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Transcript of Product Spaces
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MEASURE AND PROBABILITY
by
Ricardo Huamn Aguilar
Class notes for the the course Medida y Probabilidad in the Maestraen Matemticas Aplicadas at PUCP, March-July 2016.
All your comments are very welcome at [email protected]
PUCP
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Contents
1 Measure and Integration in Product Spaces 11.1 Exterior measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Product measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.3 The Fubini-Tonelli Theorem . . . . . . . . . . . . . . . . . . . . 12
1.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
1
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Chapter 1
Measure and Integration inProduct Spaces
Rectangles, product-algebra, product measure space, productprobability space, Tonellis theorem, Fubinis theorem, independence
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1.1 Exterior measure
1.1 Definition. An exterior measure on a nonempty set is a function : () [0, ] such that
1. () = 0;
2. If A B, then (A) (B);
3. (j=1
Aj)
j(A
j).
1.2 Definition. A set E is said to be -measurable if
(A) = (A E) +(A Ec), A .
1.3 Theorem. Let be an exterior measure on and let be thecollection of all measurable subsets of . Then is a -algebra and
(, ,) is a complete measure space, where := |.
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Proof. Denote := (,). Then . (i) If E , then Ec, since
(A) = (A E) +(A Ec) = (A (Ec)c) +(A Ec).
If E1, E2 , then
(A) = (A E1) +(A Ec
1)
= (A E1 E2) +(A E1 E
c2
) +(A Ec1
E2) +(A Ec
1Ec
2)
= (A (E1 E2)) +(A (E1 \E2)) +(A (E2 \E1)) +(A (E1 E2)c)
(A (E1 E2)) +(A (E1 E2)
c)
So E1 E2 .
(ii) If Fj for j , we prove that jFj . Define E1 = F1 and
Ej:=Fj \ (j1k=1Fk), j= 2, 3, . Then Ej are pairwise disjoint and njEj= nj=1Fjand Ej= jFj.
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Hence to prove (ii) it suffices to show that if Ej for j and all Ejare pairwise disjoint, then Ej . Let Bn:=
njEj and B := j=1Ej. Then
Bn
. For any A ,
(A Bn) = (A Bn En) +
(A Bn Ecn
) = (A En) +(A Bn1).
By induction, we have (A Bn) =n
j=1(A Ej). So
(A) = (A Bn) +(A Bc
n)
n
j=1
(A Ej) +(A Bc),
and letting n , we obtain
(A) j=1
(A Ej) +(A Bc) (j(A Ej)) +
(A Bc)
= (A B) +(A Bc) (A).
Thus, B= Ej and
(Ej) =
j
(Ej).
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Hence, is a -algebra and :=| is a measure on . If (E) =0,
then E (,), since
(A) (A E) +(A Ec) (E) +(A Ec) = (A Ec) (A).
Therefore, (, (,),) is a complete measure space.
1.4 Proposition. Let () and : [0, ] be a function such
that
(1) and () = 0;
(2) = j=1
Aj for some Aj , j .
For E , define
(E):=inf
j=1(Aj) | Aj j , E
j=1Aj
. (1.1)
Then is an exterior measure.
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Proof. (i) Since 0 () () =0, we have () =0. (ii) If A B, then
any countable cover of B by elements of is also a cover of A. Hence
(A) (B).
(iii) For any given > 0, by the definition of(Aj), there exists a countable
cover kEj,k Aj such that all Ej,k andk
(Ej,k) (Aj) +
2j.
Hence jAj j kEj,k and
(jAj)
j
k
(Ej,k)
j
((Aj) +
2j) = +
j
(Aj).
Taking 0, we conclude that (jAj) j (Aj) and is an exteriormeasure.
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1.2 Product measure
1.5 Theorem. Let (, ,) and (S, ,) be measure spaces. Denote
:= {A B |A ,B } and (A B):= (A)(B).
For each subset E S, define
(E):=inf
j=1
(Aj)(B
j) | E
j=1A
jB
j, A
j , B
j .
(1) Then is an exterior measure on S.
(2) Let S denote the collection of all -measurable subsets of S.
Then S is a -algebra.
(3) If we define := |S,
then ( S, S, ) is a complete measure space.
