On the Skorokhod Representation Theorem

0.16 MB·English

by  Jean Cortissoz

ON THE SKOROKHOD REPRESENTATION THEOREM
6
0 JEANCORTISSOZ
0
2
Abstract. In this paper we present a variant of the well known Skorokhod
n RepresentationTheorem. Inourmainresult,givenSaPolishSpace,toagiven
a continuous path α in the space of probability measures on S, we associate a
J continuous path in the space of S-valued random variables on a nonatomic

2 probability space (endowed with the topology of the convergence in proba-
2 bility). We call this associated path a lifting of α. An interesting feature of
our resultisthat wecan fixthe endpoints ofthe liftingof α, aslong astheir

] distribution correpond to the respective endpoints of α. We also discuss an
R n-dimensionalgeneralizationofthisresult.
P
.
h
t
a
m 1. Introduction
[
Let (S,d) be a complete separable metric space and (Ω,F,P) be a complete

1 non atomic probability space (here F denotes the σ-algebra where P is defined),
v and P(S) the space of probability measures on S. The Skorokhod Representation
4 Theorem states the following
2
5 Theorem 1. Suppose P , n=1,2,... and P are probability measures on S (pro-
n
1
vided with its Borel σ-algebra) such that P ⇒P (see Section 2.2, Definition 2.5).
0 n
Then there is a probability space (Ω,F,P) on which are defined S-valued random
6
0 variables Xn, n = 1,2,... and X with distributions Pn and P respectively, such
/ that lim X =X a.s.
h n→n n
t
a A stronger result is presented in [BD] The purpose of this paper is to prove a
m result in the same spirit of Theorem 1. The main result we prove in this note is,
:
v Theorem 2. Let α: [0,1]→P(S) be a continuous function (P(S) endowed with
i the topology of the weak convergence -see Section 2.2). Let X and X be random
X µ ν
variables such that law(X ) = α(0) and law(X ) = α(0). There is αˆ : [0,1] →
µ ν
r
a L0(Ω,S) continuous (L0(Ω,S) endowed with the topology of the convergence in
probability), such that αˆ(0)=X and αˆ(1)=X and law(αˆ)=α.
µ ν
A result with the same statement as our main theorem but with [0,1] replaced
by [0,1]n can be obtained using techniques similar to the ones used in this paper.
More exactly we have,
Theorem 3. Let α : [0,1]n → P(S) be a continuous function (P(S) endowed
with the topology of the weak convergence). Let X : ∂[0,1]n → L0(Ω,S) (∂ =
boundary) be a continuous function such that law(X) = α|∂[0,1]n. There is αˆ :
[0,1]n →L0(Ω,S) continuous, such that αˆ|∂[0,1]n =X and law(αˆ)=α.
We must point out that the result in [BD] may seem stronger to the results
described above. In certain sense this is true, since given α : [0,1]→ P(S) and a
1
2 JEANCORTISSOZ
representationβ : P(S)×Ω−→S (using the definitions in [BD] -notice that their
M(X) is our P(S)) by defining
αˆ(t)=β(α(t),·)
With this definition, we have that if t → t then αˆ(t ) → αˆ(t ) a.s., and
n 0 n 0
a.s. convergenceimplies convergencein probability. However,in our results we are
allowed to fix boundary values for the liftings (or representations). As a curious
consequence of Theorem 3 we obtain
Corollary 1. All the homotopy groups of L0(Ω,S) (endowed with the topology of
convergence in probability) are trivial.
EventhoughwedonotgiveadetailedargumenttoproveTheorem3,wepresent
a rough sketch of a proof in the last sections and details to be given in a further
paper. This paper is arrangedas follows: in Section 2 we collect some basic theory
asaquickreferencefortheconvenienceofthereader. InSection3wepreparesome
lemmaswe use inthe proofofTheorem2, whichwefinally provein Section4. The
sketch of a proof of Theorem 3 at the end of Section 4.
2. Some Basic Concepts
In this section we collect some important and well known definitions and facts
for the reader’s convenience. We recommend Chapter 3 of [EK] as a reference for
this section.
