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INDEPENDENT DISCRETE RV Example 1 EX: The following formulas define the behavior of conditional probabilities: €
P ( A | B ) =
P ( A , B )
P ( B )
P ( A and B ) P ( B )
P ( A ∩ B )
P ( B )
(always true) € P ( A | B ) = P ( A ) (if A and B independent) € P ( A , B ) = P ( A ) P ( B ) (if A and B independent) For the following formulas, determine whether the formula is always true when A and B are independent. a) €
P ( A | B ) P ( B | A ) = P ( A , B )
b) €
P ( A '| B ') = 1 − P ( A | B )
c) If € P ( A ) ≠ 0 and € P ( B ) ≠ 0 , then €
P ( A , B ) ≠ 0
d) For an arbitrary event, C , A is independent of €
B ∩ C.
SOL'N: a) The first equation follows by direct application of the formulas for independent events: €
P ( A | B ) P ( B | A ) = P ( A ) P ( B ) = P ( A , B )
b) Because A and B are independent, we may immediately simplify the right- hand side of the equation: €
1 − P ( A | B ) = 1 − P ( A ) = P ( A ')
Now we consider the left side of the equation: €
P ( A '| B ') =
P ( A ', B ')
P ( B ')
P ( A '∩ B ')
P ( B ')
Using the Law to Total Probability, we may relate € P ( A '∩ B ') to €
P ( B ') :
P ( A '∩ B ') + P ( A ∩ B ') = P ( B ')
or €
P ( A '∩ B ') = P ( B ') − P ( A ∩ B ')
Again using the Law of Total Probability, we may relate €
P ( A ∩ B ')
to P ( A ).
INDEPENDENT DISCRETE RV Example 1 (cont.) €
P ( A ∩ B ') + P ( A ∩ B ) = P ( A )
or €
P ( A ∩ B ') = P ( A ) − P ( A ∩ B ) = P ( A ) − P ( A ) P ( B )
Substituting into our equation for € P ( A '∩ B ') , we have the following result: €
P ( A '∩ B ') = P ( B ') − ( P ( A ) − P ( A ) P ( B )) = ( 1 − P ( B )) − ( 1 − P ( B )) P ( A )
or €
P ( A '∩ B ') = ( 1 − P ( B ))( 1 − P ( A )) = P ( A ') P ( B ')
The left side of the original equation now simplifies to €
P ( A '):
P ( A '| B ') =
P ( A ', B ')
P ( B ')
P ( A ') P ( B ')
P ( B ')
= P ( A ')
Thus, the left and right sides of the original equations are equal whenever A and B are independent: €
P ( A '| B ') = 1 − P ( A | B )
NOTE: Our derivation shows that, when A and B are independent events, we also have independence of A and B ', A ' and B , and A ' and B '. c) If € P ( A ) ≠ 0 and € P ( B ) ≠ 0 , then € P ( A , B ) = P ( A ) P ( B ) ≠ 0 follows immediately. What is less immediately obvious is that this result implies that € P ( A ∩ B ) ≠ ∅. In other words, the intersection of A and B is nonempty. Equivalently, A and B must overlap on a Venn diagram. d) For an arbitrary event, C , we investigate the independence of A and €
B ∩ C
by examining the conditional probability for A. €
P ( A | B ∩ C ) =
P ( A ∩ B ∩ C )
P ( B ∩ C )
Since € B ∩ C may be any part of B , we may consider the case where C = A ∩ B :
