Mathematics in Independent Component Analysis

Chapter 2. Neural Computation 16:1827-1850, 2004 65
1836 F. Theis
Putting this in equation 3.4 yields
0 = ( f ∂i∂j f − (∂i f )(∂j f ))(x)
= �
bkibkj((g1 ⊗···⊗gn)(g1 ⊗···⊗g ′′
k ⊗···⊗gn)
k
− (g1 ⊗···⊗g ′ k ⊗···⊗gn) 2 )(Bx)
= �
bkibkjg 2 1 ⊗···⊗g2 k−1 ⊗
k
�
gkg ′′
k
− g′2
k
for x ∈ R n . B is invertible, so the whole function is zero:
�
k
bkibkjg 2 1 ⊗···⊗g2 k−1 ⊗
�
gkg ′′
k
�
⊗ g 2 k+1 ⊗···⊗g2 n (Bx)
�
− g′2
k ⊗ g 2 k+1 ⊗···⊗g2n ≡ 0. (3.5)
Choose x ∈ R n with gk(xk) �= 0 for k = 1,...,n. Evaluating equation 3.5
at (x1,...,xl−1, y, xl+1,...,xn) for variable y ∈ R and dividing the resulting
one-dimensional equation by the constant g 2 1 (x1) ···g 2 l−1 (xl−1)g 2 l+1 (xl+1) ···
g 2 n (xn) shows
bliblj
�
glg ′′
l
�
− g′2
l
⎛
(y) =−⎝
�
k�=l
gkg
bkibkj
′′
k
g 2 k
− g′2
k
(xk)
⎞
⎠ g 2 l
(y) (3.6)
for y ∈ R. So for indices l and i �= j with bliblj �= 0, it follows from equation
3.6 that there exists a ∈ C such that gk satisfies the differential equation
ag2 l − glg ′′
l + g′2
l ≡ 0, that is, equation 3.3.
Proof of Theorem 2. i. S is assumed to have at most one gaussian or deterministic
component and existing covariance. Set X := AS.
We first show using whitening that A can be assumed to be orthogonal.
For this, we can assume S and X to have no deterministic component at all
(because arbitrary choice of the matrix coefficients of the deterministic components
does not change the covariance). Hence, by assumption, Cov(X)
is diagonal and positive definite, so let D1 be diagonal invertible with
Cov(X) = D2 1 . Similarly, let D2 be diagonal invertible with Cov(S) = D2 2 .
Set Y := D −1
1 X and T := D−1
2 S, that is, normalize X and S to covariance I.
Then
Y = D −1
1
X = D−1
1 AS = D−1
1 AD2T