2. A new inequality index for multivariate distributions

The goal of this section is to present in some details a novel inequality index, which can be applied to measure the heterogeneity of multivariate distributions [Reference Giudici, Raffinetti and Toscani22]. This index is obtained by suitably generalising a new one-dimensional index introduced in ref. [Reference Toscani35].

In what follows, for a given $1\le n \in \mathbb{N}$ , we denote by $P_s(\mathbb{R}^n)$ , $s \ge 1$ , the class of all probability measures $F = F(\textbf{x})$ on the Borel subsets of $\mathbb{R}^n$ such that

\begin{equation*} m_s(F) = \int _{\mathbb {R}} |\textbf{x}|^s dF(\textbf{x}) \lt + \infty, \end{equation*}

where, for a given column vector $\textbf{x}$ of dimension $n$ , ${\textbf{x}}^T = (x_1,x_2,\dots,x_n)$ is a point in $\mathbb{R}^n$ , and $ |\textbf{x}| =\sqrt{\textbf{x}^T \textbf{x}}$ is the modulus of the vector, i.e. the distance of the point $\textbf{x}$ from the origin of the cartesian axes.

Further, we denote by $ \tilde{P}_s (\mathbb{R}^n)$ the class of probability measures $F \in P_s(\mathbb{R}^n)$ which possess a mean value vector $\textbf{m}$ with positive components $m_k$ , $k =1,2,\dots, n$ , i.e.

\begin{equation*} {m}_k (F) = \int _{\mathbb {R}^n } x_k \, dF(\textbf{x}) \gt 0, \quad k =1,2,\dots, n, \end{equation*}

and with $P_s^+(\mathbb{R}^n)$ the subset of probability measures $F \in P_s(\mathbb{R}^n)$ such that $F(\textbf{x}) = 0$ if at least one component $x_k \le 0, k =1,2,\dots, n$ .

On the set $\tilde{P}_s(\mathbb{R}^n)$ of probability measures, we consider the set $\mathcal F_s^n$ of their $n$ -dimensional Fourier transforms, where, for $F=F(\textbf{x}) \in \tilde{P}_s(\mathbb{R}^n)$ ,

(2.1) \begin{equation} \widehat{f}(\boldsymbol{\xi }) = \int _{\mathbb{R}^n} e^{-i \textbf{x}^T\boldsymbol{\xi }} \, dF(\textbf{x}). \end{equation}

In equation (2.1), we denoted by $\boldsymbol{\xi }$ the $n$ -dimensional column vector of components $\xi _k, k =1,2,\dots, n$ .

When $n =1$ , in alternative to well-known inequality indices, for a given distribution $F \in P_s(\mathbb{R})$ , the following measure of heterogeneity was proposed in ref. [Reference Toscani35]

(2.2) \begin{equation} T(F) = \frac 1{2m} \sup _{\xi \in \mathbb{R}} \left | \left. \frac{d\widehat{f}(\xi )}{d\xi }\right |_{\xi =0} \widehat{f}(\xi ) - \frac{d\widehat{f}(\xi )}{d\xi } \right |. \end{equation}

In definition (2.2), $m\gt 0$ denotes the mean value of the distribution $F$ .

Apparently, the index (2.2) is not clearly connected with others most used indices, including the well-known Gini index [Reference Gini20, Reference Gini21], strongly related to Lorenz curve [Reference Eliazar17]. However, looking at Gini index from the Fourier transform side, an interesting relationship appears.

Let us consider a probability measure $F \in P_s^+(\mathbb{R})$ , of mean value $m\gt 0$ . As shown in ref. [Reference Toscani35], the classical Gini index

(2.3) \begin{equation} G(F) = 1 - \frac 1m\int _{\mathbb{R}_+} (1- F(x))^2\, dx. \end{equation}

can be expressed in terms of one-dimensional Fourier transform as follows:

(2.4) \begin{equation} G(F) = 1 - \frac 1{2\pi m} \int _{\mathbb{R}} \frac{|1 -\widehat{f}(\xi )|^2}{|\xi |^2} \, d\xi. \end{equation}

Expression (2.4) clarifies that the Fourier expression of the classical Gini index is a function of a certain distance between probability measures $F$ and $G$ [Reference Torregrossa and Toscani34], namely

\begin{equation*} d_2(F,G) = \int _{\mathbb {R}} \frac {|\widehat {f}(\xi )-\widehat {g}(\xi )|^2}{|\xi |^2} \, d\xi. \end{equation*}

Resorting to this analogy, in ref. [Reference Toscani35], new inequality measures have been introduced, some of them related to the supremum distance

(2.5) \begin{equation} d_\infty (F,G) = \sup _{\xi \in \mathbb{R}} \frac{|\widehat{f}(\xi )-\widehat{g}(\xi )|}{|\xi |}. \end{equation}

This type of metrics have been extensively studied in connection with the convergence to equilibrium of kinetic equations, as alternatives to more classical entropies [Reference Carrillo10, Reference Gabetta, Toscani and Wennberg19, Reference Torregrossa and Toscani34].

