# Rigid characterizations of probability measures through independence, with
applications

Thomas A. Courtade
University of California, Berkeley

(March 10, 2024)

###### Abstract

Three equivalent characterizations of probability measures through
independence criteria are given. These characterizations lead to a family of
Brascamp–Lieb-type inequalities for relative entropy, determine equilibrium
states and sharp rates of convergence for certain linear Boltzmann-type
dynamics, and unify an assortment of $L^{2}$ inequalities in probability.

##  1 Introduction

We start with the notation and definitions needed to state our main result.
Throughout, we work on the space $\mathbb{R}^{n}$, and denote the set of
$n\times n$ real matrices by $M_{n\times n}(\mathbb{R})$. Both spaces are
equipped with their usual Euclidean topologies, and we let $\|\cdot\|$ denote
the Euclidean norm on each space. For a linear subspace
$E\subset\mathbb{R}^{n}$, we let $P_{E}:\mathbb{R}^{n}\to\mathbb{R}^{n}$
denote the orthogonal projection of $\mathbb{R}^{n}$ onto $E$, represented by
a matrix in $M_{n\times n}(\mathbb{R})$. Let $\mathbf{E}$ denote the set of
all linear subspaces of $\mathbb{R}^{n}$. We equip $\mathbf{E}$ with the
coarsest topology such that the maps $E\in\mathbf{E}\mapsto P_{E}\in
M_{n\times n}(\mathbb{R})$ are continuous. For a topological space
$\mathsf{X}$, we let $P(\mathsf{X})$ denote the set of Borel probability
measures on $\mathsf{X}$. For our purposes, $\mathsf{X}$ will be either
$\mathbf{E}$, $\mathbb{R}^{n}$, or a linear subspace of $\mathbb{R}^{n}$.

Each linear subspace $E\subset\mathbb{R}^{n}$ is equipped with the induced
Euclidean topology. The orthogonal complement of $E$ with respect to the usual
inner product on $\mathbb{R}^{n}$ is denoted $E^{\perp}$. For $\mu\in
P(\mathbb{R}^{n})$, we let $\mu_{E}\in P(E)$ denote the marginal of $\mu$ on
$E$. That is, for all bounded continuous $\phi:E\to\mathbb{R}$, we have

$\int_{E}\phi(y)d\mu_{E}(y):=\int_{\mathbb{R}^{n}}(\phi\circ\pi_{E})(x)d\mu(x),$

where $\pi_{E}:\mathbb{R}^{n}\to E$ is the canonical projection of
$\mathbb{R}^{n}$ onto $E$. Generally speaking, we write $T\\#\mu$ to denote
the pushforward of $\mu$ by a measurable map $T$. For example,
$\mu_{E}=\pi_{E}\\#\mu$.

We say the measure $\mu_{E}$ is Gaussian if $\mu_{E}=N(\theta,\Sigma)$ for
some $\theta\in E$ and $\Sigma\equiv\operatorname{Cov}(\mu_{E})$ a symmetric
nonnegative form on $E$ (in particular, we allow degenerate Gaussians). The
standard Gaussian measure on $\mathbb{R}^{n}$ is always denoted by $\gamma$.

######  Definition 1.

A probability measure $\mu\in P(\mathbb{R}^{n})$ splits along $(E,E^{\perp})$
if it admits the product form $\mu=\mu_{E}\otimes\mu_{E^{\perp}}$. We say that
$\mu$ splits with respect to $\xi\in P(\mathbf{E})$ if $\mu$ splits along
$(E,E^{\perp})$, $\xi$-a.s.

For $\xi\in P(\mathbf{E})$, we define $\chi_{\xi}:\mathbb{R}^{n}\to\mathbb{R}$
by

$\chi_{\xi}(x):=\int_{\mathbf{E}}\min\\{\|P_{E}x\|,\|P_{E^{\perp}}x\|\\}d\xi(E),~{}~{}x\in\mathbb{R}^{n}.$

The zero-set of $\chi_{\xi}$ describes an important characteristic of $\xi$,
as explained below.

######  Proposition 1.

The set $\chi_{\xi}^{-1}(0)\subset\mathbb{R}^{n}$ may be uniquely written as
the union of mutually orthogonal, distinct linear subspaces
$(E_{\alpha})_{\alpha}\subset\mathbf{E}$, which we denote

$\displaystyle\chi_{\xi}^{-1}(0)=\cup_{\alpha}E_{\alpha}.$ (1)

######  Remark 1.

If the subspaces $(E_{\alpha})_{\alpha}$ are all nonzero, we do not
distinguish between the cases $\left(\cup_{\alpha}E_{\alpha}\right)$ and
$\\{0\\}\cup\left(\cup_{\alpha}E_{\alpha}\right)$ in the assertion of
uniqueness.

The proof is deferred to Section 3. Admitting it for now, Proposition 1 yields
a canonical decomposition of $\mathbb{R}^{n}$ into orthogonal subspaces,
defined below.

######  Definition 2.

Fix $\xi\in P(\mathbf{E})$. The nonzero subspaces $(E_{\alpha})_{\alpha}$
appearing in the decomposition (1) are called the independent subspaces of
$\mathbb{R}^{n}$ with respect to $\xi$. The independent decomposition of
$\mathbb{R}^{n}$ (with respect to $\xi$) is the orthogonal decomposition

$\mathbb{R}^{n}=\left(\oplus_{\alpha}E_{\alpha}\right)\oplus E_{dep},$

where $E_{dep}:=(\oplus_{\alpha}E_{\alpha})^{\perp}$ is the dependent subspace
of $\mathbb{R}^{n}$ with respect to $\xi$.

######  Definition 3.

Let $\mu\in P(\mathbb{R}^{n})$, and let $V\subset\mathbb{R}^{n}$ be a linear
subspace. We say that the covariance of $\mu_{V}$ splits with respect to
$\xi\in P(\mathbf{E})$ if

$\int_{\mathbf{E}}\|P_{E}\operatorname{Cov}(P_{V}\\#\mu)P_{E^{\perp}}\|\,d\xi(E)=0.$

The important case to consider is when $\mu_{V}$ is Gaussian. Then, for
$X\sim\mu$, the vectors $P_{E}P_{V}X$ and $P_{E^{\perp}}P_{V}X$ are
independent for $\xi$-a.e. $E\in\mathbf{E}$.

###  1.1 Characterizations of measures through independence

The following theorem provides rigid characterizations of measures that split
with respect to a given $\xi\in P(\mathbf{E})$. We defer the proof to Section
3.

######  Theorem 1.

Fix $\xi\in P(\mathbf{E})$. For $\mu\in P(\mathbb{R}^{n})$, the following are
equivalent:

  1. 1)

The measure $\mu$ splits with respect to $\xi$.

  2. 2)

In terms of the independent decomposition of $\mathbb{R}^{n}$ with respect to
$\xi$, the measure $\mu$ admits the product form

$\displaystyle\mu=\left(\otimes_{\alpha}\mu_{E_{\alpha}}\right)\otimes\mu_{E_{dep}},$
(2)

where $\mu_{E_{dep}}$ is Gaussian with covariance that splits with respect to
$\xi$.

  3. 3)

Marginal $\mu_{E_{dep}}$ satisfies
$\int_{E_{dep}}\\!\\!\\!\log(1+\|x\|)d\mu_{E_{dep}}\\!(x)\\!<\\!\infty$, and
$\mu$ is the mixture

$\displaystyle\mu=\int_{\mathbf{E}}(\mu_{E}\otimes\mu_{E^{\perp}})d\xi(E).$
(3)

A few remarks are in order.

######  Remark 2.

The marginal factors $(\mu_{E_{\alpha}})_{\alpha}$ in the decomposition (2)
are allowed to be arbitrary, unlike the marginal $\mu_{E_{dep}}$, which is
necessarily Gaussian.

######  Remark 3.

We are justified in writing the mixture (3) since the map

$\mu_{E}\otimes\mu_{E^{\perp}}(B):(E,B)\to[0,1],~{}~{}~{}~{}\mbox{Borel~{}}B\subset\mathbb{R}^{n},$

is a Markov kernel. This follows from (Borel) measurability of
$(E,x,x^{\prime})\mapsto 1_{B}(P_{E}x+P_{E^{\perp}}x^{\prime})$ for
$x,x^{\prime}\in\mathbb{R}^{n}$, Borel $B\subset{\mathbb{R}^{n}}$ and
Tonelli’s theorem.

It is unclear whether the logarithmic moment assumption in the third statement
of Theorem 1 is necessary. It may be the case that $\mu$ having the mixture
form (3) suffices to imply that $\mu_{E_{dep}}$ satisfies the moment
assumption. In view of this, the following is a natural question.

######  Question 1.

Suppose there is $\theta>0$ such that $\chi_{\xi}(x)\geq\theta\|x\|$ for all
$x\in\mathbb{R}^{n}$. If $\mu$ admits the mixture form (3), does $\mu$ satisfy
$\int_{\mathbb{R}^{n}}\log(1+\|x\|)d\mu(x)<\infty$?

An affirmative answer would allow one to remove the moment assumption from the
third statement of Theorem 1. In some simple cases that yield to ad hoc
analysis, the answer to Question 1 is affirmative. See Appendix A.

###  1.2 Relationship to other work

Theorem 1 has connections to a variety of results appearing in the literature,
many of which are discussed in detail in Section 2. At a more fundametal
level, the equivalence 1)$\Leftrightarrow$2) is related to (and, in fact,
inspired by) Valdimarsson’s characterization of extremizers [32] in the
Brascamp–Lieb inequalities [7, 8, 28]. The (geometric) Brascamp–Lieb
inequality asserts that, if subspaces $(E_{i})_{i=1}^{k}\subset\mathbf{E}$ and
positive reals $(c_{i})_{i=1}^{k}\subset(0,\infty)$ satisfy

$\displaystyle\sum_{i=1}^{k}c_{i}P_{E_{i}}=\operatorname{id}_{\mathbb{R}^{n}},$
(4)

then for any non-negative measurable $f_{i}\in L^{1}(E_{i})$, $i=1,\dots,k$,
we have

$\displaystyle\int_{\mathbb{R}^{n}}\prod^{k}_{i=1}(f_{i}\circ\pi_{E_{i}})^{c_{i}}(x)dx\leq\prod_{i=1}^{k}\left(\int_{E_{i}}f_{i}(x_{i})dx_{i}\right)^{c_{i}}.$
(5)

See, for example, [3]. Admissible $(f_{i})_{i=1}^{k}$ are said to be
extremizers if they meet (5) with equality. Roughly speaking, Valdimarsson
characterized extremal functions $(f_{i})_{i=1}^{k}$ in (5) by defining the
function

$F(x,t):=\prod_{i=1}^{k}(H_{i,t}f_{i})^{c_{i}}(\pi_{E_{i}}x),~{}~{}~{}x\in\mathbb{R}^{n},~{}t\geq
0,$

where $(H_{i,t})_{t\geq 0}$ denotes the heat semigroup acting on
$L^{1}(E_{i})$, and showing that, via monotonicity along the heat flow,
$x\mapsto F(x,t)$ must be separable in certain coordinates when the underlying
$(f_{i})_{i=1}^{k}$ are extremal. He then uses this separability to deduce
that each $f_{i}$ inherits similar separability properties to give a rigid
characterization of extremizers.

