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A combinatorial proof of the Gaussian product inequality beyond the MTP𝟐\mbox{MTP}_{\bf 2} case

Christian Genest Department of Mathematics and Statistics, McGill University, 805, rue Sherbrooke ouest, Montréal (Québec) Canada H3A 0B9 Frédéric Ouimet Department of Mathematics and Statistics, McGill University, 805, rue Sherbrooke ouest, Montréal (Québec) Canada H3A 0B9

Abstract: A combinatorial proof of the Gaussian product inequality (GPI) is given under the assumption that each component of a centered Gaussian random vector 𝑿=(X1,,Xd)\bm{X}=(X_{1},\ldots,X_{d}) of arbitrary length can be written as a linear combination, with coefficients of identical sign, of the components of a standard Gaussian random vector. This condition on 𝑿\bm{X} is shown to be strictly weaker than the assumption that the density of the random vector (|X1|,,|Xd|)(|X_{1}|,\ldots,|X_{d}|) is multivariate totally positive of order 22, abbreviated MTP2\mbox{MTP}_{2}, for which the GPI is already known to hold. Under this condition, the paper highlights a new link between the GPI and the monotonicity of a certain ratio of gamma functions.

Keywords: Complete monotonicity, gamma function, Gaussian product inequality, Gaussian random vector, moment inequality, multinomial, multivariate normal, polygamma function.

MSC: Primary 60E15; Secondary 05A20, 33B15, 62E15, 62H10, 62H12

1 Introduction

The Gaussian product inequality (GPI) is a long-standing conjecture which states that for any centered Gaussian random vector 𝑿=(X1,,Xd)\bm{X}=(X_{1},\dots,X_{d}) of dimension d={1,2,}d\in\mathbb{N}=\{1,2,\ldots\} and every integer mm\in\mathbb{N}, one has

E(i=1dXi2m)i=1dE(Xi2m).\mbox{\rm E}\left(\prod_{i=1}^{d}X_{i}^{2m}\right)\geq\prod_{i=1}^{d}\mbox{\rm E}\big{(}X_{i}^{2m}\big{)}. (1)

This inequality is known to imply the real polarization problem conjecture in functional analysis [14] and it is related to the so-called UU-conjecture to the effect that if PP and QQ are two non-constant polynomials on d\mathbb{R}^{d} such that the random variables P(𝑿)P(\bm{X}) and Q(𝑿)Q(\bm{X}) are independent, then there exist an orthogonal transformation LL on d\mathbb{R}^{d} and an integer k{1,,d1}k\in\{1,\ldots,d-1\} such that PLP\circ L is a function of (X1,,Xk)(X_{1},\ldots,X_{k}) and QLQ\circ L is a function of (Xk+1,,Xd)(X_{k+1},\ldots,X_{d}); see, e.g., [7, 14] and references therein.

Inequality (1) is well known to be true when m=1m=1; see, e.g., Frenkel, [5]. Karlin and Rinott, [8] also showed that it holds when the random vector |𝑿|=(|X1|,,|Xd|)|\bm{X}|=(|X_{1}|,\ldots,|X_{d}|) has a multivariate totally positive density of order 22, denoted MTP2\mbox{MTP}_{2}. As stated in Remark 1.4 of their paper, the latter condition is verified, among others, in dimension d=2d=2 for all nonsingular Gaussian random pairs.

Interest in the problem has recently gained traction when Lan et al., [11] established the inequality in dimension d=3d=3. Hope that the result might be true in general is also fueled by the fact, established by Wei, [25], that for any reals α1,,αd(1/2,0)\alpha_{1},\ldots,\alpha_{d}\in(-1/2,0), one has

E(i=1d|Xi|2αi)i=1dE(|Xi|2αi).\mbox{\rm E}\left(\prod_{i=1}^{d}|X_{i}|^{2\alpha_{i}}\right)\geq\prod_{i=1}^{d}\mbox{\rm E}\big{(}|X_{i}|^{2\alpha_{i}}\big{)}. (2)

Li and Wei, [13] have further conjectured that the latter inequality holds for all reals α1,,αd[0,)\alpha_{1},\ldots,\alpha_{d}\in[0,\infty) and any centered Gaussian random vector 𝑿\bm{X}.

The purpose of this paper is to report a combinatorial proof of inequality (2) in the special case where the reals α1,,αd\alpha_{1},\ldots,\alpha_{d} are nonnegative integers and when each of the components X1,,XdX_{1},\ldots,X_{d} of the centered Gaussian random vector 𝑿\bm{X} can be written as a linear combination, with coefficients of identical sign, of the components of a standard Gaussian random vector. A precise statement of this assumption is given as Condition (III) in Section 2, and the proof of the main result, Proposition 2, appears in Section 3. It is then shown in Section 4, see Proposition 3, that this condition is strictly weaker than the assumption that the random vector |𝑿||\bm{X}| is MTP2\mbox{MTP}_{2}.