Let $X$ be a random variable, of mean value $m\gt 0$ characterised by a differentiable probability measure $F \in P_s^+(\mathbb{R})$ , and denote by $f(x) = dF(x)/dx$ its probability density. In this case, in addition to the classic expression (2.3), Gini index can be expressed in alternative forms, one of which is particularly interesting to enlighten contact points with the $T$ -index defined by (2.2). This alternative expression reads

(2.6) \begin{equation} G(F) = 2 \int _{\mathbb{R}_+} (1- F(x))\left ( f(x) -\frac xmf(x)\right )\, dx. \end{equation}

Indeed,

\begin{equation*} \int _{\mathbb {R}_+} (1- F(x)f(x) \, dx = - \frac 12 \int _{\mathbb {R}_+} \frac d{dx}(1-F(x)^2\, dx = - \frac 12 \left .(1-F(x)^2\right |_0^\infty =\frac 12, \end{equation*}

while, integrating by parts

\begin{equation*} \int _{\mathbb {R}_+} (1- F(x)f(x)\frac xm \, dx = - \frac 12 \int _{\mathbb {R}_+}\frac xm\, \frac d{dx}(1-F(x)^2\, dx = \frac 1{2m} \int _{\mathbb {R}_+}(1-F(x)^2\, dx. \end{equation*}

Consequently, if we set $H(x) = 1-F(x)$ , thanks to Plancherel identity we obtain

(2.7) \begin{align} G(F) = & 2 \int _{\mathbb{R}_+} (1- F(x))\left ( f(x) -\frac xmf(x)\right )\, dx = \nonumber \\[5pt] & \frac 1{m\pi }\int _{\mathbb{R}} \overline{\widehat{H}(\xi )} \left ( \left. \frac{d\widehat{f}(\xi )}{d\xi }\right |_{\xi =0} \widehat{f}(\xi ) - \frac{d\widehat{f}(\xi )}{d\xi } \right ), \end{align}

where $\overline{\widehat{H}(\xi )}$ is the complex conjugate of the Fourier transform of $1-F(x)$ , given by

\begin{equation*} H(\xi ) = \frac {1 -\widehat {f}(\xi )}{i\xi }. \end{equation*}

Consequently, the value of the Gini index depends on the product of two different quantities, working in opposite directions. Indeed, according to (2.5), the term $\overline{\widehat{H}(\xi )}$ quantifies the distance of $\widehat{f}(\xi )$ from the state of perfect inequality, represented by the value $1$ , the Fourier transform of a Dirac delta function located in $x=0$ . On the contrary, the term

\begin{equation*} \left. \frac {d\widehat {f}(\xi )}{d\xi }\right |_{\xi =0} \widehat {f}(\xi ) - \frac {d\widehat {f}(\xi )}{d\xi }, \end{equation*}

that vanishes in correspondence to $e^{-im\xi }$ , the Fourier transform of a Dirac delta function localised in the mean value $m$ of $f(x)$ , quantifies the distance of $\widehat{f}(\xi )$ from the state of perfect equality. This clarifies both the nonlinearity of Gini index and the advantages of the choice of the inequality index (2.2) as alternative measure of the heterogeneity of the distribution. A further advantage is represented by the possibility to easily extend the measure (2.2) to higher dimensions.

Following [Reference Giudici, Raffinetti and Toscani22], we introduce on $\mathcal F_s^n$ the multivariate inequality index $T_n(F)$ , expressed by the formula

(2.8) \begin{equation} T_n(F) = \frac 1{2|\textbf{m}|} \sup _{\boldsymbol{\xi } \in \mathbb{R}^n} \left | \nabla \widehat{f}(\boldsymbol{\xi } = \textbf{0}) \widehat{f}(\boldsymbol{\xi }) - \nabla \widehat{f}(\boldsymbol{\xi }) \right |. \end{equation}

In definition (2.8), $\nabla \widehat{f}(\boldsymbol{\xi })$ denotes the gradient of the scalar function $\widehat{f}(\boldsymbol{\xi })$ . Indeed, $F \in P_s(\mathbb{R}^n)$ implies that $\widehat{f}(\boldsymbol{\xi })$ is continuously differentiable.

It is immediate to show that the functional $T_n(F)$ is invariant with respect to the scaling (dilation)

\begin{equation*} F(\textbf{x}) \to F(c\textbf{x}), \quad c \gt 0. \end{equation*}

However, as one can easily verify by direct inspection, in the multivariate setting the invariance holds under a scaling transformation like $A\textbf{x}^T$ , where $A$ is a square orthogonal matrix.

