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Unit 19: The Laws of Large Numbers Compared



            19.1 Strong Law of Large Numbers                                                      Notes


            We are now ready to give Etemadi’s proof of
            (7.1) Strong law of large numbers. Let X1, X2, ... be pairwise independent identically distributed
            random variables with E | X  | < . Let EX  =  and S  = X  + ... + X . Then S /n   a.s. as n  .
                                  i           i       n  1      n      n
            Proof : As in the proof of weak law of large numbers, we begin by truncating.
            (a) Lemma. Let Y  = K 1   and T  = Y  + ... + Y . It is sufficient to prove that T /n   a.s.
                          k   k (|X |k)  n  1     n                        n
                                 k
            Proof      P(|X | k)    P(|X | t)dt   E|X |  so P(X   Yk i.o.)  =  0.  This  shows  that
                                        
                            
                    k 1   k       0   1          1         k
                     
            |Sn(w) – Tn(w) <  a.s. for all n, from which the desired result follows.
            The second step is not so intuitive but it is an important part of this proof and the one given in
            Section 1.8.
                                  2
            (b) Lemma.      k 1 var(Y )/k   4E|X |  .
                              k
                                         1
                         
            Proof To bound the sum, we observe
                                          k
                       2
                  
            var(Y ) E(Y )   2yP(|Y | y)dy   2yP(|X | y)dy
                                   
                                                    
                 k    k    0     k        0     1
            so using Fubini’s theorem (since everything is  0 and the sum is just an integral with respect to
            counting measure on {1, 2, ....})
                           
                 2
                     2
             E(Y )/k    k  2  0   1 (y k) 2yP(|X | y)dy
                                         
                                        1
                 k
                                
                        
            k 1         k 1
             
                      
                    
                    2
                                 
            =   0    k 1 (y k) 2yP(|X | y)dy
                       
                                1
                  k 1   
                 
                         
                                
            Since E|X | =    0   P(|X | y)dy , we can complete the proof by showing
                    1         1
            (c) Lemma. If y  0 then  2y   k y k   2    4.
                                     
            Proof We being with the observation that if m  2 then
                                              
                                                  2
                                         k   2       x dx (m 1)  1
                                                  
                                                     
                                                         
                                       
                                       k m    m 1
            When y  1 the sum starts with k = [y] +1  2 so
                                         2y   k  2    2y/[y] 4
                                                        
                                            
                                           k m
            since y/[y]  2 for y  1 (the worst case being y close to 2). To cover 0  y < 1 we note that in this
            case
                                                       2 
                                       2y  k   2    2y 1    k       4
                                                 
                                         k y        k 2  
                                                      
                                          
            The first  two steps, (a) and (b) above, are standard.  Etemadi’s inspiration was that  since
                                                                          
                                                                      
                        
             
            X ,n   1, and X ,n   1,  satisfy the assumptions of the theorem of X  =  X  X ,  we can without
                                                                      n
                                                                          n
                        n
             n
                                                                 n
            loss of generality suppose X   0, As in proof of (6.8) we will prove the result first for a subsequence
                                 n
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