A probabilistic interpretation of Valdimarsson’s argument becomes apparent
through the lens of duality, where the Brascamp–Lieb inequalities correspond
to certain subadditivity properties of entropy [14]. In this case, an
inspection of the duality argument reveals that extremal $(f_{i})_{i=1}^{k}$
in (5) admit interpretation as marginal densities of a single probability
density on the ambient space $\mathbb{R}^{n}$. A probabilistic interpretation
of Valdimarsson’s result that seems to have gone unnoticed, or at least has
been underappreciated, roughly manifests as the equivalence
1)$\Leftrightarrow$2) in Theorem 1. Indeed, our definition of the independent
decomposition of $\mathbb{R}^{n}$ coincides with that defined by Valdimarsson
[32] when the probability measure $\xi$ is supported on the finite set of
points $\\{E_{1},\dots,E_{k}\\}$. We retain his terminology of “independent”
and “dependent” subspaces for consistency with that literature.

The equivalence of 3) with the other two statements in Theorem 1 does not have
an analogy in the literature on Brascamp–Lieb inequalities. However, the
mixture (3) emerges naturally in a probabilistic context, characterizing
equilibria in certain Boltzmann-like dynamics, as explained in Section 2.3.
This leads to a concrete “physical” interpretation of the geometric
Brascamp–Lieb inequalities (or, more precisely, the dual variants presented in
Section 2.2) as governing convergence to equilibrium of a stochastic process
that describes an excited particle’s velocity after it’s put in contact with
an equilibrated bath.

The rest of this paper is organized as follows. Section 2 gives a sequence of
applications of Theorem 1. Proofs of Theorem 1 and supporting results are
found in Section 3.

Acknowledgement: The author thanks Efe Aras, Pietro Caputo, and Max Fathi for
stimulating conversations. He also acknowledges NSF-CCF 1750430, the
hospitality of the Laboratoire de Probabilités, Statistique et Modélisation
(LPSM) at the Université Paris Cité, and the Invited Professor program of the
Fondation Sciences Mathématiques de Paris (FSMP).

##  2 Applications

###  2.1 Bernstein-type characterization of Gaussians

The Gaussian distribution has many surprising characterizations; for a nice
overview and pointers to further references, interested readers are referred
to the excellent monograph by Bryc [10]. One famous characterization of the
Gaussian distribution is Bernstein’s characterization via independent linear
forms [4].

######  Theorem 2.

If $X,Y$ are independent random vectors on $\mathbb{R}^{m}$, such that $(X+Y)$
and $(X-Y)$ are independent, then $X$ and $Y$ are Gaussian.

Among its other applications, Bernstein’s theorem implies Maxwell’s
characterization of centered isotropic Gaussian distributions as the unique
rotationally-invariant distributions with independent marginals. As a first
application of Theorem 1, we observe that Bernstein’s theorem follows as a
special case. Indeed, first note that we have:

######  Proposition 2.

Let $(E_{i})_{i=1}^{k}\subset\mathbf{E}$ satisfy $\cap_{i=1}^{k}(E_{i}\cup
E_{i}^{\perp})=\\{0\\}$. If $\mu\in P(\mathbb{R}^{n})$ splits along
$({E}_{i},{E}_{i}^{\perp})$ for each $i=1,\dots,k$, then $\mu$ is Gaussian.

###### Proof.

Take $\xi$ to be a uniform distribution on $\\{E_{1},\dots,E_{k}\\}$ and note
that definitions imply $\chi_{\xi}(x)>0$ for all $x\neq 0$, so that
$E_{dep}=\mathbb{R}^{n}$. Hence, the claim follows from the implication
1)$\Rightarrow$2) of Theorem 1. ∎

To see that Bernstein’s theorem follows as a consequence, consider ambient
space $\mathbb{R}^{n}$ with dimension $n=2m$, and define

$E_{1}=\\{(x,0);x\in\mathbb{R}^{m}\\},~{}~{}~{}E_{2}=\\{(x,x);x\in\mathbb{R}^{m}\\},$

where $0$ denotes the zero-vector in $\mathbb{R}^{m}$. If $X,Y$ satisfy the
hypotheses of Bernstein’s theorem, then $\mu=\operatorname{law}(X,Y)$ splits
along $(E_{i},E_{i}^{\perp})$ for $i=1,2$. Of course,
$\cap_{i=1}^{2}(E_{i}\cup E_{i}^{\perp})=\\{0\\}$, so we conclude Bernstein’s
theorem.

###  2.2 Geometric Brascamp–Lieb-type inequalities

For the applications in this section, we will assume that $\xi\in
P(\mathbf{E})$ satisfies the operator inequality

$\displaystyle\int_{\mathbf{E}}P_{E}d\xi(E)\leq(1-\lambda)\operatorname{id}_{\mathbb{R}^{n}}$
(6)

for some $\lambda\geq 0$. Obviously, there always exists such a $\lambda$ in
the interval $[0,1]$.

For two probability measures $P,Q$ defined on a common measurable space
$(\Omega,\mathcal{F})$, the relative entropy of $P$ with respect to $Q$ is
defined as

$D(P\|Q):=\begin{cases}\int_{\Omega}\log\left(\frac{dP}{dQ}\right)dP&\mbox{if
$P\ll Q$}\\\ +\infty&\mbox{otherwise.}\end{cases}$

We briefly recall the well-known facts that $(P,Q)\mapsto D(P\|Q)$ is non-
negative, convex, and weakly lower semicontinuous. Moreover, if
$T:(\Omega,\mathcal{F})\to(\Xi,\mathcal{G})$ is a measurable map between
measure spaces, we have the so-called data processing inequality
$D(T\\#P\|T\\#Q)\leq D(P\|Q)$. If $T$ is bijective with measurable inverse,
then this is an equality.

The main result of this section is the following family of (dual)
Brascamp–Lieb-type inequalities. The classical (dual, geometric) Brascamp–Lieb
inequalities correspond to the special case where $\mu=\gamma$ and $\xi$ is a
discrete measure supported on a finite number of subspaces.

######  Theorem 3.

If $\mu\in P(\mathbb{R}^{n})$ splits with respect to $\xi\in P(\mathbf{E})$
satisfying (6), then

$\displaystyle\int_{\mathbf{E}}D(\nu_{E}\|\mu_{E})d\xi(E)\leq(1-\lambda)D(\nu\|\mu),~{}~{}~{}\forall\nu\in
P(\mathbb{R}^{n}).$ (7)

######  Remark 4.

If $\xi$ is such that $E_{dep}=\\{0\\}$, then (7) is equivalent to Shearer’s
inequality for relative entropy [16]. There is no obvious way to deduce the
general case from this special case (if there were, the Brascamp–Lieb
inequalities would be derivable as a special case of the Loomis–Whitney
inequalities).

###### Proof.

We begin by observing that (7) holds when the reference measure $\mu$ is
replaced by the standard Gaussian measure $\gamma$, which splits along
$(E,E^{\perp})$ for every choice of subspace. That is, we claim that

$\displaystyle\int_{\mathbf{E}}D(\nu_{E}\|\gamma_{E})d\xi(E)\leq(1-\lambda)D(\nu\|\gamma),~{}~{}~{}\forall\nu\in
P(\mathbb{R}^{n}).$ (8)

This is a variation on the dual form of the geometric Brascamp–Lieb
inequalities, which correspond to the special case where $\xi$ is a discrete
measure supported on a finite number of points. Despite this minor difference,
Lehec’s probabilistic proof of the geometric Brascamp–Lieb inequalities (based
on a variational representation of entropy due to Föllmer [23, 24]) can be
executed nearly verbatim to conclude (8); see [27, proof of Theorem 16].

Now, we’ll use a transportation argument to conclude (7) by leveraging the
rigid product structure (2) implied by the splitting property of $\mu$. Toward
this end, let $T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ be a homeomorphism
satisfying

$\displaystyle T\circ P_{E}=P_{E}\circ T,\mbox{ $\xi$-a.s.}$ (9)

As a homeomorphism, $T^{-1}$ exists and is continuous (and therefore
measurable), and we have for $\xi$-a.e. $E$

$\displaystyle D(\nu_{E}\|(T\\#\gamma)_{E})$ $\displaystyle=D((P_{E}\circ
T)\\#(T^{-1}\\#\nu)\|(P_{E}\circ T)\\#\gamma)$ $\displaystyle=D((T\circ
P_{E})\\#(T^{-1}\\#\nu)\|(T\circ P_{E})\\#\gamma)$
$\displaystyle=D(P_{E}\\#(T^{-1}\\#\nu)\|P_{E}\\#\gamma)=D((T^{-1}\\#\nu)_{E}\|\gamma_{E}),$

where the first line is definitions, the second is the assumed commutativity,
and the third follows since relative entropy is invariant under homeomorphisms
of the measure space. Thus, integrating with respect to $\xi$ gives

$\displaystyle\int_{\mathbf{E}}D(\nu_{E}\|(T\\#\gamma)_{E})d\xi(E)$
$\displaystyle=\int_{\mathbf{E}}D((T^{-1}\\#\nu)_{E}\|\gamma_{E})d\xi(E)$
$\displaystyle\leq(1-\lambda)D(T^{-1}\\#\nu\|\gamma)$
$\displaystyle=(1-\lambda)D(\nu\|T\\#\gamma),$

where the inequality is (8) applied to the measure $\nu\leftarrow
T^{-1}\\#\nu$, and the final step is again because relative entropy is
invariant under homeomorphisms of the measure space. Hence, (7) holds whenever
$\mu$ is the pushforward of $\gamma$ by a homeomorphism
$T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ that commutes with $P_{E}$, $\xi$-a.s.

Now, let $\mu$ split with respect to $\xi$. By Theorem 1, $\mu$ admits the
product form

$\mu=\left(\otimes_{\alpha}\mu_{E_{\alpha}}\right)\otimes\mu_{E_{dep}},$

where $\mu_{E_{dep}}\in P(E_{dep})$ is Gaussian, and $\mu_{E_{\alpha}}\in
P(E_{\alpha})$. We may assume without loss of generality that $\mu_{E_{dep}}$
is centered, and therefore $\mu_{E_{dep}}=N(0,\Sigma)$, where the covariance
$\Sigma:E_{dep}\to E_{dep}$ splits with respect to $\xi$. In particular,

$\Sigma^{1/2}(E_{dep}\cap E)\subset(E_{dep}\cap
E)~{}~{}\mbox{and}~{}~{}\Sigma^{1/2}(E_{dep}\cap E^{\perp})\subset(E_{dep}\cap
E^{\perp}),~{}~{}~{}\xi-a.s.,$

due to the splitting property of $\mu$, and the fact that $\Sigma$ and its
positive semidefinite square root $\Sigma^{1/2}$ share common eigenspaces.

