Proof of Conditional Mean-Square Optimality Restricted to an Event of the Conditioning Sigma-Algebra
lemmalem:conditional-mean-square-optimality-restricted-2026aThroughout, -measurability of a real-valued function on means measurability with respect to and the Borel -algebra of the real line, and tuples are multiplied by componentwise.
Step 1: pulling the indicator inside the conditional expectation. The indicator is -measurable by claim 1 of the arithmetic lemma for measurable functions applied on , and everywhere. By claim 4 of the basic properties of conditional expectation (taking out what is known), for each the random variables and are square-integrable and is a conditional expectation of given . Set and ; then is a tuple of conditional expectations of the components of given , and its error tuple is .
Step 2: the tuple is admissible. Fix . By admissibility of there is a -measurable square-integrable random variable and an event with such that on (almost sure equality). Then is square-integrable, since pointwise and expectation is monotone by monotonicity of the integral; it coincides with on ; and is -measurable (claim 3 of the arithmetic lemma on ) and square-integrable by the same comparison. Hence is admissible in the sense of the conditional mean-square optimality lemma.
Step 3: claims 1 and 2. Apply the cited lemma to the data , , , , and the admissible tuple . Pointwise, , , and for any tuple of square-integrable random variables
because . Therefore claim 1 (orthogonal decomposition) and claim 2 (optimality) of the cited lemma for the primed data read exactly as claims 1 and 2 of the present lemma, and the integrability of all products involved is part of the setting of that lemma (each is a finite linear combination of products of two square-integrable random variables, integrable by the closure properties of the square-integrability definition; see also Step 4).
Step 4: trace form. Pointwise , each product being integrable as the product of the square-integrable random variables and ; the displayed trace formula follows from linearity of the integral applied finitely many times.
Step 5: independence of the choice. Let be any other conditional expectation of given , and set . By the uniqueness assertion of the existence and uniqueness theorem for conditional expectation, ; the set is an event, being the set where the random variable (claim 2 of the arithmetic lemma) is nonzero, it has , and off it . The event satisfies , by writing as the disjoint union of the sets and using additivity and monotonicity of the measure . Off all the agree with the . Now fix and use the polarization identity, valid pointwise because :
and the same identity with in place of . The random variables and are square-integrable and agree off the null event , so by the lemma on almost sure equality and second moments their second moments coincide. Taking expectations in the two polarization identities and using linearity of the integral gives , which is claim 3.
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Prerequisites
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