Mathematics in Independent Component Analysis

234 Chapter 16. Proc. ICA 2006, pages 917-925
for all xN∈R, because A is invertible.
Now aNN� 0, otherwise XN= aNGS G which contradicts (ii) for X. If also aNG� 0,
then by equation (3), h ′′ N is constant and therefore S N Gaussian, which again contradicts
(ii), now for S. Hence aNG= 0. By (3), aGNaGG= 0, and again aGG�0, otherwise
XG= aGNS N contradicting (ii) for S. Hence also aGN= 0 as was to show.
General proof. In order to give an idea of the main proof without getting lost in details,
we have divided it up into a sequence of lemmas; these will not be proven due
to lack of space. The characteristic function of the random vector X is defined by
�X(x) := E(exp ix⊤X), and since X is assumed to have existing covariance,�X is twice
continuously differentiable. Moreover by definition�AS(x)=�S(A ⊤x), and the characteristic
function of an independent random vector factorizes into the component characteristic
functions. So instead of using pX as in the 2-dimensional example, we use�X,
having similar properties except for the fact that the range is now complex and that the
differentiability condition can be considerably relaxed.
We will need the following lemma, which has essentially been shown in [9]; here
∇ f denotes the gradient of f and H f its Hessian.
Lemma 1. Let X∈ L2(Ω,R m ) be a random vector. Then X is Gaussian with covariance
2C if and only if it satisfies�XH�X −∇�X(∇�X) ⊤ + C�X 2≡ 0.
Note that we may assume that the covariance of S (and hence also of X) is positive
definite — otherwise, while still keeping the model, we can simply remove the subspace
of deterministic components (i.e. components of variance 0), which have to be mapped
onto each other by A. Hence we may even assume Cov(SG)=I, after whitening as
described in section 1.1. This uses the fact that the basis within the Gaussian subspace
is not unique. The same holds also for the non-Gaussian subspace, so we may choose
any BN∈ Gl(n) and BG∈ O(d−n) to get
� �
ANNBN
X= (B
AGNBN
−1
N SN)+
� �
ANGBG
(B
AGGBG
⊤ GSG). (4)
Here only orthogonal matrices BG are allowed in order for B⊤ GSG to stay decorrelated,
with SG being decorrelated.
The next lemma uses the dimension reduction model for X and S to derive an explicit
differential equation for ˆSN. The Gaussian part ˆSG in the following lemma vanishes
after application of lemma 1.
Lemma 2. For any basis BN∈ Gl(n), the non-Gaussian source characteristic function
ˆSN∈C 2 (Rn ,C) fulfills
�
ANNBN
ˆSNHˆSN −∇ˆSN(∇ˆSN) ⊤� B ⊤ NA⊤ GN + 2ANGA ⊤ GG ˆS 2 N≡ 0. (5)
Lemma 3. Let (ANN, ANG)∈Mat(n×(n+(d−n))) be an arbitrary full rank matrix.
If rank ANN < n, then we may choose coordinates BN ∈ Gl(n), BG ∈ O(d−n) and
M∈Gl(n) such that for arbitrary matrices∗∈Mat((n−1)×(n−1)),∗ ′ ∈ Mat((n−
1)×(d− n−1)):
MANNBN=
0 0
0∗
and MANGBG= 1 0
0∗ ′