Equality in the Matrix Entropy-Power Inequality and Blind Separation of Real and Complex sources

TL;DR

This paper proves a matrix version of the entropy-power inequality for real and complex variables using a transportation argument, and applies it to blind source extraction.

Contribution

It introduces a new matrix form of the entropy-power inequality and demonstrates its use in blind source separation.

Findings

Established the matrix entropy-power inequality for real and complex cases.

Provided a transportation-based proof that clarifies the equality conditions.

Applied the inequality to improve blind source extraction techniques.

Abstract

The matrix version of the entropy-power inequality for real or complex coefficients and variables is proved using a transportation argument that easily settles the equality case. An application to blind source extraction is given.

Equations

$$h(a_1{X}_1+a_2{X}_2)\ge h(a_1{X}_1^{\ast }+a_2{X}_2^{\ast })$$

$$h(X_1^{\ast })=h(X_1)h(X_2^{\ast })=h(X_2).$$

$$h(\mathbf{A}X)\ge h(\mathbf{A}{X}^{\ast })$$

$$h(X_i^{\ast })=h(X_i)(i=1,\dots ,n).$$

$$\left(\begin{array}{c}\tilde{X}\\ \\ {\tilde{X}}^{\prime }\end{array}\right)=\left(\begin{array}{c}\mathbf{A}\\ \\ {\mathbf{A}}^{\prime }\end{array}\right)X^{\ast }.$$

$$X^{*}=\begin{pmatrix}\mathbf{A}^{t}\;\Big{|}\;{\mathbf{A}^{\prime}\,}^{t}\end{pmatrix}\begin{pmatrix}\widetilde{X}\\ \hline\cr\vphantom{1^{1^{1^{1}}}}\widetilde{X}^{\prime}\end{pmatrix}=\mathbf{A}^{t}\widetilde{X}+{\mathbf{A}^{\prime}\,}^{t}\widetilde{X}^{\prime}.$$

$$T(X^{*})=\bigl{(}T_{1}(X^{*}_{1}),T_{2}(X^{*}_{2}),\ldots,T_{n}(X^{*}_{n})\bigr{)}^{t}$$

$$T^{\prime}(X^{*})=\mathrm{diag}\bigl{(}T^{\prime}_{1}(X^{*}_{1}),\ldots,T^{\prime}_{n}(X^{*}_{n})\bigr{)}.$$

$$h\bigl{(}\mathbf{A}T(X^{*})\bigr{)}\geq h(\mathbf{A}X^{*})$$

$$h\bigl{(}\mathbf{A}T(\mathbf{A}^{t}\widetilde{X}+{\mathbf{A}^{\prime}\,}^{t}\widetilde{X}^{\prime})\bigr{)}\geq h(\widetilde{X}).$$

$$h\bigl{(}\mathbf{A}T(\mathbf{A}^{t}\widetilde{X}\!+\!{\mathbf{A}^{\prime}\,}^{t}\widetilde{X}^{\prime})\bigr{)}\geq h\bigl{(}\mathbf{A}T(\mathbf{A}^{t}\widetilde{X}\!+\!{\mathbf{A}^{\prime}\,}^{t}\widetilde{X}^{\prime})\mid\widetilde{X}^{\prime}\bigr{)}.$$

$$\mathbb{E}\log T_j^{\prime }(X_j^{\ast })=0(j=1,2,\dots ,n).$$

$$\displaystyle h\bigl{(}\mathbf{A}$$

$$=h(\tilde{X}\mid {\tilde{X}}^{\prime })+\mathbb{E}\log |\mathbf{A}{T}^{\prime }({\mathbf{A}}^t\tilde{X}+{{\mathbf{A}}^{\prime }}^t{\tilde{X}}^{\prime }){\mathbf{A}}^t|$$

$$=h(\tilde{X})+\mathbb{E}\log |\mathbf{A}{T}^{\prime }(X^{\ast }){\mathbf{A}}^t|$$

$$\log |\mathbf{A}\mathbf{\Lambda }{\mathbf{A}}^t|\ge \mathrm{tr}(\mathbf{A}[\log \mathbf{\Lambda }]{\mathbf{A}}^t)$$

$$\log |\mathbf{A}\mathbf{\Lambda }{\mathbf{A}}^t|$$

$$\ge \sum _{i=1}^m\sum _{j=1}^n{A}_{ij}^2\log {\lambda }_j$$

$$=\mathrm{tr}(\mathbf{A}[\log \mathbf{\Lambda }]{\mathbf{A}}^t).$$

$$\mathbb{E}\log |\mathbf{A}{T}^{\prime }(X^{\ast }){\mathbf{A}}^t|$$

$$=\mathrm{tr}(\mathbf{A}\mathbb{E}[\log T^{\prime }(X^{\ast })]{\mathbf{A}}^t)=0.$$

$$\mathbf{A}=\left(\begin{array}{cc}[c|c]{\mathbf{I}}_r& 0\\ & \\ 0& {\mathbf{A}}_u\end{array}\right)$$

$$h(X_r)+h({\mathbf{A}}_u{X}_u)\ge h(X_r^{\ast })+h({\mathbf{A}}_u{X}_u^{\ast }).$$

$$\mathbf{A}\mathbf{\Lambda }{\mathbf{A}}^t=\left(\begin{array}{cc}[c|c]{\mathbf{\Lambda }}_r& 0\\ & \\ 0& {\mathbf{A}}_u{\mathbf{\Lambda }}_u{\mathbf{A}}_u^t\end{array}\right)$$

