Conditional expectation with sigma algebra generated by two random variables

Question

Suppose $X$ and $Y$ are independent, then we have $$E[X|\sigma(Y)]=E[X].$$ If there is another random variable $Z$, do we have the following property: $$E[X|\sigma(Y,Z)]=E[X|\sigma(Z)]?$$

Answer 1

No, you would need independence of $Y$ and $Z$ as well. Think of standard examples where pairwise independence does not imply mutual independence.

That is, $Y$ is independent of $(X,Z)$ is sufficient for $E[X\mid Y,Z]=E[X\mid Z]$, but $Y$ is independent of $X$ is not sufficient.

Answer 2

Consider a classic example.

Let $Y$ be the indicator that the first from two coin tosses shows a head, $Z$ be the indicator that the second does, and $X$ the indicator that the both tosses show the same face.

These Bernoilli($\tfrac 12$) random variables are pairwise independent but not mutually independent.

The conditional expectation that both coins show the same face, given the result of the just one toss, is not determined by that result (it is a constant).

$$\def\E{\mathop{\sf E}} \E(X\mid\sigma(Y))~=~\tfrac 12~=~\E(X)~=~\E(X\mid\sigma(Z))$$

However, the conditional expectation that both coins show the same face, given both results, is very much determined by what are those results.

$$\E(X\mid \sigma(Y,Z)) ~=~ \mathbf 1_{Y=Z}~=~X$$

This is an example of where $\E(X\mid\sigma(Y,Z))\neq \E(X\mid\sigma(Z))$ although $X\perp Y$.