Page 34 - Probability and Statistical Inference

1. Nations of Probability 11

we apply parts (ii)-(iii) from the Theorem 1.3.1 to write

(ii) ⇒ (iii) ⇒ (iv) : These are left as the Exercise 1.4.2.
(iv) ⇒ (i) : Assume that A and B are independent events. That is, in view
c
c
c
c
c
of the Theorem 1.4.1, we have P(A ∩ B ) = P(A )P(B ). Again in view of the
c
Theorem 1.4.1, we need to show that P(A ∩ B) = P(A)P(B). Now, we combine
DeMorgan’s Law from (1.2.5) as well as the parts (ii)-(iv) from the Theorem
1.3.1 to write

which is the same as P(A)P(B), the desired claim. !
Definition 1.4.3 A collection of events A , ..., A are called mutually inde-
n
1
pendent if and only if every sub-collection consists of independent events, that
is
for all 1 = i ≤ i < ... < i ≤ n and 2 ≤ k ≤ n.
1 2 k
A collection of events A , ..., A may be pairwise independent,
1
n
that is, any two events are independent according to
the Definition 1.4.2, but the whole collection of sets may not
be mutually independent. See the Example 1.4.2.
Example 1.4.2 Consider the random experiment of tossing a fair coin twice.
Let us define the following events:
A : Observe a head (H) on the first toss
1
A : Observe a head (H) on the second toss
2
A : Observe the same outcome on both tosses
3
The sample space is given by S = {HH, HT, TH, TT} with each outcome being
equally likely. Now, rewrite A = {HH, HT}, A = {HH, TH}, A = {HH, TT}.
1
3
2
Thus, we have P(A ) = P(A ) = P(A ) = 1/2. Now, P(A ∩ A ) = P(HH) = 1/4 =
3
1
2
1
2
P(A )P(A ), that is the two events A , A are independent. Similarly, one should
2
1
2
1
verify that A , A are independent, and so are also A , A . But, observe that
3
2
1
3
P(A ∩ A ∩ A ) = P(HH) = 1/4 and it is not the same as P(A )P(A )P(A ). In
2
3
3
1
1
2
other words, the three events A , A , A are not mutually independent, but they
1
2
3
are pairwise independent. !