Databases Reference
In-Depth Information
S. Tsumoto
∆
(1
,j
)=(1+
q
+
s
)
x
1
j
4
4
x
2
k
−
x
2
j
x
1
k
k
=1
k
=1
∆
(2
,j
)=(1+
p
+
r
)
x
2
j
4
4
x
1
k
−
x
1
j
x
2
k
k
=1
k
=1
∆
(3
,j
)=(
p
−
q
+
ps
−
qr
)
x
2
j
4
4
×
x
1
k
−
x
1
j
x
2
k
k
=1
k
=1
∆
(4
,j
)=(
r
−
s
+
qr
−
ps
)
4
4
×
x
1
j
x
2
k
− x
2
j
x
2
k
k
=1
k
=1
Since
p
ps
= 0 gives the only reasonable
solution
p
=
q
and
r
=
s
, the following theorem is obtained.
−
q
+
ps
−
qr
=0and
r
−
s
+
qr
−
Theorem 11.
The third and fourth rows represented by a linear combination
of first and second rows (basis) will satisfy the condition of statistical inde-
pendence if and only if p
=
q and r
=
w.
8 Pseudo-Statistical Independence
Now, we will generalize the results shown in Sect. 7. Let us consider the
m
s
rows (columns). Thus, we assume a corresponding matrix with the following
equations.
×
n
contingency table whose
r
rows (columns) are described by
n
−
⎛
⎝
⎞
⎠
x
11
x
12
··· x
1
n
x
21
x
22
···
x
2
n
.
.
.
.
.
.
x
m
1
x
m
2
···
x
mn
m−s
(
x
m−s
+
p,
1
x
m−s
+
p,
2
···
x
m−s
+
p,n
)=
k
pi
(
x
i
1
x
i
2
···
x
in
)
i
=1
×
(1
≤
s
≤
n
−
1
,
1
≤
p
≤
s
)
(22)
Then, the following theorem about
∆
(
u,v
) is obtained.
