Page 264 - DMTH404_STATISTICS

Statistics

Notes since
3
p + q = (p + q) – 3p q – 3pq < 1, pq  1 / 2 < 1
3
3
2
2
Substituting (18.4) also (18.3) we obtain
 k  3pq
P   p    2 4

 n  n 
1
Let  = 1/8 so that the above integral reads and hence
n
  k 1   1 
 P   p  1/8   3pq   3pq(1 1  x  3/2 dx)
n 1  n n  n 1 n 3/2


= 3pq(1 + 2) = 9pq <  
1/8
thus proving the strong law by exhibiting a sequence of positive numbers  = 1/n that
n
converges to zero and satisfies (13-2).
We return back to the same question: “What is the difference between the weak law and the
strong law?.

 n 
”The weak law states that for every n that is large enough, the ratio  X /n  k/n is likely
i 
 i 1 

to be near p with certain probability that tends to 1 as nincreases. However, it does not say that
k/n is boundto stay near p if the number of trials is increased. Suppose (18.1) is satisfied for a
given  in a certain number of trials n . If additional trials are conducted beyond n , the weak law
0 0
does not guarantee that the new k/n is bound to stay near p for such trials. In fact there can be
events for which k/n > p + e, for n > n in some regular manner. The probability for such an event
0
is the sum of a large number of very small probabilities, and the weak law is unable to say
anything specific about the convergence of that sum.
However, the strong law states (through (18.2)) that not onlyall such sums converge, but the
total number of all such events where k/n > p +  is in fact finite! This implies that the probability
 k 
  p    of the events as n increases becomes and remains small, since with probability
 n 
1 only finitely many violations to the above inequality takes place as n  .
Interestingly, if it possible to arrive at the same conclusion using a powerful bound known as
Bernstein’s inequality that is based on the WLLN.

Bernstein’s inequality : Note that
k
 p    k > n(p + )
n
and for any  > 0, this gives e (k – n(p+)) > 1.

Thus

 k  n  n  k n k


P   p       p q
)] k
 n  k [n(p  

n
n (k n(p  
k
  e   ))   p q n k

k

k [n(p )]  
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