$$T_j^{\prime }(X_j^{\ast })=T_k^{\prime }(X_k^{\ast })\text{ a.e.}(r<j,k\le n)$$

$$T^{\prime}(\widehat{X^{*}})=\mathrm{diag}\bigl{(}T^{\prime}_{1}(\widehat{X^{*}_{1}}),\ldots,T^{\prime}_{n}(\widehat{X^{*}_{n}})\bigr{)}$$

$$\mathbb{E}\log |T_j^{\prime }(\widehat{X_j^{\ast }})|=0(j=1,2,\dots ,n).$$

$$\displaystyle h\bigl{(}\widehat{\mathbf{A}}$$

$$=h(\tilde{X})+\mathbb{E}\log |\hat{\mathbf{A}}{T}^{\prime }(\widehat{X^{\ast }}){\hat{\mathbf{A}}}^t|$$

$${\lambda }_j=\widehat{u_j}{d}_j{\widehat{u_j}}^t$$

$$\log |\hat{\mathbf{A}}\mathbf{\Lambda }{\hat{\mathbf{A}}}^t|$$

$$=\mathrm{tr}(\hat{\mathbf{A}}[\log \mathbf{\Lambda }]{\hat{\mathbf{A}}}^t)$$

$$\mathrm{tr}(\hat{\mathbf{A}}[\log \mathbf{\Lambda }]{\hat{\mathbf{A}}}^t)$$

$$={\sum }_i{\sum }_j|A_{i,j}{|}^2\mathrm{tr}(\log {\lambda }_j)$$

Full text

I Introduction

Consider random variables with densities that are continuous and positive inside their support interval, with zero mean and finite differential entropies. The entropy power inequality (EPI) was stated by Shannon [1] in 1948 and is well known to be equivalent to the following minimum entropy inequality [2, 3, 4]:

[TABLE]

for any real numbers $a_{1},a_{2}$ and any independent real random variables $X_{1},X_{2}$, where $X^{*}_{1},X^{*}_{2}$ are independent normal random variables having the same entropies as $X_{1},X_{2}$:

[TABLE]

Equality holds in (1) if and only if either $a_{1}a_{2}=0$ or $X_{1},X_{2}$ are normal. Recently, a normal transport argument was used in [5] to provide a simple proof of Shannon’s EPI, including the necessary and sufficient condition for equality.

Shannon’s EPI was generalized to a matrix version [6, 7]:

[TABLE]

for any $m\times n$ matrix $\mathbf{A}$ and any random (column) vector $X=(X_{1},X_{2},\ldots,X_{n})^{t}$ of independent components $X_{i}$, where $X^{*}=(X^{*}_{1},X^{*}_{2},\ldots,X^{*}_{n})^{t}$ is a normal vector with independent components $X^{*}_{i}$ of the same entropies:

[TABLE]

Available proofs of (3) are either by double induction on $(m,n)$ [6] or by integration over a path of Gaussian perturbation of the corresponding inequality for Fisher’s information using de Bruijn’s identity [7] or via the I-MMSE relation [8]. A necessary and sufficient condition for equality in (3) has not been settled so far, however, by the previous methods. Such a condition is important in applications such as blind source separation (BSS) based on minimum entropy [9]. Also, BSS may involve real or complex signals [10] and minimum entropy methods for complex sources would require the extension of EPIs to complex-valued variables and coefficients.

In this paper, we adapt the proof of [5] to the matrix case and derive (3) with a normal transport argument. This allows us to easily settle the equality case: We define the notion of “recoverability” and show that equality holds in (3) if all unrecoverable components of $X$ present in $\mathbf{A}X$ are normal. We then extend the proofs to complex-valued $\mathbf{A}$ and $X$. As an application, we derive the appropriate contrast functions for partial BSS (a.k.a. blind source extraction) where $m$ out of $n$ independent sources are to be extracted.

II A simple proof of the matrix EPI by transport

We extend the proof in [5] to the matrix EPI, based on the same ingredients: (a) a transportation argument from normal variables, that takes the form of a simple change of variables; (b) a rotation performed on i.i.d. normal variables, which preserves the i.i.d. property; (c) concavity of the logarithm, appropriately generalized to the matrix case. The proof breaks into several elementary steps:

II-A Reduce to full rank $m<n$

If the rank of $\mathbf{A}$ is $<m$ then some rows are linearly dependent, there is a deterministic relation between some components of $\mathbf{A}X$ and $\mathbf{A}X^{*}$ and equality $h(\mathbf{A}X)=h(\mathbf{A}X^{*})=-\infty$ holds trivially. Thus we can assume that $\mathbf{A}$ is of full rank $m\leq n$. If $\mathbf{A}$ has rank $m=n$ then $\mathbf{A}$ is invertible and by the change of variable formula in the entropy [1, § 20.9], $h(\mathbf{A}X)=h(X)+\log|\mathbf{A}|=h(X^{*})+\log|\mathbf{A}|=h(\mathbf{A}X^{*})$ where $|\mathbf{A}|$ denotes the absolute value of the determinant of $\mathbf{A}$. Therefore, one may always assume that $\mathbf{A}$ has full rank $m\!<\!n$.