Now, let us assume $\mu$ has smooth non-vanishing density with respect to
Lebesgue measure. For each nonzero independent subspace $E_{\alpha}$, there
exists a homeomorphism $T_{\alpha}:E_{\alpha}\to E_{\alpha}$ such that
$T_{\alpha}\\#\gamma_{E_{\alpha}}=\mu_{E_{\alpha}}$ [17, Corollary 1]. On
$E_{dep}$ we define the linear map $T_{dep}:x\in E_{dep}\mapsto\Sigma^{1/2}x$;
therefore $T_{dep}\\#\gamma_{E_{dep}}=\mu_{E_{dep}}$. Since we are assuming
$\mu$ has full support, $\Sigma$ is positive definite, and therefore $T_{dep}$
is also a homeomorphism. Now, we define the operator
$T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ via the direct sum

$T=\left(\oplus_{\alpha}T_{\alpha}\right)\oplus T_{dep},$

with individual components acting on the corresponding components of the
independent decomposition of $\mathbb{R}^{n}$. By construction, $T$ is a
homeomorphism satisfying (9). Hence, (7) holds whenever $\mu$ has smooth non-
vanishing density with respect to Lebesgue measure.

The general case where $\mu$ fails to admit a non-vanishing density is handled
by localization. To this end, let $(H^{*}_{t})_{t\geq 0}$ denote the (adjoint)
heat semigroup acting on $P(\mathbb{R}^{n})$. Since $H^{*}_{t}$ operates
independently on orthogonal components, it is a consequence of the data
processing inequality that

$\displaystyle D(\nu_{E}\|\mu_{E})$ $\displaystyle\geq
D((H^{*}_{t}\nu)_{E}\|(H^{*}_{t}\mu)_{E}).$

By weak lower semicontinuity of entropy, we have the reverse inequality in the
limit as $t\downarrow 0$:

$\liminf_{t\geq 0}D((H^{*}_{t}\nu)_{E}\|(H^{*}_{t}\mu)_{E})\geq
D(\nu_{E}\|\mu_{E}).$

For every $t>0$ the measure $H^{*}_{t}\mu$ admits a smooth non-vanishing
density with respect to Lebesgue measure. Since $\mu\mapsto H^{*}_{t}\mu$
preserves the splitting property of $\mu$ for all $t\geq 0$, we have

$\displaystyle\int_{\mathbf{E}}D(\nu_{E}\|\mu_{E})d\xi(E)$
$\displaystyle\leq\liminf_{t\to
0}\int_{\mathbf{E}}D((H^{*}_{t}\nu)_{E}\|(H^{*}_{t}\mu)_{E})d\xi(E)$
$\displaystyle\leq\liminf_{t\to 0}(1-\lambda)D(H^{*}_{t}\nu\|H^{*}_{t}\mu)$
$\displaystyle\leq(1-\lambda)D(\nu\|\mu),$

where the first inequality is Fatou’s lemma, the second is (7) for reference
measure having smooth non-vanishing density, and the third inequality is data
processing. ∎

###  2.3 Linear Boltzmann-type dynamics

Consider a physical experiment where two particles of equal mass and
respective velocities $v,v_{*}\in\mathbb{R}^{n}$ undergo an elastic collision.
By conservation of energy and momentum, the particles necessarily retain
velocity components on some subspace $E\in\mathbf{E}$, and exchange on
$E^{\perp}$. That is, the post-collision velocities of the first and second
particles are, respectively:

$\displaystyle
v^{\prime}=P_{E}v+P_{E^{\perp}}v_{*},~{}~{}\mbox{and}~{}~{}v^{\prime}_{*}=P_{E}v_{*}+P_{E^{\perp}}v.$
(10)

In a Hamiltonian-based model, the subspace $E$ would be a function of the
particle positions and a suitable potential energy function. In many-particle
systems, however, it will generally be more tractable to assume a spatially
homogenous model where the subspace $E$ is drawn randomly, according to a
given distribution $\xi\in P(\mathbf{E})$. Then, the post-collision velocities
are random variables, with distributions induced by $\xi$ and the pre-
collision velocities $v,v_{*}$. We remark that the post-collision kinetic
energy of the (first) particle can be written as
$\frac{1}{2}\|v^{\prime}\|^{2}=\frac{1}{2}\|P_{E}v\|^{2}+\frac{1}{2}\|P_{E^{\perp}}v_{*}\|^{2}$.
So, when $E\sim\xi$, inequality (6) has the interpretation that, on average, a
fraction at least $\lambda$ of the total kinetic energy is exchanged through
collision.

Switching gears for a moment, let $\mu\in P(\mathbb{R}^{n})$, and define a
Markov semigroup $P_{t}=e^{t\mathcal{L}},t\geq 0,$ via the infinitesimal
generator

$\mathcal{L}f(v):=\int_{\mathbf{E}}\int_{\mathbb{R}^{n}}f(P_{E}v+P_{E^{\perp}}v_{*})d\mu(v_{*})d\xi(E)-f(v),$

on bounded measurable $f:\mathbb{R}^{n}\to\mathbb{R}$. For initial data
$\nu_{0}\in P(\mathbb{R}^{n})$, we define the evolution $(\nu_{t})_{t\geq 0}$
via duality

$\int_{\mathbb{R}^{n}}fd\nu_{t}:=\int_{\mathbb{R}^{n}}P_{t}fd\nu_{0},~{}~{}~{}f\in
C_{b}(\mathbb{R}^{n}).$

The Markov semigroup $(P_{t})_{t\geq 0}$ is related to the physical model
previously described as follows. Consider an experiment where a unit-mass
particle, excited with initial velocity $V_{0}\sim\nu_{0}$, is placed in
contact with a bath consisting of unit-mass particles, each having velocities
distributed i.i.d. according to some background distribution $\mu$. If elastic
collisions modeled by (10) occur between the excited particle and independent
particles in the bath following a rate-1 Poisson point process, with subspace
$E\sim\xi$ drawn independently for each collision, the velocity $V_{t}$ of the
excited particle at time $t$ has law $\nu_{t}$. The bath is in equilibrium
with respect to the described dynamics if, for $\nu_{0}=\mu$, we have
$\nu_{t}=\mu$ for all $t\geq 0$. The _temperature_ of the bath is
$\int_{\mathbb{R}^{n}}\|v\|^{2}d\mu(v)$.

Theorem 1 immediately provides a rigid characterization of finite-
temperature111The assumption of finite temperature is stronger than required,
since our application of Theorem 1 only requires $\mu_{E_{dep}}$ to have
finite logarithmic moment. However, all realizable physical systems have
finite temperature, so it seems more physically relevant to impose this
assumption instead of merely finite logarithmic moments on $E_{dep}$. baths in
equilibrium with respect to our dynamics.

######  Proposition 3.

A finite-temperature bath with background distribution $\mu$ is in equilibrium
with respect to the dynamics $(P_{t})_{t\geq 0}$ if and only if $\mu$ splits
with respect to $\xi$. Moreover, in this case, $\mu$ is reversible for
$(P_{t})_{t\geq 0}$.

###### Proof.

In order for the bath to be invariant to the described dynamics, we require
that $\mu$ is an invariant measure of the semigroup $(P_{t})_{t\geq 0}$. In
particular, for all $f\in C_{b}(\mathbb{R}^{n})$, we have
$\int_{\mathbb{R}^{n}}\mathcal{L}fd\mu=0$. This may be rewritten as

$\displaystyle\int_{\mathbf{E}}\int_{\mathbb{R}^{n}}fd(\mu_{E}\otimes\mu_{E^{\perp}})d\xi(E)=\int_{\mathbf{E}}\int_{\mathbb{R}^{n}}fd\mu
d\xi(E)=\int_{\mathbb{R}^{n}}fd\mu.$

In other words, the bath is in equilibrium if and only if $\mu$ is the mixture
(3). Under the finite-temperature assumption, this is equivalent to $\mu$
splitting with respect to $\xi$ by Theorem 1.

To show reversibility, we need to show
$\int_{\mathbb{R}^{n}}f\mathcal{L}gd\mu=\int_{\mathbb{R}^{n}}g\mathcal{L}fd\mu$
for, say, bounded continuous $f,g$. This is easily verified using the
characterization that $\mu$ splits along $(E,E^{\perp})$, $\xi$-a.s. ∎

Having characterized all equilibrium states for our dynamics, we now establish
quantitative rates of convergence to equilibrium.

######  Theorem 4.

Let $\lambda\geq 0$ be such that (6) holds. If $\mu$ splits with respect to
$\xi$, then

$\displaystyle D(\nu_{t}\|\mu)\leq e^{-\lambda
t}D(\nu_{0}\|\mu),~{}~{}~{}\forall t\geq 0,~{}\nu_{0}\in P(\mathbb{R}^{n}).$
(11)

Additionally, for all $f\in L^{2}(\mu)$,

$\displaystyle\operatorname{Var}_{\mu}(P_{t}f)\leq e^{-2\lambda
t}\operatorname{Var}_{\mu}(f).$ (12)

The theorem is stated in such a way that no moment assumptions on $\mu$ are
imposed. However, in view of Proposition 3, the physical interpretation of
Theorem 4 is that it governs convergence toward equilibrium for the process
that results from placing an excited particle in contact with a finite-
temperature equilibrated bath. In view of this interpretation, if the best
choice of $\lambda$ in (6) is $\lambda=0$, then our particles do not interact
in some direction, and convergence to equilibrium cannot be guaranteed.
However, if $0<\lambda\leq 1$, then (11) implies a strict trend to equilibrium
$\mu$. The respective convergence rates of $\lambda$ and $2\lambda$ in (11)
and (12) are sharp; an explanation follows the proof.

######  Remark 5.

The entropy-decay estimate (11) is equivalent to a log-Sobolev inequality with
constant $2/\lambda$ [2, Theorem 5.2.1], and the variance estimate (12) is
equivalent to a Poincaré inequality with constant $1/\lambda$ [2, Theorem
4.2.5].