Moreover, as shown in ref. [Reference Toscani35] for the one-dimensional index, $T_n$ is bounded from above by $1$ . Indeed, since for any given $F \in P_s^+(\mathbb{R}^n)$ it holds $|\widehat{f}(\boldsymbol{\xi })| \le \widehat{f}(\textbf{0}) = 1$ , and

\begin{equation*} \frac {\partial \widehat {f}(\boldsymbol {\xi })}{\partial \xi _k} = -i \int _{(\mathbb {R}_+)^n} x_k e^{-i \textbf{x}\boldsymbol {\xi }} \, dF(\textbf{x}), \quad k =1,2,\dots, n, \end{equation*}

one obtains the bound

(2.9) \begin{equation} \left | \frac{\partial \widehat{f}(\boldsymbol{\xi })}{\partial \xi _k}\right | \le \int _{(\mathbb{R}_+)^n} x_k \left |e^{-i \textbf{x}\boldsymbol{\xi }}\right | \, dF(\textbf{x}) =m_k, \end{equation}

which implies $|\nabla \widehat{f}(\boldsymbol{\xi })| \le |\nabla \widehat{f}(\boldsymbol{\xi } =\textbf{0})| = |\textbf{m}|$ .

Hence, by the triangular inequality one concludes that $T_n(F)$ satisfies the usual bounds

(2.10) \begin{equation} 0 \le T_n(F) \le 1, \end{equation}

and $T_n(F) = 0$ if and only if $\widehat{f}(\boldsymbol{\xi })$ satisfies the differential equations

\begin{equation*} \frac {\partial \widehat {f}(\boldsymbol {\xi })}{\partial \xi _k} = \left. \frac {\partial \widehat {f}(\boldsymbol {\xi })}{\partial \xi _k}\right |_{\boldsymbol {\xi } = \textbf{0} } \widehat {f}(\boldsymbol {\xi }), \quad k =1,2,\dots, n, \end{equation*}

with $\widehat{f}(\textbf{0}) = 1$ .

Thus, as in the one-dimensional case, $T_n(F)$ vanishes if and only if $\widehat{f}(\boldsymbol{\xi }) = e^{-i\textbf{m}\boldsymbol{\xi }}$ , namely if $\widehat{f}(\boldsymbol{\xi })$ is the Fourier transform of a Dirac delta function located in the point $\textbf{x} =\textbf{m}^T(F)$ . Note however that, even if the functional $F$ is defined in the whole class $\tilde{P}_s(\mathbb{R}^n)$ , the upper bound is lost if the probability measure $F \notin P_s^+(\mathbb{R}^n)$ , since in this case inequality (2.9) is no more valid.

It is remarkable that the functional $T_n(F)$ defines a measure of inequality for multivariate densities which satisfies most of the properties satisfied by its one-dimensional version $T(F)$ .

Proceeding as in the one-dimensional case, we can check that the upper bound in equation (2.10) is reached simply by evaluating the value of the multivariate index in correspondence to a multivariate random variable $\textbf{X}$ taking only the two values $\textbf{a}= (a_1,a_2, \dots, a_n)$ and $\textbf{b}= (b_1,b_2,\dots, b_n)$ in $\mathbb{R}^n$ with probabilities $1-p$ and, respectively $p$ , where $0\lt p\lt 1$ . As we will see in Section 3, this example clarifies the advantages in measuring multidimensional heterogeneity by means of this new index, with respect to the use of existing generalisations of Gini index to the multidimensional setting [Reference Andreoli and Zoli5, Reference Banerjee6]. For this reason, we will give it into details.

The Fourier transform of the distribution $F$ of $\textbf{X}$ is given by

(2.11) \begin{equation} \widehat{f}(\boldsymbol{\xi }) = (1-p)e^{-i\textbf{a}\boldsymbol{\xi }} + p e^{-i \textbf{b}\boldsymbol{\xi }}. \end{equation}

Consequently,

\begin{equation*} \nabla \widehat {f}(\boldsymbol {\xi }) = -i \left [ (1-p)\textbf{a}^T e^{-i\textbf{a}\boldsymbol {\xi }} + p\textbf{b}^T e^{-i \textbf{b}\boldsymbol {\xi }} \right ] \end{equation*}

and

\begin{equation*} \nabla \widehat {f}(\boldsymbol {\xi } = \textbf{0}) = -i \left [ (1-p)\textbf{a}^T + p\textbf{b}^T \right ], \end{equation*}

so that

\begin{equation*} \nabla \widehat {f}(\boldsymbol {\xi } = \textbf{0}) \widehat {f}(\boldsymbol {\xi }) - \nabla \widehat {f}(\boldsymbol {\xi }) = i\,p(1-p) (\textbf{b}^T -\textbf{a}^T)\left [e^{-i\textbf{a}\boldsymbol {\xi }} - e^{-i \textbf{b}\boldsymbol {\xi }}\right ]. \end{equation*}

