Weird statement in my book about (measure theoretic) conditional expectation

[AxiomOfChoice]

My book tries to illustrate the conditional expectation for a random variable $X(\omega)$ on a probability space $(\Omega,\mathscr F,P)$ by asking me to consider the sigma-algebra $\mathscr G = \{ \emptyset, \Omega \}$, $\mathscr G \subset \mathscr F$. It then argues that $E[X|\mathscr G] = E[X]$ (I'm fine with that). But it claims this should make sense, since $\mathscr G$ "gives us no information." How is this supposed to make sense? In what regard does the sigma-algebra $\mathscr G$ give us "no information" about $X$? I mean, if you know the values $X$ takes on $\mathscr G$, you know $X(\omega)$ everywhere, right?! So this obviously is the wrong interpretation (in fact, any sigma-algebra necessarily contains $\Omega$, so this interpretation would make conditional expectation useless) but I can't think of what the right one is...

 

[Eynstone]

Think of a sigma algebra as 'containing information'.Since G is the trivial sigma algebra, it contains no intrinsic information & doesn't affect the expectation.
I must admit that this terminology is vague & nearly metaphorical. It's perfectly fine if you stash this terminology if it doesn't suit your intuition.

 

[Cant or Wont]

I don't know how the book you're following sets it out.

But consider discrete random variables X,Z and the E(X|Z=z) for distinct z's and how the sigma algebra generated by Z partions Omega. So consider first the functions measurable wrt to the trivial sigma algebra. Then a richer sigma algebra, and you might get more of a feel for the idea of "information" in the sigma algebra.

Even defining your random variables, Omega etc. and doing the calculations may make the idear clearer to you.

 

[disregardthat]

But what information is hidden if G is the trivial sigma algebra?

 

[Cant or Wont]

A lot? Potentially none - X might be G measurable.

 

[micromass]

What $E[X\vert \mathcal{G}]$ means is that you know the information that X takes in $\mathcal{G}$.

So the clue is that $E[X\vert \mathcal{G}]$ is $\mathcal{G}$-measurable. In fact, it is the $\mathcal{G}$-random variable that approximates X best (and this can be made rigorous).

So $E[X\vert\mathcal{G}]$ is an approximation of X that is $\mathcal{G}$-measurable. So for any ]a,b[, we know that

$$\{E[X\vert\mathcal{G}]~\in ]a,b[\}\in \mathcal{G}\}$$

What happens if we have $\mathcal{G}=\{\emptyset,\Omega\}$, then we know that

$$\{E[X\vert\mathcal{G}]\in ]a,b[\}\in \{\emptyset,\Omega\}$$

But this places severe restrictions on $E[X\vert\mathcal{G}]$. In fact, it forces this random variable to be constant!

If we take $\mathcal{G}$ to be finer (thus to contain more sets), then we allow $$E[X\vert \mathcal{G}]$$ to take on more values. Specifically, we allow it to approximate X better.

For example, if $\mathcal{G}=\{\emptyset,\Omega, G,G^c\}$, then we must have$$\{E[X\vert\mathcal{G}]\in ]a,b[\}\in \{\emptyset,\Omega,G,G^c\}$$

This does not force our random variable to be constant. Indeed, we now allow $E[X\vert\mathcal{G}]$ to take different values on G and G^c. So our random variable is now 2-valued!

The finer we make $\mathcal{G}$, the more variable the $E[X\vert \mathcal{G}]$ can be. And the better the approximation can be!

I hope this helped.