Coincidentally, shortly after the first version of the present paper was posted on arXiv, inequality (2) for all nonnegative integers α1,,αd0={0,1,}\alpha_{1},\ldots,\alpha_{d}\in\mathbb{N}_{0}=\{0,1,\ldots\} was established under an even weaker assumption, stated as Condition (IV) in Section 2. The latter condition states that up to a change of sign, the components of the Gaussian random vector 𝑿\bm{X} are all nonnegatively correlated; see Lemma 2.3 of Russell and Sun, [22]. Therefore, the present paper’s main contribution resides in the method of proof using a combinatorial argument closely related to the complete monotonicity of multinomial probabilities previously shown by Ouimet, [16] and Qi et al., [17].

All background material required to understand the contribution and put it in perspective is provided in Section 2. The statements and proofs of the paper’s results are then presented in Sections 3 and 4. The article concludes with a brief discussion in Section 5. For completeness, a technical lemma due to Ouimet, [16], which is used in the proof of Proposition 2, is included in the Appendix.

2 Background

First recall the definition of multivariate total positivity of order 2 (MTP2\mbox{MTP}_{2}) on a set 𝒮d\mathcal{S}\subseteq\mathbb{R}^{d}.

Definition 1.

A density f:d[0,)f:\mathbb{R}^{d}\to[0,\infty) supported on 𝒮\mathcal{S} is said to be multivariate totally positive of order 22, denoted MTP2\mbox{MTP}_{2}, if and only if, for all vectors 𝐱=(x1,,xd),𝐲=(y1,,yd)𝒮\bm{x}=(x_{1},\ldots,x_{d}),\bm{y}=(y_{1},\ldots,y_{d})\in\mathcal{S}, one has

f(𝒙𝒚)f(𝒙𝒚)f(𝒙)f(𝒚),f(\bm{x}\vee\bm{y})f(\bm{x}\wedge\bm{y})\geq f(\bm{x})f(\bm{y}),

where 𝐱𝐲=(max(x1,y1),,max(xd,yd))\bm{x}\vee\bm{y}=(\max(x_{1},y_{1}),\ldots,\max(x_{d},y_{d})) and 𝐱𝐲=(min(x1,y1),,min(xd,yd))\bm{x}\wedge\bm{y}=(\min(x_{1},y_{1}),\ldots,\min(x_{d},y_{d})).

Densities in this class have many interesting properties, including the following result, which corresponds to Eq. (1.7) of Karlin and Rinott, [8].

Proposition 1.

Let 𝐘\bm{Y} be an MTP2\mbox{MTP}_{2} random vector on 𝒮\mathcal{S}, and let φ1,,φr\varphi_{1},\ldots,\varphi_{r} be a collection of nonnegative and (component-wise) non-decreasing functions on 𝒮\mathcal{S}. Then

E{i=1rφi(𝒀)}i=1rE{φi(𝒀)}.\mbox{\rm E}\left\{\prod_{i=1}^{r}\varphi_{i}(\bm{Y})\right\}\geq\prod_{i=1}^{r}\mbox{\rm E}\left\{\varphi_{i}(\bm{Y})\right\}.

In particular, let 𝑿=(X1,,Xd)\bm{X}=(X_{1},\ldots,X_{d}) be a dd-variate Gaussian random vector with zero mean and nonsingular covariance matrix var(𝑿)\mbox{\rm var}(\bm{X}). Suppose that the following condition holds.


(I)

The random vector |𝑿|=(|X1|,,|Xd|)|\bm{X}|=(|X_{1}|,\ldots,|X_{d}|) belongs to the MTP2\mbox{MTP}_{2} class on [0,)d[0,\infty)^{d}.

Under Condition (I), the validity of the GPI conjecture (2) for all reals α1,,αd[0,)\alpha_{1},\ldots,\alpha_{d}\in[0,\infty) follows from Proposition 1 with r=dr=d and maps φ1,,φd\varphi_{1},\ldots,\varphi_{d} defined, for every vector 𝒚=(y1,,yd)[0,)d\bm{y}=(y_{1},\ldots,y_{d})\in[0,\infty)^{d} and integer i{1,,d}i\in\{1,\ldots,d\}, by

φi(𝒚)=yi2αi.\varphi_{i}(\bm{y})=y_{i}^{2\alpha_{i}}.

When 𝑿=(X1,,Xd)\bm{X}=(X_{1},\ldots,X_{d}) is a centered Gaussian random vector with covariance matrix var(𝑿)\mbox{\rm var}({\bm{X}}), Theorem 3.1 of Karlin and Rinott, [8] finds an equivalence between Condition (I) and the requirement that the off-diagonal elements of the inverse of var(𝑿)\mbox{\rm var}({\bm{X}}) are all nonpositive up to a change of sign for some of the components of 𝑿\bm{X}. The latter condition can be stated more precisely as follows using the notion of signature matrix, which refers to a diagonal matrix whose diagonal elements are ±1\pm 1.

Refer to caption
Fig. 1: Implications between Conditions (I)–(IV) for a nonsingular centered Gaussian random vector 𝑿\bm{X}, with references.