2.1. Convergence in Probability.
Definition 2.1. We say that a sequence (X ) converges in probability to X, and
n
P
we denote it by X →X, if for every ǫ>0 we have
n
lim P {ω : d(X(ω),Y (ω))≥ǫ}=0
n→∞
Definition 2.2. Given X and Y random variables, define
ρ(X,Y)=inf{ǫ>0: P {ω : d(X (ω),Y (ω))≥ǫ}≤ǫ}
n
Theorem 2.1. ρ is a metric on L0(Ω,S), and given a sequence (X ) of random
n
variables and a random variable X, then
P
lim ρ(X ,X)=0 if and only if X →X.
n n
n→∞
2.2. The Space P(S) and the function law.
Definition 2.3.
P(S)={µ: µ is a probability measure on B(S)}.
Definition 2.4. Given µ ∈ P(S) and A ∈ B(S) we say that A is a set of µ-
continuity if µ(∂A)=0 (∂A is the topological boundary of A).
Asequence(µ ) inP(S)converges weaklytoµ,andwedenoteitbyµ ⇒µ,
n n∈N n
if for every set of µ-continuity A µ (A)→µ(A) as n→∞.
n
Definition 2.5. Given µ,ν ∈P(S) define
q(µ,ν)=inf{ǫ>0: µ(A)≤ν(Aǫ)+ǫ for all A⊂S closed}
ON THE SKOROKHOD REPRESENTATION THEOREM 3
Theorem 2.2. q defines a metric on P(S), and given a sequence (µ ) and a
n
probability measure µ, then
lim q(µ ,µ)=0 if and only if µ ⇒µ.
n n
n→∞
Definition 2.6. Given X ∈ L0(Ω,S), the probability measure µ defined on B(S)
by
µ(A)=P X−1(A)
is called the distribution or law of X .
(cid:0) (cid:1)
3. An important lemma
Theorem 3.1. Let (S,d) be separable, and let P,Q∈P(M). Define M(P,Q) be
the set of all P(S×S) with marginals P and Q. Then
q(P,Q)= inf inf{ǫ>0: µ{(x,y): d(x,y)≥ǫ}≤ǫ}
µ∈M(P,Q)
Proof. See [EK] (Chapter 3, Theorem 1.2) or [S] (Corollary to Theorem 10).
(cid:3)
As a corollary we get
Corollary 3.1.
q(P,Q)=inf{ρ(X,Y): law(X)=P and law(Y)=Q}.
Proof. It follows from the fact that given µ ∈ M(P,Q) there is a random
variable W ∈ L0(Ω,S×S) such that law(W) = µ. But by its very definition
W = (X,Y) where X,Y ∈L0(Ω,S) and law(X)=P and law(Y)=Q. Then all
we have to notice is that
ρ(X,Y)=inf{ǫ>0: P {ω : d(X(ω),Y (ω))≥ǫ}≤ǫ}.
(cid:3)
We will need the following “working” definitions,
Definition 3.1. δ is the measure defined by
a
1 if a∈V
δ (V)=
a 0 otherwise
(cid:26)
We say that µ is finitely supported if it can be written as
n
µ= c δ c ≥0
i ai i
i=1
X
Definition3.2. Let(A ) beapartitionofΩ. WedefinethesimpleS-valued
k k=1,...,n
random variable X = n χai as
i=1 Ai
P X(ω)=ai iff ω ∈Ai
Corollary 3.1 is used to prove
Lemma 3.1. Let µ and ν be finitely supported measures, and let ǫ>0 be such that
q(µ,ν)<ǫ.
Then, given X a random variable such that law(X)=µ there exists Y such that
law(Y)=ν and ρ(X,Y)<ǫ
4 JEANCORTISSOZ
Proof. Since µ and ν have finite support, by Theorem 3.1 we can find
m m
X′ = χaj and Y′ = χaj.
A′ B′
j j
j=1 j=1
X X
so that law(X′)=µ and law(Y′)=ν and ρ(X′,Y′)<ǫ.
