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where now, rather than as linear combinations,
Z
1
,
Z
2
may be interpreted as giving
quantifications. The constraint becomes diag(
Z
1
G
1
G
1
Z
1
+
Z
2
G
2
G
2
Z
2
)
=
2
I
. The gen-
eralization to
p
categorical variables is to minimize
p
p
G
k
Z
k
−
M
1
p
2
,
w e
M
=
G
k
Z
k
k
=
1
k
=
1
subject to the constraint
p
1
(
Z
k
G
k
G
k
Z
k
)
=
p
I
.
diag
k
=
In terms of our notation for MCA, the criterion may be written
1
trace
(
Z
LZ
)
−
p
trace
(
Z
G
GZ
)
and the constraint diag
(
Z
LZ
)
=
p
I
. This constraint says that the sum of squares of each
of the
r
columns of
GZ
is
p
. Writing
as a diagonal matrix of
r
Lagrange multipliers,
minimization gives
1
p
(
G
G
)
Z
=
LZ
LZ
−
or
G
G
(
)
Z
=
p
LZ
(
I
−
)
=
LZ
.
This is a two-sided eigenvalue problem and hence we have the usual orthogonality
conditions that
Z
LZ
and
Z
G
GZ
are both diagonal. To comply with the constraint,
we normalize so that
Z
LZ
p
I
. Note that not only diag(
Z
LZ
p
I
, as required by
the constraint, but also the off-diagonal values are zero. This has not been imposed
as a constraint but is a natural consequence of the solution. When
p
=
)
=
=
2, it turns out
that
Z
LZ
2
I
implies the separate condition
Z
1
L
1
Z
1
=
Z
2
L
2
Z
2
=
I
which is a usual
normalization in CCA, but this separability does not extend to the case when
p
>
2.
We may rewrite the two-sided eigenvalue equation as
=
L
−
1
/
2
G
G
L
−
1
/
2
L
1
/
2
Z
L
1
/
2
Z
{
(
)
}
=
so that
L
1
/
2
Z
are eigenvectors of the normalized Burt matrix
L
−
1
/
2
(
G
G
)
L
−
1
/
2
,justas
is required for MCA. The interpretation of
Z
in terms of scores supports interest in the
quantified variables
G
1
Z
1
,
G
2
Z
2
,
,
G
p
Z
p
and their mean
M
. Each of these gives
n
points that may be plotted as individual points on the MCA to give a proper biplot of
individuals/samples and variables. However, because
n
is usually large in MCA, often
we would prefer some form of density plot for the samples and, rather than
p
such plots,
would be content just with a density plot of
M
. For any eigenvalue
...
φ
with corresponding
eigenvector
z
,wehave
1
(
G
G
1
Lz
,
)
z
=
φ
that is,
1
Lz
1
Lz
=
φ
.