Therefore,

\begin{align*} T_n(F) =&\frac 1{2|\textbf{m}|} p(1-p) |\textbf{b}^T-\textbf{a}^T| \sup _{\boldsymbol{\xi } \in \mathbb{R}^n} \left | e^{-i\textbf{a}\boldsymbol{\xi }} - e^{-i \textbf{b}\boldsymbol{\xi }}\right | = \\[5pt] &\frac 1{2|\textbf{m}|} p(1-p) |\textbf{b}^T-\textbf{a}^T| \sup _{\boldsymbol{\xi } \in \mathbb{R}^n} \left | 1 - e^{-i (\textbf{b}-\textbf{a})\boldsymbol{\xi }}\right | = \frac 1{|\textbf{m}|} p(1-p) |\textbf{b}^T-\textbf{a}^T|. \end{align*}

Hence, expanding the value of the mean $\textbf{m}$ we get the formula

(2.12) \begin{equation} T_n(F) = \frac{p(1-p) |\textbf{b}^T-\textbf{a}^T|}{|(1-p)\textbf{a}^T + p\textbf{b}^T |}. \end{equation}

This expression has a structure that does not differ from the one-dimensional formula computed in [Reference Toscani35], that reads

\begin{equation*} T(F) = T_1(F) = \frac {p(1-p) |b-a|}{(1-p)a + p b}. \end{equation*}

In fact, when the mean is fixed, the value of the index does not depend on the positions of the two points $\textbf{a}$ and $\textbf{b}$ , but only on their distance.

To show that formula (2.12) can be used to reach the upper bound, let us now consider the case in which, for a given positive constant $\epsilon \ll 1$ , $p =\epsilon$ , the point $\textbf{a} =\textbf{0}$ , while $\textbf{b} = \textbf{m}/\epsilon$ is located far away, but leaving the mean value $\textbf{m}$ unchanged. In this case, $T_n(F) = 1-\epsilon$ , a value which, as $\epsilon \to 0$ converges to the upper bound expressed by the value $1$ .

As its one-dimensional version, we can further show that the inequality index $T_n$ satisfies a number of properties [Reference Giudici, Raffinetti and Toscani22].

Let $F,G\in \tilde{P}_s(\mathbb{R}^n)$ two probability measures with the same mean value, say $\textbf{m}$ . Then, for any given $\tau \in (0,1)$ it holds

(2.13) \begin{equation} T_n (\tau F + (1-\tau ) G) \le \tau \, T_n(F) + (1-\tau ) T_n(G). \end{equation}

Inequality (2.13) shows the convexity of the functional $T_n$ on the set of probability measures with the same mean.

Another interesting property characterising the inequality index $T_n$ is linked to its behaviour when evaluated on convolutions. Let $\textbf{X}$ and $\textbf{Y}$ independent multivariate random variables with probability measures in $\tilde{P}_s(\mathbb{R}^n)$ , and mean values $\textbf{m}_X$ (respectively $\textbf{m}_Y$ ). Then if $\widehat{f}(\boldsymbol{\xi })$ and $\widehat{g}(\boldsymbol{\xi })$ denote the Fourier transforms of their respective probability measures, the Fourier transform $\widehat{h}(\boldsymbol{\xi })$ of the distribution measure of the sum ${\textbf{X}}+ \textbf{Y}$ is equal to the product $\widehat{f}(\boldsymbol{\xi })\widehat{g}(\boldsymbol{\xi })$ .

Then, it can be shown that [Reference Giudici, Raffinetti and Toscani22]

(2.14) \begin{equation} T_n({\textbf{X}}+ \textbf{Y}) \le \frac{|\textbf{m}_{\textbf{X}}|}{|\textbf{m}_{\textbf{X}} +\textbf{m}_{\textbf{Y}}|} {\textbf{T}}_{\textbf{n}}({\textbf{X}}) + \frac{|\textbf{m}_{\textbf{Y}}|}{|\textbf{m}_{\textbf{X}} +\textbf{m}_{\textbf{Y}}|} {\textbf{T}}_{\textbf{n}}(\textbf{Y}). \end{equation}

It is remarkable that, at difference with the one-dimensional case, in equation (2.14), the sum of the coefficients in front of the inequality indices $T_n({\textbf{X}})$ and $T_n(\textbf{Y})$ is always greater that one. Nevertheless, one can extract from (2.14) some useful consequences.

