# 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,$\dagger$
1 Sorbonne Université CNRS  INSERM  Laboratoire d’imagerie biomédicale  LIB
F-75006  Paris  France
2 Center for Interaction Science (CIS) Centre de recherches et d’études en
sciences des interactions (Crési)  F-75006  Paris  France
3 Sorbonne Université CNRS  Laboratoire de Physique Théorique de la Matière
Condensée  LPTMC  F-75005 Paris  France
4 Institut Universitaire de France IUF  F-75005 Paris  France
* Email<EMAIL_ADDRESS>
$\dagger$ Email<EMAIL_ADDRESS>
###### Abstract

We consider Gaussian graphical models associated with an equicorrelational and
one-dimensional conditional independence graph. We show that pairwise
correlation decays exponentially 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 independence graph; equicorrelational one-dimensional connection
pattern; tridiagonal matrix; circulant matrix; Gaussian free fields.

##  1 Introduction

Let $\boldsymbol{X}=(X_{1},\dots,X_{n})$ be an $n$-dimensional variable. A
conditional independence graph on $\boldsymbol{X}$ is a graphical
representation of $\boldsymbol{X}$ which emphasizes the relationships of
conditional independence between the $X_{i}$’s (Whittaker, 1990). More
precisely, there is _no_ link between nodes $i$ and $j$ if $X_{i}$ and $X_{j}$
are conditionally independent given $\boldsymbol{X}_{[n]\setminus\\{i,j\\}}$,
denoted
$X_{i}\,\rotatebox{90.0}{$\models$}\,X_{j}|\boldsymbol{X}_{[n]\setminus\\{i,j\\}}$.
In the particular case where $\boldsymbol{X}$ is a multivariate normal
distribution, we refer to Gaussian graphical models (Uhler, 2017). Let then
$\boldsymbol{X}$ be a Gaussian graphical model characterized by its covariance
matrix $\boldsymbol{\Sigma}=(\Sigma_{ij})$, or, equivalently, its precision
(or concentration) matrix
$\boldsymbol{\Upsilon}=(\Upsilon_{ij})=\boldsymbol{\Sigma}^{-1}$. Two other
key quantities are the pairwise correlation matrix
$\boldsymbol{\Omega}=(\Omega_{ij})$, defined as
$\Omega_{ij}=\Sigma_{ij}/\sqrt{\Sigma_{ii}\Sigma_{jj}}$ for $i\neq j$ and
$\Omega_{ii}=1$, as well as the partial correlation matrix
$\boldsymbol{\Pi}=(\Pi_{ij})$, defined as
$\Pi_{ij}=-\Upsilon_{ij}/\sqrt{\Upsilon_{ii}\Upsilon_{jj}}$ for $i\neq j$ and
$\Pi_{ii}=1$. Then, for $i\neq j$, the relationship of conditional
independence
$X_{i}\,\rotatebox{90.0}{$\models$}\,X_{j}|\boldsymbol{X}_{[n]\setminus\\{i,j\\}}$
is equivalent to $\Upsilon_{ij}=0$ and $\Pi_{ij}=0$ (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, $\phi^{4}$ model) are used to
investigate the behavior of variables related through various connection
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 $i\in\mathbb{Z}^{d}$ is
associated with a real-valued variable $x_{i}$ and the corresponding
Hamiltonian is of the form

$\frac{\beta}{4d}\sum_{i,j\in\mathbb{Z}^{d}:\|i-j\|_{2}=1}(x_{i}-x_{j})^{2}+\frac{m^{2}}{2}\sum_{i\in\mathbb{Z}^{d}}x_{i}^{2},$

where $\beta\geq 0$ is the inverse temperature and $m\geq 0$ 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 $\mathbb{Z}^{d}$), 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 behavior 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 $\boldsymbol{X}$
with an equicorrelational one-dimensional connection pattern between the
$X_{i}$’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 $\tau$ that can be related to
the parameters of the one-dimensional Gaussian free field by

$\tau=\frac{\frac{\beta}{4d}}{\frac{2\beta}{4d}+\frac{m^{2}}{2}}.$ (1)

We here restrict ourselves to the case $\tau>0$ and only consider diagonally
dominant matrices, leading to $0\leq\tau<1/2$ (which corresponds to the
massive case, $m>0$).

$-n$$\tau$$-1$$\tau$$0$$\tau$1$\tau$$n$ Figure 1: A conditional independence
graph whose limit when $n\to\infty$ yields the one-dimensional Gaussian free
field.