Write X = m χaj. Since Ω is nonatomic we can find measurable sets
j=1 Aj
B ,B ,...,B such that
1 2 m
P
P (A ∩B )=P A′ ∩B′ for all i,j =1,2,...,m.
i j i j
It is clear then that Y = m(cid:0) χaj s(cid:1)atisfies ρ(X,Y)<ǫ.
j=1 Bj (cid:3)
P
Finally we have the following fundamental lemma,
Lemma 3.2. Let ǫ>0 be given and assume q(law(X),law(Y))<ǫ. Then there
is Y′ such that law(Y′)=law(Y) and ρ(X,Z)<ǫ.
Proof. We proceed by induction on the complexity of random variables. The
case when X and Y are simple is contained in corollary 3.1.
NowassumethatX isarbitraryandY simple. Letδ >0besuchthatρ(X,Y)<
ǫ−δ. Since the set of simple randomvariablesis dense in L0(Ω,S), we can choose
X′ asimplerandomvariablesuchthatρ(X,X′)<δ. Bythe inductionhypothesis,
wecanfindY′ suchthatlaw(Y′)=law(Y)andρ(X′,Y′)<ǫ−δ. Hencewehave,
ρ(X,Y′)≤ρ(X,X′)+ρ(X′,Y′)<δ+ǫ−δ =ǫ
Finally, let X and Y be arbitrary random variables and let δ >0 be such that
q(law(X),law(Y))+1000δ<ǫ.
Choose a sequence (Y ) of simple random variables converging to Y and such
n
that
1
ρ(Y ,Y )< and q(law(Y ),law(Y))<δ,
n n+1 2n+1 n
andletN besuchthat 1 <δandalsoρ(Y ,Y)<δ. Weconstructanewsequence
2N N
Y′ as follows:
j j=N,N+1,N+2,...
If j =N, choose Y′ be such that law(Y′ )=law(Y ) and
(cid:0) (cid:1) N N N
ρ(Y′ ,X)<q(law(X),law(Y))+δ
N
This can be done because
q(law(Y ),law(X)) ≤ q(law(Y ),law(Y))+q(law(Y),law(X))
N N
≤ q(law(Y),law(X))+δ.
Once we have chosen Y′ for j =N +1,...,N +M, we pick Y′ such that
j N+M+1
1
law(Y )=law Y′ and ρ Y′ ,Y′ <
N+M+1 N+M+1 N+M+1 N+M 2N+M+1
This can be done becau(cid:0)se (cid:1) (cid:0) (cid:1)
1
q law(Y ),law Y′ =q(law(Y ),law(Y ))≤ .
N+M+1 N+M N+M+1 N+M 2N+M+2
(cid:0) (cid:0) (cid:1)(cid:1)
ON THE SKOROKHOD REPRESENTATION THEOREM 5
By construction, the sequence Y′ is convergent, and for its limit Y′ it holds
j j
that law(Y′)=law(Y), and also
(cid:0) (cid:1)
ρ(X,Y′) ≤ ρ(X,Y′)+ δ
N j=1 2j
< q(law(X),law(Y))+δ+δ <ǫ.
P
(cid:3)
4. Representation Theorems
4.1. Liftings. First we give a definition we learnt from Ramiro de la Vega.
Definition 4.1. A family of measurable sets (A ) is a [0,δ]-family if it satis-
t t∈[0,δ]
fies:
(i) A ⊂A whenever s≤t,
s t
(ii) P(A )=t .
t
ThefollowinglemmawealsolearntfromdeLaVega,iswhatmakes[0,δ]-families
a useful tool.
Lemma 4.1. Let (Ω,F,P) be a complete nonatomic probability space, and let
A∈F and let δ =P(A). Then there is a [0,δ]-family (A ) such that A ⊂A
t t∈[0,δ] t
for all t.
Proof. PickanorderingofQ∩[0,1](Q: therationalnumbers)sayq ,q ,q ,...
1 2 3
. Since Ω is nonatomic we can inductively construct E ,E ,E ,... such that
q1 q2 q3
P(E )=q, and E ⊂E if q ≤r.
q q r
Finally for x∈[0,1]\Q, define
E = E
x q
q<x
[
(cid:3)
Using [0,δ]families we canintroduce the notionofa segmentjoining twosimple
random variables.