In particular, if $\textbf{X}$ and $\textbf{Y}$ belong to $P_s^+(\mathbb{R}^n)$ and $\textbf{Y}$ is a random variable that takes the value $\textbf{m}$ with positive components with probability $1$ (so that $\widehat{g}(\boldsymbol{\xi }) = e^{-i\textbf{m}\boldsymbol{\xi }}$ and $T(\textbf{Y}) = 0$ ),

(2.15) \begin{equation} T_n({\textbf{X}}+ \textbf{Y}) = \frac{|\textbf{m}_X|}{|\textbf{m}_X +\textbf{m}_Y|} T_n({\textbf{X}}) \lt T_n(\textbf{X}), \end{equation}

since in this case the length of the sum of two vectors with positive components is bigger than the length of both. It is remarkable that the same result holds even if only one component of the $\textbf{Y}$ variable is bigger than zero, while the others are not. The meaning of inequality (2.15) is clear. Since in this case $\textbf{X}+\textbf{Y}$ is nothing but $\textbf{X} +\textbf{m}$ , which corresponds to adding constant values $m_k$ to $X_k$ , for $k=1, 2,\dots, n$ , this property asserts that adding a positive constant value to one or more components to each agent decreases inequality.

Also, if the independent random variables ${\textbf{X}}_1$ and ${\textbf{X}}_2$ are distributed with the same law of $\textbf{X}$ , so that their mean values are equal, thanks to the scale property

(2.16) \begin{equation} T_n\left (\frac{{\textbf{X}}_1+{\textbf{X}}_2}2\right ) = T_n\!\left ({\textbf{X}}_1+{\textbf{X}}_2\right ) \le T_n (\textbf{X}), \end{equation}

while the mean of $({\textbf{X}}_1+{\textbf{X}}_2)/2$ is equal to the mean of $\textbf{X}$ .

A third important consequence of inequality (2.14) is related to the situation in which the random variable $\textbf{Y} =\textbf{N}$ represents a noise (of mean value $\textbf{m}\gt 0$ ) that is present when measuring the inequality index of $\textbf{X}$ . The classical choice is that the additive noise is represented by a Gaussian variable of mean $\textbf{m}$ and covariance matrix $\Sigma$ .

We have in this case

(2.17) \begin{equation} T_n({\textbf{X}}+ \textbf{N}) \le \frac{|\textbf{m}_X|}{|\textbf{m}_X +\textbf{m}|} T_n({\textbf{X}}) + \frac{|\textbf{m}|}{|\textbf{m}_X +\textbf{m}|} T_n(\textbf{ N}). \end{equation}

Hence, a precise upper bound can be obtained once we know the explicit value of the inequality index $ T_n(\textbf{N})$ . This leads to the interesting question relative to the (explicit) evaluation of the inequality index $T_n$ of a random multivariate Gaussian variable.

The evaluation is direct. The Fourier transform of the distribution function $F$ of a multivariate Gaussian variable $\textbf{N} = (N_1,N_2,\dots, N_n)$ in $\mathbb{R}^n$ , $n \gt 1$ , is given by the expression

(2.18) \begin{equation} \widehat{f}(\boldsymbol{\xi }) = \exp \left \{-i \textbf{m}^T\boldsymbol{\xi } - \frac 12 \boldsymbol{\xi }^T\Sigma \, \boldsymbol{\xi } \right \}, \end{equation}

where $\textbf{m}$ is the vector of the mean values $\langle N_k\rangle$ , and $\Sigma$ is the $n\times n$ symmetric covariance matrix, with elements

\begin{equation*} \sigma _{ij} = \langle (N_i -m_i)(N_j-m_j)\rangle. \end{equation*}

Let us first consider the simple case in which the covariance matrix $\Sigma$ is diagonal, so that $\sigma _{ij} = 0$ if $i\not =j$ , and let us set, for simplicity $\sigma _{ii} = \sigma _i$ . Then $\boldsymbol{\xi }^T\Sigma \, \boldsymbol{\xi } = \sum _{k=1}^n \sigma _k \xi _k^2$ , and

(2.19) \begin{equation} \nabla \widehat{f}(\boldsymbol{\xi }) =\left ( -i\textbf{m} - \Sigma \, \boldsymbol{\xi }\right ) \widehat{f}(\boldsymbol{\xi }), \end{equation}

where $\Sigma \, \boldsymbol{\xi }$ is the $n$ -dimensional vector with components $\sigma _k\xi _k$ , $k =1,2,\dots,n$ . Consequently, $\nabla \widehat{f}(\boldsymbol{\xi }=\textbf{0}) = -i\textbf{m}$ , and

\begin{align*} T_n(F) =& \frac 1{2|\textbf{m}|} \sup _{\boldsymbol{\xi } \in \mathbb{R}^n} \left | \Sigma \, \boldsymbol{\xi } \exp \left \{-i \textbf{m}^T\boldsymbol{\xi } - \frac 12 \boldsymbol{\xi }^T\Sigma \, \boldsymbol{\xi } \right \}\right |=\\[5pt] & \frac 1{2|\textbf{m}|} \sup _{\boldsymbol{\xi } \in \mathbb{R}^n} \left | \Sigma \, \boldsymbol{\xi } \exp \left \{ - \frac 12 \boldsymbol{\xi }^T\Sigma \, \boldsymbol{\xi } \right \}\right | \end{align*}

