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Unit 18: The Weak Law



            A stronger version of this result due to Borel and Cantellistates that the above ratio k/n tends to  Notes
            p not only in probability, but with probability 1. This is the strong law of large numbers (SLLN).
            What is the difference between the  weak law  and the  strong law? The strong  law of  large
            numbers states that if { } is a sequence of positive numbers converging to zero, then
                               n
                 k     
              P    p   n                                                            ...(18.2)
            n 1   h     
             
                                                                                k    
            From Borel-Cantelli lemma [see (2-69) Text], when (13-2) is satisfied the events  An      p   n
                                                                                h     
            can occur only for a finitenumber of indices n in an infinite sequence, or equivalently, the events
              k     
               p   n   occur infinitely often, i.e., the event k/nconverges to palmost-surely.
              h     

            Proof: To prove (18.2), we proceed as follows. Since
                                       k
                                                        4
                                          p     |k – np |     n 4
                                                            4
                                       h
            we have

                           n                        k         k       
                           (k np) p (k) 4 n   4  n   P     p      p   
                                   4
                                                4
                                           4
                                                              P
                              
                                                           
                                    n
                           
                          k 0                        n        n     
            and hence
                         n
                                 4
                          (k np) p (k)
                             
               k                 n
                         
                      
            P    p     k 0  4  4                                                       ...(13.3)
               n              n
            where
                      n      n   k  n k
                                     
            p (k) P   X     p q
                          k 
                 
             n          i
                               k
                      i 1    
                     
            By direct computation
                                       4
                                                      4
                                                      
                                        
             n                   n         n   
             (k np)4pn(k) E   X   np     E  X   p 
                          
                
                                       
                                                     
                                                 i
                                  i
                               
             
                                               
            k 0                   i 1          i 1    
                     4
                   n      n  n  n  n
                       
              E  Y i     E(YY YY )
                                      k
                                         l
                                       j
                                    i
                    i 1      i 1 k 1 j 1 l 1
                 
                                
                            
                              
                         
              n              n  n                  n  n
                                    3
                                                         2
                   4
                                                              2
               E(Y ) 4n(n 1)  E(Y )E( Y ) 3n(n 1)  E(Y )E(Y )
                                               
                    
                          
                                          
                  i
                                                              j
                                    i
                                                         i
                                        j
                               
              i 1            i 1 j 1              i 1 j 1
              
                             
                                                   
                                                     
                                   2
            = n(p  + q )pq + 3n(n – 1)(pq)  [n + 3n(n – 1)]pq
                    3
                3
            = 3n pq,                                                                 ...(18.4)
               2
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