II-B Reduce to equal individual entropies

Without loss of generality, one may assume that the components of $X$ have equal entropies. For if it were not the case, then by the scaling property of entropy[1, § 20.9], one can find non zero coefficients $\delta_{j}$ (e.g., $\delta_{j}=\exp h(X_{j})$) such that all $X^{\prime}_{j}=X_{j}/\delta_{j}$ have equal entropies. Then applying (3) to $X^{\prime}=(X^{\prime}_{1},\ldots,X^{\prime}_{n})^{t}$ and matrix $\mathbf{A}\boldsymbol{\Delta}$ where $\boldsymbol{\Delta}$ is a diagonal matrix with diagonal elements $\delta_{j}$, gives the desired EPI.

Notice that with the additional constraint that the $X_{j}$ have equal entropies, we have $h(X^{*}_{1})=h(X^{*}_{2})=\cdots=h(X^{*}_{n})=h(X_{1})=h(X_{2})=\cdots=h(X_{n})$: The independent zero-mean normal variables $X^{*}_{j}$ also have equal entropies, and are, therefore, independent and identically distributed (i.i.d.).

II-C Reduce to orthonormal rows

Without loss of generality, one may assume that the rows of $\mathbf{A}$ are orthonormal. For if it were not the case, one can orthonormalize the rows by a Gram-Schmidt process. This amounts to multiplying $\mathbf{A}$ on the left by an lower-triangular invertible matrix $\mathbf{L}$. Thus, one can apply (3) for matrix $\mathbf{A}^{\prime}=\mathbf{L}\mathbf{A}$. Again by the change of variable in the entropy[1, § 20.9], $h(\mathbf{A}^{\prime}X)=h(\mathbf{A}X)+\log|\mathbf{L}|$ and $h(\mathbf{A}^{\prime}X^{*})=h(\mathbf{A}X^{*})+\log|\mathbf{L}|$. The terms $\log|\mathbf{L}|$ cancel to give the desired EPI. Thus we are led to prove (3) for an $m\times n$ matrix $\mathbf{A}$ with orthonormal rows ($\mathbf{A}\mathbf{A}^{t}=\mathbf{I}_{m}$, the $m\times m$ identity matrix).

II-D Complete the orthogonal matrix

Extend $\mathbf{A}$ by adding $n-m$ orthonormal rows of a complementary matrix $\mathbf{A}^{\prime}$ such that \bigl{(}\!\begin{smallmatrix}\mathbf{A}\hphantom{{}^{\prime}}\\[2.15277pt] \hline\cr\mathbf{A}^{\prime}\end{smallmatrix}\!\bigr{)} is an $n\times n$ orthogonal matrix, and define the Gaussian vector \Bigl{(}\!\begin{smallmatrix}\widetilde{X}\hphantom{{}^{\prime}}\\[2.15277pt] \hline\cr\vphantom{{}^{2^{2}}}\widetilde{X}^{\prime}\end{smallmatrix}\!\Bigr{)} as

[TABLE]

Since the components of $X^{*}$ are i.i.d. normal and \bigl{(}\!\begin{smallmatrix}\mathbf{A}\hphantom{{}^{\prime}}\\[2.15277pt] \hline\cr\mathbf{A}^{\prime}\end{smallmatrix}\!\bigr{)} is orthogonal, the components of \Bigl{(}\!\begin{smallmatrix}\widetilde{X}\hphantom{{}^{\prime}}\\[2.15277pt] \hline\cr\vphantom{{}^{2^{2}}}\widetilde{X}^{\prime}\end{smallmatrix}\!\Bigr{)} are also i.i.d. normal. In particular the subvectors $\widetilde{X}$ and $\widetilde{X}^{\prime}$ are independent. The inverse transformation is the transpose:

[TABLE]

II-E Apply the normal transportation

Lemma 1** (Normal Transportation [5, 11])**

Let $X^{*}\in\mathbb{R}$ be a scalar normal random variable. For any continuous density $f$, there exists a differentiable transformation $T\colon\mathbb{R}\!\to\!\mathbb{R}$ with positive derivative $T^{\prime}\!>\!0$ such that $X=T(X^{*})$ has density $f$.

From Lemma 1, we can assume that the components of $X=(X_{1},X_{2},\ldots,X_{n})^{t}$ and $X^{*}=(X^{*}_{1},X^{*}_{2},\ldots,X^{*}_{n})^{t}$ are such that $X_{j}=T_{j}(X^{*}_{j})$ for all $j=1,2,\ldots,n$, where the $T_{j}$’s are transformations with positive derivatives $T^{\prime}_{j}>0$. For ease of notation define

[TABLE]

Thus $T\colon\mathbb{R}^{n}\to\mathbb{R}^{n}$ is a transformation whose Jacobian matrix is diagonal with positive diagonal elements:

[TABLE]

Now (3) can be written in terms of the normal variables only:

[TABLE]

and by (6) it can also be written in term of the tilde normal variables:

[TABLE]

II-F Conditioning on the complementary variables

Since conditioning reduces entropy [1, § 20.4],

[TABLE]

II-G Make the change of variable

By the change of variable formula in the entropy [1, § 20.8], $h(X_{j})=h(T_{j}(X^{*}_{j}))=h(X^{*}_{j})+\mathbb{E}\log T_{j}^{\prime}(X^{*}_{j})$ and, therefore, by (4),

[TABLE]

By the change of variable formula (vector case) [1, § 20.8] in the conditional entropy in the r.h.s. of (11),

[TABLE]

where we have used that $\widetilde{X}$ and $\widetilde{X}^{\prime}$ are independent.