###### Proof of Theorem 4.

We’ll prove the entropy decay estimate (11) first; we can assume without loss
of generality that $D(\nu_{0}\|\mu)<\infty$, else there is nothing to prove.
Toward this end, note that the Poisson timing of the collision process implies

$\displaystyle\nu_{t}=e^{-t}\nu_{0}+te^{-t}\int_{\mathbf{E}}\nu_{0E}\otimes\mu_{E^{\perp}}d\xi(E)+(1-(1+t)e^{-t})\tilde{\nu}_{t},$
(13)

where $\tilde{\nu}_{t}\in P(\mathbb{R}^{n})$ is the conditional law of the
excited particle’s velocity at time $t$, given that two or more collisions
occurred before time $t$. Thus, for all $t\geq 0$ we have

$\displaystyle D(\nu_{t}\|\mu)-D(\nu_{0}\|\mu)$
$\displaystyle\leq(e^{-t}-1)D(\nu_{0}\|\mu)+te^{-t}\int_{\mathbf{E}}D(\nu_{0E}\|\mu_{E})d\xi(E)+(1-(1+t)e^{-t})D(\tilde{\nu}_{t}\|\mu)$
$\displaystyle\leq(e^{-t}-1)D(\nu_{0}\|\mu)+te^{-t}(1-\lambda)D(\nu_{0}\|\mu)+(1-(1+t)e^{-t})D({\nu}_{0}\|\mu)$
$\displaystyle=-te^{-t}\lambda D(\nu_{0}\|\mu),$

where the first inequality follows by convexity of relative entropy applied to
the mixture (13), and the second inequality follows from an application of (3)
and the data processing inequality for relative entropy which implies
$D(\tilde{\nu}_{t}\|\mu)\leq D(\nu_{0}\|\mu)$ (this uses the fact that $\mu$
is invariant under our dynamics, so that if $\nu_{0}=\mu$, then
$\tilde{\nu}_{t}=\mu$).

Hence, from the above and the semigroup property, we have the estimate

$\displaystyle\frac{d}{dt}^{+}D(\nu_{t}\|\mu)\leq-\lambda
D(\nu_{t}\|\mu),~{}~{}~{}t\geq 0.$

An application of Grönwall’s lemma gives (11).

Inequality (11) can be linearized to obtain the variance decay inequality
(12), albeit with the suboptimal rate $\lambda$ in the exponent. The improved
rate of $2\lambda$ will be obtained by linearizing (7) directly. Toward this
end, recall that the relative entropy of $P\ll Q$ can be written as

$D(P\|Q)=\int\frac{dP}{dQ}\log\left(\frac{dP}{dQ}\right)dQ.$

Therefore, if $P$ is a perturbation of $Q$ in the sense that $dP=(1+\epsilon
f)dQ$ for a bounded measurable function $f$ and $\epsilon$ sufficiently small,
then Taylor expansion of $x\in\mathbb{R}^{+}\mapsto x\log x$ about $x=1$ gives
the local behavior of relative entropy

$D(P\|Q)=\frac{\epsilon^{2}}{2}\operatorname{Var}_{Q}(f)+o(\epsilon^{2}),$

where the first-order term is absent since $f$ necessarily satisfies $\int
fdQ=0$ for $P$ to be a probability measure. By Taylor’s theorem, we remark
that the little-$o$ term has modulus at most

$\frac{(\epsilon\|f\|_{\infty})^{3}}{6(1-\epsilon\|f\|_{\infty})^{2}}.$

So, consider a bounded measurable function $f:\mathbb{R}^{n}\to\mathbb{R}$
satisfying $\int fd\mu=0$ and define $d\mu^{\epsilon}:=(1+\epsilon f)d\mu$,
which is a valid probability measure for all $\epsilon$ sufficiently small.
Let $X\sim\mu$. For $E\in\mathbf{E}$, definitions imply

$d\mu^{\epsilon}_{E}=(1+\epsilon\mathbb{E}[f(X)|P_{E}X])d\mu_{E},$

where $\mathbb{E}[f(X)|P_{E}X]$ is the conditional expectation of $f(X)$ under
$\mu$ with respect to the $\sigma$-algebra generated by $P_{E}X$. Thus, by
linearization and (7), we have

$\displaystyle\frac{\epsilon^{2}}{2}\int_{\mathbf{E}}\operatorname{Var}(\mathbb{E}[f(X)|P_{E}X])d\xi(E)+o(\epsilon^{2})$
$\displaystyle=\int_{\mathbf{E}}D\left(\mu^{\epsilon}_{E}\big{\|}\mu_{E}\right)d\xi(E)$
$\displaystyle\leq(1-\lambda)D(\mu^{\epsilon}\|\mu)$
$\displaystyle=(1-\lambda)\frac{\epsilon^{2}}{2}\operatorname{Var}(f(X))+o(\epsilon^{2}),$

where all variances and expectations are with respect to $\mu$. Dividing
through by $\epsilon^{2}$ and letting $\epsilon$ vanish establishes the
inequality

$\displaystyle\int_{\mathbf{E}}\operatorname{Var}(\mathbb{E}[f(X)|P_{E}X])d\xi(E)\leq(1-\lambda)\operatorname{Var}(f(X))$
(14)

for all bounded measurable $f$ with $\int fd\mu=0$. Inequality (14) is
invariant to replacing $f$ by $f+c$ for $c\in\mathbb{R}$, so the zero-mean
assumption can be eliminated. Thus, a standard density argument allows us to
conclude the same holds for all $f\in L^{2}(\mu)$. Finally, using the
classical variance decomposition

$\operatorname{Var}(f(X))=\mathbb{E}[\operatorname{Var}(f(X)|P_{E}X)]+\operatorname{Var}(\mathbb{E}[f(X)|P_{E}X]),$

we may rewrite (14) in the form

$\displaystyle\operatorname{Var}_{\mu}(f)=\operatorname{Var}(f(X))$
$\displaystyle\leq\frac{1}{\lambda}\int_{\mathbf{E}}\mathbb{E}[\operatorname{Var}(f(X)|P_{E}X)]d\xi(E)$
(15)
$\displaystyle=\frac{1}{2\lambda}\int_{\mathbb{R}^{n}}\left(\mathcal{L}(f^{2})-2f\mathcal{L}(f)\right)d\mu.$

By [2, Theorem 4.2.5], this is equivalent to (12). ∎

With the proof complete, we now address optimality of the rates of convergence
in (11) and (12). To start, let $\xi\in P(\mathbf{E})$ and $\lambda\geq 0$
satisfy

$\displaystyle\int_{\mathbf{E}}P_{E}d\xi(E)=(1-\lambda)\operatorname{id}_{\mathbb{R}^{n}}.$
(16)

Now, take $\mu=\gamma$, and
$\nu_{0}=N(\theta,\operatorname{id}_{\mathbb{R}^{n}})$ for fixed
$\theta\in\mathbb{R}^{n}$. By construction, $\nu_{t}$ is the mixture

$\displaystyle\nu_{t}$ $\displaystyle=\sum_{k\geq
0}\frac{t^{k}e^{-t}}{k!}\int_{\mathbf{E}^{k}}N(P_{E_{1}}\cdots
P_{E_{k}}\theta,\operatorname{id}_{\mathbb{R}^{n}})d\xi^{\otimes
k}(E_{1},\dots,E_{k}).$

So, using the Donsker–Varadhan variational formula for entropy in combination
with the identity (due to (16))

$\int_{\mathbf{E}^{k}}\|P_{E_{1}}P_{E_{2}}\dots
P_{E_{k}}\theta\|^{2}d\xi^{\otimes
k}(E_{1},E_{2},\dots,E_{k})=(1-\lambda)^{k}\|\theta\|^{2},~{}~{}k\geq 1,$

we have for any $\beta>1$, the crude lower bound

$\displaystyle D(\nu_{t}\|\gamma)$
$\displaystyle\geq\int_{\mathbb{R}^{n}}\frac{1}{2\beta}|x|^{2}d\nu_{t}(x)-\log\left(\int_{\mathbb{R}^{n}}e^{\frac{|x|^{2}}{2\beta}}d\gamma(x)\right)$
$\displaystyle=\frac{n}{2\beta}-\frac{n}{2}\log\left(\frac{\beta}{\beta-1}\right)+\frac{1}{\beta}e^{-\lambda
t}\underbrace{D(\nu_{0}\|\gamma)}_{\frac{1}{2}\|\theta\|^{2}}.$

Hence, for any choice of $t_{0}\geq 0$ and $\epsilon>0$, we can take $\beta$
sufficiently close to 1 and $\theta$ with sufficiently large norm such that

$D(\nu_{t_{0}}\|\gamma)\geq(1-\epsilon)e^{-\lambda t_{0}}D(\nu_{0}\|\gamma).$

By an application of (11) and the semigroup property, this means we must have

$D(\nu_{t}\|\gamma)\geq(1-\epsilon)e^{-\lambda
t}D(\nu_{0}\|\gamma),~{}~{}~{}\forall t\in[0,t_{0}].$

It follows that the rate $\lambda$ in (11) cannot be improved when
$\mu=\gamma$. Using the transport map construction in the proof of Theorem 3,
this example with reference measure equal to $\gamma$ can be transported to
any reference measure $\mu$ that splits with respect to $\xi$. Therefore, the
rate $\lambda$ cannot be improved in general.

To see that the rate in (12) is sharp, assume $\mu$ splits with respect to
$\xi$, and assume for simplicity that it has finite second moments. Put
$f(x)=u\cdot x$, where $u$ is an eigenvector of $\int P_{E}d\xi(E)$ with
eigenvalue $1-\lambda$. That is,

$\displaystyle\left(\int_{\mathbf{E}}P_{E}d\xi(E)\right)u=(1-\lambda)u.$ (17)

For $X\sim\mu$, we have

$\displaystyle\int_{\mathbf{E}}\operatorname{Var}(\mathbb{E}[f(X)|P_{E}X])d\xi(E)$
$\displaystyle=\int_{\mathbf{E}}\mathbb{E}\|u\cdot P_{E}X\|^{2}d\xi(E)$
$\displaystyle=\int_{\mathbf{E}}u^{T}P_{E}\operatorname{Cov}(\mu)P_{E}u\,d\xi(E)$
$\displaystyle=u^{T}\operatorname{Cov}(\mu)\left(\int_{\mathbf{E}}P_{E}d\xi(E)\right)u$
$\displaystyle=(1-\lambda)u^{T}\operatorname{Cov}(\mu)u$
$\displaystyle=(1-\lambda)\operatorname{Var}(f(X)),$

where we used that $P_{E}$ and $\operatorname{Cov}(\mu)$ commute for
$\xi$-a.e. $E\in\mathbf{E}$ (since $\mu$ splits with respect to $\xi$), and
(17). Hence, the constant $(1-\lambda)$ is best-possible in the inequality
(14), and this is equivalent to (12).

#### Remarks on related literature

The physical model proposed in this Section can be regarded as spatially
homogeneous linear Boltzmann dynamics, where the collision kernel is replaced
by the random choice of subspace dictated by the distribution $\xi$. In this
respect, the interactions considered here are more similar in spirit to Kac’s
simplified model [26] than the full theory of the linear Boltzmann equation.
The Brascamp–Lieb inequalities on the sphere have been interpreted as
governing convergence to equilibrium in Kac’s model [15], but we are unaware
of any previous kinetic interpretation of the Euclidean Brascamp–Lieb
inequalities. In a related spirit, there is a line of recent work by P. Caputo
and collaborators that show rapid mixing of certain Markov processes is
closely connected to validity of approximate Shearer-type inequalities for
entropy (see, e.g., [11, 5, 9, 12, 13]). It is an interesting question as to
whether these results fit within the context of (yet-to-be-explored)
Brascamp–Lieb-type inequalities on the discrete spaces they consider.