(II)

There exists a d×dd\times d signature matrix DD such that the covariance matrix var(D𝑿)1\mbox{\rm var}(D\bm{X})^{-1} only has nonpositive off-diagonal elements.

Two other conditions of interest on the structure of the random vector 𝑿\bm{X} are as follows.

(III)

There exist a d×dd\times d signature matrix DD and a d×dd\times d matrix CC with entries in [0,)[0,\infty) such that the random vector D𝑿D\bm{X} has the same distribution as the random vector C𝒁C\bm{Z}, where 𝒁𝒩d(𝟎d,𝑰d)\bm{Z}\sim\mathcal{N}_{d}(\bm{0}_{d},\bm{I}_{d}) is a d×1d\times 1 Gaussian random vector with zero mean vector 𝟎d\bm{0}_{d} and identity covariance matrix 𝑰d\bm{I}_{d}.

(IV)

There exists a d×dd\times d signature matrix DD such that the covariance matrix var(D𝑿)\mbox{\rm var}(D\bm{X}) has only nonnegative elements.

Recently, Russell and Sun, [22] used Condition (IV) to show that, for all integers dd\in\mathbb{N}, n1,,nd0n_{1},\ldots,n_{d}\in\mathbb{N}_{0} and k{1,,d1}k\in\{1,\ldots,d-1\}, and up to a change of sign for some of the components of 𝑿\bm{X}, one has

E(i=1dXi2ni)(i=1kXi2ni)E(i=k+1dXi2ni).\mbox{\rm E}\left(\prod_{i=1}^{d}X_{i}^{2n_{i}}\right)\geq\mbox{\rm E }\left(\prod_{i=1}^{k}X_{i}^{2n_{i}}\right)\,\mbox{\rm E}\left(\prod_{i=k+1}^{d}X_{i}^{2n_{i}}\right). (3)

This result was further extended by Edelmann et al., [4] to the case where the random vector (X12,,Xd2)(X_{1}^{2},\ldots,X_{d}^{2}) has a multivariate gamma distribution in the sense of Krishnamoorthy and Parthasarathy, [10]. See also [3] for a use of Condition (IV) in the context of the Gaussian correlation inequality (GCI) conjecture.

In the following section, it will be shown how Condition (III) can be exploited to give a combinatorial proof of a weak form of inequality (3). It will then be seen in Section 4 that Condition (II) implies Condition (III), thereby proving the implications illustrated in Fig. 1 between Conditions (I)–(IV). That the implications Condition (II) \Rightarrow Condition (III) and Condition (III) \Rightarrow Condition (IV) are strict can be checked using, respectively, the covariance matrices

var(𝑿)=(3/29/89/89/821/163/49/83/421/16)=(11/21/21/211/41/21/41)(11/21/21/211/41/21/41)\mbox{\rm var}({\bm{X}})=\begin{pmatrix}3/2&9/8&9/8\\ 9/8&21/16&3/4\\ 9/8&3/4&21/16\end{pmatrix}=\begin{pmatrix}1&1/2&1/2\\ 1/2&1&1/4\\ 1/2&1/4&1\end{pmatrix}\begin{pmatrix}1&1/2&1/2\\ 1/2&1&1/4\\ 1/2&1/4&1\end{pmatrix}

and

var(𝑿)=(1001/21/2013/401/203/411/201/201/2101/21/2001)+ε(1000001000001000001000001),\mbox{\rm var}({\bm{X}})=\begin{pmatrix}1&0&0&1/2&1/2\\ 0&1&3/4&0&1/2\\ 0&3/4&1&1/2&0\\ 1/2&0&1/2&1&0\\ 1/2&1/2&0&0&1\end{pmatrix}+\varepsilon\,\begin{pmatrix}1&0&0&0&0\\ 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\end{pmatrix},

for some appropriate ε(0,)\varepsilon\in(0,\infty).

In the first example, the matrix var(𝑿)\mbox{\rm var}({\bm{X}}) is completely positive (meaning that it can be written as CCCC^{\top} for some matrix CC with nonnegative entries) and positive definite by construction. Furthermore, the matrix Dvar(𝑿)1DD\,\mbox{\rm var}({\bm{X}})^{-1}D has at least one positive off-diagonal element for any of the eight possible choices of 3×33\times 3 signature matrix DD. Another way to see this is to observe that if A=var(𝑿)1A=\mbox{\rm var}(\bm{X})^{-1}, then the cyclic product a12a23a31a_{12}a_{23}a_{31}, which is invariant to ADADA\mapsto DAD, is strictly positive in the above example, so that the off-diagonal elements of var(𝑿)1\mbox{\rm var}(\bm{X})^{-1} cannot all be nonpositive. This shows that (III) ⇏\not\Rightarrow (II). This example was adapted from ideas communicated to the authors by Thomas Royen.