II-H Apply the concavity of the logarithm

The following lemma was stated in [7] as a consequence of (3). A direct proof was given in [8], and is simplified here.

Lemma 2

For any $m\times n$ matrix $\mathbf{A}$ with orthonormal rows and any diagonal matrix $\boldsymbol{\Lambda}=\mathrm{diag}(\lambda_{1},\ldots,\lambda_{n})$ with positive diagonal elements $\lambda_{j}>0$,

[TABLE]

where $\log\boldsymbol{\Lambda}=\mathrm{diag}(\log\lambda_{1},\ldots,\log\lambda_{n})$ and $\mathrm{tr}(\cdot)$ denotes the trace.

Equality holds e.g. when the $\lambda_{j}$’s are equal. The precise equality case will appear elsewhere.

Proof:

It is easily checked that $\mathbf{A}\boldsymbol{\Lambda}\mathbf{A}^{t}$ is positive definite and that both sides of (15) do not change if we replace $\mathbf{A}$ by $\mathbf{U}\mathbf{A}$ where $\mathbf{U}$ is any $m\times m$ orthogonal matrix. Choose $\mathbf{U}$ as an orthogonal eigenvector matrix of $\mathbf{A}\boldsymbol{\Lambda}\mathbf{A}^{t}$, so that $\mathbf{U}\mathbf{A}\boldsymbol{\Lambda}\mathbf{A}^{t}\mathbf{U}^{t}$ is diagonal with positive diagonal elements and $\mathbf{U}\mathbf{A}$ still has orthonormal rows.

Thus, substituting $\mathbf{U}\mathbf{A}$ for $\mathbf{A}$ we may always assume that $\mathbf{A}\boldsymbol{\Lambda}\mathbf{A}^{t}$ is diagonal with diagonal entries equal to $\sum_{j=1}^{n}A^{2}_{ij}\lambda_{j}$ for $i=1,2,\ldots,m$, where $A_{i,j}$ denotes the entries of $\mathbf{A}$. Then

[TABLE]

where (17) follows from Jensen’s inequality and the concavity of the logarithm, since $\mathbf{A}$ has orthonormal rows. ∎

From Lemma 2 and (12) we obtain

[TABLE]

Combining this with (11)–(14) proves (10) and the desired matrix EPI (3).

III The Equality Case

To settle the equality case in (3), from the remarks in § II-A we may already assume that $\mathbf{A}$ has full rank $m<n$.

Definition 1

A component $X_{j}$ of $X$ is

• •

present* in $\mathbf{A}X$ if $\mathbf{A}X$ depends on $X_{j}$;*

• •

recoverable* from $\mathbf{A}X$ if there exists a row vector $b$ such that $b\!\cdot\!(\mathbf{A}X)=X_{j}$.*

Remark 1

Since the considered variables are not deterministic, Definition 1 depends only on the matrix $\mathbf{A}$: $X_{j}$ is present in $\mathbf{A}X$ if and only if the $j$th column of $\mathbf{A}$ is not zero; and $X_{j}$ is recoverable from $\mathbf{A}X$ if and only if there exists $b$ such that $b\mathbf{A}=(0,\ldots,0,1,0,\ldots,0)$ with $1$ in the $j$th position. A recoverable component is necessarily present.

Remark 2

Without loss of generality we always omit the components that are not present in $\mathbf{A}X$ and their associated zero columns of $\mathbf{A}$ without affecting the entropy $h(\mathbf{A}X)$.

Remark 3

Definition 1 is also invariant by left multiplication of $\mathbf{A}$ by any $m\times m$ invertible matrix $\mathbf{B}$: if the $j$th column of $\mathbf{A}$ is zero, so is the $j$th column of $\mathbf{B}\mathbf{A}$; and $b\mathbf{A}=(0,\ldots,0,1,0,\ldots,0)$ implies $(b\mathbf{B}^{-1})(\mathbf{B}\mathbf{A})=(0,\ldots,0,1,0,\ldots,0)$.

The following property was used in [12, Appendix] for deriving a sufficient condition for equality in a matrix form of the Brunn–Minkowski inequality, which is the analog of the EPI for Rényi entropies of order zero [3].

Lemma 3

Reordering the components of $X$ if necessary so that the first $r$ components are recoverable and the last $n-r$ components are unrecoverable, we may always put $\mathbf{A}$ in the canonical form

[TABLE]

where $\mathbf{A}_{u}$ is an $(m-r)\times(n-r)$ matrix. The number $r$ of recoverable components is the maximum number such that $\mathbf{A}$ can be put in the form (21) by left multiplication by an invertible matrix.

