Conditional expectation property: independence versus conditional independence properties

Question

My question is about those two properties of conditional expectation:

-If $\mathcal{H}$ is independent of $\sigma(\sigma(X), \mathcal{G}),$ then $$ \mathrm{E}[X \mid \sigma(\mathcal{G}, \mathcal{H})]=\mathrm{E}(X \mid \mathcal{G}), \quad \text { a.s. } $$ In particular, if $X$ is independent of $\mathcal{H},$ then $\mathrm{E}(X \mid \mathcal{H})=\mathrm{E}(X), \quad$ a.s.

-If $X, Y$ are conditionally independent given $Z$, then $P(X \in B \mid Y, Z)=P(X \in B \mid Z)$ a.s (equivalently, $\left.E\left(1_{\{X \in B\}} \mid Y, Z\right)=E\left(1_{\{X \in B\}} \mid Z\right) a.s \right)$

Is one of those properties more general than the other ? (in the sense does one imply the other)

Answer 1

The second statement is the definition of conditional independence. It is more general than "$Y$ is independent of $(X,Z)$."

Answer 2

The following is true:

If $Y$ is independent of $(X,Z)$, then $X$ and $Y$ are conditionally independent given $Z$, i.e., $$P(X\in A\mid Y,Z)=P(X\in A\mid Z).\tag{*}$$

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This is easy to see when $(X,Y,Z)$ have a joint density $f(x,y,z)$, since the hypothesis is saying $$f(x,y,z)= f(y)f(x,z).$$ This implies that $$ f(y,z)=\int_x f(x,y,z)dx\stackrel{(1)}=\int_x f(y)f(x,z)dx=f(y)f(z),$$ hence $f(y\mid z)=f(y)$ so that $f(x,y,z)=f(x,z)f(y\mid z)$ which is equivalent to $f(x\mid y,z)=f(x\mid z)$, the desired conclusion.

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For the general measure-theoretic proof, write $U:=P(X\in A\mid Z)$. By definition of conditional probability, $U$ is $\sigma(Z)$-measurable and $$ P(X\in A, Z\in C)=\int_{(Z\in C)} UdP\tag1$$ for every $C$. To prove (*), it is clear that $U$ is $\sigma(Y,Z)$ measurable, so to apply the definition of conditional probability it remains to show $$ P((X\in A)\cap F)=\int_F UdP $$ for every set $F\in \sigma(Y,Z)$. It's enough to do so for sets of the form $F:=(Y\in B, Z\in C)$. For such $F$ we have $$P(X\in A, Y\in B, Z\in C)=P(Y\in B)P(X\in A, Z\in C)\stackrel{(1)}=EI_B(Y)E\left[I_C(Z)U\right] $$ where the first equality follows from independence. Recall $U$ is $\sigma(Z)$-measurable, so by independence again this last equals $$E[I_B(Y)I_C(Z)U]=\int_{(Y\in B,Z\in C)}UdP,$$ as was to be shown.