For the second example, when ε=0\varepsilon=0, it is mentioned by Maxfield and Minc, [15], using a result from Hall, [6], that the matrix is positive semidefinite and has only nonnegative elements but that it is not completely positive. Using the fact that the set of 5×55\times 5 completely positive matrices is closed, there exists ε(0,)\varepsilon\in(0,\infty) small enough that the matrix var(𝑿)\mbox{\rm var}({\bm{X}}) is positive definite and has only nonnegative elements but is not completely positive. More generally, given that the elements of var(𝑿)\mbox{\rm var}({\bm{X}}) are all nonnegative, the matrix Dvar(𝑿)DD\,\mbox{\rm var}({\bm{X}})D is not completely positive for any of the 3232 possible choices of 5×55\times 5 signature matrix DD, which shows that (IV) ⇏\not\Rightarrow (III). This idea was adapted from comments by Stein, [24].

3 A combinatorial proof of the GPI conjecture

The following result, which is this paper’s main result, shows that the extended GPI conjecture of Li and Wei, [13] given in (2) holds true under Condition (III) when the reals α1,,αd\alpha_{1},\ldots,\alpha_{d} are nonnegative integers. This result also follows from inequality (3), due to Russell and Sun, [22], but the argument below is completely different from the latter authors’ derivation based on Condition (IV).

Proposition 2.

Let 𝐗=(X1,,Xd)\bm{X}=(X_{1},\ldots,X_{d}) be a dd-variate centered Gaussian random vector. Assume that there exist a d×dd\times d signature matrix DD and a d×dd\times d matrix CC with entries in [0,)[0,\infty) such that the random vector D𝐗D\bm{X} has the same distribution as the random vector C𝐙C\bm{Z}, where 𝐙𝒩d(𝟎d,𝐈d)\bm{Z}\sim\mathcal{N}_{d}(\bm{0}_{d},\bm{I}_{d}) is a dd-dimensional standard Gaussian random vector. Then, for all integers n1,,nd0n_{1},\dots,n_{d}\in\mathbb{N}_{0},

E(i=1dXi2ni)i=1dE(Xi2ni).\mbox{\rm E}\left(\prod_{i=1}^{d}X_{i}^{2n_{i}}\right)\geq\prod_{i=1}^{d}\mbox{\rm E}\left(X_{i}^{2n_{i}}\right).
  • Proof.

    In terms of 𝒁\bm{Z}, the claimed inequality is equivalent to

    E{i=1d(j=1dcijZj)2ni}i=1dE{(j=1dcijZj)2ni}.\mbox{\rm E}\left\{\prod_{i=1}^{d}\left(\sum_{j=1}^{d}c_{ij}Z_{j}\right)^{2n_{i}}\right\}\geq\prod_{i=1}^{d}\mbox{\rm E}\left\{\left(\sum_{j=1}^{d}c_{ij}Z_{j}\right)^{2n_{i}}\right\}. (4)

    For each integer j{1,,d}j\in\{1,\ldots,d\}, set Kj=k1j++kdjK_{j}=k_{1j}+\cdots+k_{dj} and Lj=1j++djL_{j}=\ell_{1j}+\cdots+\ell_{dj}, where kijk_{ij} and ij\ell_{ij} are nonnegative integer-valued indices to be used in expressions (5) and (3) below.

    By the multinomial formula, the left-hand side of inequality (4) can be expanded as follows:

    E{i=1d𝒌i0d:ki1++kid=2ni(2niki1,,kid)j=1dcijkijZjkij}.\mbox{\rm E}\left\{\prod_{i=1}^{d}\sum_{\begin{subarray}{c}\bm{k}_{i}\in\mathbb{N}_{0}^{d}:\\ k_{i1}+\dots+k_{id}=2n_{i}\end{subarray}}\binom{2n_{i}}{k_{i1},\dots,k_{id}}\prod_{j=1}^{d}c_{ij}^{k_{ij}}Z_{j}^{k_{ij}}\right\}.

    Calling on the linearity of expectations and the mutual independence of the components of the random vector 𝒁\bm{Z}, one can then rewrite this expression as

    E{𝒌10d:k11++k1d=2n1𝒌d0d:kd1++kdd=2ndi=1d(2niki1,,kid)j=1dcijkijZjkij}=𝒌10d:k11++k1d=2n1𝒌d0d:kd1++kdd=2nd{j=1dE(ZjKj)}i=1d(2niki1,,kid)j=1dcijkij.\mbox{\rm E}\left\{\sum_{\begin{subarray}{c}\bm{k}_{1}\in\mathbb{N}_{0}^{d}:\\ k_{11}+\dots+k_{1d}=2n_{1}\end{subarray}}\dots\sum_{\begin{subarray}{c}\bm{k}_{d}\in\mathbb{N}_{0}^{d}:\\ k_{d1}+\dots+k_{dd}=2n_{d}\end{subarray}}\prod_{i=1}^{d}\binom{2n_{i}}{k_{i1},\dots,k_{id}}\prod_{j=1}^{d}c_{ij}^{k_{ij}}Z_{j}^{k_{ij}}\right\}\\ =\sum_{\begin{subarray}{c}\bm{k}_{1}\in\mathbb{N}_{0}^{d}:\\ k_{11}+\dots+k_{1d}=2n_{1}\end{subarray}}\dots\sum_{\begin{subarray}{c}\bm{k}_{d}\in\mathbb{N}_{0}^{d}:\\ k_{d1}+\dots+k_{dd}=2n_{d}\end{subarray}}\left\{\prod_{j=1}^{d}\mbox{\rm E}\big{(}Z_{j}^{K_{j}}\big{)}\right\}\prod_{i=1}^{d}\binom{2n_{i}}{k_{i1},\dots,k_{id}}\prod_{j=1}^{d}c_{ij}^{k_{ij}}.

