HOW TO MODEL THE COVARIANCE STRUCTURE IN A
SPATIAL FRAMEWORK: VARIOGRAM OR CORRELATION
FUNCTION?

arXiv:1503.07686v3 [] 9 Dec 2016

GIOVANNI PISTONE AND GRAZIA VICARIO

Abstract. The basic Kriging’s model assumes a Gaussian distribution
with stationary mean and stationary variance. In such a setting, the joint
distribution of the spatial process is characterized by the common variance
and the correlation matrix or, equivalently, by the common variance and
the variogram matrix. We discuss in in detail the option to actually use
the variogram as a parameterization.

Kriging, Universal Kriging, Non-parametric Kriging, Elliptope, Bayes
1. Introduction
The present note is a development of the Author’s paper [7]. An application
was implemented in Vicario et al. paper [12]. In the interest of clarity we allow
here for a little overlap with the papers referred to above.
We discuss the notion of variogram as it is used in Geostatistics and we offer
some preliminary thought about the possibility of a non-parametric approach
to Universal Kriging aiming to the use of a Bayes methodology. Variograms
are due to Matheron [6], and there are many modern expositions i.e., Cressie
[3, Ch. 2], Chiles and Delfiner [2, Ch. 2], Gneiting et al. [5], Gaetan and
Guyon [4, Ch.1].
In Sec. 2 we give a brief overview of the so called Universal Kriging model
and its parameterization with the Matheron’s variogram function. In Sec. 3
we define the general variogram matrix and give a necessary and sufficient
condition for a positive variance σ 2 and a matrix Γ to be a variogram matrix
of a covariance σ 2 R, where R is a correlation matrix. In Sec. 4 we provide
some useful computations concerning the inverse variogram matrix. Sec 5 is
devoted to an interpretation of the variogram matrix as related to a projection
of the Gaussian field. In Sec. 6 we discuss the shape of the set of parameters
of the general Kriging model. A section of conclusions closes the paper.
2. Universal Krige setup
We consider a Gaussian n-vector Y , n ≥ 2, whose mean has the form
µ = µ1 and whose covariance matrix Σ = [σij ]nij=1 has constant diagonal
σii = σ 2 , i = 1, . . . , n. The assumption on the mean and the diagonal terms
is the weakest stationarity assumption, i.e. a 1-st order stationarity. We can
write Y ∼ Nn (µ1, σ 2 R), where µ is a general mean value, σ 2 the common
variance, and R = [ρij ]ni,j=1 a generic correlation matrix.
1

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PISTONE AND VICARIO

The variogram of Y is the n × n matrix Γ = [γij ]ni,j=1 whose element γij
is half the variance of the difference Yi − Yj . As the mean value is constant,
the variance of the difference is equal to the second moment of the difference.
It is expressed in terms of the common variance σ 2 and the correlations ρij ,
i, j = 1, . . . , n, as
2γij = Var (Yi − Yj ) = σ 2 (ei − ej )0 R(ei − ej ) =
σ 2 (ρii + ρjj − 2ρij ) = 2σ 2 (1 − ρij ) ,
and, in matrix form, as
Γ = σ 2 (110 − R).
The simple Gaussian model described above is commonly used in Geostatistics, when each random component Yi of the random vector Y is associated
to a location xi , i = 1, . . . , n, in a given region X, xi ∈ X, i = 1, . . . , n.
We briefly describe the most common set-up in Geostatistics. The elements
of the variogram matrix γij are assumed to be a given function γ of the distance
between two locations, γi,j = γ(d(xi , xj )). In such a case the statistical model
is characterized by the choice of a distance d(x, y), x, y ∈ X, and by the choice
of a function γ, called variogram function, defined on a real domain containing
{d(x, y)|x, y ∈ X} . The existence of a positive σ 2 and a correlation matrix
R = [ρij ] such that σ 2 (1 − ρij ) = γ(d(xi , xj )) imposes a nontrivial condition
on the function γ, see Sasvári [11] and Gneiting et al. [5] and []. Such a
model, where it is assumed that the vector of means is constant, µ = µ1 is
called universal Krige model. We do not consider the more general case of a
non-constant mean.
Krige has further qualified this model by adding assumptions on the variance function and suggesting a statistical method to estimate the value at
an untried point x0 given a set of observation Y1 , . . . , Yn at points x1 , . . . , xn .
Precisely:
(1) Krige’s modeling idea is to assume the variogram function γ to be
an increasing function on [0, ∞[, so that the variogram’s values are
increasing with the distance. Moreover, the correlation between locations is assumed to be positive. The rational is to model a variability
which increases with the distance and is bounded by a general variance:
1
Var (Yi − Yj ) = γij = γ(d(xi , xj )) = σ 2 (1 − ρij ) ≤ σ 2 .
2
The increasing function γ : R+ → R+ is assumed to be continuous
everywhere but at 0. As it is bounded at +∞, the general shape is as
in Fig. 1.
The parameters in the Krige’s universal model are unrestricted values of µ ∈ R, σ 2 > 0, and restricted values of R that are usually
estimated over a suitable parametric model.
(2) Krige’s idea is to predict the value Y0 = Yx0 at an untried location
x0 with the conditional expectation based on a plug-in estimate of
the parameters. If I = {1, . . . , n} and the locations in the model are
0≤