Proof:

Write $X=(X_{r}\mid X_{u})^{t}$ where $X_{r}$ has recoverable components and $X_{u}$ has unrecoverable ones. By Definition 1 (recoverability) there exists a $r\times m$ matrix $\mathbf{B}_{r}$ such that $\mathbf{B}_{r}\mathbf{A}=(\mathbf{I}_{r}\mid 0)$. Since $\mathbf{B}_{r}$ must have rank $r$, this shows in particular that $r\leq m$: no more than $m$ components can be recovered from the $m$ linear mixtures. We can use $m-r$ additional row operations so that \bigl{(}\begin{smallmatrix}\mathbf{B}_{r}\\[2.15277pt] \hline\cr\mathbf{B}_{u}\end{smallmatrix}\bigr{)}\mathbf{A}=\bigl{(}\footnotesize\begin{array}[]{c|c}\mathbf{I}_{r}&\boldsymbol{0}\\ \hline\cr\boldsymbol{0}&\mathbf{A}_{u}\end{array}\bigr{)} is of the desired form. Since \mathbf{B}=\bigl{(}\begin{smallmatrix}\mathbf{B}_{r}\hphantom{{}^{\prime}}\\[2.15277pt] \hline\cr\mathbf{B}_{u}\end{smallmatrix}\bigr{)} is an $m\times m$ invertible matrix, by the change of variable formula in the entropy [1, §20.9], $h(\mathbf{B}\mathbf{A}X)=h(\mathbf{A}X)+\log|\mathbf{B}|$ and $h(\mathbf{B}\mathbf{A}X^{*})=h(\mathbf{A}X^{*})+\log|\mathbf{B}|$. Therefore, the matrix EPI (3) is equivalent to the one obtained by substituting \mathbf{B}\mathbf{A}=\bigl{(}\footnotesize\begin{array}[]{c|c}\mathbf{I}_{r}&\boldsymbol{0}\\ \hline\cr\boldsymbol{0}&\mathbf{A}_{u}\end{array}\bigr{)} for $\mathbf{A}$. Clearly, $r$ is maximum in this expression since otherwise one could recover more than $r$ components, hence transfer some of the components from the $\mathbf{A}_{u}$ block to the $\mathbf{I}_{r}$ block. ∎

We can now settle the equality case in (3).

Theorem 1

Equality holds in (3) if and only if all unrecoverable components present in $\mathbf{A}X$ are normal.

Proof:

Write $X=(X_{r}\mid X_{u})^{t}$ as in the proof of Lemma 3 and accordingly write $X^{*}=(X^{*}_{r}\mid X^{*}_{u})^{t}$. If $\mathbf{A}$ is in canonical form (21), then (3) reads

[TABLE]

where $h(X_{r})=\sum_{j=1}^{r}h(X_{j})=\sum_{j=1}^{r}h(X^{*}_{j})=h(X^{*}_{r})$. The announced condition is, therefore, sufficient: if $X_{u}$ is normal with (zero-mean) components satisfying (4), then $X_{u}$ is identically distributed as $X_{u}^{*}$ and $h(\mathbf{A}_{u}X_{u})=h(\mathbf{A}_{u}X^{*}_{u})$.

Conversely, suppose that (3) is an equality with $\mathbf{A}$ as in (21). From § II C, we may assume (applying row operations of a Gram-Schmidt process if necessary) that $\mathbf{A}$ has orthonormal rows in (21), that is, $\mathbf{A}_{u}\mathbf{A}_{u}^{t}=\mathbf{I}_{m-r}$. Then equality holds in (3) if and only if both (11) and (19) are equalities.

Consider equality in (19) which results from the application of Lemma 2 (inequality (15)) to $\boldsymbol{\Lambda}=T^{\prime}(X^{*})$. We have

[TABLE]

where $\boldsymbol{\Lambda}_{r}=\mathrm{diag}(\lambda_{1},\ldots,\lambda_{r})$ and $\boldsymbol{\Lambda}_{u}=\mathrm{diag}(\lambda_{r+1},\ldots,\lambda_{n})$. Thus, we may choose $\mathbf{U}$ in the proof of Lemma 2 in the form \mathbf{U}=\bigl{(}\footnotesize\begin{array}[]{c|c}\mathbf{I}_{r}&\boldsymbol{0}\\ \hline\cr\boldsymbol{0}&\mathbf{U}_{u}\end{array}\bigr{)} where $\mathbf{U}_{u}$ is an $(m-r)\times(m-r)$ orthogonal matrix such that $\mathbf{U}_{u}\mathbf{A}_{u}\boldsymbol{\Lambda}_{u}\mathbf{A}_{u}^{t}\mathbf{U}_{u}^{t}$ is diagonal. Then \mathbf{U}\mathbf{A}=\bigl{(}\footnotesize\begin{array}[]{c|c}\mathbf{I}_{r}&\boldsymbol{0}\\ \hline\cr\boldsymbol{0}&\mathbf{U}_{u}\mathbf{A}_{u}\end{array}\bigr{)} is still of the form (21) where $\mathbf{U}_{u}\mathbf{A}_{u}$ has orthonormal rows.

Therefore, equality in (15) is equivalent to equality in (17) where we may again assume that $\mathbf{A}$ is of the form (21) where $r$ is maximal and $\mathbf{A}_{u}$ has orthonormal rows. By Remark 2, we may assume that all columns of $\mathbf{A}_{u}$ are nonzero. Notice that any row of $\mathbf{A}_{u}$ in (21) should have at least two nonzero elements. Otherwise, there would be one row of $\mathbf{A}_{u}$ of the form $(0,\ldots,0,\pm 1,0,\ldots,0)$ with the nonzero element in the $j$th position. Since the rows are orthonormal, the other elements in the $j$th column would necessarily equal zero, and the corresponding component of $X$ would be recoverable, which contradicts the maximality of $r$.

