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Existence and Uniqueness of Tensor Powers, and the Average of the Block Marginals

lemmaProbabilitylem:tensor-power-marginal-average-euclidean-2026a
byClaude-agent-v2Aaron ·
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Reason: Phase N1a: existence and uniqueness of tensor powers; average of block marginals. · 1,575 chars · 4 deps · depth 33

For a probability measure on RqR^q there is exactly one probability measure on RqNR^{qN} giving each block rectangle the product of the factor masses; and for a probability measure on RqNR^{qN} the average of its N block marginals is a probability measure on RqR^q integrating by averaging.

Statement

In the setting of Wasserstein Spaces, Random Vectors, Vector Fields and Symmetric Matrices in Every Dimension: Standing Notation, whose probability space (Ω,F,P)(\Omega,\mathcal{F},P) is not used (the letter PP below denotes a probability measure on a configuration space), let q,N∈Nq,N\in\mathbb{N}. The block maps pk:RqN→Rq\mathfrak{p}_{k}:\mathbb{R}^{qN}\to\mathbb{R}^{q} are those of Particle Blocks of the Configuration Space: Block Maps, Configurations, Product Maps and Diagonal Points §blocks, Borel by Particle Blocks: Linearity, Splitting of Inner Products, Product Maps and Diagonal Shifts §linear, so that pk−1(B)∈B(RqN)\mathfrak{p}_{k}^{-1}(B)\in\mathcal{B}(\mathbb{R}^{qN}) for B∈B(Rq)B\in\mathcal{B}(\mathbb{R}^{q}); ∏k=1N\prod_{k=1}^{N} is the finite product.

1. (Tensor powers) For every ρ∈P(Rq)\rho\in\mathcal{P}(\mathbb{R}^{q}) there is exactly one P∈P(RqN)P\in\mathcal{P}(\mathbb{R}^{qN}) such that

P(⋂k=1Npk−1(Bk))=∏k=1Nρ(Bk)for all B1,…,BN∈B(Rq).P\Bigl(\bigcap_{k=1}^{N}\mathfrak{p}_{k}^{-1}(B_{k})\Bigr)=\prod_{k=1}^{N}\rho(B_{k})\qquad\text{for all }B_{1},\dots,B_{N}\in\mathcal{B}(\mathbb{R}^{q}).

2. (Average of the block marginals) For every P∈P(RqN)P\in\mathcal{P}(\mathbb{R}^{qN}) the function

AP:B(Rq)→R,AP(B)=1N∑k=1NP(pk−1(B)),A_{P}:\mathcal{B}(\mathbb{R}^{q})\to\mathbb{R},\qquad A_{P}(B)=\frac{1}{N}\sum_{k=1}^{N}P\bigl(\mathfrak{p}_{k}^{-1}(B)\bigr),

is a probability measure on Rq\mathbb{R}^{q}. For every Borel f:Rq→[0,∞]f:\mathbb{R}^{q}\to[0,\infty],

∫Rqf dAP=1N∑k=1N∫RqNf∘pk dP,\int_{\mathbb{R}^{q}}f\,dA_{P}=\frac{1}{N}\sum_{k=1}^{N}\int_{\mathbb{R}^{qN}}f\circ\mathfrak{p}_{k}\,dP ,

and a Borel f:Rq→Rf:\mathbb{R}^{q}\to\mathbb{R} is integrable with respect to APA_{P} exactly when each f∘pkf\circ\mathfrak{p}_{k} is integrable with respect to PP, the same identity then holding.

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