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arXiv:2011.14614v1 [math.ST] 30 Nov 2020
Exponential decay of pairwise correlation in Gaussian graphical
models with an equicorrelational one-dimensional connection
pattern
Guillaume Marrelec1,2,*, Alain Giron1,2,* and Laura Messio3,4,†
1Sorbonne Universit´e, CNRS, INSERM, Laboratoire d’imagerie biom´edicale, LIB, F-75006, Paris, France
2Center for Interaction Science (CIS), Centre de recherches et d’´etudes en sciences des interactions (Cr´esi),
F-75006, Paris, France
3Sorbonne Universit´e, CNRS, Laboratoire de Physique Th´eorique de la Mati`ere Condens´ee, LPTMC, F-75005
Paris, France
4Institut Universitaire de France, IUF, F-75005 Paris, France
*Email: firstname.lastname@inserm.fr
†Email: laura.messio@lptmc.jussieu.fr
Abstract
We consider Gaussian graphical models associated with an equicorrelational and one-
dimensional conditional independence graph. We show that pairwise correlation decays ex-
ponentially as a function of distance. We also provide a limit when the number of variables
tend to infinity and quantify the difference between the finite and infinite cases.
Keywords: Gaussian graphical model; multivariate normal distributions; conditional inde-
pendence graph; equicorrelational one-dimensional connection pattern; tridiagonal matrix;
circulant matrix; Gaussian free fields.
1 Introduction
Let X= (X1,...,Xn) be an n-dimensional variable. A conditional independence graph on Xis
a graphical representation of Xwhich emphasizes the relationships of conditional independence
between the Xi’s (Whittaker, 1990). More precisely, there is no link between nodes iand jif
Xiand Xjare conditionally independent given X[n]\{i,j}, denoted Xi
|=
Xj|X[n]\{i,j}. In the
particular case where Xis a multivariate normal distribution, we refer to Gaussian graphical
models (Uhler, 2017). Let then Xbe a Gaussian graphical model characterized by its covariance
matrix Σ= (Σij ), or, equivalently, its precision (or concentration) matrix Υ= (Υij ) = Σ−1.
Two other key quantities are the pairwise correlation matrix Ω= (Ωij ), defined as Ωij =
Σij/pΣiiΣj j for i6=jand Ωii = 1, as well as the partial correlation matrix Π= (Πij ),
defined as Πij =−Υij /pΥiiΥjj for i6=jand Πii = 1. Then, for i6=j, the relationship of
conditional independence Xi
|=
Xj|X[n]\{i,j}is equivalent to Υij = 0 and Πij = 0 (Whittaker,
1990, Chap. 6).
Our interest in Gaussian graphical models originates from statistical mechanics, where the
Ising model and its various extensions (Potts model, XY model, Heisenberg model, n-vector
model, φ4model) are used to investigate the behavior of variables related through various con-
nection patterns. One extension of the Ising model to continuous real variables with noncompact
support is the so-called Gaussian free field model (Friedli and Velenik, 2017, Chap. 8). In this
case, each vertex i∈Zdis associated with a real-valued variable xiand the corresponding
Hamiltonian is of the form
β
4dX
i,j∈Zd:ki−jk2=1
(xi−xj)2+m2
2X
i∈Zd
x2
i,
1
where β≥0 is the inverse temperature and m≥0 is the mass. In massive models (m > 0),
pairwise correlation is known to decrease exponentially with distance (Friedli and Velenik, 2017,
Prop. 8.30).
While this result is shown in the “thermodynamic limit”, that is, for an infinite-dimensional
variable (i.e., on Zd), we are here interested in the finite case. The reasons for this interest are
twofold. First, a main way to approach statistical mechanics is through simulations, which only
deal with finite case scenarios. It is therefore important to understand what the expected behav-
ior of the system should be in such cases. Does pairwise correlation also decay exponentially?
Also, we would like to gain a sense of how convergence from the finite to the infinite case occurs
through some results regarding the speed of convergence.
