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(repeated as appropriate) and ( I - H ) X the matrix of deviations from the means, that is,
the within-group variation. First we note that the decomposition
X
=
HX
+ (
I - H
)
X
(4.11)
satisfies X X = X HX + X ( I - H ) X , due to the idempotency of H , showing that T =
B + W has Type B orthogonality. However, Type A orthogonality does not hold for the
decomposition (4.11) since HXX ( I - H ) = 0.
In the following, we are interested in the accuracy of the recovery of both the
group means and the within-group values, and also to what extent Type A and Type B
orthogonality are valid. To define between-group predictivity we proceed similarly to the
discussion of measures of fit in Chapter 3 where we have a decomposition in terms of
the PCA of the canonical means and the fitted values
XL + ( X -
X ) L ,
XL =
(4.12)
where the fitted values X are given by (4.6). We would expect (4.12) to satisfy both
Type A and Type B orthogonality. The centring matrix C can easily be incorporated by
multiplying (4.12) from the left by C 1 / 2 . Note that due to the diagonality of N and
the idempotency of I - K 1 11 ,wehavethat C 1 / 2 is equal to N 1 / 2 for the weighted
case and, for the unweighted case, either to I - K 1 11 or simply I . Thus, for Type B
orthogonality,
L X CXL = L X C XL + L ( X
X ) C ( X
X ) L ,
(4.13)
and for Type A orthogonality,
1
/
2
/
1
2
= C 1 / 2 XLL X C
X ) LL ( X
X ) C 1 / 2
C 1 / 2 XLL X C
+ C 1 / 2
( X
.
(4.14)
Although (4.13) and (4.14) include the centring matrix C , they essentially reproduce
(3.12) and (3.14). To verify that they continue to be valid, we need to show that the
corresponding cross product terms vanish. That this is so for Type B orthogonality follows
from
L X C
X
L M 1 JM X CX
MJM 1
(
X
)
L
=
L
= L M 1 J [ M X CXM ] ( I J ) M 1 L
= L M 1 J ( I J ) M 1 L = 0 ,
(
I
)
( X CX ) M = WM
since the term in square brackets simplify on recalling that
and
L WL = M WM = I . In the case of Type A orthogonality, we have
C 1 / 2 XLL ( X
X ) C 1 / 2
= C 1 / 2 XMJM 1 LL ( I M 1 JM ) X C 1 / 2
=
M X C 1 / 2
C 1 / 2 XMJ [ M 1 LL M 1 ]
(
I
J
)
= C 1 / 2 XMJ ( I J ) M X C 1 / 2
= 0 ,
since W 1
= LL = MM .
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