Page 266 - DMTH404_STATISTICS

Statistics

Notes 
To compute the probability of the event  y , note that its complement is given by
n

n m
c
   
  y n    y c n
 n m  n m


and using Eq. (2-68) Text,
2 /4


m


 c  c 2 /4 2e

P   y n   P(y )   2e  n  2 .
n

 n m  n m n m 1 e  /4



This gives
2 /4



m
      2e
P   y n    1 P   y n    1  2  1 as m  


 n m    n m  1 e  /4


or,
 k 
P p     p   , for all n  m   1 as m  .

 n 
Thus k /n is bound to stay near p for all large enough n, in probability,a conclusion already
reached by the SLLN.
Discussion: Let Thus if we toss a fair coin 1,000 times, from the weak law
 k 1  1
P    0.01  .

 n 2  40
Thus on the average 39 out of 40 such events each with 1000 or more trials will satisfy the
 k 1 
inequality    0.1  or, it is quite possible that one out of 40 such events may not satisfy it.
 n 2 
As a result if we continue the coin tossing experiment for an additional 1000 moretrials, with k
representing the total number of successes up to the current trial n, for n = 1000  2000, it is quite
possible that for few such n the above inequality may be violated. This is still consistent with the
weak law, but “not so often” says the strong law. According to the strong law such violations can
occur only a finite number of times each with a finite probability in an infinite sequence of trials,
and hence almost always the above inequality will be satisfied, i.e., the sample space of
k/n coincides with that of p as n  .
Next we look at an experiment to confirm the strong law:
Example: 2n red cards and 2n black cards (all distinct) are shuffled together to form a
single deck, and then split into half. What is the probability that each half will contain n red and
n black cards?

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