Page 26 - DMTH404_STATISTICS

Statistics

Notes Boole’s inequality : If A , A , . . . , A are N events, then
1 2 N

N


N
P  A j   P(A )

  j
 J 1  j 1


Proof : By P5, the result is true for N = 2. Assume that it is true for N  r, and observe that
A A ...  A is the same as B  A , where B = A  A  . . . Ar. Then by P5,
1 2 r+1 r+1 1 2
r

r 1

P  A   P(B  A ) P(B) P(A )   P(A ) P(A ),






 j   r 1 r 1 j r 1
 J 1  j 1


where the last inequality is a consequence of the induction hypothesis. Hence, if Boole’s inequality
holds for N  r, it holds for N = r + 1 and hence for all N  2.
A similar induction argument yields
P7 : If A , A , . . . , A are pair wise disjoint events, i.e., if A  A = , i  j, then
1 2 n i j

n
P  A   P(A ) P(A ) ... P(A ) ...(3)



 j   1 2 n
 J 1 

We sometimes refer to the relation (3) as the Property of finite additivity.
We can generalise P7 to apply to an’infinite sequence of events.
P8 : If (A , n  1 ) is a sequence of pair wise disjoint events, then
n




P  A j   P(A )

  j
 J 1  j 1


P8 is called the -additivity pioperty.
In the general theory of probability, which covers non-discrete sample spaces as well, -additivity
and therefore finite additivity is included as an axiom to be satisfied by probabilities of events.
We now discuss some examples based on the above properties.
Example 3: Let us check whether the probabilities P(A) and P(B) are consistently defined
in the following cases.
(i) P(A) = 0.3 P(B) = 0.4, P(A  B) = 0.4
(ii) P(A) = 0.3 P(B) = 0.4, P(A  B) = 0.8
Here we have to see whether P1, P2, P3, P5 and P6 are satisfied or not. P4, P7 and P8 do not apply
here since we are considering only two sets. In both the cases P(A) and P(B) are not consistently
defined. Since A  B  A, by P6. P(A  B)  P(A). In case (i), P(A  B) = 0.4 > 0.3 = P(A), which is
impossible. Similar is the’situatmn with case (ii). Moreover, note that case (ii) also violates
P1 and P2. Recall that by P2,
P (A  B) = P(A) + P(B) – P(A  B)
but P(A) + P(B) – P(A  B) = 0.3 + 0.4 – 0.8 = – 0.1 which is impossible.

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