Under these assumptions, we show that $\Omega_{ij}^{(n)}$, the pairwise
correlation between any two variables $X_{i}^{(n)}$ and $X_{j}^{(n)}$,
decreases exponentially with the distance $|j-i|$ between variables, with a
rate given by

$\lambda=\arg\cosh\left(\frac{1}{2\tau}\right).$ (2)

More specifically, we show the following theorem.

###### Theorem 1

Let $\boldsymbol{X}^{(n)}$ be a Gaussian graphical model with conditional
independence graph given by Figure 1. Then the following results yield:

  * •

$0<\Omega_{ij}^{(n)}<e^{-|j-i|\lambda}$ for all $n$;

  * •

$\Omega_{ij}^{(n)}\to e^{-|j-i|\lambda}$ when $n\to\infty$;

  * •

The absolute error $\Omega_{ij}^{(n)}-e^{-|j-i|\lambda}$ is
$O\left[e^{-2(n+1)\lambda}\right]$ when $n\to\infty$;

  * •

The relative error $\Omega_{ij}^{(n)}e^{|j-i|\lambda}-1$ is equal to

$-\left\\{\sinh[2\max(i,j)\lambda]-\sinh[2\min(i,j)\lambda]\right\\}e^{-2(n+1)\lambda}+o\left[e^{-2(n+1)\lambda}\right]$

when $n\to\infty$.

Here, $O(\cdot)$ and $o(\cdot)$ are the usual big-O and little-o
Bachmann–Landau notations, respectively, with

$u_{n}=O(v_{n})\qquad\Leftrightarrow\qquad\exists
n_{0},c\quad|u_{n}|<c|v_{n}|\ \forall n>n_{0}$

and

$u_{n}=o(v_{n})\qquad\Leftrightarrow\qquad\frac{u_{n}}{v_{n}}\stackrel{{\scriptstyle
n\to\infty}}{{\to}}0.$

##  2 Proof of Theorem

We start by expressing pairwise correlation in the case of the simpler model
of an $n$-dimensional Gaussian graphical model $\boldsymbol{Y}^{(n)}$ with
conditional independence graph given by Figure 2. We then relate the pairwise
correlations for both models and derive the results for
$\boldsymbol{X}^{(n)}$.

1$\tau$2$\tau$3$\tau$4$\tau$$n$ Figure 2: Conditional independence graph of
the Gaussian graphical model $\boldsymbol{Y}^{(n)}$.

###  2.1 Partial correlation matrix

Assume that $\boldsymbol{Y}^{(n)}$ 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

$\Pi_{ij}^{(n)}=\left\\{\begin{array}[]{cl}1&\mbox{if $i=j$}\\\ \tau&\mbox{if
$|j-i|=1$}\\\ 0&\mbox{otherwise.}\end{array}\right.$ (3)

###  2.2 From partial to pairwise correlation

Letting $\boldsymbol{I}_{n}$ be the $n$-by-$n$ identity matrix and setting
$\boldsymbol{\Upsilon}^{(n)}=2\boldsymbol{I}_{n}-\boldsymbol{\Pi}^{(n)}$, the
(pairwise) correlation matrix $\boldsymbol{\Psi}^{(n)}=(\Psi_{ij}^{(n)})$
corresponding to the distribution can be obtained in two steps:

  1. 1.

Invert $\boldsymbol{\Upsilon}^{(n)}$ to obtain
$\boldsymbol{\Sigma}^{(n)}={\boldsymbol{\Upsilon}^{(n)}}^{-1}$;

  2. 2.

Decompose $\boldsymbol{\Sigma}^{(n)}=(\Sigma_{ij}^{(n)})$ using the
correlation transform:

$\boldsymbol{\Sigma}^{(n)}=\boldsymbol{\Delta}^{(n)}\boldsymbol{\Psi}^{(n)}\boldsymbol{\Delta}^{(n)},$

where $\boldsymbol{\Delta}^{(n)}=(\Delta_{ij}^{(n)})$ is a diagonal matrix
with $\Delta_{ii}^{(n)}=\sqrt{\Sigma_{ii}^{(n)}}$.