MODELLING COVARIANCES WITH VARIOGRAMS

3

1,4

Sill

1,2

variogram

1
0,8
0,6
0,4

Range

0,2

Nugget effect
0
0

1

2
distance

3

4

Figure 1. A general variogram function is 0 at 0, can have
jump at 0 which is called Nugget, has a finite limit at +∞
named Sill. The Range is a length such that the value is equal
to the limit value for any practical purpose.
x0 , x1 , . . . , xn , the regression value is
Y\
0−µ =

Σ0,I Σ−1
I,I (YI



− µ1I ),

ΣI,I ΣI,0
with Σ =
Σ0,I σ 2 .



The set of data that give the same prediction is an affine plane in Rn .
The variance of the prediction is σ02 − Σ0,I Σ−1
I,I ΣI,0 .
In this paper we do not follow this approach, but we adopt a general nonparametric attitude, where µ is real number, σ 2 is a positive real number, R
is a positive definite matrix with unit diagonal, possibly with positive entries.
The variogram matrix Γ is not restricted and we do not enforce the existence
of any special form.
3. The variogram matrix
Our plan now is to express the Krige’s computations in term of Matheron’s
variogram matrix Γ. One good reason to use Γ as a basic parameter is because
its empirical estimator is unbiased.
Let us discuss in some detail the basic transformation of matrix parameters
(1)

Γ = σ 2 (110 − R) = σ 2 110 − Σ .

We note that Γ = 0 if, and only if, R = 110 , such extreme case being always
excluded in the following. In fact, in most cases we will assume det R 6= 0.
The entries of Γ are nonnegative and bounded by 2σ 2 because the correlations are bounded between −1 and 1. If all the correlations are nonnegative,
the the entries of Γ are bounded by σ 2 .

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PISTONE AND VICARIO

The difference between the covariance matrix and the variogram matrix is
a matrix of rank 1, Γ + Σ = σ 2 110 . Let us remark moreover that
1 0
1
11 =
(Γ + Σ)
n
nσ 2
is the orthogonal projector on the space of constant vectors Span (1).
We denote by S= the cone of nonnegative definite matrices with constant
diagonal, by S=1 the convex set of correlation matrices, and by V the cone of
variogram matrices. We have the following characterization of V.
(2)

Proposition 1. A nonzero matrix Γ is the variogram matrix of some covariance matrix of the form Σ = σ 2 R, with σ 2 > 0 and R a correlation matrix, if,
and only if, the three following conditions hold
(1) Γ is symmetric, and has zero diagonal;
(2) Γ is conditionally negative definite, i.e. wΓw ≤ 0 if w0 1 = 0;
(3) sup {x0 Γx|x0 1 = 1} ≤ σ 2 .
Proof. Assume Γ = σ 2 (110 − R) 6= 0 with R correlation matrix and σ 2 > 0.
Condition 1 follows from the definition. If we write a generic vector as x =
w + α1 with w0 1 = 0, we have
n2 σ 2 α2 = x0 Γx + x0 Σx .
In particular, Condition 2 follows because α = 0 implies x0 Γx = −x0 Σx.
Finally, if x0 1 = 1, that is nα = 1, we have x0 Γx − σ 2 = x0 Σx ≥ 0 and
Condition 3 follows.
Viceversa, consider the matrix 110 − σ −2 Γ. It is symmetric, with unit
diagonal. We need only to show it is positive definite:
x0 (110 − σ −2 Γ)x = (x0 1)2 − σ −2 x0 Γx =
0

2

(x 1) (1 − σ

−2

because of Condition 3.