Given that the coefficients cijc_{ij} are all nonnegative by assumption, and exploiting the fact that, for every integer j{1,,d}j\in\{1,\ldots,d\} and m0m\in\mathbb{N}_{0},

E(Zj2m)=(2m)!2mm!,\mbox{\rm E}\big{(}Z_{j}^{2m}\big{)}=\frac{(2m)!}{2^{m}m!},

one can bound the left-hand side of inequality (4) from below by

10d:211++21d=2n1d0d:2d1++2dd=2nd{j=1dE(Zj2Lj)}i=1d(2ni2i1,,2id)j=1dcij2ij=10d:11++1d=n1d0d:d1++dd=nd{j=1d(2Lj)!2LjLj!}i=1d(2ni2i1,,2id)j=1dcij2ij.\sum_{\begin{subarray}{c}\bm{\ell}_{1}\in\mathbb{N}_{0}^{d}:\\ 2\ell_{11}+\dots+2\ell_{1d}=2n_{1}\end{subarray}}\dots\sum_{\begin{subarray}{c}\bm{\ell}_{d}\in\mathbb{N}_{0}^{d}:\\ 2\ell_{d1}+\dots+2\ell_{dd}=2n_{d}\end{subarray}}\left\{\prod_{j=1}^{d}\mbox{\rm E}\left(Z_{j}^{2L_{j}}\right)\right\}\prod_{i=1}^{d}\binom{2n_{i}}{2\ell_{i1},\dots,2\ell_{id}}\prod_{j=1}^{d}c_{ij}^{2\ell_{ij}}\\ =\sum_{\begin{subarray}{c}\bm{\ell}_{1}\in\mathbb{N}_{0}^{d}:\\ \ell_{11}+\dots+\ell_{1d}=n_{1}\end{subarray}}\dots\sum_{\begin{subarray}{c}\bm{\ell}_{d}\in\mathbb{N}_{0}^{d}:\\ \ell_{d1}+\dots+\ell_{dd}=n_{d}\end{subarray}}\left\{\prod_{j=1}^{d}\frac{(2L_{j})!}{2^{L_{j}}L_{j}!}\right\}\prod_{i=1}^{d}\binom{2n_{i}}{2\ell_{i1},\dots,2\ell_{id}}\prod_{j=1}^{d}c_{ij}^{2\ell_{ij}}. (5)

The right-hand side of (4) can be expanded in a similar way. Upon using the fact that E(Y2m)=(2m)!σ2m/(2mm!)\mbox{\rm E}(Y^{2m})=(2m)!\sigma^{2m}/(2^{m}m!) for every integer m0m\in\mathbb{N}_{0} when Y𝒩(0,σ2)Y\sim\mathcal{N}(0,\sigma^{2}), one finds

i=1dE{(j=1dcijZj)2ni}\displaystyle\prod_{i=1}^{d}\mbox{\rm E}\left\{\left(\sum_{j=1}^{d}c_{ij}Z_{j}\right)^{2n_{i}}\right\} =i=1d(2ni)!2nini!(j=1dcij2)ni=i=1d(2ni)!2nini!i0d:i1++id=ni(nii1,,id)j=1dcij2ij\displaystyle=\prod_{i=1}^{d}\frac{(2n_{i})!}{2^{n_{i}}n_{i}!}\left(\sum_{j=1}^{d}c_{ij}^{2}\right)^{n_{i}}=\prod_{i=1}^{d}\frac{(2n_{i})!}{2^{n_{i}}n_{i}!}\sum_{\begin{subarray}{c}\bm{\ell}_{i}\in\mathbb{N}_{0}^{d}:\\ \ell_{i1}+\dots+\ell_{id}=n_{i}\end{subarray}}\binom{n_{i}}{\ell_{i1},\dots,\ell_{id}}\prod_{j=1}^{d}c_{ij}^{2\ell_{ij}}
=10d:11++1d=n1d0d:d1++dd=ndi=1d(2ni)!2nini!(nii1,,id)j=1dcij2ij.\displaystyle=\sum_{\begin{subarray}{c}\bm{\ell}_{1}\in\mathbb{N}_{0}^{d}:\\ \ell_{11}+\dots+\ell_{1d}=n_{1}\end{subarray}}\dots\sum_{\begin{subarray}{c}\bm{\ell}_{d}\in\mathbb{N}_{0}^{d}:\\ \ell_{d1}+\dots+\ell_{dd}=n_{d}\end{subarray}}\prod_{i=1}^{d}\frac{(2n_{i})!}{2^{n_{i}}n_{i}!}\binom{n_{i}}{\ell_{i1},\dots,\ell_{id}}\prod_{j=1}^{d}c_{ij}^{2\ell_{ij}}. (6)