###  2.3 Expression of $\Psi_{ij}^{(n)}$

If $\boldsymbol{\Pi}^{(n)}$ has the form of Equation (3), then
$\boldsymbol{\Upsilon}^{(n)}$ is also a tridiagonal matrix with off-diagonal
element equal to $-\tau$. Defining $\lambda$ as in Equation (2) and applying
results from Hu and O’Connell (1996), we obtain that

$\Sigma_{ij}^{(n)}=\frac{1}{\tau}\,\frac{\cosh[(n+1-|j-i|)\lambda]-\cosh[(n+1-i-j)\lambda]}{2\sinh(\lambda)\,\sinh[(n+1)\lambda]}.$

Using a basic identity of hyperbolic functions (Gradshteryn and Ryzhik, 2007,
§1.314)

$\cosh(x)-\cosh(y)=2\sinh\left(\frac{x+y}{2}\right)\sinh\left(\frac{x-y}{2}\right),$

we obtain

$\Sigma_{ij}^{(n)}=\frac{1}{\tau}\,\frac{\sinh\left[\frac{2(n+1)-i-j-|j-i|}{2}\lambda\right]\,\sinh\left[\frac{i+j-|j-i|}{2}\lambda\right]}{\sinh(\lambda)\,\sinh[(n+1)\lambda]}.$

In particular, the diagonal elements read

$\Sigma_{ii}^{(n)}=\frac{1}{\tau}\,\frac{\sinh\left[(n+1-i)\lambda\right]\,\sinh\left(i\lambda\right)}{\sinh(\lambda)\,\sinh[(n+1)\lambda]}.$

This leads to the following expression for the correlation coefficient

$\Psi_{ij}^{(n)}=\frac{\sinh\left[\frac{2(n+1)-i-j-|j-i|}{2}\lambda\right]\,\sinh\left[\frac{i+j-|j-i|}{2}\lambda\right]}{\sqrt{\sinh\left[(n+1-i)\lambda\right]\,\sinh\left(i\lambda\right)}\sqrt{\sinh\left[(n+1-j)\lambda\right]\,\sinh\left(j\lambda\right)}}.$

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
$\Psi_{ij}^{(n)}=\Psi_{ji}^{(n)}$. So, if $j>i$, the previous result can be
simplified to yield

$\displaystyle\Psi_{ij}^{(n)}$ $\displaystyle=$
$\displaystyle\frac{\sinh\left[(n+1-j)\lambda\right]\,\sinh\left(i\lambda\right)}{\sqrt{\sinh\left[(n+1-i)\lambda\right]\,\sinh\left(i\lambda\right)}\sqrt{\sinh\left[(n+1-j)\lambda\right]\,\sinh\left(j\lambda\right)}}$
(4) $\displaystyle=$
$\displaystyle\sqrt{\frac{\sinh\left[(n+1-j)\lambda\right]\,\sinh\left(i\lambda\right)}{\sinh\left[(n+1-i)\lambda\right]\,\sinh\left(j\lambda\right)}}$
$\displaystyle=$ $\displaystyle
e^{-\lambda(j-i)}\sqrt{\frac{1-e^{-2[(n+1-j)]\lambda}}{1-e^{-2[(n+1-i)]\lambda}}\frac{1-e^{-2i\lambda}}{1-e^{-2j\lambda}}}.$

###  2.4 Connection between $\boldsymbol{Y}^{(n)}$ and $\boldsymbol{X}^{(n)}$

Gaussian free fields can be obtained as the limit when $n\to\infty$ of a
$(2n+1)$-dimensional variables
$\boldsymbol{X}^{(n)}=(X_{-n},\dots,X_{-1},X_{0},X_{1},\dots,X_{n})$ with a
conditional independence graph given by Figure 1. Results regarding this model
can be derived from the previous model and calculations by replacing $n$ with
$2n+1$ and considering pairwise correlations of the form
$\Omega_{ij}^{(n)}\equiv\Psi_{n+1+i,n+1+j}^{(2n+1)}$. In this perspective,
Equation (4) leads to, for $i<j$,

$\Omega_{ij}^{(n)}=e^{-(j-i)\lambda}\sqrt{\frac{1-e^{-2[(n+1-j)]\lambda}}{1-e^{-2[(n+1-i)]\lambda}}\frac{1-e^{-2(n+1+i)\lambda}}{1-e^{-2(n+1+j)\lambda}}}.$
(5)

###  2.5 Bounds

From Equation (5), it is straightforward to see that $\Omega_{ij}^{(n)}$ is
always strictly positive. Also, since $u\mapsto 1-e^{-2(n+1-u)\lambda}$ is a
strictly increasing function of $u$, and $u\mapsto 1-e^{-2(n+1-u)\lambda}$ a
strictly decreasing function of $u$, we obtain for $i<j$

$\sqrt{\frac{1-e^{-2[(n+1-j)]\lambda}}{1-e^{-2[(n+1-i)]\lambda}}}<1\qquad\mbox{and}\qquad\sqrt{\frac{1-e^{-2(n+1+i)\lambda}}{1-e^{-2(n+1+j)\lambda}}}<1,$

so that

$0<\Omega_{ij}^{(n)}<e^{-(j-i)\lambda}$

for all $n$.

