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Conditional treatment-indicator factorization

Statement

Assume the Axiom of Choice (AC) for the conditional-expectation existence convention used here. Let $(Ω, F, P)$ be a probability space, let $G \subseteq F$ be a sub- $σ$ -algebra, let $A$ be a random element, and let $C$ be a measurable set in its state space. If $Z$ is a real integrable random variable and $Z ⊥ ⁣ ⁣ ⁣ ⊥ A ∣ G$ , then, with $I = 1_{{A \in C}}$ , $E [I Z ∣ G] = E [I ∣ G] E [Z ∣ G] almost surely .$ The conclusion holds for every integrable $Z$ , not only bounded outcomes.

Facts & Assumptions

Given: AC, a probability space, $G \subseteq F$ , random elements $A, Z$ , a measurable treatment set $C$ , $Z \in L^{1} (P)$ , and $Z ⊥ ⁣ ⁣ ⁣ ⊥ A ∣ G$ .

[A1]

AC is assumed for existence of the conditional-expectation classes; $Z$ is integrable and conditionally independent of $A$ given $G$ .

[F1]

Conditional independence means the bounded-product conditional-expectation identity for every pair of bounded measurable functions. (Conditional independence given a sigma-algebra)

[F2]

Under AC, $E [X ∣ G]$ denotes the almost-sure class of integrable $G$ -measurable conditional-expectation versions. (Conditional expectation as an ae class)

[F3]

Conditional expectation preserves order and constants, preserves expectation, and satisfies $E ∣ E [X ∣ G] ∣ \leq E ∣ X ∣$ for integrable $X$ . (Basic algebra and order properties of conditional expectation)

[F4]

If measurable $Z_{n} \to Z$ almost everywhere and $∣ Z_{n} ∣ \leq ∣ Z ∣$ for integrable $Z$ , then $Z_{n} \to Z$ in $L^{1}$ . (Dominated convergence)

Proof

1.1givenA1F2F3

Set $I = 1_{{A \in C}}$ , $e = E [I ∣ G]$ , and $U = E [Z ∣ G]$ . These conditional-expectation classes exist under AC, and $0 \leq e \leq 1$ almost surely by positivity and preservation of constants.

2.1A1F1step 1.1

For each integer $n \geq 1$ , put $Z_{n} = \max (- n, \min (Z, n))$ . The functions $z \mapsto \max (- n, \min (z, n))$ are bounded and measurable, so [F1] applied to $1_{C} (A)$ and this truncation of $Z$ gives $E [I Z_{n} ∣ G] = e E [Z_{n} ∣ G]$ .

2.2F3F4step 1.1

By [F4], $Z_{n} \to Z$ in $L^{1}$ . Since $∣ I ∣ \leq 1$ , $I Z_{n} \to I Z$ in $L^{1}$ ; [F3] then gives $E [I Z_{n} ∣ G] \to E [I Z ∣ G]$ in $L^{1}$ and $e E [Z_{n} ∣ G] \to e U$ in $L^{1}$ , using $0 \leq e \leq 1$ .

3.1step 2.1step 2.2∎

The two sides of step 2.1 are equal for every $n$ and converge in $L^{1}$ to the two sides claimed in the Statement. Uniqueness of limits in $L^{1}$ gives $E [I Z ∣ G] = e U$ almost surely.

Depends on

Used by

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Sources

Hernán and Robins, Causal Inference: What If, updated 2026 author-hosted edition, Chapter 2, Technical Point 2.3 (standard reference, not scraped)
The Mathematics of Indirect Treatment Comparisons, Chapter 10, Lemma 10.3 and Theorem 10.3 proof (standard reference, not scraped)