0 

1
1
x Γ
x ≥0
x0 1
x0 1


The lower bound imposed on σ 2 means that the parameterization with σ 2 ,
carrying one degree of freedom, and with Γ, carrying n(n − 1)/2 degrees of
freedom, has a drawback in that the two parameters are not independently
defined on a product set. Note the relation 10 Γ1 = σ 2 (n2 − 10 R1) that we are
going to discuss in Sec. 4 below.
In conclusion, there is a one-to-one transformation of parameters
(σ 2 , Γ) ↔ (σ 2 , R) ↔ Σ
with σ 2 ∈ R> , Γ ∈ V, R ∈ S=1 , Σ ∈ S= , namely:
(1) The mapping from Σ ∈ S= to the couple (σ 2 , Γ) ∈ R> × V factors as

−2
1
1
Tr (Σ)
Σ) = (σ 2 , R) ∈]0, ∞[×R
S= 3 Σ 7→ ( Tr (Σ) ,
n
n

MODELLING COVARIANCES WITH VARIOGRAMS

5

and
]0, ∞[×R 3 (σ 2 , R) 7→ (σ 2 , σ 2 (110 − R)) =

(σ 2 , Γ) ∈ (σ 2 , Γ) Γ ∈ V, sup {x0 Γx|x0 1 = 1} ≤ σ 2 .
(2) Inverse is
 2
(σ , Γ) Γ ∈ V, sup {x0 Γx|x0 1 = 1} ≤ σ 2 3 (σ 2 , Γ) 7→
σ 2 110 − Γ = Σ ∈ S=
4. Inverse variogram matrix Γ−1
The equations two above, both based on the definition of the variogram
matrix as Γ = σ 2 (110 − R), provide a simple connection between the parameterization based on the covariance matrix Σ and the parameterization based
on the couple σ 2 and Γ. However, we want to spell out the computation of
an other key statistical parameter, namely the concentration matrix Σ−1 . We
begin by recalling a well known equation in matrix algebra [8]. We review the
result in detail as we need an exact statement of the conditions under which
is true in our case.
Proposition 2 (Sherman-Morrison formula). Assume the matrix A is invertible. The matrix 110 − A is invertible if, and only if, 10 A−1 1 6= 1. In such a
case,
det(110 − A) = (−1)n (1 − 10 A−1 1) det A ,
(110 − A)−1 = −A−1 − (1 − 10 A−1 1)−1 A−1 110 A−1 .
Proof. The multi-linear expansion of det(110 − A) is written in terms of the
adjoints (−A)ij of each element (−A)ij by
n X
n
X
det(11 − A) = det (−A) +
(−A)ij
0

j=1 i=1
n

n−1

= (−1) det A − (−1)

n
X

Aij

i,j=1

= (−1)n det A − (−1)n−1 1(adj A)10 .
As det A 6= 0, we can factor-out (−1)n (det A) to get
det(110 − A) = (−1)n (det A)(1 − 10 A−1 1)
and the statement about the determinant follows. The inversion formula is
directly checked.

We are concerned with the invertibility of Γ = σ 2 (110 − R), hence we need
to discuss the condition 10 R−1 1 6= 1.
Proposition 3. Let R be a correlation matrix and assume det R 6= 0. Let
λj > 0, j = 1, . . . , n, be the spectrum of R and let uj be a set of unit eigenvectors. It can be proved:

6

PISTONE AND VICARIO

P
Q
(1) Tr R = nj=1 λj = n and det R = nj=1 λj ≤ 1, with equality if, and
only if, R P
= I.
(2) Tr R−1 = nj=1 λ−1
j ≥ n with equality if, and only if, R = I.
0 −1
(3) 1 R 1 6= 1.
P
Q
Proof.
(1) n = Tr R = nj=1 λj . From det R = nj=1 λj , as the arithmetic
mean is larger than the the geometric mean,
! n1
Pn
n
Y
1
j=1 λj
≥
λj
1=
= (det R) n ,
n
j=1
with equality if, and only if the λj ’s are all equal, hence equal to 1,
which happens if R = I.
(2) The geometric mean is larger or equal than the harmonic mean, hence
! n1
!−1
n
n
Y
X
1
1 ≥ (det R) n =
λj
≥n
λ−1
,
j
j=1

j=1

P
with equality if, and only if, λj = 1, j = 1, . . . , n. It follows n1 nj=1 λ−1
j ≥
1.
P
0
(3) We derive a contradiction from 1 = 10 R−1 1. As R−1 = nj=1 λ−1
j uj uj
Pn
and j=1 (10 uj )2 = k1k2 = n2 ,
0

