Using sums-of-squares to prove Gaussian product inequalities

Abstract

Abstract The long-standing Gaussian product inequality (GPI) conjecture states that E [ ∏ j = 1 n ∣ X j ∣ y j ] ≥ ∏ j = 1 n E [ ∣ X j ∣ y j ] E\left[{\prod }_{j=1}^{n}{| {X}_{j}| }^{{y}_{j}}]\ge {\prod }_{j=1}^{n}E\left[{| {X}_{j}| }^{{y}_{j}}] for any centered Gaussian random vector ( X 1 , … , X n ) \left({X}_{1},\ldots ,{X}_{n}) and any non-negative real numbers y j {y}_{j} , j = 1 , … , n j=1,\ldots ,n . In this study, we describe a computational algorithm involving sums-of-squares representations of multivariate polynomials that can be used to resolve the GPI conjecture. To exhibit the power of the novel method, we apply it to prove new four- and five-dimensional GPIs: E [ X 1 2 m X 2 2 X 3 2 X 4 2 ] ≥ E [ X 1 2 m ] E [ X 2 2 ] E [ X 3 2 ] E [ X 4 2 ] E\left[{X}_{1}^{2m}{X}_{2}^{2}{X}_{3}^{2}{X}_{4}^{2}]\ge E\left[{X}_{1}^{2m}]E\left[{X}_{2}^{2}]E\left[{X}_{3}^{2}]E\left[{X}_{4}^{2}] for any m ∈ N m\in {\mathbb{N}} , and E [ ∣ X 1 ∣ y X 2 2 X 3 2 X 4 2 X 5 2 ] ≥ E [ ∣ X 1 ∣ y ] E [ X 2 2 ] E [ X 3 2 ] E [ X 4 2 ] E [ X 5 2 ] E\left[{| {X}_{1}| }^{y}{X}_{2}^{2}{X}_{3}^{2}{X}_{4}^{2}{X}_{5}^{2}]\ge E\left[{| {X}_{1}| }^{y}]E\left[{X}_{2}^{2}]E\left[{X}_{3}^{2}]E\left[{X}_{4}^{2}]E\left[{X}_{5}^{2}] for any y ≥ 1 10 y\ge \frac{1}{10} .