In the present study, we focus on the unidimensional case (d= 1) and consider the particular
case of a (finite) Gaussian graphical model on Xwith an equicorrelational one-dimensional con-
nection pattern between the Xi’s, as represented in Figure 1. Such a conditional independence
graph entails that the Gaussian graphical model has a tridiagonal partial correlation matrix
with an off-diagonal element τthat can be related to the parameters of the one-dimensional
Gaussian free field by
τ=
β
4d
2β
4d+m2
2
.(1)
We here restrict ourselves to the case τ > 0 and only consider diagonally dominant matrices,
leading to 0 ≤τ < 1/2 (which corresponds to the massive case, m > 0).
−n
τ
−1
τ
0
τ
1
τ
n
Figure 1: A conditional independence graph whose limit when n→ ∞ yields the one-dimensional
Gaussian free field.
Under these assumptions, we show that Ω(n)
ij , the pairwise correlation between any two
variables X(n)
iand X(n)
j, decreases exponentially with the distance |j−i|between variables,
with a rate given by
λ= arg cosh 1
2τ.(2)
More specifically, we show the following theorem.
Theorem 1 Let X(n)be a Gaussian graphical model with conditional independence graph given
by Figure 1. Then the following results yield:
•0<Ω(n)
ij < e−|j−i|λfor all n;
•Ω(n)
ij →e−|j−i|λwhen n→ ∞;
•The absolute error Ω(n)
ij −e−|j−i|λis Oe−2(n+1)λwhen n→ ∞;
•The relative error Ω(n)
ij e|j−i|λ−1is equal to
−{sinh[2 max(i, j)λ]−sinh[2 min(i, j )λ]}e−2(n+1)λ+ohe−2(n+1)λi
when n→ ∞.
Here, O(·) and o(·) are the usual big-O and little-o Bachmann–Landau notations, respectively,
with
un=O(vn)⇔ ∃n0, c |un|< c|vn| ∀n > n0
and
un=o(vn)⇔un
vn
n→∞
→0.
2
2 Proof of Theorem
We start by expressing pairwise correlation in the case of the simpler model of an n-dimensional
Gaussian graphical model Y(n)with conditional independence graph given by Figure 2. We
then relate the pairwise correlations for both models and derive the results for X(n).
1
τ
2
τ
3
τ
4
τ
n
Figure 2: Conditional independence graph of the Gaussian graphical model Y(n).
2.1 Partial correlation matrix
Assume that Y(n)is a Gaussian graphical model with conditional independence graph given by
Figure 2. The corresponding partial correlation matrix is then given by the following n-by-n
symmetric tridiagonal matrix
Π(n)
ij =
1 if i=j
τif |j−i|= 1
0 otherwise.
(3)
2.2 From partial to pairwise correlation
Letting Inbe the n-by-nidentity matrix and setting Υ(n)= 2In−Π(n), the (pairwise) corre-
lation matrix Ψ(n)= (Ψ(n)
ij ) corresponding to the distribution can be obtained in two steps:
1. Invert Υ(n)to obtain Σ(n)=Υ(n)−1;
2. Decompose Σ(n)= (Σ(n)
ij ) using the correlation transform:
Σ(n)=∆(n)Ψ(n)∆(n),
where ∆(n)= (∆(n)
ij ) is a diagonal matrix with ∆(n)
ii =qΣ(n)
ii .
2.3 Expression of Ψ(n)
ij
If Π(n)has the form of Equation (3), then Υ(n)is also a tridiagonal matrix with off-diagonal
element equal to −τ. Defining λas in Equation (2) and applying results from Hu and O’Connell
(1996), we obtain that
Σ(n)
ij =1
τ
cosh[(n+ 1 − |j−i|)λ]−cosh[(n+ 1 −i−j)λ]
2 sinh(λ) sinh[(n+ 1)λ].
Using a basic identity of hyperbolic functions (Gradshteryn and Ryzhik, 2007, §1.314)
cosh(x)−cosh(y) = 2 sinh x+y
2sinh x−y
2,
we obtain
Σ(n)
ij =1
τ
sinh h2(n+1)−i−j−|j−i|
2λisinh hi+j−|j−i|
2λi
sinh(λ) sinh[(n+ 1)λ].
3
In particular, the diagonal elements read
Σ(n)
ii =1
τ
sinh [(n+ 1 −i)λ] sinh (iλ)
sinh(λ) sinh[(n+ 1)λ].
