Independence and Negation

Two events A,B are said to be statistically independent if and only 
                  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

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 

      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


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

            Intersection(A,B) = -------

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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