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Appendix
11
For proving theorems 1 and 2 we firstly assume that the conditions of SP (arrived
at in section 2.1) are satisfied. That is,
C
≡
(
C
1
&
C
2
&
C
3
)
ʸ
=(
A
1
−
B
1
)+(
A
2
−
B
2
)+(
ʲ
−
ʱ
)
>
0
Further, we stipulate the following definitions:
a = (members of A in partition 1)/(total members of A)
b = (members of B in partition 1)/(total members of B)
ʱ
=
aA
1
+
A
2
(1
−
a
)
b
)
A
1
,
A
2
,
B
1
and
B
2
have the same meanings defined in section 2.1. We have
defined
ʱ
and
ʲ
differently than what we had done in section 2.1 only to ease
the proofs of the following theorems; otherwise the two sets of definitions are
mathematically equivalent. To take an example, in Table 1 (Type I SP),
A
1
=
180
/
200,
A
2
= 100
/
300,
B
1
= 480
/
600,
B
2
=10
/
100,
a
= 200
/
500,
b
= 600
/
700.
Hence,
ʱ
= (180/500) + (100/500) = 280/500 = 56% and
ʲ
= (480/700) +
(10/700) = 490/700 = 70%.
Theorem 1.
Simpson's paradox results only if A
1
ʲ
=
bB
1
+
B
2
(1
−
=
A
2
.
Proof: Let us assume that
A
1
=
A
2
. Then,
ʱ
=
aA
1
+
A
2
(1
a
)=
aA
1
+
A
2
−
aA
2
=
A
1
=
A
2
. Given this, there are three possible scenarios. (I)
B
1
>B
2
,or
(II)
B
1
<B
2
or (III)
B
1
=
B
2
.
(I) If
B
1
>B
2
,then[
B
1
b
+
B
1
(1
−
−
b
)]
>
[
B
1
b
+
B
2
(1
−
b
)]. Therefore,
B
1
>ʲ
.
Yet, if
A
1
≥
B
1
,and
ʱ
=
A
1
,then
ʱ>ʲ
, which contradicts the assumption
that
ʲ
ʱ
. Therefore, if
A
1
=
A
2
, then it can't be that
B
1
>B
2
.
(II) If
B
1
<B
2
,then[
B
1
b
+
B
2
(1
≥
−
b
)
<
[
B
2
b
+
B
2
(1
−
b
)] =
B
2
. Therefore,
ʲ<
B
2
.Yet,
A
2
≥
B
2
,
A
1
≥
B
2
,and
ʱ
≥
B
2
>ʲ
. This contradicts the assumption
that
ʲ
ʱ
. Therefore, if
A
1
=
A
2
, then it can't be the case that
B
1
<B
2
.
(III) If
B
1
=
B
2
,then
ʲ
=
bB
1
+
B
2
(1
≥
−
b
)=
bB
1
+
B
1
(1
−
b
)=
B
1
.Giventhat
A
1
≥
ʱ
. Therefore,
ʲ
=
ʱ
.Since
A
1
=
A
2
=
ʱ
,and
B
1
=
B
2
=
ʲ
,itmustbethat
A
1
=
B
1
,
A
2
=
B
2
,
and
ʱ
=
ʲ
.That
ʱ
=
ʲ
contradicts the assumption that out case is paradoxical,
characterized by the reversal which we don't find here. Therefore, if
A
1
=
A
2
,it
can't be the case that
B
1
=
B
2
. Therefore,
A
1
=
A
2
. Without
A
1
=
A
2
, Simpson's
paradox cannot occur.
Theorem 2.
Simpson's paradox arises only if B
1
B
1
,
A
1
=
ʱ
,and
B
1
=
ʲ
,then
ʱ
≥
ʲ
.Yet,byassumption,
ʲ
≥
=
B
2
.
Proof: Let us assume that
B
1
=
B
2
.Then
ʲ
=
bB
1
+
B
2
(1
−
b
)=
bB
1
+
B
1
(1
−
b
)=
B
1
=
B
2
.Giventhat
A
1
>A
2
,itistruethat[
aA
1
+
A
2
(1
−
a
)]
>
[
aA
2
+
A
2
(1
−
a
)]. Given that
A
1
>A
2
, it follows that [
aA
1
+
A
2
(1
−
a
)]
>
[
aA
2
+
A
2
(1
−
a
)]. Therefore,
ʱ>A
2
.Yet,
A
2
≥
B
2
=
ʲ
.So
ʱ>ʲ
, which contradicts the
assumption. Therefore,
B
1
=
B
2
. Without
B
1
=
B
2
, Simpson's paradox cannot
occur.
11
We are indebted to Davin Nelson for the following proofs.