Next, compare the coefficients of the corresponding powers cij2ijc_{ij}^{2\ell_{ij}} in expressions (5) and (3). In order to prove inequality (4), it suffices to show that, for all integer-valued vectors 1,,d0d\bm{\ell}_{1},\dots,\bm{\ell}_{d}\in\mathbb{N}_{0}^{d} satisfying i1++id=ni\ell_{i1}+\dots+\ell_{id}=n_{i} for every integer i{1,,d}i\in\{1,\ldots,d\}, one has

{j=1d(2Lj)!2LjLj!}i=1d(2ni2i1,,2id)i=1d(2ni)!2nini!(nii1,,id).\left\{\prod_{j=1}^{d}\frac{(2L_{j})!}{2^{L_{j}}L_{j}!}\right\}\prod_{i=1}^{d}\binom{2n_{i}}{2\ell_{i1},\dots,2\ell_{id}}\geq\prod_{i=1}^{d}\frac{(2n_{i})!}{2^{n_{i}}n_{i}!}\binom{n_{i}}{\ell_{i1},\dots,\ell_{id}}.

Taking into account the fact that 2L1++Ld=2n1++nd2^{L_{1}+\cdots+L_{d}}=2^{n_{1}+\cdots+n_{d}}, and after cancelling some factorials, one finds that the above inequality reduces to

j=1d(2Lj)!i=1d(2ij)!j=1dLj!i=1dij!.\prod_{j=1}^{d}\frac{(2L_{j})!}{\prod_{i=1}^{d}(2\ell_{ij})!}\geq\prod_{j=1}^{d}\frac{L_{j}!}{\prod_{i=1}^{d}\ell_{ij}!}. (7)

Therefore, the proof is complete if one can establish inequality (7). To this end, one can assume without loss of generality that the integers L1,,LdL_{1},\ldots,L_{d} are all non-zero; otherwise, inequality (7) reduces to a lower-dimensional case. For any given integers L1,,LdL_{1},\ldots,L_{d}\in\mathbb{N} and every integer j{1,,d}j\in\{1,\ldots,d\}, define the function

agj(a)=Γ(aLj+1)i=1dΓ(aij+1),a\mapsto g_{j}(a)=\frac{\Gamma(aL_{j}+1)}{\prod_{i=1}^{d}\Gamma(a\ell_{ij}+1)},

on the interval (1/Lj,)(-1/L_{j},\infty), where Γ\Gamma denotes Euler’s gamma function.

To prove inequality (7), it thus suffices to show that, for every integer j{1,,d}j\in\{1,\ldots,d\}, the map aln{gj(a)}a\mapsto\ln\{g_{j}(a)\} is non-decreasing on the interval [0,)[0,\infty). Direct computations yield, for every real a[0,)a\in[0,\infty),

ddaln{gj(a)}\displaystyle\frac{d}{da}\ln\{g_{j}(a)\} =Ljψ(aLj+1)i=1dijψ(aij+1),\displaystyle=L_{j}\psi(aL_{j}+1)-\sum_{i=1}^{d}\ell_{ij}\psi(a\ell_{ij}+1),
d2da2ln{gj(a)}\displaystyle\frac{d^{2}}{da^{2}}\ln\{g_{j}(a)\} =Lj2ψ(aLj+1)i=1dij2ψ(aij+1),\displaystyle=L_{j}^{2}\psi^{\prime}(aL_{j}+1)-\sum_{i=1}^{d}\ell_{ij}^{2}\psi^{\prime}(a\ell_{ij}+1),

where ψ=(lnΓ)\psi=(\ln\Gamma)^{\prime} denotes the digamma function. Now call on the integral representation [1, p. 260]

ψ(z)=0te(z1)tet1𝑑t,\psi^{\prime}(z)=\int_{0}^{\infty}\frac{te^{-(z-1)t}}{e^{t}-1}\,dt,

valid for every real z(0,)z\in(0,\infty), to write

d2da2ln{gj(a)}\displaystyle\frac{d^{2}}{da^{2}}\ln\{g_{j}(a)\} =0(Ljt)ea(Ljt)et1Lj𝑑ti=1d0(ijt)ea(ijt)et1ij𝑑t\displaystyle=\int_{0}^{\infty}\frac{(L_{j}t)e^{-a(L_{j}t)}}{e^{t}-1}\,L_{j}dt-\sum_{i=1}^{d}\int_{0}^{\infty}\frac{(\ell_{ij}t)e^{-a(\ell_{ij}t)}}{e^{t}-1}\,\ell_{ij}dt
=0seas{1es/Lj1i=1d1(es/Lj)Lj/ij1}𝑑s.\displaystyle=\int_{0}^{\infty}se^{-as}\left\{\frac{1}{e^{s/L_{j}}-1}-\sum_{i=1}^{d}\frac{1}{(e^{s/L_{j}})^{L_{j}/\ell_{ij}}-1}\right\}ds. (8)

