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Unit 3: Conditional Probability and Independence Baye’s Theorem

Let’s try to understand this through an example. Notes

Example 20: An unbiased coin is tossed thx& times. Let A, denote the event that a head
turns up on the j-th toss, j = 1,2,3. Let’s see if A , A and A are independent.
1 2 3
Since the coin is unbiased, we assign the same probability, 1/8, to each of the eight possible
outcomes.
Check that
P(A ) = P(A ) = P(A ) = 1/2
1 2 3
P(A  A ) = P(A  A ) = P(A  A ) = 1/4, and
1 2 2 3 3 1
P(A  A  A ) = 1/8.
1 2 3
Thus, all the four equations in (15) are satisfied and the events A , A , A are mutually independent.
1 2 3
We have seen that the last condition in (15) alone is not enough, since it does not guarantee the
independence of pairs of events.
Similarly, the first three equations of (15) alone are not sufficient to guarantee that all the four
conditions required for mutual independence would be satisfied. To see this, consider the
following example.

Example 21: An unbiased die is rolled twice. Let A denote the event “odd face on the
1
first roll”, A denote the event “odd face on the second roll” and A denote the event that the total
2 3
score is odd. With the classical assignment of probability 1/36 to each of the sample points, you
can easily check that
P(A ) = P(A ) = P(A ) = 18/36 = 1/2, and that
1 2 3
P(A  A ) = P(A  A ) = P(A  A ) = 9/36 = 1/4.
1 2 2 3 3 1
Thus, the first three equations in (15) are satisfied. But the last one is not valid. The reason for it
is that P(A  A  A ) is zero (Do you agree ?), and P(A ), P(A ), P(A ) are all positive.
1 2 3 1 2 3
If the first three equationsof (15) are satisfied, we say that A , A and A are pairwise independent.
1 2 3
Example 21 shows that pairwise independence does not guarantee mutual independence.
Now we are sure you can define the concept of independence of n events. Does your definition
agree with Definition 6?
Definition 6 : The n events A , A , , . . , A corresponding to the same random experiment are
1 2 n
mutually independent if for all r = 2. . . ., n, 1  i < i < . . . < i  n, the product rule holds.
j 2 r
r

P(A  ... A )  P(A ) ...(17)
1 i r i j 1 j i

 n 
Since r of the n events can be chosen in   ways, (17) represents
r
 
n
 n     n  n

     ...    = 2 – n – 1
2
3
n
     
conditions.
Try to write Definition 6 for n = 3 and see if it matches Definition 5.

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