Question about conditional expectation w.r.t $\sigma$-algebra

Question

I know the definition of the conditional expectation regarding two random variables. Now I encountered the folllowing

Definition: Let $\mathcal{G} \subseteq \mathcal{F}$ be a sub-$\sigma$-algebra of $\mathcal{F}$. We say Z is the conditional expectation of $X$ given $\mathcal{G}$, denoted as $E[X|\mathcal{G}]$ if :

1. $E[|Z|] < \infty$
2. $Z$ is $\mathcal{G}$ measurable
3. for all $G \in \mathcal{G}$: $E[Z 1_G]=E[X 1_G]$.

I can't comprehend that definition.

I know the (very simple) Definition of conditional expectation of $X$ given $Y=y$, i.e. $E[X|Y=y]$.

And I also know that we can view the conditional expectation $E[X|Y]$ as a function in $y$.

But how can one understand the conditional expectation of $E[X|\mathcal{G}]$?

I assume that $E[X|\mathcal{G}]$ is a generalization of $E[X|Y]$, but how does it generalize it?

For example $E[X|Y]$ is the generalization of $E[X|Y=y]$, since $E[X|Y=y]$ is a constant that we get for a specific $y \in \mathbb{R}$. But if we take $\omega \in \Omega$ such that $Y(\omega)=y$, then for this $\omega$ we have $E[X|Y](\omega)=E[X|Y=y]$. Thus I interpret E[X|Y] as taking all values of $E[X|Y=y]$ and writing it as a function.

But I do not see how the "$\sigma$"-algebra version $E[X|\mathcal{G}]$ does extend the existing $E[X|Y]$ conditional expectation. And I am also confused by the notation $E[X|\mathcal{G}]$. Up so far, the conditional expectation was not "conditioned" on a $\sigma$-algebra, (it was conditioned on a value $Y=y$ and on a r.v. $Y$). Also in the above Definition, we use a $\sigma$-algbra but still have a random variable $Z$. Why is that so?

I would be happy if someone could explain this to me.

Answer 1

Firstly you can think of conditioning on a random variable to be some function of that random variable $$\mathbb{E}[X|Y]=f(Y).$$

This is equivalent to the definition of conditioning with respect to a sigma algebra generated by $Y$ is an your definition. If we take the sigma algebra generated by $Y$, call it $\mathcal{F}_Y$, then

$$\mathbb{E}[X|\mathcal{F}_Y]=\mathbb{E}[X|Y].$$

Now remember that if $X$ and $Y$ have finite variance, the conditional expectation with respect to a sigma algebra can be defined as the orthogonal projection onto the space of functions measurable with respect to the $\sigma$-algebra. This projection is going to be a random variable which is measurable with respect to $\mathcal{F}_Y$ and indeed will be a function of $Y$, namely $\mathbb{E}[X|Y]=f(Y)$, by virtue of being $\mathcal{F}_Y$-measurable.

This theorem doesn't say how you actually compute the conditional expectation in practice. One way to compute them is if you know the joint and conditional densities of your random variables, then there is a formula from undergrad probability. But to extend probability theory to the general case where the random variables don't necessarily have densities, there is a more abstract definition for what the conditional expectation means but it doesn't say how you might find this.