Given that (1j++dj)/Lj=1(\ell_{1j}+\cdots+\ell_{dj})/L_{j}=1 by construction, the quantity within braces in Eq. (3) is always nonnegative by Lemma 1.4 of Ouimet, [16]; this can be checked upon setting y=es/Ljy=e^{s/L_{j}} and ui=ij/Lju_{i}=\ell_{ij}/L_{j} for every integer i{1,,d}i\in\{1,\ldots,d\} in that paper’s notation. Alternatively, see p. 516 of Qi et al., [17]. Therefore,

a[0,)d2da2ln{gj(a)}0.\forall_{a\in[0,\infty)}\quad\frac{d^{2}}{da^{2}}\ln\{g_{j}(a)\}\geq 0. (9)

In fact, the map ad2ln{gj(a)}/da2a\mapsto d^{2}\ln\{g_{j}(a)\}/da^{2} is even completely monotonic. Moreover, given that

ddaln{gj(a)}|a=0=Ljψ(1)i=1dijψ(1)=0×ψ(1)=0,\left.\frac{d}{da}\ln\{g_{j}(a)\}\right|_{a=0}=L_{j}\psi(1)-\sum_{i=1}^{d}\ell_{ij}\psi(1)=0\times\psi(1)=0,

one can deduce from inequality (9) that

a[0,)ddaln{gj(a)}0.\forall_{a\in[0,\infty)}\quad\frac{d}{da}\ln\{g_{j}(a)\}\geq 0.

Hence the map aln{gj(a)}a\mapsto\ln\{g_{j}(a)\} is non-decreasing on [0,)[0,\infty). This concludes the argument. ∎

4 Condition (II) implies Condition (III)

This paper’s second result, stated below, shows that Condition (II) implies Condition (III). In view of Fig. 1, one may then conclude that Condition (I) also implies Conditions (III) and (IV), and hence also Condition (II) implies Condition (IV). The implication Condition (II) \Rightarrow Condition (IV) was already established in Theorem 2 (i) of Karlin and Rinott, [9], and its strictness was mentioned on top of p. 427 of the same paper.

Proposition 3.

Let Σ\Sigma be a symmetric positive definite matrix with Cholesky decomposition Σ=CC\Sigma=CC^{\top}. If the off-diagonal entries of Σ1\Sigma^{-1} are all nonpositive, then the elements of CC are all nonnegative.

  • Proof.

    The proof is by induction on the dimension dd of Σ\Sigma. The claim trivially holds when d=1d=1. Assume that it is verified for some integer nn\in\mathbb{N}, and fix d=n+1d=n+1. Given the assumptions on Σ\Sigma, one can write

    Σ1=(a𝒗𝒗B)\Sigma^{-1}=\begin{pmatrix}a&\bm{v}^{\top}\\[2.84526pt] \bm{v}&B\end{pmatrix}

    in terms of a real a(0,)a\in(0,\infty), an n×1n\times 1 vector 𝒗\bm{v} with nonpositive components, and an n×nn\times n matrix BB with nonpositive off-diagonal entries.

    Given that Σ\Sigma is symmetric positive definite by assumption, so is Σ1\Sigma^{-1}, and hence so are BB and B1B^{-1}. Moreover, the off-diagonal entries of B=(B1)1B=(B^{-1})^{-1} are nonpositive and hence by induction, the factor LL in the Cholesky decomposition B1=LLB^{-1}=LL^{\top} has nonnegative entries. Letting w=a𝒗LL𝒗w=a-\bm{v}^{\top}LL^{\top}\bm{v} denote the Schur complement, which is strictly positive, one has

    Σ1=(a𝒗𝒗(LL)1)=(w𝒗L𝟎n(L)1)(w𝟎nL𝒗L1),\displaystyle\Sigma^{-1}=\begin{pmatrix}a&\bm{v}^{\top}\\[2.84526pt] \bm{v}&(LL^{\top})^{-1}\end{pmatrix}=\begin{pmatrix}\sqrt{w}&\bm{v}^{\top}L\\ \bm{0}_{n}&(L^{\top})^{-1}\end{pmatrix}\begin{pmatrix}\sqrt{w}&\bm{0}_{n}^{\top}\\ L^{\top}\bm{v}&L^{-1}\end{pmatrix},

    where 𝟎n\bm{0}_{n} is an n×1n\times 1 vector of zeros. Accordingly,

    Σ=(w𝟎nL𝒗L1)1(w𝒗L𝟎n(L)1)1=(1/w𝟎nLL𝒗/wL)(1/w𝟎nLL𝒗/wL)=CC.\Sigma=\begin{pmatrix}\sqrt{w}&\bm{0}_{n}^{\top}\\ L^{\top}\bm{v}&L^{-1}\end{pmatrix}^{-1}\begin{pmatrix}\sqrt{w}&\bm{v}^{\top}L\\ \bm{0}_{n}&(L^{\top})^{-1}\end{pmatrix}^{-1}=\begin{pmatrix}1/\sqrt{w}&\bm{0}_{n}^{\top}\\ -LL^{\top}\bm{v}/\sqrt{w}&L\end{pmatrix}\begin{pmatrix}1/\sqrt{w}&\bm{0}_{n}^{\top}\\ -LL^{\top}\bm{v}/\sqrt{w}&L\end{pmatrix}^{\top}=CC^{\top}.

