58.15 The Convergence Of Sums Of Symmetric Random Variables
It turns out that when random variables have symmetric distributions, some
remarkable things can be said about infinite sums of these random variables.
Conditions are given here that enable one to conclude the convergence of the
sequence of partial sums from the convergence of some subsequence of partial
The following lemma is like an earlier result but is proved here for convenience.
Definition 58.15.1 Let X be a random variable. ℒ
μ means λX
= μ. This
is called the law of X. It is the same as saying the distribution measure of X is μ.
Lemma 58.15.2 Let
be a probability space and let X
: Ω → E be a random
variable, where E is a real separable Banach space. Also let ℒ
μ, a probability
measure defined on ℬ
, the Borel sets of E. Suppose h
: E → ℝ is in L1
nonnegative and Borel measurable. Then
Proof: First suppose A is a Borel set in E. Then
Thus for nonnegative simple Borel measurable functions s,
Now approximating with an increasing sequence of nonnegative simple functions and
using the monotone convergence theorem, the desired formula holds for nonnegative
Borel measurable functions h.
If h is Borel measurable and in L1
then you can consider the formula for the
positive and negative parts and get the result in this case also. This proves the
Here is a simple definition and lemma about random variables whose distribution is
Definition 58.15.3 Let X be a random variable defined on a probability space,
having values in a Banach space, E. Then it has a symmetric distribution if
whenever A is a Borel set,
In terms of the distribution,
It is good to observe that if X,Y are independent random variables defined on a
such that each has symmetric distribution, then
has symmetric distribution. Here is why. Let A
be a Borel set in E.
Then by Theorem
on Page 6394
By induction, it follows that if you have n
independent random variables each having
symmetric distribution, then their sum has symmetric distribution.
Here is a simple lemma about random variables having symmetric distributions. It
will depend on Lemma 58.15.2 on Page 6399.
Lemma 58.15.4 Let X ≡
and Y be random variables defined on a
such that Xi,i
,n and Y have values in E a
separable Banach space. Thus X has values in En. Suppose also that
independent and that Y has symmetric distribution. Then if A ∈ℬ
, it follows
You can also change the inequalities in the obvious way, < to ≤ , > or ≥.
Proof: Denote by λX and λY the distribution measures for X and Y respectively.
Since the random variables are independent, the distribution for the random variable,
is λX × λY
where this denotes product measure. Since the
Banach space is separable, the Borel sets are contained in the product measurable sets.
Then by symmetry of the distribution of Y
This proves the lemma. Other cases are similar.
Now here is a really interesting lemma.
Lemma 58.15.5 Let E be a real separable Banach space. Assume ξ1,
independent random variables having values in E, a separable Banach space which have
symmetric distributions. Also let Sk
i=1kξi. Then for any r >
Proof: First of all,
I need to estimate the second of these terms. Let
Thus Ak consists of those ω where
for the first time at k.
and the sets in the above union are disjoint. Consider Aj ∩
in this set, it follows
is of the form
for some Borel set, A. Then letting Y = ∑
i=j+1Nξi in Lemma 58.15.4 and Xi = ξi,
and so, referring to 58.15.27
, this has shown
It follows that
and using 58.15.26, this proves the lemma.
This interesting lemma will now be used to prove the following which concludes a
sequence of partial sums converges given a subsequence of the sequence of partial sums
Lemma 58.15.6 Let
be a sequence of independent random variables having
values in a separable real Banach space, E whose distributions are symmetric.
Letting Sk ≡∑
converges a.e. Also suppose that for every
m > nk,
Then in fact,
and off a set of measure zero, the convergence of Sk to S is uniform.
Proof: Let nk ≤ l ≤ m. Then by Lemma 58.15.5
In using this lemma, you could renumber the ζi so that the sum
where ξj = ζj+nk.
Then using 58.15.29,
fails to converge then
must be in infinitely many of the sets,
each of which has measure no more than 2−
. Thus ω must be in a set of measure
zero. This proves the lemma.