INTRODUCTION TO STATISTICS AND ECONOMETRICS
CONDITIONAL PROBABILITY AND INDEPENDENCE
1.2.2 Axioms of Conditional Probability
The concept of conditional probability is intuitively easy to understand. For example, it makes sense to talk about the conditional probability that number one will show in a roll of a die given that an odd number has occurred. In the frequency interpretation, this conditional probability can be regarded as the limit of the ratio of the number of times one occurs to the number of times an odd number occurs. In general we shall consider the “conditional probability of A given B,” denoted by P(A B), for any pair of events A and В in a sample space, provided P(B) > 0, and establish axioms that govern the calculation of conditional probabilities.
Axioms of Conditional Probability
(In the following axioms it is assumed that P(B) >0.)
|
(1) |
P(A 1 |
B) > |
0 for |
any |
event A. |
|
|
(2) |
P(A | |
B) = |
1 for |
any |
event A I |
) B. |
|
(3) |
If (A |
n B), |
і = 1 |
, 2, . |
. . , are mutually exclusive, then |
|
|
P(A і |
и a2 |
и. . . |
1 В) |
= P(Aj | |
B) + P(A21 B) +_______ |
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|
(4) |
if £: |
J H and В |
D G |
and P{G) |
Ф 0, then |
|
|
P(H |
И) _ |
P{H) |
||||
|
P(G | |
В) |
P(G) |
Axioms (1), (2), and (3) are analogous to the corresponding axioms of probability. They mean that we can treat conditional probability just like probability by regarding В as the sample space. Axiom (4) is justified by observing that the relative frequency of H versus G remains the same before and after В is known to have happened. Using the four axioms of conditional probability, we can prove
THEOREM 2.4.1 P(A В) = P(A П B)/P(B) for any pair of events A and В such that P(B) > 0.
Proof. From axiom (3) we have (2.4.1) P(A І В) = P(A П В I В) + P(A П В B),
where В denotes the complement of B. But from axioms (2) and (3) we can easily deduce that P(C В) = 0 if С П В = 0. Therefore we can eliminate the last term of (2.4.1) to obtain
(2.4.2) P(AB) = P(A П BB).
The theorem follows by putting H = А П В and G = В in axiom (4) and noting P(B I Б) = 1 because of axiom (2). □
The reason axiom (1) was not used in the above proof is that axiom (1) follows from the other three axioms. Thus we have proved that (2), (3), and (4) imply Theorem 2.4.1. It is easy to show the converse. Therefore we may postulate either axioms (2), (3), and (4) or, more simply, Theorem 2.4.1 as the only axiom of conditional probability. Most textbooks adopt the latter approach.
If the conditioning event В consists of simple events with equal probability, Theorem 2.4.1 shows P(A В) — n{A П B)/n(B). Therefore, the counting techniques of Section 2.3 may be used to calculate a conditional probability in this case.