−1

1=1R 1=

n
X

0
2
λ−1
j (1 uj )

n
X
=n
(λj )−1 θj ,
2

j=1

where θj = (10 u)2 /n2 ≥ 0 and
λ 7→ λ−1 we obtain

j=1

Pn

j=1 θj

= 1. From the convexity of

n
n
X
X
−1
2
1=n
(λj ) θj ≥ n
λj θj
2

j=1

!−1
,

j=1

hence the contradiction
n
1 X
1
1
1≤ 2
λj θj ≤ 2 max {λj |j = 1, . . . , n} ≤ .
n j=1
n
n

From Proposition 3 we have immediately the following result of interest.
Proposition 4. Assume the correlation matrix Σ = σ 2 R ∈ S= is invertible.
It follows that Γ = σ 2 (110 − R) is invertible, with
(3)

Γ−1 = −Σ−1 − (σ −2 − 10 Σ−1 1)−1 Σ−1 110 Σ−1

and
(4)

Σ−1 = −Γ−1 − (σ −2 − 10 Γ−1 1)−1 Γ−1 110 Γ−1

MODELLING COVARIANCES WITH VARIOGRAMS

7

Proof. From the assumption it follows det R 6= 0 so that
σ −2 − 10 Σ−1 1 = σ −2 (1 − 10 R1) 6= 0 ,
hence the conclusion.



We can now analyze the likelihood of the Gaussian model N(µ1, σ 2 R) in
terms of the variogram.
First, we compute the determinant of the correlation matrix


det σ 2 R = det σ 2 110 − Γ

= σ 2n det 110 − σ −2 Γ


 
= σ 2n det −σ −2 Γ + 10 adj −σ −2 Γ 1
= det (−Γ) − σ 2 10 adj (−Γ) 1 .
Second, we compute the quadratic form of the concentration matrix

y 0 Σ−1 y = y 0 −Γ−1 − (σ −2 − 10 Γ−1 1)−1 Γ−1 110 Γ−1 y
= −y 0 Γ−1 y − (σ −2 − 10 Γ−1 1)−1 (y 0 Γ−1 1)2 .
Third, we compute the log-likelihood with µ = 0:

log p y σ 2 , Γ =

n
1
1
− log (2π) − log det 110 − σ −2 Γ − 2 y 0 (110 − σ −2 Γ)−1 y =
2
2
2σ

n
1
− log (2π) − log det (−Γ) − σ 2 10 adj (−Γ) 1
2
2
1
1
+ y 0 Γ−1 y + (σ 2 − 10 Γ−1 1)−1 (y 0 Γ−1 1)2 .
2
2
Here are the essentials of the computations leading to a maximum likelihood
estimation of Γ are the following. In the direction of a generic symmetric
matrix with zero diagonal H,


dH (Γ 7→ log det 110 − σ −2 Γ ) = Tr (σ 2 110 − Γ)−1 H
and
dH (Γ 7→ y 0 (110 − σ −2 Γ)−1 y)
= σ 2 Tr (σ 2 110 − Γ)−1 yy 0 (σ 2 110 − Γ)−1 H



.

so that the normal equations for Γ reduce to the condition
−(σ 2 110 − Γ)−1 + (σ 2 110 − Γ)−1 yy 0 (σ 2 110 − Γ)−1

is diagonal .

The approach with parameters σ 2 , Γ is feasible in principle, but it does not
appear promising in term of ease of computation. We will see in Sec. 5 below
a different, possibly better, approach.

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PISTONE AND VICARIO

5. Projecting on Span (1)⊥
We now change our point of view to consider the same problem from a
different angle suggested by the observation the variogram does not change if
we change the general mean µ. In fact, we can associate the variogram with
the state space description of the Gaussian vector.
The following proposition is a new characterization of the variogram matrix
in our setting.
Proposition 5.
(1) The matrix Γ is a variogram matrix of a covariance
matrix Σ ∈ V+ if, and only if, the matrix

0 

1 0
1 0
(5)
Σ0 = − I − 11 Γ I − 11
n
n
is symmetric, positive definite and with constant diagonal.
(2) If Y0 ∼ Nn (0, Σ0 ), then its variogram is Γ and Y0 is supported by
Span (1)⊥ .
Proof.

