Doubt on the definition of conditional expectation and a property

Question

I'm reviewing conditional expectation, which I've forgotten, since I've never really used it after I have studied it. What I'm in trouble with is the expectation w.r.t. a $\sigma$-algebra. So, given a probability space $(\Omega,\mathcal{A}, P )$, let $Y$ be a random variable, $\mathcal{G}$ a sub-$\sigma$-algebra of $\mathcal{A}$, $E[Y|\mathcal{G}]$ is that $ \mathcal{G} $-measurable random variable $Z$ such that: $$ \int_{G}ZdP =\int_{G}YdP \ \ \forall G \in \mathcal{G}$$ Now I know this is a definition but I cannot understand really well in which sense $ E[Y|\mathcal{G}] $ is a random variable. I'll try to explain my doubt: if I consider a $\mathcal{G}$-measurable random variable $X$, and take $E[Y|X]$, it's clear to me the randomicity of E[Y|X] since it is explicit here the dependence on $X$, so whenever I run an experiment and $X$ assumes a value $ x$, $E[Y|X]$ assumes the value $E[Y|x]$. While for $E[Y|\mathcal{G}]$, this dependence is not directly specified in the definition. When I look at the notation $E[Y|\mathcal{G}]$, I translate in "the expectation of $Y$, given we are in the sigma algebra $\mathcal{G}$", that is "given one of the events of $\mathcal{G}$ is occured" or maybe "will occur". I think maybe my problem is I cannot find an explicit expression for $E[Y|\mathcal{G}]$ as a function of a $\mathcal{G}$-measurable random variable. A related problem is why $ Y \ \ \mathcal{G} $- measurable $\implies E[Y|\mathcal{G}] = Y $. I know it's a bit confused question, I hope someone can clarify this.

Answer 1

The starting random variable $Y$ is $\mathcal A$-measurable, but it may not be $\mathcal G$-measurable. Intuitively, $\mathbb E\left[Y\mid\mathcal G\right]$ is a random variable which as close as possible to $Y$ and also $\mathcal G$-measurable. The vector projection explained in these notes formalizes this point of view.

Indeed, when $\mathcal G$ is generated by a random variable $X$, then $\mathbb E\left[Y\mid\mathcal G\right]$ can be written as a function of $X$ (Doob's theorem).

If $Y$ is $\mathcal G$-measurable, then $\int_{G}ZdP =\int_{G}YdP \ \ \forall G \in \mathcal{G}$ is satisfied for $Z=Y$ and since $Y$ is $\mathcal G$-measurable, it satisfies the other requirement in the definition.