Definition 4.2. Let
m m
X = χaj and Y = χaj
Aj Bj
j=1 j=1
X X
be two simple random variables. Define
E =A ∩B and e =P (E ).
ij i j ij ij
and let [E ] be a [0,e ] family of E . A segment α : [a,b] → L0(Ω,S)
ij t ij ij X,Y
joining X and Y is defined as
(cid:0) (cid:1)
m m m
α (t)= χai + χai
X,Y Xi=1 Eii∪ k=1,k6=i[Eki](bt−−aa)eki! Xi=1j=X1,j6=i Eij\[Eij](bt−−aa)eij
S
We describe some important properties of these segments.
6 JEANCORTISSOZ
Proposition4.1. αˆ :=α thusdefinedisacontinuousfunctionwithα (a)=
X,Y X,Y
X and α (b)=Y. Moreover,
X,Y
b−t t−a
α(t):=law(αˆ(t))= law(X)+ law(Y).
b−a b−a
(cid:18) (cid:19) (cid:18) (cid:19)
Proof. First we show that α is continuous. It is an immediate consequence
X,Y
of the following inequality. Let ǫ>0 be given and s≤t, then we have,
P {ω : d(αˆ(t),αˆ(s))≥ǫ} ≤ P [E ]
{(i,k):d(ai,ak)≥ǫ} ik (t−s)eik
≤ P(t−s) ...eik ≤t−(cid:16)s. (cid:17)
Let us show that law(αˆ) = α. To make things easier, we will assume a = 0
P
and b = 1. Then all we must show is that the coefficient of δ in law(αˆ) is
ai
(1−t)P (A )+tP(B ). Let’s fix i=1. Then the sought coefficient is given by
i i
P(E )+ m P [E ] + m P E \[E ]
11 j=2 j1 tej1 j=2 1j 1j tm1j
=Pe11+ (cid:16)mj=2tmj1+(cid:17) Pmj=2(m1j(cid:16)−tm1j) (cid:17)
=P (A )+t(P(B )−e )−t m e
=P (A1P)+t(P(B1)−Pe11)−t jm=2e1j
=P1(A )+tP1(B )−11t mPej=2 1j
1 1 j=P1 1j
=P (A )+tP(B )−tP(A ).
1 1 1
P
(cid:3)
The notion of segments can be generalized to the concept of a “poligonal”.
Definition4.3. Wecallβ : [0,1]−→P(S)apoligonalwithverticesatµ ,µ ,...,µ ,µ
0 1 n n+1
if there is a partition 0 = t ,t ,...,t ,t = 1 of [0,1] such that β restricted to
0 1 n n+1
[t ,t ] is given by
i i+1
t −t t−t
β(t)= i+1 µ + i µ .
t −t i t −t i+1
(cid:18) i+1 i(cid:19) (cid:18) i+1 i(cid:19)
An easy consequence of proposition 4.1 is the following fact about poligonals,
Proposition 4.2. Let α: [0,1]−→P(S) be a poligonal with vertices at measures
of finite support, and let α(0) = µ and α(1) = ν. Given X and X such that
µ ν
law(X ) = µ and law(X ) = ν, there is a lifting αˆ : [0,1] −→ L0(Ω,S) (i.e.,
µ ν
law(αˆ)=α), such that αˆ(0)=X and αˆ(1)=X .
µ ν
Also poligonalsare dense inthe space ofcontinuousmaps fromthe unit interval
to P(S). Before we write and prove the exact statement of this fact we will need
the following observation.
Lemma 4.2. Let µ,ν ∈P(S). For t∈[0,1] we have
q(ν,tµ+(1−t)ν)≤q(ν,µ).
Proof. Let ǫ>0 be suchthat µ(A)≤ν(Aǫ)+ǫ for all A⊂S closed. Then we
have,
tµ(A)+(1−t)ν(A) ≤ tν(Aǫ)+tǫ+(1−t)ν(Aǫ)+(1−ǫ)ǫ
= ν(Aǫ)+ǫ
and from this the statement of the lemma follows.