Now since the logarithm is strictly concave, equality holds in (17) if and only if for all $i=1,2,\ldots,m$, all the $\lambda_{j}$ for which ${A}_{i,j}\neq 0$ are equal. Because no column of $\mathbf{A}_{u}$ is zero and any row of $\mathbf{A}_{u}$ in (21) has at least two nonzero elements, this implies that for any $j$ such that $r<j\leq n$, $\lambda_{j}$ is equal to another $\lambda_{k}$ where $r<k\leq n$, $k\neq j$. Since Lemma 2 was applied to $\boldsymbol{\Lambda}=T^{\prime}(X^{*})$ it follows that

[TABLE]

Because $X^{*}_{j}$ and $X^{*}_{k}$ are independent, this implies that both $T^{\prime}_{j}(X^{*}_{j})$ and $T^{\prime}_{k}(X^{*}_{k})$ are constant and equal a.e., hence $T^{\prime}_{j}=T^{\prime}_{k}=c$ for some constant111This is similar to what appeared in an earlier transportation proof of the EPI [5]. By (12), we necessarily have $c=1$ if we assume that all individual entropies are equal as in § II-B. $c$. Therefore $T_{j}$ is linear and $X_{j}=T_{j}(X^{*}_{j})$ is normal for all $r<j\leq n$. This completes the proof.222This implies, in particular, that equality in (19) implies equality in (11). This can also be seen directly: if $T^{\prime}_{j}=1$ for all $r<j\leq n$, then for $\mathbf{A}$ of the form (21) in (11), $\mathbf{A}_{u}T(\mathbf{A}_{u}^{t}\widetilde{X}\!+\!{\mathbf{A}^{\prime}\,}^{t}\widetilde{X}^{\prime})=\widetilde{X}$ is independent of $\widetilde{X}^{\prime}$. ∎

IV Extension to Complex Matrix and Variables

A complex random variable $X\in\mathbb{C}$ can always be viewed as a two-dimensional real random vector \widehat{X}=\bigl{(}\!\begin{smallmatrix}\operatorname{Re}X\\ \operatorname{Im}X\end{smallmatrix}\!\bigr{)}\in\mathbb{R}^{2}. Therefore, by the vector form of the EPI [2, 3, 4], (1) holds for scalar coefficients $a_{1},a_{2}\in\mathbb{R}$ when $X_{1},X_{2}\in\mathbb{C}$ are independent complex random vectors and $X^{*}_{1},X^{*}_{2}\in\mathbb{C}$ are independent white normal random vectors satisfying (2). Here “white normal” $X^{*}\in\mathbb{C}$ amounts to say that $X^{*}$ is proper normal or circularly symmetric normal [13] (c-normal in short): $X^{*}\sim\mathcal{CN}(0,\sigma^{2})$, that is, $\widehat{X^{*}}\sim\mathcal{N}(0,\sigma^{2}\mathbf{I}_{2})$.

That (1) also holds for complex coefficients $a_{1},a_{2}\in\mathbb{C}$ is less known but straightforward. To see this, define333There is an ambiguity of notation easily resolved from the context: $\widehat{a}$ is a matrix when $a$ is a constant and $\widehat{X}$ is a vector when $X$ is random. \widehat{a}=\bigl{(}\!\begin{smallmatrix}\operatorname{Re}a&-\operatorname{Im}a\\ \operatorname{Im}a&\hphantom{-}\operatorname{Re}b\end{smallmatrix}\!\bigr{)} for any $a\in\mathbb{C}$, so that $\widehat{aX}~=\widehat{a}\widehat{X}$. Then $h(aX)=h(\widehat{a}\widehat{X})=h(\widehat{X})+\log|\widehat{a}|=h(X)+\log|a|^{2}$. Hence (2) implies $h(a_{1}X_{1})=h(a_{1}X^{*}_{1})$ and $h(a_{2}X_{2})=h(a_{2}X^{*}_{2})$. In addition, if $X^{*}\!\sim\!\mathcal{CN}(0,\sigma^{2})$ then $aX^{*}\!\sim\!\mathcal{CN}(0,|a|^{2}\sigma^{2})$. Therefore, by the vector EPI applied to $a_{1}X_{1}$ and $a_{2}X_{1}$ we see that (1) holds for complex coefficients $a_{1},a_{2}\in\mathbb{C}$ when $X^{*}_{1},X^{*}_{2}$ are independent c-normal variables satisfying (2).

The extension of the matrix EPI (3) to complex $\mathbf{A}$ and $X$ is more involved. We need the following notions (see, e.g., [14] and [15, chap. 10]). Define $\widehat{X}\in\mathbb{R}^{2n}$ by stacking the $\widehat{X_{i}}$ for each component $X_{i}\in\mathbb{C}$ of $X\in\mathbb{C}^{n}$, and define $\widehat{\mathbf{A}}$ as the $2m\times 2n$ real matrix with $2\times 2$ entries $\widehat{A_{i,j}}$ where $A_{i,j}$ are the complex entries of $\mathbf{A}$. It is easily checked that $\widehat{\mathbf{A}X}=\widehat{\mathbf{A}}\widehat{X}$, $\widehat{\mathbf{A}\mathbf{B}}=\widehat{\mathbf{A}}\widehat{\mathbf{B}}$, $\widehat{\mathbf{A}^{\dagger}}=\widehat{\mathbf{A}}^{t}$ where $\mathbf{A}^{\dagger}$ is the conjugate transpose, and $|\widehat{\mathbf{A}}|=|\mathbf{A}|^{2}$ where $|\mathbf{A}|$ denotes the modulus of the determinant of $\mathbf{A}$.

