6Ergodic theory

II Probability and Measure

6 Ergodic theory
We are now going to study a new topic ergodic theory. This is the study
the “long run behaviour” of system under the evolution of some Θ. Due to time
constraints, we will not do much with it. We are going to prove two ergodic
theorems that tell us what happens in the long run, and this will be useful when
we prove our strong law of large numbers at the end of the course.
The general settings is that we have a measure space (
E, E, µ
) and a measur-
able map Θ :
E E
that is measure preserving, i.e.
µ
(
A
) =
µ
1
(
A
)) for all
A E.
Example.
Take (
E, E, µ
) = ([0
,
1)
, B
([0
,
1))
, Lebesgue
). For each
a
[0
,
1), we
can define
Θ
a
(x) = x + a mod 1.
By what we’ve done earlier in the course, we know this translation map preserves
the Lebesgue measure on [0, 1).
Our goal is to try to understand the “long run averages” of the system when
we apply Θ many times. One particular quantity we are going to look at is the
following:
Let f be measurable. We define
S
n
(f) = f + f Θ + ··· + f Θ
n1
.
We want to know what is the long run behaviour of
S
n
(f)
n
as n .
The ergodic theorems are going to give us the answer in a certain special
case. Finally, we will apply this in a particular case to get the strong law of
large numbers.
Definition
(Invariant subset)
.
We say
A E
is invariant for Θ if
A
= Θ
1
(
A
).
Definition
(Invariant function)
.
A measurable function
f
is invariant if
f
=
f Θ.
Definition (E
Θ
). We write
E
Θ
= {A E : A is invariant}.
It is easy to show that
E
Θ
is a
σ
-algebra, and
f
:
E R
is invariant iff it is
E
Θ
measurable.
Definition
(Ergodic)
.
We say Θ is ergodic if
A E
Θ
implies
µ
(
A
) = 0 or
µ(A
C
) = 0.
Example.
For the translation map on [0
,
1), we have Θ
a
is ergodic iff
a
is
irrational. Proof is left on example sheet 4.
Proposition.
If
f
is integrable and Θ is measure-preserving. Then
f
Θ is
integrable and
Z
f Θdµ =
Z
E
fdµ.
It turns out that if Θ is ergodic, then there aren’t that many invariant
functions.
Proposition. If Θ is ergodic and f is invariant, then there exists a constant c
such that f = c a.e.
The proofs of these are left as an exercise on example sheet 4.
We are now going to spend a little bit of time studying a particular example,
because this will be needed to prove the strong law of large numbers.
Example
(Bernoulli shifts)
.
Let
m
be a probability distribution on
R
. Then
there exists an iid sequence
Y
1
, Y
2
, ···
with law
m
. Recall we constructed this
in a really funny way. Now we are going to build it in a more natural way.
We let
E
=
R
N
be the set of all real sequences (
x
n
). We define the
σ
-algebra
E
to be the
σ
-algebra generated by the projections
X
n
(
x
) =
x
n
. In other
words, this is the smallest
σ
-algebra such that all these functions are measurable.
Alternatively, this is the σ-algebra generated by the π-system
A =
(
Y
nN
A
n
, A
n
B for all n and A
n
= R eventually
)
.
Finally, to define the measure µ, we let
Y = (Y
1
, Y
2
, ···) : Ω E
where
Y
i
are iid random variables defined earlier, and is the sample space of
the Y
i
.
Then
Y
is a measurable map because each of the
Y
i
’s is a random variable.
We let µ = P Y
1
.
By the independence of Y
i
’s, we have that
µ(A) =
Y
nN
m(A
n
)
for any
A = A
1
× A
2
× ··· × A
n
× R × ··· × R.
Note that the product is eventually 1, so it is really a finite product.
This (
E, E, µ
) is known as the canonical space associated with the sequence
of iid random variables with law m.
Finally, we need to define Θ. We define Θ : E E by
Θ(x) = Θ(x
1
, x
2
, x
3
, ···) = (x
2
, x
3
, x
4
, ···).
This is known as the shift map.
Why do we care about this? Later, we are going to look at the function
f(x) = f (x
1
, x
2
, ···) = x
1
.
Then we have
S
n
(f) = f + f Θ + ··· + f Θ
n1
= x
1
+ ··· + x
n
.
So
S
n
(f)
n
will the average of the first
n
things. So ergodic theory will tell us
about the long-run behaviour of the average.
Theorem. The shift map Θ is an ergodic, measure preserving transformation.
Proof. It is an exercise to show that Θ is measurable and measure preserving.
To show that Θ is ergodic, recall the definition of the tail σ-algebra
T
n
= σ(X
m
: m n + 1), T =
\
n
T
n
.
Suppose that A
Q
nN
A
n
A. Then
Θ
n
(A) = {X
n+k
A
k
for all k} T
n
.
Since
T
n
is a
σ
-algebra, we and Θ
n
(
A
)
T
N
for all
A A
and
σ
(
A
) =
E
, we
know Θ
n
(A) T
N
for all A E.
So if A E
Θ
, i.e. A = Θ
1
(A), then A T
N
for all N . So A T .
From the Kolmogorov 0-1 law, we know either
µ
[
A
] = 1 or
µ
[
A
] = 0. So
done.