sigma_2 = \{\emptyset, [0,1)\}$, both sigma algebras. Let $X(\omega) = 4$ if $\omega\in[0,1/2)$ and $X(\omega)=5$ if $\omega\in[1/2,1)$. Now, $X$ is $\sigma_1$-measurable, and $E(X \mid \sigma_2) = 4.5$ for $\omega \in[0,1)$. The conditional expectation takes a value that $X$ did not take at all! This is what typically happens, and also usually the conditional expectation with respect to an algebra is not a constant function.
The conditional expectation is what you get when your knowledge about $\omega$ is restricted (and so you have to average over your ignorance).
To the modified question, I say that your intuition is correct. Think of the $\sigma$-algebra as your knowledge, and sub-$\sigma$-algebras as representing the picture as seen by someone with less knowledge. A martingale, for example, is then a function (the path of a particle across all time, perhaps) about which you are gradually gaining knowledge.