(cid:3)
Nowwearereadytostateandprovethefollowingdensitypropertyofpoligonals
ON THE SKOROKHOD REPRESENTATION THEOREM 7
Lemma 4.3. Given α : [0,1] −→ P(S) and ǫ > 0 there is a poligonal β with
vertices at measures of finite support such that
sup q(α(t),β(t))<ǫ
t∈[0,1]
Proof. Let ǫ > 0 be given. By the uniform continuity of α, we can find δ > 0
suchthatwhenever|s−t|<δwehaveq(α(t),α(s))< ǫ. LetN >0bebigenough
5
sothat 1 <δ,anddefineapartitionoftheinterval[0,1]byt = i i=0,1,...,N.
N i N
For each i pick a finitely supported measure µ such that q(µ ,α(t )) ≤ ǫ. Let β
i i i 5
be the poligonal defined by the segments β : [t ,t ]−→P(S) with endpoints µ
i i+1 i
and µ . For t∈[t ,t ] we have,
i+1 i i+1
q(α(t),β(t)) ≤ q(α(t),α(t ))+q(α(t ),µ )+q(µ ,β(t))
i i i i
by Lemma 4.2
≤ q(α(t),α(t ))+q(α(t ),µ )+q(µ ,µ )
i i i i i+1
≤ q(α(t),α(t ))+q(α(t ),µ )+q(µ ,α(t ))
i i i i i
+q(α(t ),α(t ))+q(α(t ),µ )
i i+1 i+1 i+1
≤ ǫ + ǫ + ǫ + ǫ + ǫ + ǫ =ǫ.
5 5 5 5 5 5
(cid:3)
4.2. Proof of the Main Theorem. We are almost ready to prove the Main
Theorem of this paper. The following fact will be used in its proof.
Lemma 4.4. Let α : [0,1] → P(S) and let ǫ > 0 be given. Let β be an arbitrary
poligonal with vertices at measures of finite support and such that
sup q(α(t),β(t))<ǫ
t∈[0,1]
Then, given any continuous lifting αˆ of α, there is a lifting βˆ of β such that
sup ρ αˆ(t),βˆ(t) <5ǫ.
t∈[0,1]
(cid:16) (cid:17)
For the proof of this lemma we need the following observation,
Lemma 4.5. Let
m m
X = χaj and X = χaj
µ Aj ν Bj
j=1 j=1
X X
be such that law(X ) = µ and law(X ) = ν are finitely supported measures. Let
µ ν
αˆ :=α be as in Definition 4.2. Then we have
Xµ,Xν
ρ(X ,αˆ(t))≤ρ(X ,X ).
µ µ ν
Proof. (We use the notation of Definition 4.2) Given ǫ>0 we have
P{ω : d(X (ω),X (ω))≥ǫ} = P(E )
µ ν {(i,j):d(ai,aj)≥ǫ} ij
≥ P [E ]
P{(i,j):d(ai,aj)≥ǫ} ij teij
= PP {ω : d(Xµ(ω,αˆ(t(cid:16))))≥ǫ}.(cid:17)
The conclusion of the lemma follows.
(cid:3)
Proof of Lemma 4.4. Let αˆ be a continuous lifting of α. Take a partition
0=t <t <···<t =1 of the unit interval, in such a way that
0 1 n+1
ρ(X ,X )<ǫ where X =αˆ(t )
i i+1 i i
8 JEANCORTISSOZ
Choose Y for i = 0,1,...,n+1 so that law(Y ) = β(t ) and ρ(X ,Y ) < ǫ.
i i i i i
Then we have
ρ(Y ,Y )≤ρ(Y ,X )+ρ(X ,X )+ρ(X ,Y )<3ǫ.
i i+1 i i i i+1 i+1 i+1
Construct a lifting βˆ of β, such that βˆ restricted to the segment [t ,t ] is a
i i+1
lifting of β : [t ,t ]−→P(S) with βˆ(t )=Y as givenby Definition 4.2. Then βˆ
i i+1 i i
is continuous and for t ≤t<t we have
i i+1
ρ αˆ(t),βˆ(t) ≤ ρ(αˆ(t),X )+ρ(X ,Y )+ρ Y ,βˆ(t)
i i i i
(cid:16) (cid:17) by Lemma 4.5 (cid:16) (cid:17)
≤ ρ(αˆ(t),X )+ρ(X ,Y )+ρ(Y ,Y )
i i i i i+1
≤ ǫ+ǫ+3ǫ=5ǫ.
