Page 46 - DMTH404_STATISTICS

Statistics

Notes We now proceed to study some implications of independence of two events A and A .
1 2
Recall that
c
P(A ) = P(A  A ) + P(A  A ).
1 1 2 1 2
Then
c
P(A  A ) = P(A ) – P(A  A )
1 2 1 1 2
Now, if A and A are independent, we get
1 2
c
P(A  A ) = P(A ) {1 – P(A )}
1 2 1 2
c
= P(A ) P( A )
1 2
c
Thus, the independence of A andA implies that of A and A . Now interchange the roles of
1 2 1 2
c
A and A What do you get? We get that if A and A are independent, then so are A and A . The
1 2 1 2 2 2
c
c
c
independence of A and A then implies the independence of A and A too.
1 2 1 2
Now here is an interesting fact.
If A is an almost sure event, then A and another event B are independent.
C
Let us see how . Since A is an alm ost sure event, P(A ) = 1. H ence P(A ) = 0 and therefore,
C
P(A  B) = 0. In particular,
P(B) = P(A  B) + P(A  B) = P(A  B).
c
One consequence of this is that
P(A  B) = I.P(B) = P(A) P(B),
which implies that A and B are independent.
Can you prove a similar result for a null event ? You can check that if A is a null event, then A and
any other event B are independent.
Now, can we extend the definition of independence of two events to that of the independence of
three events? The obvious way seems to be to call A , A , A , independent if P(A  A  A )
1 2 3 1 2 3
= P(A )P(A )P(A ). But this does not work. Because if 3 events are independgnt, we would expect
1 2 3
any two of them also to be independent. But this is not ensured by the condition above. To
appreciate this, consider the case when A = A = A, 0 < P (A) < 1, and P(A ) = 0. Then P(A  A )
1 2 3 1 2
2
= P(A)  P(A ) P(A ) = P[(A)] .
1 2
Thus, A and A are not independent, but P(A  A  A ) = P(A ) P(A ) P(A ) is satisfied. So, to get
1 2 1 2 3 1 2 3
around this problem we add some more conditions and get the following definition
Definition 5 : Three events A , A and A corresponding to the same random experiment are said
1 2 3
to be stochastically or mutually independent if
P(A  A ) = p(A ) P(A )
1 2 1 2
P(A  A ) = P(A ) P(A ) ...(15)
2 3 2 3
and P(A  A  A ) = P(A ) P(A ) P(A ).
1 2 3 1 2 3

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