###  2.6 Asymptotics

We can now provide the limit of $\Omega_{ij}^{(n)}$ when $n\to\infty$. Using
the fact that $1-e^{-2[(n+1-u)]\lambda}$ tends to 1 when $n\to\infty$ for a
given $u$, Equation (5) leads to

$\Omega_{ij}^{(n)}\stackrel{{\scriptstyle
n\to\infty}}{{\to}}e^{-(j-i)\lambda}.$ (6)

Besides, using the following Taylor expansion for $u\to 0$,

$(1+u)^{k}=1+ku+o(u),$ (7)

we can express $\Omega_{ij}^{(n)}/e^{-(j-i)\lambda}$ as

$\displaystyle\Omega_{ij}^{(n)}e^{(j-i)\lambda}$ $\displaystyle=$
$\displaystyle\left[1-e^{-2(n+1-j)\lambda}\right]^{\frac{1}{2}}\left[1-e^{-2(n+1-i)\lambda}\right]^{-\frac{1}{2}}$
$\displaystyle\qquad\times\left[1-e^{-2(n+1+i)\lambda}\right]^{\frac{1}{2}}\left[1-e^{-2(n+1+j)\lambda}\right]^{-\frac{1}{2}}$
$\displaystyle=$
$\displaystyle\left\\{1-\frac{1}{2}e^{-2(n+1-j)\lambda}+o\left[e^{-2(n+1)\lambda}\right]\right\\}$
$\displaystyle\qquad\times\left\\{1+\frac{1}{2}e^{-2(n+1-i)\lambda}+o\left[e^{-2(n+1)\lambda}\right]\right\\}$
$\displaystyle\times\left\\{1-\frac{1}{2}e^{-2(n+1+i)\lambda}+o\left[e^{-2(n+1)\lambda}\right]\right\\}$
$\displaystyle\qquad\times\left\\{1+\frac{1}{2}e^{-2(n+1+j)\lambda}+o\left[e^{-2(n+1)\lambda}\right]\right\\}$
$\displaystyle=$ $\displaystyle
1-\frac{1}{2}\left[e^{2j\lambda}-e^{2i\lambda}+e^{-2i\lambda}-e^{-2j\lambda}\right]e^{-2(n+1)\lambda}+o\left[e^{-2(n+1)\lambda}\right]$
$\displaystyle=$ $\displaystyle
1-\left[\sinh(2j\lambda)-\sinh(2i\lambda)\right]e^{-2(n+1)\lambda}+o\left[e^{-2(n+1)\lambda}\right].$

We therefore have that

$\Omega_{ij}^{(n)}e^{(j-i)\lambda}=1+O\left[e^{-2(n+1)\lambda}\right],$

so that

$\Omega_{ij}^{(n)}-e^{-(j-i)\lambda}=e^{-(j-i)\lambda}\left[\Omega_{ij}^{(n)}e^{(j-i)\lambda}-1\right]=O\left[e^{-2(n+1)\lambda}\right].$

###  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 $\Omega_{ij}^{(n)}=\Omega_{ji}^{(n)}$.
The most general results can therefore be expressed by replacing $i$ with
$\min(i,j)$, $j$ with $\max(i,j)$, and $j-i$ with $|j-i|$, leading to

  * •

Bounds: $0<\Omega_{ij}^{(n)}<e^{-|j-i|\lambda}$ for all $n$;

  * •

Limit: $\Omega_{ij}^{(n)}\to e^{-|j-i|\lambda}$ when $n\to\infty$;

  * •

Asymptotic expansion: the absolute error $\Omega_{ij}^{(n)}-e^{-|j-i|\lambda}$
is $O\left[e^{-2(n+1)\lambda}\right]$, and the relative error is given by

$\displaystyle\Omega_{ij}^{(n)}e^{(j-i)\lambda}-1$ $\displaystyle=$
$\displaystyle-\left\\{\sinh[2\max(i,j)\lambda]-\sinh[2\min(i,j)\lambda]\right\\}e^{-2(n+1)\lambda}$
$\displaystyle\qquad+o\left[e^{-2(n+1)\lambda}\right].$

##  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 $\lambda$ of Equation (2). We also provided bounds for pairwise correlation
as well as asymptotic expansions of the absolute and relative errors.

