Gaussian product inequality for absolute moments
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For any centered Gaussian vector, including singular covariance, and positive real exponents αᵢ, must E[∏ᵢ|Xᵢ|^αᵢ]≥∏ᵢE[|Xᵢ|^αᵢ]? This is the strong Gaussian product inequality for absolute moments.
Why it matters and what counts as progress
Why it matters: The strong inequality asks whether arbitrary centered Gaussian dependence can only increase every positive absolute product moment.
What counts as progress: Prove the inequality beyond known exponent or covariance classes, determine equality or establish sharper correlation-dependent bounds.
How the problem developed
- 2008 · Conjecture [1]
Peter E. Frenkel Formulated the Gaussian product inequality for centered Gaussian vectors and equal positive even exponents.
Wenbo V. Li and Ang Wei Proposed the corresponding inequality for arbitrary positive exponents, which later became the stronger form addressed by the 2026 paper.
- 27 Jul 2026 · Reported claim [3]
ChatGPT 5.6 Sol and Dylan Greaves, with Frédéric Ouimet reworking the proof and equality characterization The paper proves the strong inequality for every centered Gaussian vector, including degenerate vectors, and all positive exponents; with positive marginal variances it characterizes equality by independence.
AI contribution
A proof of the strong Gaussian product inequality conjecture
For any centered Gaussian vector, including singular covariance, and positive real exponents αᵢ, the expected product of |Xᵢ|^αᵢ is at least the product of the corresponding marginal expectations.
Sources
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