This leads to the following expression for the correlation coefficient
Ψ(n)
ij =
sinh h2(n+1)−i−j−|j−i|
2λisinh hi+j−|j−i|
2λi
psinh [(n+ 1 −i)λ] sinh (iλ)psinh [(n+ 1 −j)λ] sinh (jλ).
In the following, we will restrict our attention to i < j without loss of generality. For j < i,
we can then use the symmetry identity Ψ(n)
ij = Ψ(n)
ji . So, if j > i, the previous result can be
simplified to yield
Ψ(n)
ij =sinh [(n+ 1 −j)λ] sinh (iλ)
psinh [(n+ 1 −i)λ] sinh (iλ)psinh [(n+ 1 −j)λ] sinh (jλ)
=ssinh [(n+ 1 −j)λ] sinh (iλ)
sinh [(n+ 1 −i)λ] sinh (jλ)
=e−λ(j−i)s1−e−2[(n+1−j)]λ
1−e−2[(n+1−i)]λ
1−e−2iλ
1−e−2jλ .(4)
2.4 Connection between Y(n)and X(n)
Gaussian free fields can be obtained as the limit when n→ ∞ of a (2n+1)-dimensional variables
X(n)= (X−n,...,X−1, X0, X1,...,Xn) with a conditional independence graph given by Fig-
ure 1. Results regarding this model can be derived from the previous model and calculations by
replacing nwith 2n+ 1 and considering pairwise correlations of the form Ω(n)
ij ≡Ψ(2n+1)
n+1+i,n+1+j.
In this perspective, Equation (4) leads to, for i < j,
Ω(n)
ij =e−(j−i)λs1−e−2[(n+1−j)]λ
1−e−2[(n+1−i)]λ
1−e−2(n+1+i)λ
1−e−2(n+1+j)λ.(5)
2.5 Bounds
From Equation (5), it is straightforward to see that Ω(n)
ij is always strictly positive. Also, since
u7→ 1−e−2(n+1−u)λis a strictly increasing function of u, and u7→ 1−e−2(n+1−u)λa strictly
decreasing function of u, we obtain for i < j
s1−e−2[(n+1−j)]λ
1−e−2[(n+1−i)]λ<1 and s1−e−2(n+1+i)λ
1−e−2(n+1+j)λ<1,
so that
0<Ω(n)
ij < e−(j−i)λ
for all n.
2.6 Asymptotics
We can now provide the limit of Ω(n)
ij when n→ ∞. Using the fact that 1 −e−2[(n+1−u)]λtends
to 1 when n→ ∞ for a given u, Equation (5) leads to
Ω(n)
ij
n→∞
→e−(j−i)λ.(6)
4
Besides, using the following Taylor expansion for u→0,
(1 + u)k= 1 + ku +o(u),(7)
we can express Ω(n)
ij /e−(j−i)λas
Ω(n)
ij e(j−i)λ=h1−e−2(n+1−j)λi1
2h1−e−2(n+1−i)λi−1
2
×h1−e−2(n+1+i)λi1
2h1−e−2(n+1+j)λi−1
2
=1−1
2e−2(n+1−j)λ+ohe−2(n+1)λi
×1 + 1
2e−2(n+1−i)λ+ohe−2(n+1)λi
×1−1
2e−2(n+1+i)λ+ohe−2(n+1)λi
×1 + 1
2e−2(n+1+j)λ+ohe−2(n+1)λi
= 1 −1
2he2jλ −e2iλ +e−2iλ −e−2jλie−2(n+1)λ+ohe−2(n+1)λi
= 1 −[sinh(2jλ)−sinh(2iλ)] e−2(n+1)λ+ohe−2(n+1)λi.
We therefore have that
Ω(n)
ij e(j−i)λ= 1 + Ohe−2(n+1)λi,
so that
Ω(n)
ij −e−(j−i)λ=e−(j−i)λhΩ(n)
ij e(j−i)λ−1i=Ohe−2(n+1)λi.