These results are in line with what is known about the one-dimensional
Gaussian free field. Indeed, setting $\beta=1$, pairwise correlation is known
to be of the form $\exp(-\xi_{m}|j-i|)$ with (Friedli and Velenik, 2017, Th.
8.33)

$\xi_{m}=\ln(1+m^{2}+\sqrt{2m^{2}+m^{4}}).$

Using the relationship between $\tau$ and $(\beta,m)$ 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 $\xi_{m}$ corresponds to our
$\lambda$.

Another quantity of interest is $\alpha=e^{-\lambda}$, which can be expressed
using again the expression of $\arg\cosh$ in terms of logarithm (Gradshteryn
and Ryzhik, 2007, §1.622), leading to

$\frac{1}{\alpha}=\frac{1+\sqrt{1-{4\tau^{2}}}}{2\tau},$

or equivalently

$\alpha=\frac{1-\sqrt{1-{4\tau^{2}}}}{2\tau}.$ (8)

From the definition, it is obvious that $\alpha\in[0,1)$, and that pairwise
correlation decreases as $\alpha^{|j-i|}$. $\alpha$ appears 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
$\boldsymbol{Y}^{(n)}$ with conditional independence graph of Figure 2 that
was used to derive our main results. The corresponding results are given in
Appendix A. They are more complex due to the proximity of the boundary point 0
to $i$ and $j$.

Another finite pattern of conditional independence that would lead to one-
dimensional Gaussian free fields is the one given in Figure 3. In this case,
the partial correlation matrix is symmetric circulant and it can be shown that
the pairwise correlation still decays exponentially with the same rate
$\lambda$ (see Appendix B). However, we were not able to provide bounds nor an
asymptotic expansion in that particular case.

1234$n$ 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 known as total correlation (Watanabe,
1960), multivariate constraint (Garner, 1962), $\delta$ (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.

##  Appendix A Results for $\boldsymbol{Y}^{(n)}$

###  A.1 Bounds

We start from Equation (4) of the manuscript. From this equation, it is
obvious that $\Psi_{ij}^{(n)}>0$. Since $u\mapsto 1-e^{-2u\lambda}$ is a
strictly increasing function of $u$ and $u\mapsto 1-e^{-2[(n+1-u)]\lambda}$ a
strictly decreasing function of $u$, the term in the square root is smaller
than one for $i<j$ and

$\Psi_{ij}^{(n)}<e^{-(j-i)\lambda}.$

We therefore still have that pairwise correlation decreases exponentially with
distance.

###  A.2 Limit

We can now provide the limit of $\Psi_{ij}^{(n)}$ when $n\to\infty$. Still
from Equation (4) of the manuscript, we have

$\Psi_{ij}^{(n)}\stackrel{{\scriptstyle
n\to\infty}}{{\to}}e^{-(j-i)\lambda}\,\sqrt{\frac{1-e^{-2i\lambda}}{1-e^{-2j\lambda}}}\equiv\Psi_{ij}^{(\infty)}.$
(9)

Note that, in this case, $\Psi_{ij}^{(\infty)}$ is a function of both $i$ and
$j$ that cannot be expressed as a function of $j-i$ (the distance between $i$
and $j$) only. An upper bound for $\Psi_{ij}^{(\infty)}$ is given by

$\Psi_{ij}^{(\infty)}<e^{-(j-i)\lambda}.$

Also, still from Equation (4) of the manuscript, we have

$\Psi_{ij}^{(n)}=\Psi_{ij}^{(\infty)}\sqrt{\frac{1-e^{-2(n+1-j)\lambda}}{1-e^{-2(n+1-i)\lambda}}}.$
(10)

Since $u\mapsto 1-e^{-2(n+1-u)\lambda}$ is a strictly decreasing function of
$u$, we obtain for $i<j$

$\sqrt{\frac{1-e^{-2(n+1-j)\lambda}}{1-e^{-2(n+1-i)\lambda}}}<1,$

so that

$\Psi_{ij}^{(n)}<\Psi_{ij}^{(\infty)}$

for all $n$.