By definition,

\begin{equation*} \left | \Sigma \, \boldsymbol {\xi } \exp \left \{ - \frac 12 \boldsymbol {\xi }^T\Sigma \, \boldsymbol {\xi } \right \}\right |^2 = \left (\sum _{k=1}^n \sigma _k^2\xi _k^2 \right )\exp \left \{ -\sum _{k=1}^n \sigma _k \xi _k^2 \right \}. \end{equation*}

On the other hand,

\begin{equation*} \sum _{k=1}^n \sigma _k^2\xi _k^2 \le \max _{1\le i \le n}\sigma _i \sum _{k=1}^n \sigma _k\xi _k^2, \end{equation*}

while

\begin{equation*} \sup _{{\boldsymbol {\xi }} \in \mathbb {R}^n} \left (\sum _{k=1}^n \sigma _k\xi _k^2 \right )\exp \left \{ -\frac 12 \sum _{k=1}^n \sigma _k \xi _k^2 \right \} = \frac 1{e}. \end{equation*}

Therefore, we conclude with the expression

(2.20) \begin{equation} T_n(F) = \frac 1{2 \sqrt e} \frac 1{|\textbf{m}|} \sqrt{\max _{1\le i \le n}\sigma _i}. \end{equation}

We outline that in the one-dimensional case the inequality index (2.20) coincides with a quantity proportional to the coefficient of variation, with constant of proportionality equal to $1/(2\sqrt e)$ . It is remarkable that this constant is independent of the dimension.

The general case in which the covariance matrix is not diagonal can be easily treated in the same way, by observing that, since the matrix $\Sigma$ is symmetric and positive definite, we can diagonalise it. To this extent, let us set $\boldsymbol{\xi } = A\boldsymbol{\eta }$ , where $A$ is an orthogonal matrix, so that $A^TA =A^{-1}A = I$ .

Then

\begin{equation*} \boldsymbol {\xi }^T\Sigma \, \boldsymbol {\xi } = (A\boldsymbol {\eta })^T \Sigma A\boldsymbol {\eta } = \boldsymbol {\eta }^T A^T\Sigma A\, \boldsymbol {\eta }. \end{equation*}

Now, let us choose the orthogonal matrix $A$ in such a way that the matrix $D= A^T\Sigma A$ is diagonal, and the elements $\lambda _k$ on the diagonal, $k=1,2,\dots,n$ , are the positive eigenvalues of the matrix $\Sigma$ .

In this case,

\begin{equation*} \boldsymbol {\xi }^T\Sigma \, \boldsymbol {\xi } = \boldsymbol {\eta }^T D \, \boldsymbol {\eta }. \end{equation*}

In addition,

\begin{equation*} \textbf{m}^T\boldsymbol {\xi } = \textbf{m}^T A\, \boldsymbol {\eta } = (A^T\textbf{m})^T \boldsymbol {\eta }, \end{equation*}

where, since the matrix $A$ is orthogonal,

\begin{equation*} (A^T\textbf{m})^T (A^T\textbf{m}) = \textbf{m}^T AA^T \textbf{m} )= \textbf{m}^T I \textbf{m} = \textbf{m}^T \textbf{m}. \end{equation*}

Moreover, the gradient of the Fourier transform, since the matrix $\Sigma$ is symmetric, is expressed by formula (2.19). Then, since the matrix $A$ is orthogonal,

\begin{equation*} \left | \Sigma \, \boldsymbol {\xi }\right |^2 = \left | \Sigma A\, \boldsymbol {\eta }\right |^2 = \left | A^T \Sigma A\, \boldsymbol {\eta }\right |^2 = \left | D\, \boldsymbol {\eta }\right |^2= \sum _{k=1}^n \lambda _k^2\eta _k^2. \end{equation*}

Now, taking the supremum over all values of $\boldsymbol{\xi }$ is the same as to take the supremum over $\boldsymbol{\eta }$ . Therefore, for the multivariate Gaussian distribution, one obtains the expression

(2.21) \begin{equation} T_n(N) = \frac 1{2\sqrt e} \frac 1{|\textbf{m}|} \sqrt{\max _{1\le i \le n}\lambda _i}, \end{equation}

where the $\lambda _k$ ’s are the positive eigenvalues of the covariance matrix of the multivariate Gaussian distribution.