(1) If Γ = σ 2 (110 − R) is the variogram matrix of Σ = σ 2 R, then
from Eq. (5) we have

0 

1 0
1 0
2
Σ0 = σ I − 11 R I − 11 ,
n
n
which is indeed positive definite. Let us show that the diagonal elements of Σ0 are constant.

0 

1 0
1 0
2 0
(Σ0 )ii = σ ei I − 11 R I − 11 ei
n
n
0 


1
1
2
= σ ei − 1 R ei − 1
n
n


2
1 0
2
0
= σ ei Rei − ei R1 + 2 1 R1
n
n


1 0
= σ2
1 R1 − 1
n2
Viceversa, assume Σ0 is a covariance matrix. As ei −ej ∈ Span (1)⊥ ,
the variogram of Σ0 has elements

(ei − ej )0 Σ0 (ei − ej ) =

0


1 0
1 0
0
(ei − ej ) I − 11 (−Γ) I − 11 (ei − ej ) =
n
n
0
− (ei − ej ) Γ(ei − ej ) = −γii − γjj + 2γij = 2γij .
0

(2) As 10 (ei − ej ) = 0, then 10 I − n1 110 (−Γ) I − n1 110 1 = 0, hence
the distribution of Y0 is supported by the space Span (1)⊥ .


MODELLING COVARIANCES WITH VARIOGRAMS

9

It is possible to split every Gaussian Y with covariance matrix Σ ∈ S=
according the splitting Rn = Span (1) ⊕ Span (1)⊥ . The corresponding projections split the Gaussian process Y into two components, one with the covariance as in Proposition 5, the other proportional to the empirical mean.
Note that the two components have a singular covariance matrix.
Proposition 6. Let Y ∼ Nn (µ,
Σ), Σ = σ 2 R ∈ S+ with variogram Γ =

σ 2 (110 − R). Let Yb = I − n1 110 Y ∼ Nn (0, Σ0 ) be the projection of Y onto
Span (1)⊥ so that we can write Y = Yb + Y , where each component of Y is the
empirical mean n1 10 Y .
(1) The distribution of Yb depends on the variogram only,
0 



1 0
1 0
b
,
Y ∼ Nn 0, − I − 11 Γ I − 11
n
n
and the variogram matrix of Yb is Γ.
(2) The distribution of 1 10 Y , conditionally to Yb is Gaussian with mean µ.
n

Proof.

0 
 
0 

1 0
1 0
1 0
1 0
− I − 11 Γ I − 11 = I − 11 Σ I − 11 .
n
n
n
n

This suggests the following empirical estimation algorithm.
(1) Project the independent sample data y1 , .P
. . , yN onto Span (1)⊥ by
subtracting the empirical mean y+ = n1
yi , to get yb1 = y1 −
b
y+ , . . . , yN = yN − y+ . Use the empirical estimator of the variogram
matrix on the projected data.
(2) Estimate µ with the empirical mean.
In the same spirit, we could suggest the simulation of a random variable
with variogram matrix Γ by the generation of Nn (0, Σ0 ) data in Span (1)⊥ .
Both suggestions will be further discussed in future work.
6. Elliptope
We now turn to the geometrical description of the set of variograms. From
the basic equation Γ = σ 2 (110 −R) it follows that the set of variogram matrices
is an affine image of the set of correlation matrices in the space of symmetric
matrices. The set of correlation matrices is a convex bounded set whose
geometrical shape has been studied in a number of papers, i.e. [10], [10], [9].
Such a shape is of central interest in a non parametric Bayesian approach
to the statistics of the universal Kriging model. It appears also in convex
optimization, where it has been called elliptope.
Let us discuss the case n = 3. All principal minors of R are nonnegative,