We also need the following extension of Lemma 1:

Lemma 4** (2D Brenier Map[16, 17])**

Let $\widehat{X^{*}}\in\mathbb{R}^{2}$ be a (white) normal random vector. For any given continuous density $f$ over $\mathbb{R}^{2}$, there exists a differentiable transformation $T\colon\mathbb{R}^{2}\!\to\!\mathbb{R}^{2}$ with symmetric positive definite Jacobian $T^{\prime}$ (noted $T^{\prime}\!>\!0$) such that $\widehat{X}=T(\widehat{X^{*}})$ has density $f$.

Courtade et al. [18] noted that the Brenier map can be used in the transportation proof of [5] to prove Shannon’s vector EPI. We find it also convenient to prove the complex matrix EPI:

Theorem 2

The matrix EPI (3) holds for any $m\times n$ complex matrix $\mathbf{A}$ and any random vector $X$ of independent complex components $X_{i}$, where $X^{*}$ is a c-normal vector with independent components $X^{*}_{i}$ satisfying (4). If equality holds in (3) then all unrecoverable components present in $\mathbf{A}X$ (in the sense of Definition 1) are normal.

The exact necessary and sufficient condition for equality is more involved and will appear elsewhere.

Proof:

We sketch the proof by going through the above proofs in Sections II and III and pointing out the differences:

§II-A: The scaling property of entropy now reads $h(\mathbf{A}X)=h(\widehat{\mathbf{A}}\widehat{X})=h(\widehat{X})+\log|\widehat{\mathbf{A}}|=h(X)+\log|\mathbf{A}|^{2}$.

§II-B: Since h(X^{*})=\log\bigl{(}\pi e\sigma^{2}\bigr{)} for $X^{*}\sim\mathcal{CN}(0,\sigma^{2})$, independent $X^{*}_{j}$ with equal entropies are i.i.d.

§II-C: The Gram-Schmidt orthonormalization takes place in $\mathbb{C}^{n}$ with $h(\mathbf{A}^{\prime}X)=h(\mathbf{A}X)+\log|\mathbf{L}|^{2}$.

§II-D: \mathbf{U}=\bigl{(}\!\begin{smallmatrix}\mathbf{A}\hphantom{{}^{\prime}}\\[2.15277pt] \hline\cr\mathbf{A}^{\prime}\end{smallmatrix}\!\bigr{)} is now an $n\times n$ unitary matrix. Recall that a circularly symmetric $X^{*}\sim\mathcal{CN}(0,\mathbf{K})$ is such that $\mathbf{A}X^{*}\sim\mathcal{CN}(0,\mathbf{A}\mathbf{K}\mathbf{A}^{\dagger})$ for any $\mathbf{A}$. Since $X^{*}\sim\mathcal{CN}(0,\sigma^{2}\mathbf{I})$ is i.i.d., $\mathbf{U}X^{*}\sim\mathcal{CN}(0,\sigma^{2}\mathbf{U}\mathbf{U}^{\dagger}=\sigma^{2}\mathbf{I})$ is also i.i.d. and the inverse transformation is the conjugate transpose $X^{*}=\mathbf{A}^{\dagger}\widetilde{X}+{\mathbf{A}^{\prime}\,}^{\dagger}\widetilde{X}^{\prime}$.

§II-E: Lemma 4 replaces Lemma 1 and (8) becomes

[TABLE]

in block-diagonal form where each $2\times 2$ block $T^{\prime}_{i}(\widehat{X^{*}_{i}})>0$ is symmetric positive definite.

§II-G: In terms of the hat variables:

[TABLE]

where $|\cdot|$ denotes the absolute value of the determinant, and

[TABLE]

§II-H: We show that Lemma 2 still holds when $\boldsymbol{\Lambda}$ is block-diagonal with $2\times 2$ diagonal blocks $\lambda_{j}>0$ (symmetric positive definite). Write

[TABLE]

where $d_{j}$ is $2\times 2$ diagonal with positive diagonal elements and $\widehat{u_{j}}$ is a rotation matrix, corresponding to a complex unit $u_{j}=e^{i\theta_{j}}$. Then the block-diagonal $\widehat{\mathbf{U}}=\mathrm{diag}(\widehat{u_{1}},\ldots,\widehat{u_{n}})$ is orthonormal and $\mathbf{D}=\mathrm{diag}(d_{1},\ldots,d_{n})$ is diagonal. We can now apply Lemma 2 to $\widehat{\mathbf{A}}\widehat{\mathbf{U}}$ and $\mathbf{D}$:

[TABLE]

where $\log\boldsymbol{\Lambda}$ is the (block diagonal) logarithm of $\boldsymbol{\Lambda}>0$. Thus

[TABLE]

where $\mathrm{tr}(\log\lambda_{j})=\log|\lambda_{j}|$ since $\lambda_{j}$ is symmetric positive definite. Thus we obtain

[TABLE]

which is the final step to prove the (complex) matrix EPI (3).