###  A.3 Asymptotic behavior

Using the Taylor expansion of Equation (7) of the manuscript, we can express
$\Psi_{ij}^{(n)}/\Psi_{ij}^{(\infty)}$ as

$\displaystyle\frac{\Psi_{ij}^{(n)}}{\Psi_{ij}^{(\infty)}}$ $\displaystyle=$
$\displaystyle\left[1-e^{-2(n+1-j)\lambda}\right]^{\frac{1}{2}}\left[1-e^{-2(n+1-i)\lambda}\right]^{-\frac{1}{2}}$
(11) $\displaystyle=$
$\displaystyle\left\\{1-\frac{1}{2}e^{-2(n+1-j)\lambda}+o\left[e^{-2(n+1)\lambda}\right]\right\\}$
$\displaystyle\qquad\times\left\\{1+\frac{1}{2}e^{-2(n+1-i)\lambda}+o\left[e^{-2(n+1)\lambda}\right]\right\\}$
$\displaystyle=$ $\displaystyle
1-\frac{e^{2j\lambda}-e^{2i\lambda}}{2}e^{-2(n+1)\lambda}+o\left[e^{-2(n+1)\lambda}\right].$

This result directly entails that

$\Psi_{ij}^{(n)}-\Psi_{ij}^{(\infty)}=\Psi_{ij}^{(\infty)}\left[\frac{\Psi_{ij}^{(n)}}{\Psi_{ij}^{(\infty)}}-1\right]=O\left[e^{-2(n+1)\lambda}\right].$

###  A.4 General results

All previous results were proved for $i<j$. The case $j<i$ is solved by using
the symmetry identity $\Psi_{ij}^{(n)}=\Psi_{ji}^{(n)}$, so that $i$, $j$ and
$j-i$ are replaced with $\min(i,j)$, $\max(i,j)$ and $|j-i|$, respectively. In
the end, setting

$\Psi_{ij}^{(\infty)}=e^{-|j-i|\lambda}\,\sqrt{\frac{1-e^{-2\min(i,j)\lambda}}{1-e^{-2\max(i,j)\lambda}}},$
(12)

we obtain the following results:

  * •

Bounds: $0<\Psi_{ij}^{(n)}<\Psi_{ij}^{(\infty)}<e^{-|j-i|\lambda}$;

  * •

Limit: $\Psi_{ij}^{(n)}\to\Psi_{ij}^{(\infty)}$ as $n\to\infty$;

  * •

Asymptotic expansion:

$\frac{\Psi_{ij}^{(n)}}{\Psi_{ij}^{(\infty)}}=1-\frac{1}{2}\left[e^{2\max(i,j)\lambda}-e^{2\min(i,j)\lambda}\right]e^{-2(n+1)\lambda}+o\left[e^{-2(n+1)\lambda}\right]$

and

$\Psi_{ij}^{(n)}-\Psi_{ij}^{(\infty)}=O\left[e^{-2(n+1)\lambda}\right].$

##  Appendix B Circulant partial correlation matrix

###  B.1 Model

Assume that $\boldsymbol{X}^{(n)}$ 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-$n$ symmetric circulant matrix

$\Pi_{ij}^{(n)}=\left\\{\begin{array}[]{cl}1&\mbox{if $i=j$}\\\ \tau&\mbox{if
$|j-i|\in\\{1,n-1\\}$}\\\ 0&\mbox{otherwise.}\end{array}\right.$ (13)

It can be expressed in the general form of circulant matrices as

$\boldsymbol{\Pi}^{(n)}=\mathrm{circ}\left[c_{0}^{(n)},c_{1}^{(n)},\cdots,c_{n-1}^{(n)}\right]$

with $c_{0}^{(n)}=1$, $c_{1}^{(n)}=c_{n-1}^{(n)}=\tau$ and 0 otherwise.

