2020
DOI: 10.1016/j.jmaa.2020.123858
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The three-dimensional Gaussian product inequality

Abstract: We prove the 3-dimensional Gaussian product inequality, i.e., for any real-valued centered Gaussian random vector (X, Y, Z) and m ∈ N, it holds thatOur proof is based on some improved inequalities on multi-term products involving 2-dimensional Gaussian random vectors. The improved inequalities are derived using the Gaussian hypergeometric functions and have independent interest. As by-products, several new combinatorial identities and inequalities are obtained. MSC: Primary 60E15; Secondary 62H12

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Cited by 14 publications
(10 citation statements)
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“…This conjecture was shown to be true for d = 2 (see Theorem 2.1 in [2]), for d = 3 (see Theorem 1.1 in [5]), and whenever |X| := (|X 1 |, |X 2 |, . .…”
Section: Introductionmentioning
confidence: 95%
See 1 more Smart Citation
“…This conjecture was shown to be true for d = 2 (see Theorem 2.1 in [2]), for d = 3 (see Theorem 1.1 in [5]), and whenever |X| := (|X 1 |, |X 2 |, . .…”
Section: Introductionmentioning
confidence: 95%
“…The Gaussian product inequality (GPI) conjecture has gained traction lately following the proof of the three-dimensional case by Lan et al [5] and the proof of the closely related Gaussian correlation inequality by Royen [11] (in fact, Royen proved an extension involving Gamma random variables), see also Lata la & Matlak [6].…”
Section: Introductionmentioning
confidence: 99%
“…The weak form of the GPI was established by Frenkel [4] for n 1 = • • • = n d = 2 and arbitrary d, and by Lan, et al [11] for d = 3 and integer exponents n 1 , n 2 , and n 3 with equality between at least two exponents. Genest and Ouimet [5] developed recently a novel and far-reaching approach to the GPI, proving (1.2) for arbitrary d with nonnegative even integer exponents n j when the covariance matrix Σ is completely positive, i.e., Σ = CC ′ where C = (c ij ) is a d×d matrix with c ij ≥ 0 for all i, j = 1, .…”
Section: Introductionmentioning
confidence: 97%
“…In [4,Theorem 3.2], Lan et al used the Gaussian hypergeometric functions to prove the following inequality: for any m 1 , m 2 ∈ N and any centered Gaussian random vector (X 1 , X 2 , X 3 ),…”
Section: Introductionmentioning
confidence: 99%
“…Note that the assumption X 2 and X 3 have the same exponent m 2 plays an essential role in the proof of [4,Theorem 3.2]. A natural question is whether the three-dimensional GPI still holds when the exponents of X 1 , X 2 , X 3 are all different.…”
Section: Introductionmentioning
confidence: 99%