2.7 General results
All results were proved for i < j. As mentioned earlier, the case j < i can be solved by using
the symmetry identity Ω(n)
ij = Ω(n)
ji . The most general results can therefore be expressed by
replacing iwith min(i, j), jwith max(i, j), and j−iwith |j−i|, leading to
•Bounds: 0 <Ω(n)
ij < e−|j−i|λfor all n;
•Limit: Ω(n)
ij →e−|j−i|λwhen n→ ∞;
•Asymptotic expansion: the absolute error Ω(n)
ij −e−|j−i|λis Oe−2(n+1)λ, and the relative
error is given by
Ω(n)
ij e(j−i)λ−1 = −{sinh[2 max(i, j)λ]−sinh[2 min(i, j)λ]}e−2(n+1)λ
+ohe−2(n+1)λi.
3 Discussion
In the present manuscript, we considered a (finite-dimensional) Gaussian graphical model with
the conditional independence graph depicted in Figure 1. We proved that the pairwise correlation
decays exponentially at a rate given by λof Equation (2). We also provided bounds for pairwise
correlation as well as asymptotic expansions of the absolute and relative errors.
5
These results are in line with what is known about the one-dimensional Gaussian free field.
Indeed, setting β= 1, pairwise correlation is known to be of the form exp(−ξm|j−i|) with
(Friedli and Velenik, 2017, Th. 8.33)
ξm= ln(1 + m2+p2m2+m4).
Using the relationship between τand (β , m) of Equation (1) as well as the expression of arg cosh
in terms of logarithm (Gradshteryn and Ryzhik, 2007, §1.622), it can be shown that ξmcorre-
sponds to our λ.
Another quantity of interest is α=e−λ, which can be expressed using again the expression
of arg cosh in terms of logarithm (Gradshteryn and Ryzhik, 2007, §1.622), leading to
1
α=1 + √1−4τ2
2τ,
or equivalently
α=1−√1−4τ2
2τ.(8)
From the definition, it is obvious that α∈[0,1), and that pairwise correlation decreases as α|j−i|.
αappears naturally in the case where the Gaussian graphical model has a partial correlation
matrix that is circulant instead of tridiagonal (see below).
One could wonder what the results are for the Gaussian graphical model Y(n)with condi-
tional independence graph of Figure 2 that was used to derive our main results. The corre-
sponding results are given in Appendix A. They are more complex due to the proximity of the
boundary point 0 to iand j.
Another finite pattern of conditional independence that would lead to one-dimensional Gaus-
sian free fields is the one given in Figure 3. In this case, the partial correlation matrix is sym-
metric circulant and it can be shown that the pairwise correlation still decays exponentially
with the same rate λ(see Appendix B). However, we were not able to provide bounds nor an
asymptotic expansion in that particular case.
1
2
3
4
n
Figure 3: Another instance of conditional independence graph, which corresponds to a symmetric
circulant partial correlation matrix.
Our results show that pairwise correlation in (finite-dimensional) Gaussian graphical models
behave in a manner very similar to one-dimensional (infinite-dimensional) Gaussian free fields,
the difference between both cases decreasing exponentially with n. As a consequence, computer
simulations can be trusted to provide precise approximations for the behavior of one-dimensional
Gaussian free fields.
Beyond pairwise correlation, a measure that we think would be relevant to quantify the global
level of dependence within the system is a multivariate generalization of mutual information
6
known as total correlation (Watanabe, 1960), multivariate constraint (Garner, 1962), δ(Joe,
1989), or multiinformation (Studeny, 1998). In the case of multivariate normal distributions,
this measure has a simple expression in terms of the covariance matrix. While we were not able
to provide the closed form expression in the case of tridiagonal nor circulant partial correlation
matrices, we believe that such expressions might be helpful to understand the global behavior
of the system.
Now that we have solved the case d= 1, we would like to investigate more general cases with
more complex connectivity patterns, still in the case of a finite n. Note that a major advantage
of multivariate normal distributions is that their structures of conditional independence can
be read off their precision matrices. For instance, moving from a one-dimensional to a two-
dimensional model simply implies to change from a tridiagonal partial correlation matrix to
a partial correlation matrix with more non-zero bands. More complex connectivity patterns
with specific features (e.g., random or small world) simply translate into different patterns
in the precision matrix which can then be investigated either analytically or through computer
simulations. And, again, multiinformation has a simple expression that could provide interesting
insight into the global behavior of the system.
A Results for Y(n)
A.1 Bounds
We start from Equation (4) of the manuscript. From this equation, it is obvious that Ψ(n)
ij >0.