###  B.2 Pairwise correlation

In this case,
$\boldsymbol{\Upsilon}^{(n)}=2\boldsymbol{I}_{n}-\boldsymbol{\Pi}^{(n)}$ is a
symmetric circulant matrix as well with

$\boldsymbol{\Upsilon}^{(n)}=\mathrm{circ}(1,-\tau,0,\dots,0,-\tau).$

For $n\geq 2$, the $n$ eigenvalues of $\boldsymbol{\Upsilon}^{(n)}$ are given
by (Chen, 1987; Chao, 1988)

$\mu_{k}^{(n)}=1-2\tau\cos(k\theta_{n}),\qquad k=0,\dots,n-1,$ (14)

where we set $\theta_{n}=2\pi/n$. Note that we have $\mu_{0}^{(n)}=1-2\tau$;
for $k\geq 1$, $\mu_{n-k}^{(n)}=\mu_{k}^{(n)}$ ; for $n$ even, we also have
$\mu_{\frac{n}{2}}^{(n)}=1+2\tau$. Let
$\boldsymbol{Q}^{(n)}=n(\boldsymbol{\Upsilon}^{(n)})^{-1}$, so that
$\boldsymbol{\Sigma}^{(n)}=(\boldsymbol{\Upsilon}^{(n)})^{-1}=\frac{1}{n}\boldsymbol{Q}^{(n)}$.
Then $\boldsymbol{Q}^{(n)}$ is also a symmetric circulant matrix,

$\boldsymbol{Q}^{(n)}=\mathrm{circ}\left[q_{0}^{(n)},q_{1}^{(n)},\dots,q_{n-1}^{(n)}\right],$

with (Chao, 1988)

$q_{k}^{(n)}=\sum_{j=0}^{n-1}\frac{e^{-ijk\theta_{n}}}{\mu_{j}^{(n)}}=\sum_{j=0}^{n-1}\frac{e^{-ijk\theta_{n}}}{1-2\tau\cos(j\theta_{n})}.$
(15)

In particular, we have for the diagonal term ($k=0$)

$q_{0}^{(n)}=\sum_{j=0}^{n-1}\frac{1}{\mu_{j}^{(n)}}=\sum_{j=0}^{n-1}\frac{1}{1-2\tau\cos(j\theta_{n})}.$
(16)

Since $\boldsymbol{Q}^{(n)}$ is a symmetric circulant matrix, so is
$\boldsymbol{\Sigma}^{(n)}$,

$\boldsymbol{\Sigma}^{(n)}=\mathrm{circ}\left[\sigma_{0}^{(n)},\sigma_{1}^{(n)},\dots,\sigma_{n-1}^{(n)}\right],$

with

$\sigma_{k}^{(n)}=\frac{q_{k}^{(n)}}{n}.$ (17)

Finally, the correlation matrix $\boldsymbol{\Omega}^{(n)}$ is also a
symmetric circulant matrix,

$\boldsymbol{\Omega}=(\Omega_{ij}^{(n)})=\mathrm{circ}\left[1,\omega_{1}^{(n)},\dots,\omega_{n-1}^{(n)}\right]$

with

$\omega_{k}^{(n)}=\frac{\sigma_{k}^{(n)}}{\sqrt{[\sigma_{0}^{(n)}]^{2}}}=\frac{\sigma_{k}^{(n)}}{\sigma_{0}^{(n)}}=\frac{q_{k}^{(n)}}{q_{0}^{(n)}},$
(18)

with $q_{k}^{(n)}$ and $q_{0}^{(n)}$ given by Equations (15) and (16),
respectively.

###  B.3 Riemannian sum

Let $h_{k}$ be the function that maps any $x\in[0,2\pi]$ to

$h_{k}(x)=\frac{e^{-ikx}}{1-2\tau\cos(x)}.$ (19)

Setting $x_{j}^{(n)}=j\theta_{n}$ for $j=0,\dots,n$, we have

$0=x_{0}^{(n)}<x_{1}^{(n)}<\dots<x_{n}^{(n)}=2\pi.$

We now define

$S_{k}^{(n)}=\sum_{j=0}^{n-1}h_{k}[x_{j}^{(n)}][x_{j+1}^{(n)}-x_{j}^{(n)}]=\theta_{n}\sum_{j=0}^{n-1}\frac{e^{-ijk\theta_{n}}}{1-2\tau\cos(j\theta_{n})}=\theta_{n}q_{k}^{(n)}.$
(20)

By construction, $S_{k}^{(n)}$ is a left Riemann sum that converges to

$S_{k}^{(n)}\stackrel{{\scriptstyle
n\to\infty}}{{\to}}I_{k}=\int_{0}^{2\pi}h_{k}(x)\,\mathrm{d}x.$