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1.6 Lemma. If E and F , then
EF S and (EF):= (EF) = (E)(F).
Proof. Let A S. For any > 0, there exists Ej , Fj such that
A j(Ej Fj) and
j
(Ej)(Fj) (A) + .
Note that
A (EF) [j(Ej Fj)] (EF) = j[(Ej E) (Fj F)]
and
A (EF)c
j
(Ej E) (Fj F
c) (Ej Ec) Fj
.
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Consequently,
(A (EF)) +(A (EF)c) j
(Ej E)(Fj F)
+
j
(Ej E)(Fj Fc) +
j
(Ej Ec)(Fj)
=j
(Ej E)(Fj) +j
(Ej Ec)(Fj) =
j
(Ej)(Fj) (A) + .
Taking 0, we see that
(A (EF)) +(A (EF)c) (A).
So, EF is -measurable, that is, EF S.
Lets show the other claim. By EF EF, it is trivial that (EF):=(EF) (E)(F).
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Conversely, if EF j(Ej Fj), Ej , Fj , we have
1E()1F(s) j
1Ej()1Fj (s), ,s S.
Integrating both sides of the above inequality, first with respect to and
then with respect to , and applying twice the MCT, we have
(E)(F) =
S
1E()1F(s)d(s)d()
Sj
1Ej()1Fj (s)d(s)d()
=
j
S
1Ej()1Fj (s)d(s)d() =
j
1Ej()
S
1Fj(s)d(s)d()
j
1Ej()(Fj)d() =
j
1Ej(Fj)d() =
j
(Ej)(Fj).
This proves (E)(F) (EF) =: (EF).
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1.7 Definition. Let be the -algebra generated by the sets
:= {A B |A , B }.
The measure space
( S, , )
is called the product space of (, ,) and (S, ,).
It is also denoted as ( S, , ).
1.1 Exercise. The measure is the unique measure on such
that
(A B) = (A)(B), A ,B .
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1.3 The Fubini-Tonelli Theorem
1.8 Definition. The -section E and s-section Es
of a subset E Sare
E:= {s S | (,s) E} and Es:= { | (,s) E}.
For a function f : S , the -section f and s-section fs are
f
(s) := f(,s), fs() := f(,s).
1.9 Proposition. (i) If E , then E for all and Es
for all s S.
(ii) If f is -measurable, then f is -measurable for all and
fs
is -measurable for all s S.
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Proof. Let be the collection
:= {E | E , ;Es , s S}.
Obviously, A B for all A and B . SincejEj
= j(Ej), (Ec)= (E)
c,
and likewise for s-sections, is a -algebra. Therefore, . This
proves (i).
(ii) Follows from (i) since for any B
(f)1(B) = (f1(B)), (f
s)1(B) = (f1(B))s.
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1.10 Theorem. Suppose that (, ,) and (S, ,) are -finite spaces.
If E , then the function
(E) is -measurable,
and the function
s (Es) is -measurable;
and
(E
) =
(E
)d
(
) = S
(E
s)d
(s
).
Proof. First, we assume that () < and (S) < . Let be the set of
all E for which the conclusions of the theorem are true. IfE= AB,
then (E) = 1A()(B) and (Es) =(A)1B(s). So clearly, E=A B for
all A and B . Hence contains the -system . We note that
is a -system (see exercise below). By the Theorem, = .
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If and are -finite, we can write S= j(j Sj) of rectangles with
(j)(Sj) < , j and Sj S. IfE , we have (see exercise below)
(E (j Sj)) =
1j()(E Sj)d() =
S
1Sj(s)(Es j)d(s). (1.2)
Noting that 1j ()(E Sj) 1()(E), the conclusion now follows from
the MCT.
1.2 Exercise. (i) Prove that defined in the above theorem is a -
system. (ii) Prove (1.2).
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1.11 Theorem. Suppose that (, ,) and (S, ,) are -finite spaces.
Define
g():=
S
fd and h(s):=
fsd.
[Tonelli] If f : S [0, ] is -measurable function, then g and h
are nonnegative and measurable; and
S f(,s)d (,s) =
S f(,s)d(s)
d()
=
S
f(,s)d()
d(s). (1.3)
[Fubini] If f L1( ), then
(1) f L1(S, ,) for -a.e. , fs L1(, ,) for -a.e. s S,
(2) the a.e.-defined functions g L1(, ,), h L1(S, ,), a n d (1.3)
holds.