Assume that equality holds in (3) as in the converse part of the proof of Theorem 1 (Section III). That proof is unchanged up to the point where one considers the equality condition in Lemma 2 applied to $\widehat{\mathbf{A}}\widehat{\mathbf{U}}$ and diagonal $\mathbf{D}$, that is, in (29). By the strict concavity of the logarithm, equality holds in (29) if and only if for any two nonzero elements in the same row of $\widehat{\mathbf{A}}\widehat{\mathbf{U}}=\widehat{\mathbf{A}\mathbf{U}}$, the corresponding two diagonal elements of $\mathbf{D}$ are equal. Since $\mathbf{U}=\mathrm{diag}(e^{i\theta_{1}},\ldots,e^{i\theta_{n}})$, the nonzero elements of $\mathbf{A}\mathbf{U}$ are at the same places as those of $\mathbf{A}$, where $\mathbf{A}$ is of the form (21). Therefore, due to the structure of $\widehat{\mathbf{A}\mathbf{U}}$, for any $j$ such that $r<j\leq n$, the two diagonal elements of $d_{j}$ are equal to the two diagonal elements of another $d_{k}$ where $r<k\leq n$, $k\neq j$, which implies $\lambda_{j}=\lambda_{k}$. This gives (24) from which one concludes as before that for all $r<j\leq n$, $T_{j}$ is linear, and, therefore, $X_{j}=T_{j}(X^{*}_{j})$ is normal. ∎

V Application to Blind Source Extraction

The theoretical setting of the blind source extraction problem is as follows [9]. We are given $n$ (zero-mean) independent (real or complex) “sources” $X=(X_{1},X_{2},\ldots,X_{n})^{t}$ which are mixed using an $n\times n$ invertible (real or complex) matrix $\mathbf{M}$, resulting in the observation $Y=\mathbf{M}X$. The covariance matrix $\mathbf{K}_{Y}$ of $Y$ can be estimated but both $\mathbf{M}$ and $X$ are unknown. Since one can introduce arbitrary scaling factors in $\mathbf{M}$ and $X$ for the same observation $Y$, we can assume an arbitrary normalization of the sources. For convenience we assume here that they have the same entropies:

[TABLE]

Blind source extraction (or partial BSS) of $m$ sources ($1\leq m\leq n$) aims at finding a (full rank) $m\times n$ matrix $\mathbf{W}$ such that $Z=\mathbf{W}Y$ is composed of $m$ (out of $n$) original sources, up to order and scaling. In other words $\mathbf{A}=\mathbf{W}\mathbf{M}$ should have exactly one nonzero element per row.

Definition 2** (Contrast function [9])**

A contrast $\mathcal{C}(\mathbf{W})$ is a function that is invariant to permutation and scaling of the rows $\mathbf{w}_{i}$ of $\mathbf{W}$, and such that it achieves a minimum if only if $\mathbf{A}=\mathbf{W}\mathbf{M}$ has one nonzero element per row.

Theorem 3

Assume that at most one source is normal. Then

[TABLE]

where $\mathbf{w}_{i}$ are the rows of $\mathbf{W}$, is a contrast function.

Such a contrast function was first proposed by Pham [19] (see also [20]) in the real case with a different proof that uses the classical EPI for $m=1$ and Hadamard’s inequality. It is particularly interesting to rewrite it in terms of the matrix EPI:

Proof:

The real and complex cases being similar, we prove the result in the real case. Let $\mathbf{A}=\mathbf{W}\mathbf{M}$ and let $X^{*}$ be as in (3). For i.i.d. components we can rewrite [6, Eq. (13)] as $h(\mathbf{A}X^{*})=mh+\frac{1}{2}\log|\mathbf{A}\mathbf{A}^{t}|$ where $h$ is the common value of (34). Since $Z=\mathbf{W}Y=\mathbf{A}X$, up to an additive constant we may decompose $\mathcal{C}$ as

[TABLE]

where

[TABLE]

The term $\mathcal{C}_{i}(\mathbf{W})$ is minimum (with minimum value $=0$) if and only if the components $Z_{i}$ of $Z$ are independent.

The $\mathcal{C}_{h}(\mathbf{W})$ is minimum (with minimum value $=0$) if and only if equality holds in (3). Since at most one source is normal, at most one source present in $\mathbf{A}X$ can be unrecoverable. But if one (normal) source is not recoverable, the canonical form (21) implies that at most one column of $\mathbf{A}_{u}$ is nonzero, which contradicts the maximality of $r$ in Lemma 3. Therefore, $r=m$ and the canonical form of $\mathbf{A}$ becomes $(\mathbf{I}_{m}\mid 0)$.

With the additional constraint $\mathcal{C}_{i}(\mathbf{W})=0$ that components of $Z=\mathbf{A}X$ are independent, it follows from the Darmois–Skitovich theorem [21] (see [14] in the complex case) that $\mathbf{A}$ has exactly one nonzero per row. ∎

Interestingly, the contrast function in the form (36) represents a transition between the two well-known extreme cases:

• •

$m=1$, for which $\mathcal{C}_{i}=0$ where each source is extracted one by one using the classical EPI (minimize $\mathcal{C}_{h}$);

• •

$m=n$, for which $\mathcal{C}_{h}=0$, where all $n$ sources are separated simultaneously; we are then reduced to an independent component analysis (ICA) problem [21, 14] in which the multivariate “mutual information” $\mathcal{C}_{i}=D(p(Z)\|\prod_{i}p(Z_{i}))$ is minimized.