###  2.4 Inequalities in probability

It’s well-known that the Brascamp–Lieb inequalities (5) imply many classical
analytic and geometric inequalities (e.g., the Hölder, sharp Young, and
Loomis–Whitney inequalities). None of these applications require
Valdimarsson’s characterization of extremizers. However, by incorporating the
latter circle of ideas into the picture through their manifestation in Theorem
1, we eventually arrived at the probabilistic inequalities (14) and (15),
which we summarize below. In this Section, we’ll explain how they generalize a
variety of familiar inequalities in probability.

######  Theorem 5.

Let $\lambda\geq 0$ be such that (6) holds, and let $\mu$ split with respect
to $\xi$. If $X\sim\mu$, then for all $f\in L^{2}(\mu)$

$\displaystyle\int_{\mathbf{E}}\operatorname{Var}(\mathbb{E}[f(X)|P_{E}X])d\xi(E)\leq(1-\lambda)\operatorname{Var}(f(X))$
(18)

and, equivalently,

$\displaystyle\operatorname{Var}(f(X))$
$\displaystyle\leq\frac{1}{\lambda}\int_{\mathbf{E}}\mathbb{E}[\operatorname{Var}(f(X)|P_{E}X)]d\xi(E).$
(19)

Since (18) and (19) follow from linearizing the Brascamp–Lieb-type
inequalities (7), we find it appropriate to refer to them as _linearized
Brascamp–Lieb inequalities_. They were described by the author in the extended
abstract [20], along with the following few examples which we repeat here (see
also [19], in the context of linearizing Shearer’s inequality).

Since the standard Gaussian measure $\gamma$ splits with respect to any
$\xi\in P(\mathbf{E})$, we immediately obtain a family of variance
inequalities for $\gamma$.

######  Example 1.

If $X\sim\gamma$, then (18) and (19) hold for every $\xi\in P(\mathbf{E})$ and
$\lambda\geq 0$ satisfying (6).

In general, if one starts with a random vector $X$ with given independence
structure, it may be possible to construct a measure $\xi\in P(\mathbf{E})$
such that the hypotheses of Theorem 5 are satisfied. This approach handily
recovers an assortment of classical $L^{2}$ inequalities, as demonstrated in
the following few examples.

######  Example 2.

(Efron–Stein inequality [22, 31]) Let $X=(X_{i})_{i=1}^{k}$ be a random vector
with independent components $(X_{i})_{i=1}^{k}$, and define

$X^{(i)}=(X_{1},\dots,X_{i-1},X_{i+1},\dots X_{k}).$

For any measurable $f$ with $\operatorname{Var}(f(X))<\infty$,

$\displaystyle\operatorname{Var}(f(X))\leq\sum_{i=1}^{k}\mathbb{E}[\operatorname{Var}(f(X)|X^{(i)})].$
(20)

###### Proof.

We can assume $X$ takes values in $\mathbb{R}^{n}$, and choose $E_{i}$ such
that $X_{i}$ is the component of $X$ in $E_{i}^{\perp}$. This implies the
orthogonal decomposition $\mathbb{R}^{n}={\oplus}_{1\leq i\leq
k}E^{\perp}_{i}$, which yields

$\frac{1}{k-1}\sum_{i=1}^{k}P_{E_{i}}=\operatorname{id}_{\mathbb{R}^{n}}.$

By the independence hypothesis, the law of $X$ splits along
$(E_{i},E_{i}^{\perp})$ for each $i=1,\dots,k$, and therefore (20) follows
from (19) by taking $\xi$ equal to the uniform distribution on
$\\{E_{1},\dots,E_{k}\\}\subset\mathbf{E}$. ∎

We remark that the Efron–Stein inequality is often interpreted as a spectral
gap inequality for the Gibbs sampling procedure where coordinates of a vector
$X$ are resampled, uniformly at random. From this perspective, (19) may be
thought of as the same, except where the component of $X$ lying in subspace
$E^{\perp}$ is resampled according to the distribution $E\sim\xi$.

######  Example 3.

(Dembo–Kagan–Shepp inequality [21]) Let $(X_{i})_{i\geq 1}$ be a sequence of
i.i.d. random vectors, and define the cumulative sum
$S_{n}=\sum_{j=1}^{n}X_{j}$. For any measurable $g$,

$\displaystyle\operatorname{Var}(\mathbb{E}[g(S_{n})|S_{m}])\leq\frac{m}{n}\operatorname{Var}(g(S_{n})),~{}~{}~{}~{}n\geq
m\geq 1.$ (21)

###### Proof.

For simplicity of notation, we’ll assume each $X_{i}$ is one-dimensional.
Consider the random vector $X=(X_{1},\dots,X_{n})$ taking values in
$\mathbb{R}^{n}$, with $X_{j}$ the projection of $X$ along natural basis
vector $e_{j}$, $j=1,\dots,n$. Take $(E_{i})_{i=1}^{k}$ be an enumeration of
all $k={n\choose m}$ subspaces of $\mathbb{R}^{n}$, each equal to the linear
span of exactly $m$ natural basis vectors. By construction, the law of $X$
splits along $(E_{i},E_{i}^{\perp})$ for each $i=1,\dots,k$, and

$\sum_{i=1}^{k}P_{E_{i}}\xi(\\{E_{i}\\})=\frac{m}{n}\operatorname{id}_{\mathbb{R}^{n}}.$

for $\xi$ the uniform measure on the discrete set $\\{E_{1},\dots,E_{k}\\}$.
By symmetry, $\mathbb{E}[g(S_{n})|P_{E_{i}}X]$ are equal in law for each
$i=1,\dots,k$. So, an application of (18) with $f(X)=g(S_{n})$ gives

$\displaystyle\operatorname{Var}(\mathbb{E}[g(S_{n})|X_{1},\dots,X_{m}])\leq\frac{m}{n}\operatorname{Var}(g(S_{n})),~{}~{}~{}~{}n\geq
m\geq 1.$

The claim follows since $S_{m}$ is a sufficient statistic of
$(X_{1},\dots,X_{m})$ for $S_{n}$. ∎

######  Remark 6.

In the special case where $(X_{i})_{i\geq 1}$ are i.i.d. $\gamma$, inequality
(21) yields the optimal estimate for contraction of the Ornstein–Uhlenbeck
semigroup in $L^{2}(\gamma)$. This is equivalent to the Gaussian Poincaré
inequality.

A straightforward Corollary of Theorem Theorem 5 is the following improvement
of Jensen’s inequality by a factor of $(1-\lambda)$, under the structural
assumption that $\mu$ splits with respect to $\xi$.

######  Corollary 1.

Let $\mu$ split with respect to $\xi$, and $\lambda\geq 0$ satisfy (6). If
$(\psi_{E})_{E\in\mathbf{E}}$ is a collection of functions from
$\mathbb{R}^{n}$ to $\mathbb{R}$ such that

$(E,x)\in\mathbf{E}\times\mathbb{R}^{n}\mapsto(\psi_{E}\circ
P_{E})(x)\in\mathbb{R}$

is measurable, then

$\displaystyle\operatorname{Var}_{\mu}\left(\int_{\mathbf{E}}(\psi_{E}\circ
P_{E})d\xi(E)\right)\leq(1-\lambda)\int_{\mathbf{E}}\operatorname{Var}_{\mu}\left(\psi_{E}\circ
P_{E}\right)d\xi(E).$ (22)

###### Proof.

Put $f(x):=\int_{\mathbf{E}}(\psi_{E}\circ P_{E})(x)d\xi(E)$, and let
$X\sim\mu$. Applying Cauchy-Schwarz twice followed by (18), we have

$\displaystyle\operatorname{Var}(f(X))$
$\displaystyle=\int_{\mathbf{E}}\operatorname{Cov}(f(X),(\psi_{E}\circ
P_{E})(X))d\xi(E)$
$\displaystyle=\int_{\mathbf{E}}\operatorname{Cov}(\mathbb{E}[f(X)|P_{E}X],(\psi_{E}\circ
P_{E})(X))d\xi(E)$
$\displaystyle\leq\int_{\mathbf{E}}\operatorname{Var}\left(\mathbb{E}[f(X)|P_{E}X]\right)^{1/2}\operatorname{Var}\left((\psi_{E}\circ
P_{E})(X)\right)^{1/2}d\xi(E)$
$\displaystyle\leq\left(\int_{\mathbf{E}}\operatorname{Var}\left(\mathbb{E}[f(X)|P_{E}X]\right)d\xi(E)\right)^{1/2}\left(\int_{\mathbf{E}}\operatorname{Var}\left((\psi_{E}\circ
P_{E})(X)\right)d\xi(E)\right)^{1/2}$
$\displaystyle\leq(1-\lambda)^{1/2}\operatorname{Var}\left(f(X)\right)^{1/2}\left(\int_{\mathbf{E}}\operatorname{Var}\left((\psi_{E}\circ
P_{E})(X)\right)d\xi(E)\right)^{1/2}.$

∎

As an example, we recover an inequality due to Madiman and Barron [29], which
is itself a generalization of a classical result on $U$-statistics due to
Hoeffding [25]. To state it, recall that $\mathcal{T}\subset 2^{[n]}$ is
called an $r$-cover of $[n]:=\\{1,\dots,n\\}$ if each element of $[n]$ is
contained in at most $r$ members of $\mathcal{T}$.

######  Example 4 (Madiman–Barron inequality).

Let $X=(X_{m})_{m=1}^{n}$ be a collection of $n$ independent random random
variables, let $(S_{i})_{i=1}^{k}\subset 2^{[n]}$ be an $r$-cover of $[n]$,
and define $X_{S_{i}}:=(X_{m})_{m\in S_{i}}$. For any real-valued measurable
$\psi_{i}:X_{S_{i}}\mapsto\psi_{i}(X_{S_{i}})$,

$\displaystyle\operatorname{Var}\left(\sum_{i=1}^{k}\psi_{i}(X_{S_{i}})\right)\leq
r\sum_{i=1}^{k}\operatorname{Var}\left(\psi_{i}(X_{S_{i}})\right).$ (23)

###### Proof.

Let $E_{i}=\operatorname{span}\\{(e_{j})_{j\in S_{i}}\\}$, where $e_{j}$ is
the $j$th natural basis vector of $\mathbb{R}^{n}$. The random vector
$X=(X_{1},\dots,X_{n})$ takes values in $\mathbb{R}^{n}$, and has law which
splits along $(E_{i},E_{i}^{\perp})$ for all $i=1,\dots,k$. Letting $\xi$ be
the uniform measure on $\\{E_{1},\dots,E_{k}\\}$, we have

$\sum_{i=1}^{k}P_{E_{i}}\xi(\\{E_{i}\\})\leq\frac{r}{k}\operatorname{id}_{\mathbb{R}^{n}}.$

Hence, the claim follows from (22). ∎

######  Remark 7.

A weighted version of (23) also appears in [29], which can be obtained by
letting $\xi$ be an arbitrary (non-uniform) distribution on
$\\{E_{1},\dots,E_{k}\\}$.

