Digital Signal Processing Reference
In-Depth Information
4
5
3
4
2
3
1
2
0
1
1
0
2
1
3
2
4
3
5
6
4
6
4
2
0
2
4
6
4
3
2
1
0
1
2
3
4
5
Figure 4.6
Kurtosis maximization, second example: Source and mixture scatterplots. A two-
dimensional Laplacian distribution (super-Gaussian) with 20000 samples was
chosen, again mixed by a rotation of 30 degrees.
(lemma 3.7), we therefore get
kurt(
y
)=kurt(
q
1
s
1
)+kurt(
q
2
s
2
)=
q
1
kurt(
s
1
)+
q
2
kurt(
s
2
)
.
By normalization, we can assume
E
(
s
1
)=
E
(
s
2
)=
E
(
y
2
)=1,so
q
1
+
q
2
= 1, which means that
q
lies on the circle
q
S
1
.
∈
The question is: What are the maxima of
S
1
−→ R
q
1
kurt(
s
1
)+
q
2
kurt(
s
2
)
q
→|
|
R
2
can be quickly solved
using Lagrange multipliers. Using the function without absolute values,
we can take derivatives and get two equations:
This maximization on a smooth submanifold of
4
q
i
kurt(
s
i
)+2
λq
i
=0
for
λ
∈ R
,
i
=1
,
2. So
2
q
1
kurt(
s
1
)=
2
q
2
kurt(
s
2
)
λ
=
−
−
or
q
1
=0or
q
2
= 0 (assuming that the kurtoses are not zero). Obviously
only the latter two equations correspond to maxima, so from
q
S
1
we
∈
get solutions
q
∈{±
e
1
,
±
e
2
}
with the
e
i
denoting the unit vectors. And this is exactly what we
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