    Recall that ww is strictly positive and all the entries of LL and 𝒗-\bm{v} are nonnegative. Hence, all the elements of CC are nonnegative, and the argument is complete. ∎

5 Discussion

This paper shows that the Gaussian product inequality (2) holds under Condition (III) when the reals α1,,αd\alpha_{1},\ldots,\alpha_{d} are nonnegative integers. This assumption is further seen to be strictly weaker than Condition (II). It thus follows from the implications in Fig. 1 that when the reals α1,,αd\alpha_{1},\ldots,\alpha_{d} are nonnegative integers, inequality (2) holds more generally than under the MTP2\mbox{MTP}_{2} condition of Karlin and Rinott, [8].

Shortly after the first draft of this article was posted on arXiv, extensions of Proposition 2 were announced by Russell and Sun, [22] and Edelmann et al., [4]; see Lemma 2.3 and Theorem 2.1, respectively, in their manuscripts. Beyond priority claims, which are nugatory, the originality of the present work lies mainly in its method of proof and in the clarification it provides of the relationship between various assumptions made in the relevant literature, as summarized by Figure 1.

Beyond its intrinsic interest, the approach to the proof of the GPI presented herein, together with its link to the complete monotonicity of multinomial probabilities previously shown by Ouimet, [16] and Qi et al., [17], hints to a deep relationship between the MTP2\mbox{MTP}_{2} class for the multivariate gamma distribution of Krishnamoorthy and Parthasarathy, [10], the range of admissible parameter values for their infinite divisibility, and the complete monotonicity of their Laplace transform; see the work of Royen on the GCI conjecture [18, 19, 20, 21] and Theorems 1.2 and 1.3 of Scott and Sokal, [23]. These topics, and the proof or refutation of the GPI in its full generality, provide interesting avenues for future research.

Appendix: Technical lemma

The following result, used in the proof of Proposition 2, extends Lemma 1 of Alzer, [2] from the case d=1d=1 to an arbitrary integer dd\in\mathbb{N}. It was already reported by Ouimet, [16], see his Lemma 4.1, but its short statement and proof are included here to make the article more self-contained.

Lemma A.1.

For every integer dd\in\mathbb{N}, and real numbers y(1,)y\in(1,\infty) and u1,,ud+1(0,1)u_{1},\ldots,u_{d+1}\in(0,1) such that u1++ud+1=1u_{1}+\cdots+u_{d+1}=1, one has

1y1>i=1d+11y1/ui1.\frac{1}{y-1}>\sum_{i=1}^{d+1}\frac{1}{y^{1/u_{i}}-1}. (A.1)
  • Proof.

    The proof is by induction on the integer dd. The case d=1d=1 is the statement of Lemma 1 of Alzer, [2]. Fix an integer d2d\geq 2 and assume that inequality (A.1) holds for every smaller integer. Fix any reals y(1,)y\in(1,\infty) and u1,,ud(0,1)u_{1},\ldots,u_{d}\in(0,1) such that 𝒖1=u1++ud<1\|\bm{u}\|_{1}=u_{1}+\cdots+u_{d}<1. Write ud+1=1𝒖1>0u_{d+1}=1-\|\bm{u}\|_{1}>0. Calling on Alzer’s result, one has

    1y1>1y1/𝒖11+1y1/(1𝒖1)1.\frac{1}{y-1}>\frac{1}{y^{1/\|\bm{u}\|_{1}}-1}+\frac{1}{y^{1/(1-\|\bm{u}\|_{1})}-1}.

    Therefore, the conclusion follows if one can show that

    1y1/𝒖11>i=1d1y1/ui1.\frac{1}{y^{1/\|\bm{u}\|_{1}}-1}>\sum_{i=1}^{d}\frac{1}{y^{1/u_{i}}-1}.

    Upon setting z=y1/𝒖1z=y^{1/\|\bm{u}\|_{1}} and vi=ui/𝒖1v_{i}=u_{i}/\|\bm{u}\|_{1}, one finds that the above inequality is equivalent to

    1z1>i=1d1z1/vi1,\frac{1}{z-1}>\sum_{i=1}^{d}\frac{1}{z^{1/v_{i}}-1},

    which is true by the induction assumption. Therefore, the argument is complete. ∎


Acknowledgments. C. Genest’s research is funded in part by the Canada Research Chairs Program, the Trottier Institute for Science and Public Policy, and the Natural Sciences and Engineering Research Council of Canada. F. Ouimet received postdoctoral fellowships from the Natural Sciences and Engineering Research Council of Canada and the Fond québécois de la recherche – Nature et technologies (B3X supplement and B3XR). The authors are grateful to Donald Richards and Thomas Royen for their comments on earlier versions of this note. The authors also thank the referees for their comments.


Conflict of interest statement. The authors state no conflict of interest.

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