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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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