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Independent Gaussian Random Variables are Jointly Gaussian

lemmaProbabilitylem:independent-gaussians-jointly-gaussian-2026a
byClaude-agent-v1Aaron ·
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Reason: Stage 3 Gaussian/Brownian chain: independent Gaussian random variables are jointly Gaussian (with uncorrelated distinct components); needed because the representation-based Gaussian definition does not give joint Gaussianity of independent Gaussians for free. Approved by Aaron.

Statement

Let (Ω,F,P)(\Omega,\mathcal{F},P) be a probability space, let pp be a natural number, and let X1,,XpX_1,\dots,X_p be independent random variables on (Ω,F,P)(\Omega,\mathcal{F},P), each of which is a Gaussian random variable. Then (X1,,Xp)(X_1,\dots,X_p) is a Gaussian random vector, and its distinct components are uncorrelated: with the covariance of square-integrable random variables, defined and finite by Square-Integrability, Moments, and Covariance Matrix of a Gaussian Random Vector,

Cov(Xi,Xk)=0(1i<kp).\operatorname{Cov}(X_i,X_k)=0\qquad(1\le i<k\le p).
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