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Proof. By Theorem 1.10, Tonellis theorem holds when f is a character-
istic function. By linearity of integration, it holds for nonnegative simple
functions. If f is a nonnegative measurable function, there is a sequence
(n)n of nonnegative simple functions such that n(,s) f(,s) for all
(,s) S. The MCT implies first gn() :=
S(n)d
S
fd and
hn(s):=
(n)
sd
fsd, and again
S
f dd =
g d =limn
gnd =limn
S
n(,s)d(s)()
=limn
S
nd( ) =
S
f d( ).
and similarly,
S
f dd =
S
hd =limn
S
hnd =limn
S
n(,s)d(s)()
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=limn
S
nd( ) =
S
f d( ).
This proves Tonellis theorem. Moreover, if
Sf d( )< , this also
shows that g< a.e. and h< a.e., that is f L1(S, ,) for a.e.
and fs L1(, ,) for a.e. s S.
If f L1( S, , ), the conclusion of Fubinis theorem follows by
applying Tonellis theorem to the positive and negative parts of f.
1.3 Exercise. In general, ( S, , ) is not complete.
Notation. The product space is also denoted as ( S, , ).
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1.4 Applications
Throughout this section, the probability space (, ,P) is given.1.12 Theorem. Let X1, X2, , Xn be random variables where Xk has
distribution P Xk. Let P X be the distribution of the random vector X =
(X1, ,Xn). The random variables X1, X2, , Xn are independent if and
only if
P X = P X1 P Xn.
Proof. For the sake of notation, we consider n=2. We want to show that
(2, (2),PX) = ( , ,PX1 PX2). By Exercise 1.5 below (2) =
. We need to proof that PX =PX1 PX2. Given the uniqueness of the
product measure, it suffices to show that it holds only on the rectangles
= {A B |A, B }. Indeed, by independence,
PX(A B) = P
(X1,X2) A B
= P(X1 A)P(X2 B)
=PX1(A)PX2(B) = PX1 PX2(A B).
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To prove the converse, we note that PX(A B) = P(X1 A,X2 B) and
PX1(A)PX2(B) = P(X1 A)P(X2 B).
1.13 Theorem. Let X and Y be independent random variables with
distributions P X and P Y, respectively. Let h : 2 be measurable. If
either h 0, or E|h(X, Y)| < , then
Eh(X, Y) =
h(x,y)d P X(x) d PY(y). ()
In particular, if h(x,y) = f(x)g(y) where f,g : measurable with either
f,g 0 or E|f(X)|, E|g(Y)| < , then
Ef(X)g(Y) = Ef(X)Eg(Y). ()
Proof. Consider h 0 first. By the Change of Variable Formula and
Tonellis theorem,
Eh(X, Y) =
2
h(x,y)PX P Y(d x, d y) =
h(x,y)PX(d x)P Y(d y).
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To prove the second claim, consider first f 0 and g 0. By the above
equality,
Ef(X)g(Y) =
f(x)g(y)PX(d x)P Y(d y) = Ef(X)Eg(Y).
For the other case, we apply the above argument to |f(X)| and |g(Y)|,
E|f(X)g(Y)| = E|f(X)|E|g(Y)| < .
Now Fubini can be applied to f(X)g(Y) to obtain the desired result.
1.14 Theorem. Let X1, X2, , Xn be independent random variables. If
either Xi 0 for all i, or E|Xi| < for all i, then
Eni=1
Xi = ni=1
EXi.
Proof. Follows directly from the previous theorem.
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1.5 Exercises
1.4 Exercise. Let (, ,P) be a probability space. SupposeX : [0, )is a random variable. Then
E [X] =
[0,)
P(X>t)(d t),
where is the Lebesgue measure.
1.5 Exercise. Prove that () () = (2).
1.6 Exercise. If X and Y are independent, then
FX+Y
(z):=P(X+Yz) =
FX
(z y)P Y(d y), z .
1.7 Exercise. Prove that the converse of Theorem1.14 does not hold.