Conditional Expectation and Conditional Independence

Question

Suppose we have 3 $\sigma$ algebras A, B and C such that A is independent of C. The random variable X is measurable with respect to $\sigma$(B,C), the $\sigma$ algebra generated by B and C. Is it true that E[X|B]=E[X|$\sigma$(A,B)]? Is there anything else we can say about E[X|$\sigma$(A,B)]? What if $B\subseteq A$?

The idea I'm trying to formalize is that if X depends on B and C measurable functions, then for some suitable larger sigma algebra B' that contains B, E[X|B'] is B-measurable. Or equivalently if X depends on an A-measurable random variable Y and a sigma algebra C, E[X|A] is a function of Y. I would've thought there'd be a nice geometric interpretaion of this by viewing the conditional expectation as orthogonal projection, but I'm getting confused by all the separate details flying around. Thanks in advance.

Answer 1

(Assume that) X is measurable with respect to σ(σ(Y),C) for some random variable Y, (that) Y is A-measurable, and (that) A is independent of C. Is it true that E[X|A] can be written as a function of Y?

Yes, by hypothesis, $X=F(Y,Z)$ for some measurable function $F$ and some $C$-measurable random variable $Z$, then:

One has $E(X\mid A)=G(Y)$ where, for every $y$, $G(y)=E(F(y,Z))$.

To show this, it suffices to check that $E(F(Y,Z)T)=E(G(Y)T)$ for every $A$-measurable random variable $T$.

To check this identity, note that, since $Z$ is independent of $(Y,T)$, $E(F(Y,Z)T)=E(H(Y,T))$ where, for every $(y,t)$, $H(y,t)=E(F(y,Z)t)$. But $H(y,t)=E(F(y,Z))t=G(y)t$, which proves the result.

Key intermediate step: If $(U,V)$ is independent and $K$ measurable, then $E(K(U,V))=E(L(U))$ where, for every $u$, $L(u)=E(K(u,V))$.