Can we characterize conditional independence by conditional probability?

Question

Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. Let $(S, d)$ be a Polish space and $\mathcal S$ its Borel $\sigma$-algebra.

Let $\mathcal A$ be a sub-$\sigma$-algebra of $\mathcal F$. Let $T:\Omega \to S$ be measurable. A map $\nu: S \times \mathcal A \to [0, 1]$ is called the regular conditional probability of $\mathcal A$ given $T$ if

• for every $s \in S$, we have $\nu(s, \cdot)$ is a probability measure on $\mathcal A$,
• for every $A \in \mathcal A$, we have $\nu(\cdot, A)$ is $\mathcal S$-measurable.
• for every $A \in \mathcal A$ and $B \in \mathcal S$, $$ \mathbb P [A \cap T^{-1} (B)] = \int_B \nu(s, A) \ \mathrm d (T_\sharp \mathbb P) (s), $$ where $T_\sharp \mathbb P$ is the push-forward of $\mathbb P$ by $T$.

Let $X, Y, Z:\Omega \to S$ be random variables. We say that $X$ and $Y$ are conditionally independent given $Z$ if $$ \mathbb P [X \in A, Y\in B |Z] = \mathbb P [X \in A |Z] \cdot \mathbb P [Y\in B |Z] \quad \text{a.s.} \quad \forall A, B \in \mathcal S. $$

Let $\mathcal A := \sigma (X, Y)$. Let $\nu: S \times \mathcal A \to [0, 1]$ be the conditional probability of $\mathcal A$ given $Z$.

Can we characterize the conditional independence of $X,Y$ given $Z$ by the map $\nu$?

---

Update: Now let $\mathcal A := \mathcal F$ and fix $s_0 \in S$. We have $$ \mathbb P [A \cap \{T=s_0\}] = \int_{\{s_0\}} \nu(s, A) \ \mathrm d (T_\sharp \mathbb P) (s) = v(s_0, A) \mathbb P [T=s_0]. $$

Hence $\nu (s_0, A) = \mathbb P[A | T=s_0]$. If $A = \{T=s_0\}$ then $\mathbb P[A | T=s_0]=1$. Hence $\nu (s_0, \{T=s_0\}) = 1$, i.e., $T=s_0$ under $\nu (s_0, \cdot)$.

Answer 1

We actually have such a characterization. We need the following lemma, i.e.,

Lemma Let $A \in \mathcal A$. Then $\nu (T, A)$ is a version of $\mathbb P[A | T]$.

Proof Clearly, $\nu (T, A)$ is $\sigma(T)$-measurable. Let $B \in \mathcal S$ and $C := T^{-1} (B)$. We need to prove $\mathbb E [ \nu (T, A) 1_C] = \mathbb P[A \cap C]$. Indeed, $$ \mathbb E [ \nu (T, A) 1_C] = \int_{C} \nu (T(\omega), A) \ \mathrm d \mathbb P (\omega) = \int_B \nu (s, A) \ \mathrm d (T_\sharp \mathbb P) (s) = \mathbb P [A \cap B], $$

where the second equality above follows from change-of-variables formula.

Let $A, B \in \mathcal S$. By above Lemma, $$ \begin{align} \mathbb P [X \in A, Y\in B |Z] &= \nu (Z, X^{-1} (A) \cap Y^{-1} (B)) \quad \text{a.s.}, \\ \mathbb P [X \in A |Z] &= \nu (Z, X^{-1} (A)) \quad \text{a.s.}, \\ \mathbb P [Y\in B |Z] &= \nu (Z, Y^{-1} (B)) \quad \text{a.s.} \end{align} $$

It follows that $X$ and $Y$ are conditionally independent given $Z$ if and only if $$ \nu (Z, X^{-1} (A) \cap Y^{-1} (B)) = \nu (Z, X^{-1} (A)) \cdot \nu (Z, Y^{-1} (B)) \quad \text{a.s.} $$ for all $A, B \in \mathcal S$.