The examples presented in this section have many interesting consequences of
their own. For example, applications of the Efron–Stein inequality are
ubiquitous in statistics [6], and the Dembo–Kagan–Shepp inequality and the
Madiman–Barron inequalities give simple proofs [18, 29] of the monotonicity of
entropy along the central limit theorem originally due to Arstein, Ball,
Barthe, and Naor [1], and its generalizations. Also, (18) in the special case
where $E_{dep}=\\{0\\}$ was used by the author to establish generalized
subadditivity estimates for Poincaré constants of convolution measures [19].

##  3 Proof of Theorem 1

The proof of Theorem 1 proceeds in two steps. First, we treat the case where
$\xi$ is a discrete measure supported on a finite set of subspaces. This
avoids some of the subtleties that need to be resolved in the second step,
where we only assume that $\mu$ splits with respect to a measure $\xi\in
P(\mathbf{E})$.

To get started, we prove Proposition 1 to establish that the independent
decomposition of $\mathbb{R}^{n}$ with respect to $\xi\in P(\mathbf{E})$ is
well defined.

###### Proof of Proposition 1.

Note that $0\in\chi_{\xi}^{-1}(0)$. If $\chi_{\xi}^{-1}(0)=\\{0\\}$, then the
claim is trivial. Hence, we can assume without loss of generality that
$\\{0\\}\subsetneq\chi_{\xi}^{-1}(0)$. It will suffice to show that
$\chi_{\xi}^{-1}(0)$ is the union of mutually orthogonal subspaces; uniqueness
follows immediately. Toward this end, define a binary relation $\sim$ on
$V:=\chi_{\xi}^{-1}(0)\setminus\\{0\\}$ via:

$x\sim y~{}~{}~{}\Leftrightarrow~{}~{}~{}\chi(x-y)=0,~{}~{}x,y\in V.$

The relation $\sim$ is obviously reflexive and symmetric. Now, note that
$x\sim y$ implies

$x,y,(x-y)\in(E\cup E^{\perp}),~{}~{}~{}\xi-a.s.$

Since $x,y\in V$ themselves, this is only possible if the vectors $x,y$ are
both in $E$, or are both in $E^{\perp}$, $\xi$-a.s. In other words, for
$x,y\in V$,

$x\sim y~{}~{}\Leftrightarrow~{}~{}x,y\in E~{}\mbox{or}~{}x,y\in
E^{\perp},~{}\xi-a.s.$

Since $V$ excludes the zero-vector, $\sim$ is transitive, and is therefore an
equivalence relation on $V$. By similar logic, if $x,y\in V$ and $x\not\sim
y$, then $x,y$ are orthogonal in $\mathbb{R}^{n}$. As a result, $\sim$
partitions $V$ into at most $n$ equivalence classes, which we enumerate as
$V=\cup_{\alpha}V_{\alpha}$, and $V_{\alpha}\perp V_{\alpha^{\prime}}$ for
distinct classes. To conclude, we define $E_{\alpha}=\\{0\\}\cup V_{\alpha}$,
which is a linear subspace of $\mathbb{R}^{n}$ by definitions. ∎

###  3.1 Special case of Theorem 1 when $\xi$ is a discrete measure

Assume that $\xi$ is a discrete measure, supported on $k$ points
$\\{\tilde{E}_{1},\dots,\tilde{E}_{k}\\}\subset\mathbf{E}$. With this
assumption in place, the assertion that $\mu$ splits with respect to $\xi$
simplifies to saying that $\mu$ splits along
$(\tilde{E}_{i},\tilde{E}_{i}^{\perp})$ for each $i=1,\dots,k$. Moreover, in
reference to the independent decomposition of $\mathbb{R}^{n}$ with respect to
$\xi$, the independent subspaces $(E_{\alpha})_{\alpha}$ are explicitly
identified via

$\cup_{\alpha}E_{\alpha}=\cap_{i=1}^{k}(\tilde{E}_{i}\cup\tilde{E}_{i}^{\perp}).$

That is, each independent subspace is of the form
$E_{\alpha}=\cap_{i=1}^{k}\tilde{E}_{i}^{\alpha_{i}}$, where
$\alpha_{i}\in\\{o,\perp\\}$ and $E^{o}:=E$. In this context, a preliminary
version of Theorem 1 is the following statement. The goal of this subsection
is to prove it.

######  Theorem 6.

If $\mu\in P(E)$ splits along $(\tilde{E}_{i},\tilde{E}_{i}^{\perp})$ for each
$i=1,\dots,k$, then $\mu$ takes the product form

$\displaystyle\mu=\left(\otimes_{\alpha}\mu_{E_{\alpha}}\right)\otimes\mu_{E_{dep}},$
(24)

with $\mu_{E_{dep}}\in P(E_{dep})$ a Gaussian measure.

Theorem 6 is a probabilistic manifestation of Valdimarsson’s core argument in
[32]. In our proof, we’ll point out the key steps that follow Valdimarsson’s
analysis.

We first establish the case where there are no nontrivial independent
subspaces, which has been stated already as Proposition 2. We will require the
following two lemmas.

######  Lemma 1.

Under the assumptions of Proposition 2,
$\int_{\mathbb{R}^{n}}\|x\|d\mu(x)<\infty$.

###### Proof.

First, we claim that there is some $\theta>0$ such that

$\max_{1\leq i\leq
k}\min\\{\|P_{\tilde{E}_{i}}x\|,\|P_{\tilde{E}_{i}^{\perp}}x\|\\}\geq\theta\|x\|,~{}~{}\forall
x\in\mathbb{R}^{n}.$

Suppose not. By homogeneity, there exists a sequence $(u_{n})_{n\geq
1}\subset\mathbb{R}^{n}$ with norm $\|u_{n}\|=1$ such that

$\min\\{\|P_{\tilde{E}_{i}}u_{n}\|,\|P_{\tilde{E}_{i}^{\perp}}u_{n}\|\\}\to
0,~{}~{}~{}\forall i=1,\dots,k.$

By continuity and compactness, there is a unit vector $u\in\mathbb{R}^{n}$
such that

$\min\\{\|P_{\tilde{E}_{i}}u\|,\|P_{\tilde{E}_{i}^{\perp}}u\|\\}=0.$

Hence $0\neq u\in\cap_{i=1}^{k}(\tilde{E}_{i}\cup\tilde{E}_{i}^{\perp})$,
which is the desired contradiction.

Therefore, for $X\sim\mu$ and all $t\geq 0$,

$\displaystyle\Pr\\{\|X\|\geq t\\}$
$\displaystyle\leq\Pr\left(\cup_{i=1}^{k}\\{\min\\{\|P_{\tilde{E}_{i}}X\|,\|P_{\tilde{E}_{i}^{\perp}}X\|\\}\geq\theta
t\\}\right)$
$\displaystyle\leq\sum_{i=1}^{k}\Pr\left(\\{\|P_{\tilde{E}_{i}}X\|\geq\theta
t\\}\cap\\{\|P_{\tilde{E}_{i}^{\perp}}X\|\geq\theta t\\}\right)$
$\displaystyle=\sum_{i=1}^{k}\Pr\\{\|P_{\tilde{E}_{i}}X\|\geq\theta
t\\}\Pr\\{\|P_{\tilde{E}_{i}^{\perp}}X\|\geq\theta t\\}$ $\displaystyle\leq
k\left(\Pr\\{\|X\|\geq\theta t\\}\right)^{2}.$

The key step is the third line, where we used the assumption that $\mu$ splits
along $(\tilde{E}_{i},\tilde{E}_{i}^{\perp})$ for each $i$. By Bryc, Theorem
1.3.5, this self-majorization property of the complementary distribution
function ensures that there are finite $M,\alpha,\beta>0$ (depending only on
$k,\theta$) such that

$\Pr\\{\|X\|\geq t\\}\leq M\exp(-\beta t^{\alpha}),~{}~{}~{}t\geq 0.$

This guarantees $\mathbb{E}[\|X\|]<\infty$. ∎

######  Lemma 2 ([30, Proposition 2]).

Let $X$ be a random vector on $\mathbb{R}^{n}$ with $\mathbb{E}\|X\|<\infty$,
independent from $Z\sim N(0,\operatorname{id}_{\mathbb{R}^{n}})$. The random
vector $X+Z$ has a non-vanishing density $f$ which has derivatives of all
orders, and satisfes

$\|\nabla\log f(x)\|\leq C(1+\|x\|),~{}~{}~{}\forall x\in\mathbb{R}^{n}$

for some finite constant $C$ depending only on $\mathbb{E}\|X\|$.

###### Proof of Proposition 2.

Since the splitting property is stable under convolutions with isotropic
Gaussians, we can assume $\mu$ is of the form $\mu_{0}*\gamma$ for some
probability measure $\mu_{0}\in P(\mathbb{R}^{n})$. By considering
characteristic functions, showing $\mu$ is Gaussian establishes the same for
$\mu_{0}$.

Let $f$ denote the density of $\mu$. By Lemma 2 and our regularizing
assumption, $f$ is non-vanishing and has derivatives of all orders. In
combination with Lemma 1, we further have the linear growth estimate

$\|\nabla\log f(x)\|\leq C(1+\|x\|),~{}~{}~{}x\in\mathbb{R}^{n}$

for some $C$, depending only on the moments of $\mu$. Now, just as in [32],
this justifies taking the Fourier transform of $s:x\mapsto\nabla\log f(x)$ as
a tempered distribution; we denote this Fourier transform by $\hat{s}$. By the
splitting property, $\hat{s}$ is supported on $(H_{i}+H_{i}^{\perp})$ for each
$i$, where $H_{i}$ is the complexification of $\tilde{E}_{i}$. Taking
intersections over $i=1,\dots,k$, we find that $\hat{s}$ is supported at the
origin, and therefore $s=\nabla\log f$ is polynomial. By the linear growth
condition, we conclude $\nabla\log f$ must be affine, and therefore $\log f$
is quadratic. Hence, $f$ is a Gaussian density, as desired. ∎

We’ll need the following lemma, which is a restatement of the classical fact
that a bivariate function with vanishing mixed partial derivatives is
separable in its variables. It is taken for granted in [32, p. 268], but
finding no suitable reference, we opt to give its short proof here.

######  Lemma 3.

Let $V$ be a linear subspace of $\mathbb{R}^{n}$, and let
$\phi:\mathbb{R}^{n}\to\mathbb{R}$ be twice-differentiable. The following are
equivalent:

  1. 1)

It holds that $\nabla^{2}\phi(x)V\subset V$ for every $x\in\mathbb{R}^{n}$.

