Does measurability imply conditional independence (and vice versa)?

Question

Consider a probability triple $(\Omega,\Sigma,P)$, two $\sigma$-subalgebras $\mathcal{G},\mathcal{H} \subseteq \Sigma$, and a random variable $X$.

Suppose that $X$ is $\sigma(\mathcal{G},\mathcal{H})$-measurble, do we have the following statement?

$$X \,\text{is}\, \mathcal{G}\text{-measurable} \, \Leftrightarrow X\perp \mathcal{H}|\mathcal{G}$$

Notations:

• $X\perp \mathcal{H}|\mathcal{G}$ means $\sigma(X)\perp \mathcal{H}|\mathcal{G}$, where $\sigma(X)$ is the $\sigma$-algebra generated by $X$.

• $\sigma(\mathcal{H},\mathcal{G})$ is the smallest $\sigma$-algebra that contains $\mathcal{H}$ and $\mathcal{G}$.

Intuitively, LHS means as long as we have the information from $\mathcal{G}$, information from $\mathcal{H}$ does not matter anymore, which coincides with that of RHS.

Any proof or pointer is welcomed.

Answer 1

$X\perp \mathcal{H}|\mathcal{G}\Rightarrow X \,\text{is}\, \mathcal{G}\text{-measurable}:$ $$X=E[X\mid \mathcal{H},\mathcal{G}]=E[X\mid \mathcal{G}]$$ $X \,\text{is}\, \mathcal{G}\text{-measurable} \, \Rightarrow X\perp \mathcal{H}|\mathcal{G}:$ $$P[(X\in A) \cap B \mid \mathcal{G}]=E[I_A(X)I_B \mid \mathcal{G}]=I_A(X).E[I_B \mid \mathcal{G}]=P[X\in A \mid\mathcal{G}].P[B \mid\mathcal{G}]$$

Answer 2

After some searching, I find the answer to my question is YES. Below, I share the proof in case someone needs it in the future.

Prop. 13, Sec. 3.2, pg. 137, Probability Theory with Applications (M. Rao) states that:

Let $\mathcal{B},\mathcal{B}_1,\mathcal{B}_2$ be $\sigma$-subalgebras from $(\Omega,\Sigma,P)$, then, the following are equivalent:

(i) $\mathcal{B}_1 \perp \mathcal{B}_2 | \mathcal{B}$.

(ii) $P(B_1|\sigma(\mathcal{B},\mathcal{B_2}))=P(B_1|\mathcal{B})$ a.e. (almost everywhere) $B_1\in \mathcal{B}_1$.

(iii) $P(B_2|\sigma(\mathcal{B},\mathcal{B_1}))=P(B_2|\mathcal{B})$ a.e. $B_2\in \mathcal{B}_2$.

(iv) For all $X:\Omega\mapsto \mathbb{R}$ that is $\mathcal{B}_1$-measurable, $E(X|\sigma(\mathcal{B},\mathcal{B}_2))=E(X|\mathcal{B})$ a.e.

Now, let $\mathcal{B}_1:=\sigma(X)$, $\mathcal{B}_2:=\mathcal{H}$, and $\mathcal{B}:=\mathcal{G}$, by (i) $\Leftrightarrow$ (iv), we have:

$$X\perp \mathcal{H} | \mathcal{G} \Leftrightarrow E(X|\sigma(\mathcal{H},\mathcal{G}))=E(X|\mathcal{G}).$$

Given the precondition that $X$ is $\sigma(\mathcal{H},\mathcal{G})$-measurable, we have:

$$X=E(X|\sigma(\mathcal{H},\mathcal{G})).$$

Therefore, we have:

$$X\perp \mathcal{H} | \mathcal{G} \Leftrightarrow E(X|\mathcal{G})=X.$$

which is equivalent to saying that:

$$X\perp \mathcal{H} | \mathcal{G} \Leftrightarrow X \, \text{is} \, \mathcal{G} \, \text{measurable}.$$