Going back to formula (2.17), in the presence of a additive noise represented by a Gaussian variable of mean $\textbf{m}$ and covariance matrix $\Sigma$ one has the bound

(2.22) \begin{equation} T_n({\textbf{X}}+ \textbf{N}) \le \frac{|\textbf{m}_X|}{|\textbf{m}_X +\textbf{m}|} T_n({\textbf{X}}) + \frac 1{2\sqrt e|\textbf{m}_X +\textbf{m}|}\sqrt{\max _{1\le i \le n}\lambda _i} \end{equation}

It is interesting to remark that formula (2.22) continues to hold in presence of a centred Gaussian random noise of covariance matrix $\Sigma$ , and in this case

(2.23) \begin{equation} T_n({\textbf{X}}+ \textbf{N}) \le T_n({\textbf{X}}) + \frac 1{2\sqrt e|\textbf{m}_X| }\sqrt{\max _{1\le i \le n}\lambda _i}. \end{equation}

3. About the Gini-type index for multivariate distributions

Sections 2 has been devoted to the definition of a new multivariate inequality index, to its main properties, and to its evaluation in correspondence to a multivariate Gaussian distribution. This analysis takes a great advantage from the possibility to express the index in terms of a multidimensional Fourier transform.

It is therefore fair to ask whether the use of the Fourier transform can also bring advantage in the definition of a multivariate Gini index. As a matter of fact, the extension of the classical Gini index to measure inequality in multivariate distributions has shown numerous attempts, as certified by the references of the recent paper [Reference Banerjee6], in which the author is motivated by the objective of designing a multidimensional Gini index of inequality, to quantify of standard of living, that would satisfy all of a number of reasonable properties. This in reason of the fact that, as noticed in the introduction of [Reference Banerjee6], many existing multidimensional inequality indices of Gini type proposed by economists from time to time have remained elusive in this respect.

To better understand the difficulties that appear when trying to build a multidimensional generalisation of the Gini index, following the line considered in this paper, we will build a multivariate version of the index obtained by resorting to its Fourier one-dimensional transform. Indeed, the Fourier expression of the one-dimensional Gini index considered in ref. [Reference Toscani35] appears ready to be extended to higher dimensions.

As shown in Section 2, for any probability measure $F\in P_s^+(\mathbb{R})$ of mean value $m\gt 0$ , Gini index has a simple expression in Fourier transform, given by (2.4).

Considering that the value zero in equation (2.4) is obtained when $\widehat{f}(\xi ) = e^{-im\xi }$ , Gini index can be fruitfully rewritten as

(3.1) \begin{equation} G(F) = \frac 1{2\pi m} \left [ \int _{\mathbb{R}} \frac{|1 - e^{-im\xi } |^2}{\xi ^2} \, d\xi - \int _{\mathbb{R}} \frac{|1 -\widehat{f}(\xi )|^2}{\xi ^2} \, d\xi \right ]. \end{equation}

Expression (3.1) can be easily extended to measure the inequality of a multivariate distribution $F\in P_s^+(\mathbb{R}^n)$ , $n \gt 1$ by setting

(3.2) \begin{equation} G_n(F) = \frac{\mu _n}{|\textbf{m}|} \left [ \int _{\mathbb{R}^n} \frac{|1 - e^{-i\textbf{m}\boldsymbol{\xi }} |^2}{|\boldsymbol{\xi }|^{n+1}} \, d\boldsymbol{\xi } - \int _{\mathbb{R}^n} \frac{|1 -\widehat{f}(\boldsymbol{\xi })|^2}{|\boldsymbol{\xi }|^{n+1}} \, d\boldsymbol{\xi }\right ], \end{equation}

where the constant $\mu _n$ is such that

(3.3) \begin{equation} \frac 1{\mu _n}= \frac 1{|\textbf{m}|} \int _{\mathbb{R}^n} \frac{|1 - e^{-i\textbf{m}\boldsymbol{\xi }} |^2}{|\boldsymbol{\xi }|^{n+1}} \, d\boldsymbol{\xi }. \end{equation}

Evaluating the integral on the right-hand side by resorting to a $n$ -dimensional spherical coordinate system, one realises that the value of the constant $\mu _n$ does not depend on the vector $\textbf{m}$ and equals

\begin{equation*} \mu _n =\Gamma \left (\frac {n-1}2 +1 \right )\sqrt {(2\pi )^{n-1}}, \end{equation*}

where $\Gamma (\!\cdot\!)$ denotes as usual the Gamma function. Formula (3.2) is valid for all values of $n \in \mathbb{N}$ , including $n =1$ , that consistently gives $\mu _1=1$ .

Resorting to (3.3), we can express the multivariate Gini-type index (3.2) in the (simpler) form

(3.4) \begin{equation} G_n(F) = 1- \frac{\mu _n}{|\textbf{m}|} \int _{\mathbb{R}^n} \frac{|1 -\widehat{f}(\boldsymbol{\xi })|^2}{|\boldsymbol{\xi }|^{n+1}} \, d\boldsymbol{\xi }. \end{equation}

The same idea can be applied to recover an expression for a multivariate Pietra index [Reference Pietra32, Reference Toscani35]. However, despite their eventual theoretical interest, does not seem that this type of expressions, if compared to the multivariate $T_n$ index considered in this paper, share good properties.

