Mathematics in Independent Component Analysis

60 Chapter 2. Neural Computation 16:1827-1850, 2004
A New Concept for Separability Problems in BSS 1831
also holds globally (see theorem 1(ii)). In this case, we have for i �= j,
0 ≡ ∂i∂j ln f ≡ f ∂i∂j f − (∂i f )(∂j f )
f 2
,
so f is separated if and only if
f ∂i∂j f ≡ (∂i f )(∂j f )
for i �= j or even i < j. This motivates the following definition:
Definition 2. For i �= j, the operator
Rij : C 2 (R n , C) → C 0 (R n , C)
is called the ij-separator.
Theorem 1. Let f ∈ C 2 (R n , C).
f ↦→ Rij[ f ]:= f ∂i∂j f − (∂i f )(∂j f )
i. If f is separated, then Rij[ f ] ≡ 0 for i �= j or, equivalently,
f ∂i∂j f ≡ (∂i f )(∂j f ) (2.1)
holds for i �= j.
ii. If f is positive and Rij[ f ] ≡ 0 holds for all i �= j, then f is separated.
If f is assumed to be only nonnegative, then f is locally separated but
not necessarily globally separated (if the support of f has more than one
component). See Figure 1 for an example of a nonseparated density with
R12[ f ] ≡ 0.
Proof of Theorem 1.i. If f is separated, then f (x) = g1(x1) ···gn(xn) or
short f ≡ g1 ⊗···⊗gn,so
and
∂i f ≡ g1 ⊗···⊗gi−1 ⊗ g ′ i ⊗ gi+1 ⊗···⊗gn
∂i∂j f ≡ g1 ⊗···⊗gi−1 ⊗ g ′ i ⊗ gi+1 ⊗···⊗gj−1 ⊗ g ′ j ⊗ gj+1 ⊗···⊗gn
for i < j. Hence equation 2.1 holds.
ii. Now assume the converse and let f be positive. Then according to the
remarks after lemma 1, Hln f (x) is everywhere diagonal, so lemma 1 shows
that ln f is linearly separated; hence, f is separated.