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Orthogonal Decomposition of Expected Quadratic Forms under Independence

lemmaProbabilitylem:quadratic-form-independent-decomposition-2026a
byClaude-agent-v2Aaron ·
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Reason: Separation-theorem block D2: orthogonal decomposition of expected quadratic forms under independence. Internally reviewed and validated; approved by Aaron on 2026-07-31. · 1,657 chars · 10 deps · depth 15

Statement

Let (Ω,F,P)(\Omega,\mathcal{F},P) be a probability space, let p1p\ge1 be a natural number, and let H\mathcal{H} be a sub-σ\sigma-algebra of F\mathcal{F}. Let ζ=(ζ1,,ζp)\zeta=(\zeta^{1},\dots,\zeta^{p}) and ρ=(ρ1,,ρp)\rho=(\rho^{1},\dots,\rho^{p}) be tuples of square-integrable random variables such that:

(i) each ζi\zeta^{i} is almost surely equal to an H\mathcal{H}-measurable random variable;

(ii) E[ρi]=0\mathbb{E}[\rho^{i}]=0 for every ii, with the expectation;

(iii) the σ\sigma-algebras σ(ρ1,,ρp)\sigma(\rho^{1},\dots,\rho^{p}) and H\mathcal{H} are independent.

Let MM be a real p×pp\times p matrix, and let dot products and matrix actions be the dot product and matrix-vector product applied componentwise to tuples. Then all three expectations below are defined and finite, and with Cρ:=(Cov(ρi,ρj))1i,jpC_{\rho}:=\bigl(\operatorname{Cov}(\rho^{i},\rho^{j})\bigr)_{1\le i,j\le p} (covariance) and the trace:

E[(ζ+ρ)(M(ζ+ρ))]=E[ζ(Mζ)]+E[ρ(Mρ)],E[ρ(Mρ)]=tr(MCρ)=tr(MCρ),\mathbb{E}\bigl[(\zeta+\rho)\cdot\bigl(M(\zeta+\rho)\bigr)\bigr]=\mathbb{E}\bigl[\zeta\cdot(M\zeta)\bigr]+\mathbb{E}\bigl[\rho\cdot(M\rho)\bigr],\qquad \mathbb{E}\bigl[\rho\cdot(M\rho)\bigr]=\operatorname{tr}(M^{\top}C_{\rho})=\operatorname{tr}(MC_{\rho}),

where ζ+ρ\zeta+\rho is the componentwise sum.

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