## Independence and Negation

Two events A,B are said to be statistically independent if and only
if
P(A and B) = P(A)P(B)
Intuitively it's clear that if A and B are independent events, then
A and B' (the negation of B) must also be independent. To demonstrate
this formally, notice that we have
P(B) + P(B') = 1 and P(B|A) + P(B'|A) = 1
Therefore, on the assumption that A and B are independent, so (1)
applies, and recalling that P(A and B) = P(A) P(B|A) for any two
events A and B, we have P(B) = P(B|A), from which it follows (in
view of the two preceding identities) that P(B') = P(B'|A), and
so P(A and B') = P(A) P(B'), which confirms that A and B' are
independent.
A more direct approach, pointed out by Noel Vaillant, avoiding the
use of conditional probabilities, is to note that
P(A) = P(A and (B or B')) = P(A and B) + P(A and B')
Since, by assumption, P(A and B) = P(A)P(B), it follows immediately
that
P(A and B') = P(A) - P(A)P(B) = P(A)P(B')
We can also consider this as a geometrical proposition. Suppose
A and B are overlapping regions of an overall region S as shown
below.
Let a,b,c,d denote the areas of the enclosed regions as shown in
this figure. Regarding the total region S as the "sample space",
the overall content of this space is a+b+c+d, and we can use this
to normalize each of the other areas. Hence, the normalized areas
of the various regions are
a+b+c+d a+c b+c
S = ------- A = ------- B = -------
a+b+c+d a+b+c+d a+b+c+d
c
Intersection(A,B) = -------
a+b+c+d
The condition corresponding to "independence of A and B" is that the
normalized area of the intersection of A and B equals the product of
the normalized areas of A and B, as follows
c / a+c \ / b+c \
------- = ( ------- )( ------- )
a+b+c+d \ a+b+c+d / \ a+b+c+d /
This reduces to the simple condition
ab = cd
We can also evaluate the product of the normalized areas of A and
the S-complement of B as follows
/ a+c \ / a+d \ a^2 + ac + ad + cd
( ------- )( ------- ) = ------------------
\ a+b+c+d / \ a+b+c+d / (a+b+c+d)^2
Now, under the condition of "independence" of A and B, we can
substitute ab for cd in the right-hand expression, which gives
a^2 + ac + ad + ab a
------------------ = -------
(a+b+c+d)^2 a+b+c+d
This, as expected, is the normalized area of the intersection of A
with the S-complement of B, implying that these two regions also
satisfiy the mutual "independence" condition.

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