  2. 2)

It holds that

$\phi(x)+\phi(0)=\phi(P_{V}x)+\phi(P_{V^{\perp}}x),~{}~{}\forall
x\in\mathbb{R}^{n}.$

###### Proof.

The direction $2)\Rightarrow 1)$ is trivial, so we only need to establish the
reverse implication. To that end, decompose $x=x_{\parallel}+x_{\perp}$, where
$x_{\parallel}=P_{V}x$ and $x_{\perp}=P_{V^{\perp}}x$. Let $v\in V$. For
convenience, we’ll write $\langle\cdot,\cdot\rangle$ to denote the usual
scalar product on $\mathbb{R}^{n}$. For some $t\in(0,1)$ we claim

$\displaystyle\langle v,\nabla\phi(x)\rangle$ $\displaystyle\equiv\langle
v,\nabla\phi(x_{\parallel},x_{\perp})\rangle$ $\displaystyle=\langle
v,\nabla\phi(x_{\parallel},0)\rangle+\langle x_{\perp},\nabla(\langle
v,\nabla\phi(x_{\parallel},tx_{\perp})\rangle)\rangle$ $\displaystyle=\langle
v,\nabla\phi(x_{\parallel},0)\rangle+\langle
x_{\perp},\nabla^{2}\phi(x_{\parallel},tx_{\perp})v\rangle$
$\displaystyle=\langle v,\nabla\phi(x_{\parallel},0)\rangle$
$\displaystyle=\langle v,P_{V}\nabla\phi(P_{V}x)\rangle.$

Indeed, the second line holds for some $t\in(0,1)$ by the mean value theorem,
the third follows by definition of the Hessian, the fourth follows by
orthogonality because $V$ is an invariant subspace of $\nabla^{2}\phi$, and
the last because $\phi(x_{\parallel},0)\equiv\phi(P_{V}x)$, $v\in V$ and
$P_{V}$ is self-adjoint. Since $\nabla^{2}\phi(x)$ is a symmetric matrix with
invariant subspace $V$, it follows that $V^{\perp}$ is also an invariant
subspace of $\nabla^{2}\phi(x)$. Hence, by a symmetric argument, we also have

$\langle v^{\prime},\nabla\phi(x)\rangle=\langle
v^{\prime},P_{V^{\perp}}\nabla\phi(P_{V^{\perp}}x)\rangle,~{}~{}\forall
v^{\prime}\in V^{\perp}.$

Putting things together, this implies for all $v\in V,v^{\prime}\in V^{\perp}$
that

$\displaystyle\langle v+v^{\prime},\nabla\phi(x)\rangle$
$\displaystyle=\langle
v+v^{\prime},P_{V}\nabla\phi(P_{V}x)+P_{V^{\perp}}\nabla\phi(P_{V^{\perp}}x)\rangle$
$\displaystyle=\langle v+v^{\prime},\nabla(\phi\circ
P_{V})(x)+\nabla(\phi\circ P_{V^{\perp}})(x)\rangle,$

where the second line followed from the chain rule and self-adjointness of the
individual projections. Since $v+v^{\prime}$ can be taken to be an arbitrary
vector in $E$, we must have $\phi=\phi\circ P_{V}+\phi\circ P_{V^{\perp}}+C$
for some constant $C$. Evaluating at $x=0$ reveals $C=-\phi(0)$. ∎

###### Proof of Theorem 6.

Since the splitting property is stable under convolutions with isotropic
Gaussians, we can assume $\mu$ is of the form $\mu_{0}*\gamma$ for some
probability measure $\mu_{0}\in P(\mathbb{R}^{n})$. Showing $\mu$ is of the
required form shows the same for the initial data $\mu_{0}$.

Let $f$ denote the density of $\mu$. By our regularizing assumption, $f$ is
non-vanishing and has derivatives of all orders. We now follow [32, Proof of
Lemma 13]. Since $\log f$ is separable in the variables $P_{\tilde{E}_{i}}x$
and $P_{\tilde{E}^{\perp}_{i}}x$ for each $i$ by the splitting property, it
follows from Lemma 3 that

$\nabla^{2}\log
f(x)\tilde{E}^{\alpha_{i}}_{i}\subset\tilde{E}^{\alpha_{i}}_{i},~{}~{}~{}\forall
x\in\mathbb{R}^{n},\alpha_{i}\in\\{o,\perp\\},i=1,\dots,k.$

Taking intersections, every independent subspace $E_{\alpha}$ satisfies

$\nabla^{2}\log f(x)E_{\alpha}\subset E_{\alpha},~{}~{}~{}\forall
x\in\mathbb{R}^{n},$

and therefore

$\nabla^{2}\log f(x)E_{ind}\subset E_{ind},~{}~{}~{}\forall
x\in\mathbb{R}^{n}.$

By Lemma 3, this implies that $\log f$ is of the form

$\log f(x)+\log f(0)=\log f(P_{E_{ind}}x)+\log
f(P_{E_{dep}}x),~{}~{}~{}\forall x\in\mathbb{R}^{n}.$

In other words, $\mu$ splits along $(E_{ind},E_{dep})$, and we can write
$\mu=\mu_{E_{ind}}\otimes\mu_{E_{dep}}$.

A straightforward induction shows that $\mu_{E_{ind}}$ has product form
$\otimes_{\alpha}\mu_{E_{\alpha}}$. It remains to show $\mu_{E_{dep}}$ is
Gaussian. Toward this end, since $E_{\alpha}\subset\tilde{E}^{\alpha_{i}}_{i}$
for each $i$ and suitable choice of ${\alpha_{i}}\in\\{o,\perp\\}$, it follows
that $E_{dep}$ admits the orthogonal decomposition

$E_{dep}=(E_{dep}\cap\tilde{E}_{i})\oplus(E_{dep}\cap\tilde{E}^{\perp}_{i}),~{}~{}\mbox{for
each $i=1,\dots,k$.}$

Hence, by the splitting property of $\mu$, we have that $\mu_{E_{dep}}$ splits
along $(E_{dep}\cap\tilde{E}_{i},E_{dep}\cap\tilde{E}^{\perp}_{i})$ for each
$i=1,\dots,k$, and

$\cap_{i=1}^{k}((E_{dep}\cap\tilde{E}_{i})\cup(E_{dep}\cap\tilde{E}^{\perp}_{i}))=E_{dep}\cap\left(\cap_{i=1}^{k}(\tilde{E}_{i}\cup\tilde{E}^{\perp}_{i})\right)=E_{dep}\cap
E_{ind}=\\{0\\}.$

An application of Proposition 2 completes the proof. ∎

###  3.2 Proof of Theorem 1 in the general case

###### Proof of Theorem 1.

2)$\Rightarrow$1): If $E_{dep}=\mathbb{R}^{n}$, then $\mu$ is a Gaussian
measure with covariance that splits with respect to $\xi$, and therefore $\mu$
splits along $(E,E^{\perp})$, $\xi$-a.s. (see the remarks following Definition
3). So, assume $E_{dep}\subsetneq\mathbb{R}^{n}$, and let
$B_{\alpha}=\\{u_{1},\dots,u_{\dim(E_{\alpha})}\\}$ be a basis for a nonzero
independent subspace $E_{\alpha}$. By definition of $\chi_{\xi}$, for each
$i=1,\dots,\dim(E_{\alpha})$, there exists a set
$\mathbf{E}_{i,\alpha}\subset\mathbf{E}$ of full $\xi$-measure such that

$u_{i}\in\cap_{E\in\mathbf{E}_{i,\alpha}}(E\cup E^{\perp}).$

Since $\dim(E_{\alpha})$ is finite, we have that
$\mathbf{E}_{\alpha}:=\cap_{i=1}^{\dim(E_{\alpha})}\mathbf{E}_{i,\alpha}$ is
another set of full $\xi$-measure, and
$B_{\alpha}\subset\cap_{E\in\mathbf{E}_{\alpha}}(E\cup E^{\perp})$. This
implies

$E_{\alpha}\subset\cap_{E\in\mathbf{E}_{\alpha}}(E\cup E^{\perp}).$

Since there are at most $n$ distinct nonzero independent subspaces, the set
$\mathbf{E}_{ind}:=\cap_{\alpha}\mathbf{E}_{\alpha}$ also has
$\xi(\mathbf{E}_{ind})=1$, and has the property that exactly one of the
following is true:

$\displaystyle
P_{E}P_{E_{\alpha}}=P_{E_{\alpha}}~{}(\mbox{and~{}}P_{E_{\perp}}P_{E_{\alpha}}=0),~{}\mbox{or}~{}P_{E}P_{E_{\alpha}}=0~{}(\mbox{and~{}}P_{E_{\perp}}P_{E_{\alpha}}=P_{E_{\alpha}}).$
(25)

Finally, since $\mu_{E_{dep}}$ is Gaussian with covariance that splits with
respect to $\xi$, there is a set $\mathbf{E}_{dep}\subset\mathbf{E}$ such that
$\xi(\mathbf{E}_{dep})=1$ and $\mu_{E_{dep}}\otimes\delta_{0}$ splits along
$(E,E^{\perp})$ for all $E\in\mathbf{E}_{dep}$, where $\delta_{0}$ is the
centered dirac mass on $E_{ind}$. This follows by Definition 3 and the
following remarks.

Therefore, collecting definitions, if $\mu$ admits the product form (2), then
it splits along $(E,E^{\perp})$ for every
$E\in\mathbf{E}_{ind}\cap\mathbf{E}_{dep}$. Indeed, this is most easily seen
by letting $X\sim\mu$ and writing

$P_{E}X=P_{E}P_{E_{dep}}X+\sum_{\alpha}P_{E}P_{E_{\alpha}}X,\mbox{~{}and~{}}P_{E^{\perp}}X=P_{E^{\perp}}P_{E_{dep}}X+\sum_{\alpha}P_{E^{\perp}}P_{E_{\alpha}}X.$

Assuming $E\in\mathbf{E}_{ind}\cap\mathbf{E}_{dep}$, the Gaussian factors
$P_{E}P_{E_{dep}}X$ and $P_{E^{\perp}}P_{E_{dep}}X$ are independent as just
discussed. Moreover, the factors $P_{E}P_{E_{\alpha}}X$ and
$P_{E^{\perp}}P_{E_{\alpha}}X$ are independent by (25) for each $\alpha$. Due
to the assumed product form (2), we conclude that $\mu$ splits along
$(E,E^{\perp})$ for every $E\in\mathbf{E}_{ind}\cap\mathbf{E}_{dep}$. Since
$\xi(\mathbf{E}_{ind}\cap\mathbf{E}_{dep})=1$, the claim follows.

