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Statistics

Notes and then use monotonicity to control the values in between. This time however, we let  > 1, and
n
k(n) = [a ]. Chebyshev’s inequality implies that if > 0
 
 P(|T k(n)  ET k(n) | k(n))  2  var(T k(n) )/k(n) 2
n 1 n 1


 k(n)
=   2  k(n)  2  var(Y )
m
n 1 m 1



m 
=   2  var(Y ) k(n)  2
m 1 n:k(n) m


where we have used Fubini’s theorem to interchange the two summations (everything is  0).
n
n
n
Now k(n) = [ ] and [ ]   /2 for n  1, so summing the geometric series and noting that the
first term is  m -2
 [ n  2 4    2n  4(1    2  1  2
] 
) m
n
n
n:  m n:  m
Combining our computations shows
 
2
 P(|T k(n)  ET k(n) | k(n)) 4(1    2  1  2  E(Y )m  2  

) 
m
n 1 m 1


by (b). Since  is arbitrary (T – ET )/k(n)  0. The dominated convergence theorem implies
k(n) k(n)
EY  EX as k  , so ET /k(n)  Ex and we have shown T /k(n)  EX a.s. To handle the
k 1 k(n) 1 k(n) 1
intermediate values, we observe that if k(n)  m < k(n + 1)
T T T

k(n) m k(n 1)
 
k(n 1) m k(n)

n
(here we use Y  0), so recalling k(n) = [ ] we have k(n + 1)/k(n)   and
i
1
EX  lim inf Tm/m  lim sup T /m   EX 1
1
m
 n m 
Since  > 1 is arbitrary the proof is complete.
The next result shows that the strong law holds whenever EX exists.
i


(7.2) Theorem. Let X , X , ... be i.i.d. with EX =  and EX < . If S = X + ... + X then Sn/n  
1 2 i i n 1 n
a.s.
M
M
M
M
M
M
Proof Let M > 0 and X = Xi  M. The X are i.i.d with E| X | <  so if S = X  ... X then

i i i i i n
M
M
M
(7.1) implies S /n  EX . Since X  X it follows that
n i i i
M
lim inf S /n  lim S /n  EX M
n
n
i
n n
M 

M

The monotone convergence theorem implies E(X )  EX   as M, so EX  E(X )
i
i
i
i
M 
S /n   which imlies the desired result.
 E(X )   and we have lim inf n n
i
The rest of this section is devoted to applications of the strong law of large numbers.
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