###  B.4 Computation of integral

We therefore need to compute $I_{k}$. Using Euler’s formula

$e^{ix}=\cos(x)+i\sin(x),$

we obtain

$I_{k}=\int_{0}^{2\pi}\frac{e^{-ikx}}{1-\tau\left(e^{ix}+e^{-ix}\right)}\,\mathrm{d}x.$

Performing the parameter change $z=e^{ix}$, we can now write this integral as
a contour integral on the unit circle

$\displaystyle I_{k}$ $\displaystyle=$
$\displaystyle\oint_{|z|=1}\frac{\overline{z}^{k}}{1-\tau(z+z^{-1})}\,\frac{\mathrm{d}z}{iz}$
$\displaystyle=$
$\displaystyle\frac{1}{i}\oint_{|z|=1}\frac{\overline{z}^{k}}{-\tau
z^{2}+z-\tau}\,\mathrm{d}z.$

The roots of $-\tau z^{2}+z-\tau$ are given by

$\alpha=\frac{1-\sqrt{1-4\tau^{2}}}{2\tau}$

and $1/\alpha$. The integral therefore yields

$I_{k}=-\frac{1}{i\tau}\oint_{|z|=1}\frac{\overline{z}^{k}}{\left(z-\alpha\right)\left(z-\frac{1}{\alpha}\right)}\,\mathrm{d}z.$

Factoring the fraction yields

$\frac{1}{\left(z-\alpha\right)\left(z-\frac{1}{\alpha}\right)}=\frac{u}{z-\alpha}-\frac{u}{z-\frac{1}{\alpha}}$

with

$u=\frac{\alpha}{\alpha^{2}-1}=-\frac{\tau}{\sqrt{1-4\tau^{2}}}.$

We therefore have for the integral

$\displaystyle I_{k}$ $\displaystyle=$
$\displaystyle\frac{1}{i\sqrt{1-4\tau^{2}}}\oint_{|z|=1}\left(\frac{\overline{z}^{k}}{z-\alpha}-\frac{\overline{z}^{k}}{z-\frac{1}{\alpha}}\right)\,\mathrm{d}z.$

$1/\alpha$ is outside the unit circle, so that

$\oint_{|z|=1}\frac{\overline{z}^{k}}{z-\frac{1}{\alpha}}\,\mathrm{d}z=0.$

For the other other integral, we need to compute the residual of
$f(z)=\overline{z}^{k}/(z-\alpha)$ at $z=\alpha$. Since it is a simple pole,
we have

$\mathrm{Res}_{z=\alpha}f(z)=\lim_{z\to\alpha}(z-\alpha)f(z)=\lim_{z\to\alpha}\overline{z}^{k}=\alpha^{k},$

since $\alpha\in\mathbb{R}$. We are then then led to

$\oint_{|z|=1}\frac{\overline{z}^{k}}{z-\alpha}\,\mathrm{d}z=2i\pi\alpha^{k}$

and, finally,

$I_{k}=\frac{2\pi\alpha^{k}}{\sqrt{1-4\tau^{2}}}.$ (21)

In particular, we have for $k=0$

$I_{0}=\frac{2\pi}{\sqrt{1-4\tau^{2}}}.$ (22)

###  B.5 Asymptotic approximation

Now that we computed $I_{k}$, we can go back to the pairwise correlation.
Since the sum $S_{k}^{(n)}$ of Equation (20) converges toward the integral
$I_{k}$, we have for $\sigma_{k}^{(n)}$, using Equations (17) and (20),

$\sigma_{k}^{(n)}=\frac{q_{k}^{(n)}}{n}=\frac{S_{k}^{(n)}}{n\theta_{n}}=\frac{S_{k}^{(n)}}{2\pi}\stackrel{{\scriptstyle
n\to\infty}}{{\to}}\frac{I_{k}}{2\pi}=\frac{\alpha^{k}}{\sqrt{1-4\tau^{2}}}$
(23)

and for $\omega_{k}^{(n)}$, using Equations (18) and (20),

$\omega_{k}^{(n)}=\frac{q_{k}^{(n)}}{q_{0}^{(n)}}=\frac{S_{k}^{(n)}}{S_{0}^{(n)}}\stackrel{{\scriptstyle
n\to\infty}}{{\to}}\frac{I_{k}}{I_{0}}=\alpha^{k}.$ (24)

Since $k=|j-i|$, we can conclude that

$\Omega_{ij}^{(n)}\stackrel{{\scriptstyle n\to\infty}}{{\to}}\alpha^{|j-i|}.$

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