1 x y
det (R) = det x 1 z  = 1 − x2 − y 2 − z 2 + 2xyz ≥ 0
y z 1

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PISTONE AND VICARIO

Figure 2. The
3-elliptope

Figure
3. Algebraic
variety

and 1 − x2 , 1 − y 2 , 1 − z 2 ≥ 0. The three last inequalities define the cube
Q = [−1, +1]3 while the equation 1 − x2 − y 2 − z 2 + 2xyz = 0 is a cubic
algebraic variety whose intersection with the cube Q is the border of the
elliptope. All horizontal section z = c, −1 ≤ c ≤ 1, of the elliptope are the
interior of the ellipses
1 − x2 − y 2 + 2cxy ≥ c2
Same for other sections. See Fig.s 2 and 3.
Various proposals of apriory distribution on the elliptope exist, see for example [1].
Going on with the discussion of our example, the volume is easily computed
and the uniform apriori is defined. Simulation is feasible for example by
rejection method. An other option is to write R = A0 A where the columns of A
are unit vectors. This gives an other possible apriori starting from independent
unit vectors. Simulation is feasible, for example starting with independent
standard Gaussians.
An interesting option is the Cholesky representation. A symmetric matrix
A is positive definite if there exists an upper triangular matrix
 0 

t1
t11 t12 t13
T = t02  =  0 t22 t23  , tii ≥ 0
t03
0 0 t33
such that



t211
t11 t12
t11 t13
t212 + t222
t12 t13 + t22 t23 
A = T 0 T = [ti · tj ]ij = t11 t12
t11 t13 t12 t13 + t22 t23 t213 + t223 + t233
Moreover, t11 t22 t33 6= 0 ⇔ T is unique and invertible if, and only if, A is
invertible. It is an identifiable parameterization for non singular matrices.

MODELLING COVARIANCES WITH VARIOGRAMS

11

In the case of a the correlation matrix R = T 0 T with
 0  p

t13
t1
1 − t212 − t213 p t12
i−1
+
T = t02  = 
0
1 − t223 t23  , ti ∈ 0 × Sn−1+1
0
t3
0
0
1
It follows
p
p


2
2
2
2
1
t
−
t
1
−
t
1 − tp
12
13
12
12 − t13 t13
p
R = p1 − t212 − t213 t12
t12 t13 + 1 − t223 t23 
p1
1
1 − t212 − t213 t13 t12 t13 + 1 − t223 t23
and
det (R) = (1 − t212 − t213 )(1 − t223 )
7. Conclusion
In the last decades Kriging models have been recommended not only for the
original application, but spatial noisy data in general. Thanks to the availability of comprehensive computing facilities and the recent progresses in software
development, the numerical simulation of technologically complex systems has
become an attractive alternative option to the physical experimentation. The
most popular meta-model used when dealing with Computer Experiments
(CE) is the Kriging model. The accuracy of this model strongly depends on
the detection of the correlation structure of the responses. In the Bayesian
approach, where the posterior distribution of a prediction Krige’s Y0 given the
training set (Y1 , . . . , Yn ) requires less uncertainty as possible on the correlation function, the use of the variogram as a parameter should be preferred
because it does not demand a parametric approach as the correlation estimation does. The authors proved in a previous paper [7] the equivalence between
variogram and spatial correlation function for stationary and intrinsically stationary processes. Here the study has been devoted to the characterization
of matrices which are admissible variograms in the case of 1-stationarity. We
expect these findings will allow for the imputation of an apriori distribution
on the set of variogram matrices, as Bayesians do for the correlation in the
Kriging modernization.
Acknowledgments
An early version of this paper was presented at the 4th Stochastic Modeling Techniques and Data Analysis International Conference, June 1–4, 2016,
University of Malta, Valletta, Malta, with the title Bayes and Krige: Generalities. The Authors thank both Guillaume Kon Kam King (CCA, Moncalieri)
and Luigi Malagò (CCA, Moncalieri) for suggesting relevant references. We
thank Emilio Musso (DISMA, Politecnico di Torino) for help with the elliptopes picture. G. Pistone is supported by de Castro Statistics, Collegio Carlo
Alberto, Montalieri, and he is a member of GNAFA-INDAM.

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PISTONE AND VICARIO

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G.Pistone: de Castro Statistics, Collegio Carlo Alberto, Via Real Collegio 30, 10024 Moncalieri, Italy (E-mail: giovanni.pistone@carloalberto.org).
G. Vicario: DISMA Luigi Lagrange, Politecnico di Torino, Corso Duca degli
Abruzzi 24, 10124 Torino, Italy (E-mail: grazia.vicario@polito.it)