(cid:3)
Proof of Theorem 2. Take a sequence of poligonals (α ) with vertices at
n n∈N
measures of finite support, and such that
1
α →α and sup (q(α (t),α (t)))< .
n n n+1 5n+1
t∈[0,1]
By proposition 4.2 and lemma 4.4, we can lift this sequence to a sequence (αˆ )
n
so that αˆ (0)→X , αˆ (1)→X and
n µ n ν
1
sup(ρ(α (t),α (t)))< .
n n+1 5n
It is clear by construction that (αˆ ) is a convergent sequence. Let αˆ be its
n
limit. Then, since the convergence is uniform, αˆ is continuous, and because law is
a continuous function, law(αˆ)=α. This finishes the proof.
(cid:3)
4.3. On Theorem 3. Here we say a couple of words on how to approach a proof
for Theorem 3, with further details to be given in a further paper. First, we use a
special family of functions to approximate continous maps. Let
g : [0,1]n −→P(S)
(µ1,µ2,...,µn)
as follows. First we define,
g =(1−t )µ +t µ
(µ1,µ2) 1 1 1 2
and then inductively
g =(1−t )g +t µ
(µ1,µ2,...,µn,µn+1) n+1 (µ1,µ2,...,µn) n+1 n+1
Weusethefunctionsg (ornaturalvariationsofthem)toapproximate
(µ1,µ2,...,µn)
continuous functions from [0,1]n to P(S) in the same way we use polygonals to
approximate continuous functions from [0,1] to P(S). Therefore we must learn
how to lift these functions. We use the following procedure inductive procedure.
Let
gˆ (t ,...,t )= χai
(µ1,...,µn) 1 n Ai(t1,...,tn)
be a lift of g . Given X = χXai a lift of µ , we construct a
(µ1,...,µn) n+1 Bi n+1
→
lifting of g as follows (we adopt the notation t:= (t ,...,t ), and
(µ1,...,µn,µn+1) P 1 n
ON THE SKOROKHOD REPRESENTATION THEOREM 9
→
µ:=(µ ,...,µ )). Firstforeachjchoosea[0,m ]-familyofB (herem =P (B )).
1 n j j j j
→ → → →
For fixed t construct a [0,e t ]-family of E t =A t ∩B , by taking
ij ij i j
→ (cid:16) →(cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
Eij t =Eij t ∩[Bj] → .
γ sup s:P Eij t ∩[Bj]s =γ
h (cid:16) (cid:17)i (cid:16) (cid:17) n (cid:16) (cid:16) (cid:17) (cid:17) o
Then define,
gˆ(cid:16)→µ,µn+1(cid:17)(cid:16)→t,tn+1(cid:17) = PiχaEiii →t ∪ k=1,k6=i[Eki]tn+1·eki(→t)!
+ m (cid:16) (cid:17)m S χai .
i=1 j=1,j6=i →
Eij t \[Eij]tn+1·eij(→t)
P P (cid:16) (cid:17)
Of course we must show that the lifting thus defined is continuous. Once we
haveprovedthat this lifting is continuous,we approximatethe givenfunction with
prescribed boundary values using functions in this family of liftings by choosing
the vertices of the approximation wisely, and then taking a uniform limit of the
approximations,as it was done for the case of the unit interval.
References
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rem,Proc.Amer.Math.Soc.89(1983), 691–692.
[EK] S.EthierandT.Kurtz,MarkovProcesses: CharacterizationandConvergence,WileySeries
inProbabilityandMathematical Statistics,JohnWileyandSons,1986.
[S] V.Strassen,TheExistenceofProbability Measureswithgivenmarginals,Ann.Math.Statist.
36,1965.