Since u7→ 1−e−2uλ is a strictly increasing function of uand u7→ 1−e−2[(n+1−u)]λa strictly
decreasing function of u, the term in the square root is smaller than one for i < j and
Ψ(n)
ij < e−(j−i)λ.
We therefore still have that pairwise correlation decreases exponentially with distance.
A.2 Limit
We can now provide the limit of Ψ(n)
ij when n→ ∞. Still from Equation (4) of the manuscript,
we have
Ψ(n)
ij
n→∞
→e−(j−i)λs1−e−2iλ
1−e−2jλ ≡Ψ(∞)
ij .(9)
Note that, in this case, Ψ(∞)
ij is a function of both iand jthat cannot be expressed as a function
of j−i(the distance between iand j) only. An upper bound for Ψ(∞)
ij is given by
Ψ(∞)
ij < e−(j−i)λ.
Also, still from Equation (4) of the manuscript, we have
Ψ(n)
ij = Ψ(∞)
ij s1−e−2(n+1−j)λ
1−e−2(n+1−i)λ.(10)
Since u7→ 1−e−2(n+1−u)λis a strictly decreasing function of u, we obtain for i < j
s1−e−2(n+1−j)λ
1−e−2(n+1−i)λ<1,
so that
Ψ(n)
ij <Ψ(∞)
ij
for all n.
7
A.3 Asymptotic behavior
Using the Taylor expansion of Equation (7) of the manuscript, we can express Ψ(n)
ij /Ψ(∞)
ij as
Ψ(n)
ij
Ψ(∞)
ij
=h1−e−2(n+1−j)λi1
2h1−e−2(n+1−i)λi−1
2
=1−1
2e−2(n+1−j)λ+ohe−2(n+1)λi
×1 + 1
2e−2(n+1−i)λ+ohe−2(n+1)λi
= 1 −e2jλ −e2iλ
2e−2(n+1)λ+ohe−2(n+1)λi.(11)
This result directly entails that
Ψ(n)
ij −Ψ(∞)
ij = Ψ(∞)
ij "Ψ(n)
ij
Ψ(∞)
ij −1#=Ohe−2(n+1)λi.
A.4 General results
All previous results were proved for i < j. The case j < i is solved by using the symmetry
identity Ψ(n)
ij = Ψ(n)
ji , so that i,jand j−iare replaced with min(i, j), max(i, j) and |j−i|,
respectively. In the end, setting
Ψ(∞)
ij =e−|j−i|λs1−e−2 min(i,j)λ
1−e−2 max(i,j)λ,(12)
we obtain the following results:
•Bounds: 0 <Ψ(n)
ij <Ψ(∞)
ij < e−|j−i|λ;
•Limit: Ψ(n)
ij →Ψ(∞)
ij as n→ ∞;
•Asymptotic expansion:
Ψ(n)
ij
Ψ(∞)
ij
= 1 −1
2he2 max(i,j)λ−e2 min(i,j)λie−2(n+1)λ+ohe−2(n+1)λi
and
Ψ(n)
ij −Ψ(∞)
ij =Ohe−2(n+1)λi.
B Circulant partial correlation matrix
B.1 Model
Assume that X(n)is a Gaussian graphical model with a conditional independence graph given
by Figure 3 of the manuscript. The corresponding partial correlation matrix is then given by
the following n-by-nsymmetric circulant matrix
Π(n)
ij =
1 if i=j
τif |j−i| ∈ {1, n −1}
0 otherwise.
(13)
8
It can be expressed in the general form of circulant matrices as
Π(n)= circ hc(n)
0, c(n)
1,···, c(n)
n−1i
with c(n)
0= 1, c(n)
1=c(n)
n−1=τand 0 otherwise.
B.2 Pairwise correlation
In this case, Υ(n)= 2In−Π(n)is a symmetric circulant matrix as well with
Υ(n)= circ(1,−τ, 0,...,0,−τ).
