Gaussian product inequality conjecture
Statement
Let be a centered Gaussian vector, not necessarily nondegenerate. Then, for every , Moreover, if for every , then equality holds if and only if are independent.
Context
The GPI conjecture has a real two-decade literature across probability.
People
Attempts
No person has examined this. There is 1 attempt here and 1 machine check recorded against it. A machine check is a judgement recorded by a tool: no account is credited for it, nobody has put their name to it, and it is not verification by a person. Saying whether the mathematics holds is the most useful thing anybody can do on this page.
The AI provided a complete and correct solution without the characterization of equality in terms of independence (but only because the equality case was not in the original prompt by Dylan Greaves).
Reviews
1 machine checkNo person has reviewed this attempt. 1 machine check below — a machine check is not human verification.
Machine check · not human verification
machine: correctscope Lean formalization of the result
The prompt and output are available at https://chatgpt.com/share/6a5ea69b-1648-83e8-80b1-014ae0b1003c. This early version of the proof was formalized in Lean using Codex; see https://github.com/dylgre/gaussian-product-inequality. The proof has also been checked by ChatGPT 5.6 Sol (Pro), Gemini 3.1 Pro (Extended Thinking), Grok 4.5 (Expert), and Frédéric Ouimet.
No ProbXiv account is credited for this check. Nobody has put their name to it, so it carries no personal accountability and does not count as verification by a person.
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