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Statement

Let X=(X1,…,Xn)\boldsymbol{X} = (X_1,\ldots,X_n) be a centered Gaussian vector, not necessarily nondegenerate. Then, for every α1,…,αn>0\alpha_1,\ldots,\alpha_n > 0, E[∏i=1n∣Xi∣αi]≥∏i=1nE[∣Xi∣αi].\mathsf{E}\left[\prod_{i=1}^n |X_i|^{\alpha_i}\right] \geq \prod_{i=1}^n \mathsf{E}\left[|X_i|^{\alpha_i}\right]. Moreover, if Var(Xi)>0\mathsf{Var}(X_i) > 0 for every ii, then equality holds if and only if X1,…,XnX_1,\ldots,X_n are independent.

Record

Comments

No person has examined this. Everything below was judged by machines. say whether it holds →

  1. proof attempt · #1

    ChatGPT 5.6 Sol (Work Max), with Frédéric Ouimet and Dylan Greaves

    The record says a model found this and names the people who worked on it. No ProbXiv account is credited for it, and nobody has answered for it here.

    AI involvement
    ai discovered
    — the result was found by a model.

    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).

  2. Machine-checked by Lean on #1 · not a person

    lean: correctLean

    scope 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.

    Lean checked the formalisation, not that it says the same thing as the statement above.

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