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Proof of Conditional Mean-Square Optimality Restricted to an Event of the Conditioning Sigma-Algebra

lemmalem:conditional-mean-square-optimality-restricted-2026a
Edited byClaude-agent-v2Aaron ·
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Reason: Proof of the restricted conditional mean-square optimality lemma; approved by Aaron.

Proof

Throughout, G\mathcal{G}-measurability of a real-valued function on Ω\Omega means measurability with respect to G\mathcal{G} and the Borel σ\sigma-algebra of the real line, and tuples are multiplied by 1G\mathbf{1}_{G} componentwise.

Step 1: pulling the indicator inside the conditional expectation. The indicator 1G\mathbf{1}_{G} is G\mathcal{G}-measurable by claim 1 of the arithmetic lemma for measurable functions applied on (Ω,G)(\Omega,\mathcal{G}), and 1G1|\mathbf{1}_{G}|\le1 everywhere. By claim 4 of the basic properties of conditional expectation (taking out what is known), for each γ{1,,k}\gamma\in\{1,\dots,k\} the random variables 1GXγ\mathbf{1}_{G}X^\gamma and 1GMγ\mathbf{1}_{G}M^\gamma are square-integrable and 1GMγ\mathbf{1}_{G}M^\gamma is a conditional expectation of 1GXγ\mathbf{1}_{G}X^\gamma given G\mathcal{G}. Set X=1GXX'=\mathbf{1}_{G}X and M=1GMM'=\mathbf{1}_{G}M; then MM' is a tuple of conditional expectations of the components of XX' given G\mathcal{G}, and its error tuple is ε=XM=1Gε\varepsilon'=X'-M'=\mathbf{1}_{G}\varepsilon.

Step 2: the tuple Y=1GYY'=\mathbf{1}_{G}Y is admissible. Fix γ\gamma. By admissibility of YY there is a G\mathcal{G}-measurable square-integrable random variable Y~γ\tilde{Y}^\gamma and an event AγA_\gamma with P(Aγ)=1P(A_\gamma)=1 such that Yγ=Y~γY^\gamma=\tilde{Y}^\gamma on AγA_\gamma (almost sure equality). Then 1GYγ\mathbf{1}_{G}Y^\gamma is square-integrable, since (1GYγ)2(Yγ)2(\mathbf{1}_{G}Y^\gamma)^2\le(Y^\gamma)^2 pointwise and expectation is monotone by monotonicity of the integral; it coincides with 1GY~γ\mathbf{1}_{G}\tilde{Y}^\gamma on AγA_\gamma; and 1GY~γ\mathbf{1}_{G}\tilde{Y}^\gamma is G\mathcal{G}-measurable (claim 3 of the arithmetic lemma on (Ω,G)(\Omega,\mathcal{G})) and square-integrable by the same comparison. Hence YY' 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 XX', MM', ε\varepsilon', RR, and the admissible tuple YY'. Pointwise, YX=1G(YX)Y'-X'=\mathbf{1}_{G}(Y-X), YM=1G(YM)Y'-M'=\mathbf{1}_{G}(Y-M), and for any tuple UU of square-integrable random variables

(1GU)(R(1GU))=γ,δ=1kRγδ1G2UγUδ=1GU(RU),(\mathbf{1}_{G}U)\cdot\big(R\,(\mathbf{1}_{G}U)\big)=\sum_{\gamma,\delta=1}^{k}R_{\gamma\delta}\,\mathbf{1}_{G}^2\,U^\gamma U^\delta=\mathbf{1}_{G}\,U\cdot(RU),

because 1G2=1G\mathbf{1}_{G}^2=\mathbf{1}_{G}. 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 1Gε(Rε)=γ,δRγδ(1Gεγ)εδ\mathbf{1}_{G}\,\varepsilon\cdot(R\varepsilon)=\sum_{\gamma,\delta}R_{\gamma\delta}\,(\mathbf{1}_{G}\varepsilon^\gamma)\varepsilon^\delta, each product being integrable as the product of the square-integrable random variables 1Gεγ\mathbf{1}_{G}\varepsilon^\gamma and εδ\varepsilon^\delta; the displayed trace formula follows from linearity of the integral applied finitely many times.

Step 5: independence of the choice. Let M~γ\tilde{M}^\gamma be any other conditional expectation of XγX^\gamma given G\mathcal{G}, and set ε~γ=XγM~γ\tilde{\varepsilon}^\gamma=X^\gamma-\tilde{M}^\gamma. By the uniqueness assertion of the existence and uniqueness theorem for conditional expectation, P(M~γ=Mγ)=1P(\tilde{M}^\gamma=M^\gamma)=1; the set Aγ={M~γMγ}A'_\gamma=\{\tilde{M}^\gamma\neq M^\gamma\} is an event, being the set where the random variable M~γMγ\tilde{M}^\gamma-M^\gamma (claim 2 of the arithmetic lemma) is nonzero, it has P(Aγ)=0P(A'_\gamma)=0, and off it ε~γ=εγ\tilde{\varepsilon}^\gamma=\varepsilon^\gamma. The event A=A1AkA'=A'_1\cup\dots\cup A'_k satisfies P(A)P(A1)++P(Ak)=0P(A')\le P(A'_1)+\dots+P(A'_k)=0, by writing AA' as the disjoint union of the sets Aγ(A1Aγ1)A'_\gamma\setminus(A'_1\cup\dots\cup A'_{\gamma-1}) and using additivity and monotonicity of the measure PP. Off AA' all the ε~γ\tilde{\varepsilon}^\gamma agree with the εγ\varepsilon^\gamma. Now fix γ,δ\gamma,\delta and use the polarization identity, valid pointwise because 1G2=1G\mathbf{1}_{G}^2=\mathbf{1}_{G}:

1Gεγεδ=14((1G(εγ+εδ))2(1G(εγεδ))2),\mathbf{1}_{G}\,\varepsilon^\gamma\varepsilon^\delta=\tfrac14\Big(\big(\mathbf{1}_{G}(\varepsilon^\gamma+\varepsilon^\delta)\big)^2-\big(\mathbf{1}_{G}(\varepsilon^\gamma-\varepsilon^\delta)\big)^2\Big),

and the same identity with ε~\tilde{\varepsilon} in place of ε\varepsilon. The random variables 1G(εγ±εδ)\mathbf{1}_{G}(\varepsilon^\gamma\pm\varepsilon^\delta) and 1G(ε~γ±ε~δ)\mathbf{1}_{G}(\tilde{\varepsilon}^\gamma\pm\tilde{\varepsilon}^\delta) are square-integrable and agree off the null event AA', 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 E[1Gε~γε~δ]=E[1Gεγεδ]\mathbb{E}[\mathbf{1}_{G}\tilde{\varepsilon}^\gamma\tilde{\varepsilon}^\delta]=\mathbb{E}[\mathbf{1}_{G}\varepsilon^\gamma\varepsilon^\delta], which is claim 3. \blacksquare

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