Random variable independent of $\sigma$-algebra and conditional expectation What does it mean to say that a random variable is independent of a sigma-algebra, and why then does this imply that $E(RV|  \sigma) = RV$?. I have no clue what this independence stuff is about (couldn't find it using google either), and surely $E(RV|  \sigma) = RV$ seems like the logical definition of it?
 A: A random variable $X$ is independent of a $\sigma$-algebra $\mathcal{G}$ if the $\sigma$-algebras $\sigma(X)$ and $\mathcal{G}$ are independent. Recall that $\sigma(X)$ consists of all sets of the form $\{X\in B\}$, where $B$ is a Borel subset of $\mathbb{R}$.
If $X$ has finite expectation and is independent of $\mathcal{G}$, then in fact
$$ \mathbb{E}[X|\mathcal{G}]=\mathbb{E}[X]$$
On the other hand, if $X$ is $\mathcal{G}$-measurable (and has finite expectation) then
$$ \mathbb{E}[X|\mathcal{G}]=X $$
This fits with the intuitive description of conditional expectation as an average based on the information contained in $\mathcal{G}$.
A: As @carmichael561 remarks, independence between random variable $X$ and sigma-algebra $\cal G$ amounts to the assertion
$$
P(X\in B\mid A)=P(X\in B)\tag1$$
for every Borel $B$ and every $A\in\cal G$. In particular it implies
$$
E(X\mid A)=E(X).\tag2
$$
for every $A\in\cal G$. Regarding your question whether (2) can serve as the definition, the answer is that (2) is not equivalent to (1). It can be proved that condition (1) implies the desirable property
$$
E(h(X)\mid A) = E(h(X))\tag3
$$
for all functions $h$ (assuming $h$ measurable and $h(X)$ integrable), but you can't get to (3) from (2) alone.
Here's an example where (2) does not imply (1). On the four-point set $\Omega=\{a,b,c,d\}$ let $P$ put mass $\frac12$ on the set $A:=\{a\}$ and mass $\frac16$ on the remaining three singletons. Define $X$ by $X(a)=X(b)=0$, $X(c)=1$, $X(d)=-1$. Take $\cal G$ to be the sigma-algebra generated by set $A$. You can check that $X$ satisfies condition (2) for every set in $\cal G$. However, $P(X=0\mid A)=1\ne \frac23=P(X=0)$, so condition (1) is not satisfied. Moreover, $E(X^2\mid A)=0\ne \frac13=E(X^2)$, so condition (3) is not satisfied.