For n≥2, the neigenvalues of Υ(n)are given by (Chen, 1987; Chao, 1988)
µ(n)
k= 1 −2τcos(kθn), k = 0,...,n−1,(14)
where we set θn= 2π/n. Note that we have µ(n)
0= 1 −2τ; for k≥1, µ(n)
n−k=µ(n)
k; for neven,
we also have µ(n)
n
2= 1 + 2τ. Let Q(n)=n(Υ(n))−1, so that Σ(n)= (Υ(n))−1=1
nQ(n). Then
Q(n)is also a symmetric circulant matrix,
Q(n)= circ hq(n)
0, q(n)
1,...,q(n)
n−1i,
with (Chao, 1988)
q(n)
k=
n−1
X
j=0
e−ijkθn
µ(n)
j
=
n−1
X
j=0
e−ijkθn
1−2τcos(jθn).(15)
In particular, we have for the diagonal term (k= 0)
q(n)
0=
n−1
X
j=0
1
µ(n)
j
=
n−1
X
j=0
1
1−2τcos(jθn).(16)
Since Q(n)is a symmetric circulant matrix, so is Σ(n),
Σ(n)= circ hσ(n)
0, σ(n)
1,...,σ(n)
n−1i,
with
σ(n)
k=q(n)
k
n.(17)
Finally, the correlation matrix Ω(n)is also a symmetric circulant matrix,
Ω= (Ω(n)
ij ) = circ h1, ω(n)
1,...,ω(n)
n−1i
with
ω(n)
k=σ(n)
k
q[σ(n)
0]2
=σ(n)
k
σ(n)
0
=q(n)
k
q(n)
0
,(18)
with q(n)
kand q(n)
0given by Equations (15) and (16), respectively.
9
B.3 Riemannian sum
Let hkbe the function that maps any x∈[0,2π] to
hk(x) = e−ikx
1−2τcos(x).(19)
Setting x(n)
j=jθnfor j= 0,...,n, we have
0 = x(n)
0< x(n)
1<···< x(n)
n= 2π.
We now define
S(n)
k=
n−1
X
j=0
hk[x(n)
j][x(n)
j+1 −x(n)
j] = θn
n−1
X
j=0
e−ijkθn
1−2τcos(jθn)=θnq(n)
k.(20)
By construction, S(n)
kis a left Riemann sum that converges to
S(n)
k
n→∞
→Ik=Z2π
0
hk(x) dx.
B.4 Computation of integral
We therefore need to compute Ik. Using Euler’s formula
eix = cos(x) + isin(x),
we obtain
Ik=Z2π
0
e−ikx
1−τ(eix +e−ix)dx.
Performing the parameter change z=eix, we can now write this integral as a contour integral
on the unit circle
Ik=I|z|=1
zk
1−τ(z+z−1)
dz
iz
=1
iI|z|=1
zk
−τz2+z−τdz.
The roots of −τ z2+z−τare given by
α=1−√1−4τ2
2τ
and 1/α. The integral therefore yields
Ik=−1
iτ I|z|=1
zk
(z−α)z−1
αdz.
Factoring the fraction yields
1
(z−α)z−1
α=u
z−α−u
z−1
α
with
u=α
α2−1=−τ
√1−4τ2.
10
We therefore have for the integral
Ik=1
i√1−4τ2I|z|=1 zk
z−α−zk
z−1
α!dz.
1/α is outside the unit circle, so that
I|z|=1
zk
z−1
α
dz= 0.
For the other other integral, we need to compute the residual of f(z) = zk/(z−α) at z=α.
Since it is a simple pole, we have
Resz=αf(z) = lim
z→α(z−α)f(z) = lim
z→αzk=αk,
since α∈R. We are then then led to
I|z|=1
zk
z−αdz= 2iπαk
and, finally,
Ik=2παk
√1−4τ2.(21)
In particular, we have for k= 0
I0=2π
√1−4τ2.(22)
B.5 Asymptotic approximation
Now that we computed Ik, we can go back to the pairwise correlation. Since the sum S(n)
kof
Equation (20) converges toward the integral Ik, we have for σ(n)
k, using Equations (17) and (20),
σ(n)
k=q(n)
k
n=S(n)
k
nθn
=S(n)
k
2π
n→∞
→Ik
2π=αk
√1−4τ2(23)
and for ω(n)
k, using Equations (18) and (20),
ω(n)
k=q(n)
k
q(n)
0
=S(n)
k
S(n)
0
n→∞
→Ik
I0
=αk.(24)
Since k=|j−i|, we can conclude that
Ω(n)
ij
n→∞
→α|j−i|.
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