1)$\Rightarrow$2): We can proceed in the same manner as above to identify a
set $\mathbf{E}_{ind}$ of full $\xi$-measure such that

$\cup_{\alpha}E_{\alpha}\subset\cap_{E\in\mathbf{E}_{ind}}(E\cup E^{\perp}).$

Now, by hypothesis, there is a set $\mathbf{E}_{split}$ of full $\xi$-measure
such that $\mu$ splits along $(E,E^{\perp})$ for every
$E\in\mathbf{E}_{split}$. Since $\mathbf{E}_{ind}\cap\mathbf{E}_{split}$ has
full measure, we have

$\chi_{\xi}(x)=\int_{\mathbf{E}_{ind}\cap\mathbf{E}_{split}}\min\\{\|P_{E}x\|,\|P_{E^{\perp}}x\|\\}d\xi(E),~{}~{}x\in\mathbb{R}^{n},$

so that we conclude

$\cup_{\alpha}E_{\alpha}\subset\cap_{E\in\mathbf{E}_{ind}}(E\cup
E^{\perp})\subset\cap_{E\in\mathbf{E}_{ind}\cap\mathbf{E}_{split}}(E\cup
E^{\perp})\subset\chi_{\xi}^{-1}(0).$

By virtue of Proposition 1, we conclude

$\displaystyle\cup_{\alpha}E_{\alpha}=\cap_{E\in\mathbf{E}_{ind}\cap\mathbf{E}_{split}}(E\cup
E^{\perp}).$ (26)

Since everything is finite dimensional, there is a finite collection of
subspaces
$(\tilde{E}_{i})_{i=1}^{k}\subset\mathbf{E}_{ind}\cap\mathbf{E}_{split}$ such
that

$\displaystyle\cup_{\alpha}E_{\alpha}=\cap_{i=1}^{k}(\tilde{E}_{i}\cup\tilde{E}_{i}^{\perp})$
(27)

Indeed, we can construct $(\tilde{E}_{i})_{i=1}^{k}$ by a greedy algorithm,
where $\tilde{E}_{1}$ is selected arbitrarily from
$\mathbf{E}_{ind}\cap\mathbf{E}_{split}$, and $E_{m+1}$ for $m\geq 1$ is
selected from $\mathbf{E}_{ind}\cap\mathbf{E}_{split}$ so that

$\dim(\operatorname{sp}(\cap_{i=1}^{m+1}(\tilde{E}_{i}\cup\tilde{E}_{i}^{\perp})))<\dim(\operatorname{sp}(\cap_{i=1}^{m}(\tilde{E}_{i}\cup\tilde{E}_{i}^{\perp}))),$

unless if this is impossible, in which case we terminate the algorithm and set
$k=m$. Until termination, this procedure reduces dimension by a positive
integer number at each step, and we always have

$\cup_{\alpha}E_{\alpha}\subset\cap_{i=1}^{m}(\tilde{E}_{i}\cup\tilde{E}_{i}^{\perp}),~{}~{}\mbox{for
each~{}$m\geq 1$}.$

So, in view of (26), the procedure must terminate at some point with a
collection $(\tilde{E}_{i})_{i=1}^{k}$ satisfying (27).

Since $(\tilde{E}_{i})_{i=1}^{k}\subset\mathbf{E}_{split}$ by construction,
$\mu$ splits along $(\tilde{E}_{i},\tilde{E}_{i}^{\perp})$ for each
$i=1,\dots,k$. So, an application of Theorem 6 shows that $\mu$ admits the
product form (2), with $\mu_{E_{dep}}$ necessarily being a Gaussian measure.
Since $\mu$ splits along $(E,E^{\perp})$ $\xi$-a.s., and also along
$(E_{dep},E_{dep}^{\perp})$, it follows immediately that $\mu_{E_{dep}}$ has
covariance that splits with respect to $\xi$.

1) & 2) $\Rightarrow$3): This is trivial. The assumption that $\mu$ splits
along $(E,E^{\perp})$, $\xi$-a.s., ensures that (3) holds. Since 2) guarantees
that $\mu_{E_{dep}}$ is Gaussian, it has finite second moments.

3)$\Rightarrow$1): Suppose $\mu$ is of the form (3). Without loss of
generality, we can amplify the assumption that $\mu_{E_{dep}}$ has finite
logarithmic moments to the stronger assumption that $\mu$ has finite
logarithmic moments. This can be accomplished by contracting the measure $\mu$
along the directions in $E_{ind}$ (in fact, we could assume without loss of
generality that $\mu_{E_{ind}}$ is compactly supported). By definition of the
independent subspaces, both the mixture property (3) and the splitting
property 1) are invariant to such transformations.

Next, defining $\tilde{\mu}=\mu*\gamma$, the fact that
$\gamma=\gamma_{E}\otimes\gamma_{E^{\perp}}$ for every $E\in\mathbf{E}$
implies we also have

$\displaystyle\tilde{\mu}=\int_{\mathbf{E}}(\tilde{\mu}_{E}\otimes\tilde{\mu}_{E^{\perp}})d\xi(E).$
(28)

If $\tilde{\mu}$ splits along some $(E,E^{\perp})$, then so does $\mu$.
Therefore, it suffices to show that $\tilde{\mu}$ splits with respect to
$\xi$. Together with the assumption of finite logarithmic moments, this
regularization via convolution implies finiteness of the Shannon entropy
$|h(\tilde{\mu})|<\infty$, where Shannon entropy of a measure
$\nu\ll\operatorname{Leb}$ is defined as

$h({\nu}):=-\int\log\left(\frac{d{\nu}}{dx}\right)d{\nu}=-\int\frac{d\nu}{dx}\log\left(\frac{d{\nu}}{dx}\right)d{x},$

provided the integral exists in the Lebesgue sense. With this in hand, we
write

$\displaystyle h(\tilde{\mu})$
$\displaystyle\geq\int_{\mathbf{E}}h(\tilde{\mu}_{E}\otimes\tilde{\mu}_{E^{\perp}})d\xi(E)$
(29)
$\displaystyle=\int_{\mathbf{E}}\left(h(\tilde{\mu}_{E})+h(\tilde{\mu}_{E^{\perp}})\right)d\xi(E)$
(30) $\displaystyle\geq h(\tilde{\mu}).$ (31)

In the above, (29) is (strict) concavity of the function $t\mapsto-t\log t$ on
$t\geq 0$; (30) is because entropy is additive on product measures; (31) is
subadditivity of entropy. Thus, we have equality throughout, and we conclude
that Jensen’s inequality (29) must hold with equality since $h(\tilde{\mu})$
is finite. This occurs only if we have

$\frac{d(\tilde{\mu}_{E}\otimes\tilde{\mu}_{E^{\perp}})}{dx}=\frac{d\tilde{\mu}}{dx}~{}a.e.,~{}~{}\xi-a.s.$

This implies $\tilde{\mu}=\tilde{\mu}_{E}\otimes\tilde{\mu}_{E^{\perp}}$,
$\xi$-a.s., as desired. ∎

##  Appendix A Remarks on Question 1

The following computation demonstrates simple instances where the answer to
Question 1 is affirmative. However, it is unclear how to extend the argument
for this special case to the general setting. As we did in Section 2.1,
consider an ambient space $\mathbb{R}^{n}$ with dimension $n=2m$, and define

$E_{1}=\\{(x,0);x\in\mathbb{R}^{m}\\},~{}~{}~{}E_{2}=\\{(x,x);x\in\mathbb{R}^{m}\\},$

where $0$ denotes the zero-vector in $\mathbb{R}^{m}$. For $\mu\in
P(\mathbb{R}^{n})$, define for convenience:

$\mu_{1}:=\mu_{E_{1}},~{}~{}\mu_{2}:=\mu_{E^{\perp}_{1}},~{}~{}\mu_{+}:=\mu_{E_{2}},~{}~{}\mu_{-}:=\mu_{E_{2}^{\perp}},$

which are all probability measures on $\mathbb{R}^{m}$. Now, let
$(X_{1},X_{2})\sim\mu_{1}\otimes\mu_{2}$, and
$(X_{+},X_{-})\sim\mu_{+}\otimes\mu_{-}$. If $\mu$ is representable as the
mixture

$\displaystyle\mu=p\mu_{1}\otimes\mu_{2}+(1-p)\mu_{+}\otimes\mu_{1}$ (32)

for some $p\in(0,1)$, then computing the $E_{1}$ and $E_{1}^{\perp}$-marginals
from (32) reveals

$X_{1}\overset{law}{=}\frac{X_{+}+X_{-}}{\sqrt{2}},~{}~{}X_{2}\overset{law}{=}\frac{X_{+}-X_{-}}{\sqrt{2}}.$

Similarly, computing the $E_{2}$ and $E_{2}^{\perp}$-marginals gives

$X_{+}\overset{law}{=}\frac{X_{1}+X_{2}}{\sqrt{2}},~{}~{}X_{-}\overset{law}{=}\frac{X_{1}-X_{2}}{\sqrt{2}}.$

Thus, if $(X^{\prime}_{1},X_{2}^{\prime})$ is an independent copy of
$(X_{1},X_{2})$, we have

$X_{1}\overset{law}{=}\frac{X_{1}+X_{2}+X_{1}^{\prime}-X_{2}^{\prime}}{2},~{}~{}X_{2}\overset{law}{=}\frac{X_{1}+X_{2}-X_{1}^{\prime}+X_{2}^{\prime}}{2}$

Now, let $Z$ be equal in distribution to
$X_{1}+X_{2}-X_{1}^{\prime}-X_{2}^{\prime}$, and let $(Z_{i})_{i=1}^{4}$ be
independent copies of $Z$. By the above, we have

$Z\overset{law}{=}\frac{Z_{1}+Z_{2}+Z_{3}+Z_{4}}{2}.$

Hence, $Z$ is a stable law, and noting the scale parameter involved, it must
be Gaussian. By an application of Cramér’s decomposition theorem, $X_{1}$ and
$X_{2}$ must be Gaussian. In particular, $\mu$ has finite logarithmic moment.

Similar computations work when $E_{2}=\\{(ax,bx);x\in\mathbb{R}^{m}\\}$ for
nonzero $a,b\in\mathbb{R}$. However, the calculations become unwieldy in more
general settings, such as if $\mu$ is a nontrivial mixture of three distinct
nontrivial products of marginals.

Given that Question 1 only requires us to establish finiteness of logarithmic
moments (rather than the stronger statement that $\mu$ is Gaussian), a more
promising approach may be to show that if $\mu$ admits mixture form (3), then
the tail probabilities $\mu(\\{x:\|x\|^{2}\geq t\\})$ can be nontrivially
bounded to conclude finiteness of the logarithmic moment (e.g., as was done in
the proof of Lemma 1). For example, it suffices to show that there are $c,C<1$
and $t_{0}\geq 0$ such that

$\displaystyle\mu(\\{x:\|x\|^{2}>t\\})\leq
C\mu(\\{x:\|x\|^{2}>ct\\}),~{}~{}~{}\forall t\geq t_{0}.$ (33)

Indeed, for $X\sim\mu$, we have $\mathbb{E}[\log(1+\|X\|)]<\infty$ if and only
if $\mathbb{E}[\log(1+\|X\|^{2})]<\infty$. Now, by the tail sum formula and a
change of variables,

$\mathbb{E}[\log(1+\|X\|^{2})]=\int_{0}^{\infty}\frac{\mu(\\{x:\|x\|^{2}>t\\})}{1+t}dt.$

A simple induction shows that $t\mapsto\frac{1}{1+t}\mu(\\{x:\|x\|^{2}>t\\})$
is integrable if (33) holds. For now, however, Question 1 remains unanswered.

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