Measure, Integration & Real Analysis, 2021a

Chapter 12 Probability Measures 397
Proof Suppose j ∈ Z + . Let f 1 , f 2 ,...be the sequence of simple functions converging
pointwise to X j as constructed in the proof of 2.89. The Dominated Convergence
Theorem (3.31) implies that EX j = lim n→∞ Ef n . Because of how each f n is constructed,
each Ef n depends only on n and the numbers P(c ≤ X j < d) for c < d.
However,
(
P(c ≤ X j < d) = lim P ( X j ≤ d − 1 ) (
m→∞
m − P Xj ≤ c −
m
1 ) )
for c < d. Because {X k } k∈Γ is an identically distributed family, the numbers above
on the right are independent of j. Thus EX j = EX k for all j, k ∈ Z + .
Apply the result from the paragraph above to the identically distributed family
{X k 2 } k∈Γ and use 12.20 to conclude that σ(X j )=σ(X k ) for all j, k ∈ Γ.
The next result has the nicely intuitive interpretation that if we repeat a random
process many times, then the probability that the average of our results differs from
our expected average by more than any fixed positive number ε has limit 0 as we
increase the number of repetitions of the process.
12.38 Weak Law of Large Numbers
Suppose (Ω, F, P) is a probability space and {X k } k∈Z + is an i.i.d. family of
random variables in L 2 (P), each with expectation μ. Then
for all ε > 0.
(∣ ∣∣
lim P ( n
1n
) ∣ ) ∣∣
n→∞ ∑ X k − μ ≥ ε = 0
k=1
Proof Because the random variables {X k } k∈Z + all have the same expectation and
same standard deviation, by 12.37 there exist μ ∈ R and s ∈ [0, ∞) such that
for all k ∈ Z + . Thus
( 1
12.39 E
n
EX k = μ and σ(X k )=s
n )
∑ X k = μ and σ 2( 1
n
k=1
n )
∑ X k = 1 n )
n
k=1
2 σ2( ∑ X k = s2 n ,
k=1
where the last equality follows from 12.22 (this is where we use the independent part
of the hypothesis).
Now suppose ε > 0. In the special case where s = 0, all the X k are almost surely
equal to the same constant function and the desired result clearly holds. Thus we
assume s > 0. Let t = √ nε/s and apply Chebyshev’s inequality (12.21) with this
value of t to the random variable
n 1 ∑n k=1 X k, using 12.39 to get
(∣ ∣∣ ( n
1 ) ∣ ) ∣∣
P
n
∑ X k − μ ≥ ε ≤ s2
nε
k=1
2 .
Taking the limit as n → ∞ of both sides of the inequality above gives the desired
result.
Measure, Integration & Real Analysis, by Sheldon Axler