The problems that appear when passing from the one-dimensional version (2.4) to its natural multivariate version (3.4) can be easily understood by evaluating the value of the Gini index $G_n$ , $n \gt 1$ , in correspondence to the multivariate random variable $\textbf{X}$ taking only two values, introduced in Section 2. This variable is characterised by the Fourier transform (2.11), so that, to compute the value of Gini index, as expressed by formula (3.4), we need to evaluate the integral

\begin{equation*} I_n(F)= \frac {\mu _n}{|\textbf{m}|} \int _{\mathbb {R}^n} \frac {|1 - (1-p)e^{-i\textbf{a}\boldsymbol {\xi }} + p e^{-i \textbf{b}\boldsymbol {\xi }}|^2}{|\boldsymbol {\xi }|^{n+1}} \, d\boldsymbol {\xi }, \end{equation*}

where $|\textbf{m}| = |(1-p)\textbf{a}^T + p\textbf{b}^T |$ . It is immediate to show that the integral $I_n$ can be split into three terms, i.e.

(3.5) \begin{align} I_n(F) &= \frac{\mu _n}{|\textbf{m}|} \int _{\mathbb{R}^n} \frac{2(1-p) (1 - \cos \textbf{a} \boldsymbol{\xi })}{|\boldsymbol{\xi }|^{n+1}} \, d\boldsymbol{\xi } + \nonumber \\[5pt] & \frac{\mu _n}{|\textbf{m}|} \int _{\mathbb{R}^n} \frac{2p (1 - \cos \textbf{b} \boldsymbol{\xi })}{|\boldsymbol{\xi }|^{n+1}} \, d\boldsymbol{\xi } + \frac{\mu _n}{|\textbf{m}|} \int _{\mathbb{R}^n} \frac{2p(1-p) (1 - \cos\!(\textbf{b}-\textbf{a}) \boldsymbol{\xi })}{|\boldsymbol{\xi }|^{n+1}} \, d\boldsymbol{\xi }. \end{align}

The three integrals on the right-hand side of (3.5) can be easily evaluated by resorting to a $n$ -dimensional spherical coordinate system to give

\begin{equation*} I_n(F) = \frac {1}{|\textbf{m}|}\left [ (1-p)|\textbf{a}^T| + p|\textbf{b}^T| - p(1-p)|(\textbf{b}^T - \textbf{a}^T |\right ], \end{equation*}

so that

(3.6) \begin{align} G_n(F) &= 1 -\frac{1}{|\textbf{m}|}\left [ (1-p)|\textbf{a}^T| + p|\textbf{b}^T| - p(1-p)|(\textbf{b}^T - \textbf{a}^T |\right ]=\nonumber \\[5pt] & \frac{|(1-p)\textbf{a}^T + p\textbf{b}^T | - \left [ (1-p)|\textbf{a}^T| + p|\textbf{b}^T|\right ] + p(1-p)|(\textbf{b}^T - \textbf{a}^T |}{|(1-p)\textbf{a}^T + p\textbf{b}^T |} \le \nonumber \\[5pt] & \frac{ p(1-p)|(\textbf{b}^T - \textbf{a}^T |}{|(1-p)\textbf{a}^T + p\textbf{b}^T |} = T_n(F). \end{align}

Hence, at difference with the $T_n$ multivariate index, the multivariate Gini index $G_n$ does not coincide, even in the case of a simple two-valued distribution, with the $n$ -dimensional extension of the univariate index. In other words, while the unidimensional indices depend only on the modulus of the difference between the two values assumed by the random variable, in the multivariate case only the $T_n$ index retains this property, while the multivariate Gini index $G_n$ , apparently derived from a natural extension of the one-dimensional index by maintaining the scaling property, does not.

In fact, the additional term appearing on the numerator of formula (3.6) is given by

\begin{equation*} |(1-p)\textbf{a}^T + p\textbf{b}^T | - \left [ (1-p)|\textbf{a}^T| + p|\textbf{b}^T|\right ], \end{equation*}

even in the presence of two vectors with positive components, in dimension $n \gt 1$ is dependent on the position of the points $\textbf{a}$ and $\textbf{b}$ on the space $\mathbb{R}^n$ , and it is equal to zero if and only if the two vectors $\textbf{a}$ and $\textbf{b}$ are parallel. This unpleasant fact shows that even the passage to Fourier transform does not allow a simple extension of the Gini index to multivariate distributions.

On the contrary, the presence of heavy difficulties in defining a easy to treat inequality index able to measure multivariate distributions characterises the $T_n$ index introduced in this paper as a good candidate for future applications.