Product space for two processes with independent increments under nonlinear expectations

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by  Qiang Gao

Product space for two processes with independent increments under
nonlinear expectations
Qiang Gao ∗ Mingshang Hu † Xiaojun Ji ‡ Guomin Liu §
6 January 22, 2016
1
0
2
Abstract. In this paper, we consider the product space for two processes with independent increments
n
a undernonlinearexpectations. Byintroducingadiscretizationmethod, weconstructanonlinearexpectation
J
under which the given two processes can be seen as a new process with independent increments.
1
2 Key words. G-expectation, Nonlinear expectation, Distribution, Independence, Tightness

] AMS subject classifications. 60H10, 60E05
R
P
. 1 Introduction
h
t
a
m Peng [6, 7] introduced the notions of distribution and independence under nonlinear expectation spaces.
[ Under sublinear case, Peng [10] obtained the corresponding central limit theorem for a sequence of i.i.d.

1 random vectors. The limit distribution is called G-normal distribution. Based on this distribution, Peng
v
[8, 9] gavethe definition of G-Brownianmotion, which is a kind of process with stationary and independent
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0 increments, and then discussed the Itoˆ stochastic analysis with respect to G-Brownian motion.
5
It is well-known that the existence for a sequence of i.i.d. random vectors is important for central
5
0 limit theorem. In the nonlinear case, Peng [12] introduced the product space technique to construct a
.
1 sequence of i.i.d. random vectors. But this product space technique does not hold in the continuous time
0
6 case. More precisely, let (Mt)t≥0 and (Nt)t≥0 be two d-dimensional processes with independent increments
1 defined respectively on nonlinear expectation spaces (Ω ,H ,Eˆ ) and (Ω ,H ,Eˆ ), we want to construct a
1 1 1 2 2 2
:
v 2d-dimensional process (M˜ ,N˜ ) with independent increments defined on a nonlinear expectation space
t t t≥0
i
X (Ω,H,Eˆ) such that (M˜ ) =d (M ) and (N˜ ) =d (N ) . Usually, set Ω=Ω ×Ω , M˜ (ω)=M (ω ),
t t≥0 t t≥0 t t≥0 t t≥0 1 2 t t 1
ar N˜t(ω) = Nt(ω2) for each ω = (ω1,ω2) ∈ Ω, t ≥ 0. If we use Peng’s product space technique, then we can
only get a 2d-dimensional process (M˜ ,N˜ ) such that (M˜ ) is independent from (N˜ ) or (N˜ ) is
t t t≥0 t t≥0 t t≥0 t t≥0
∗SchoolofMathematics,Shandong University,Jinan,Shandong250100, PRChina. [email protected]
†QiluInstitute of Finance, Shandong University,Jinan, Shandong 250100, PRChina. [email protected]. Research
supportedbyNSF(No. 11201262 and11301068) andShandong Province(No.BS2013SF020 andZR2014AP005)
‡SchoolofMathematics,Shandong University,Jinan,Shandong250100, PRChina. [email protected]
§QiluInstituteofFinance,ShandongUniversity,Jinan,Shandong 250100, PRChina. [email protected]. Gao, Hu,Ji
andLiu’sresearchwaspartiallysupportedbyNSF(No. 10921101) andbythe111Project(No. B12023)
1
independent from (M˜ ) . Different from linear expectation case, the independence is not mutual under
t t≥0
nonlinear case (see [4]). So this (M˜ ,N˜ ) is not a process with independent increments.
t t t≥0
In this paper, we introduce a discretization method, which can overcome the problem of independence.
More precisely, for each given D = {i2−n : i ≥ 0}, we can construct a nonlinear expectation Eˆn under
n
which (M˜ ,N˜ ) posesses independent increments. But Eˆn, n ≥ 1, are not consistent, i.e., the values of
t t t∈Dn
the same random variable under Eˆn are not equal. Fortunately, we can prove that the limit of Eˆn exisits
by using the notion of tightness, which was introduced by Peng in [13] to prove central limit theorem under
sublinear case. Denote the limit of Eˆn by Eˆ, we show that (M˜ ,N˜ ) is the process with independent
t t t≥0
increments under Eˆ.
This paper is organized as follows: In Section 2, we recall some basic notions and results of nonlinear
expectations. The main theorem is stated and proved in Section 3.
2 Preliminaries
We present some basic notions and results of nonlinear and sublinear expectations in this section. More
details can be found in [1-20].
Let Ω be a given nonempty set and H be a linear space of real-valued functions on Ω such that if
X ,...,X ∈ H, then ϕ(X ,X ,...,X ) ∈ H for each ϕ ∈ C (Rd), where C (Rd) denotes the set of
1 d 1 2 d b.Lip b.Lip
all bounded and Lipschitz functions on Rd. H is considered as the space of random variables. Similarly,
{X =(X ,...,X ):X ∈H,i≤d} denotes the space of d-dimensional random vectors.
1 d i
Definition 2.1 AsublinearexpectationEˆ onHisafunctionalEˆ :H→Rsatisfyingthefollowingproperties:
for each X,Y ∈H,
(i) Monotonicity: Eˆ[X]≥Eˆ[Y] if X ≥Y;
(ii) Constant preserving: Eˆ[c]=c for c∈R;
(iii) Sub-additivity: Eˆ[X+Y]≤Eˆ[X]+Eˆ[Y];
(iv) Positive homogeneity: Eˆ[λX]=λEˆ[X] for λ≥0.
The triple (Ω,H,Eˆ) is called a sublinear expectation space. If (i) and (ii) are satisfied, Eˆ is called a
nonlinear expectation and the triple (Ω,H,Eˆ) is called a nonlinear expectation space.
Let (Ω,H,Eˆ) be a nonlinear (resp. sublinear) expectation space. For each given d-dimensional random
vector X, we define a functional on C (Rd) by
b.Lip
Fˆ [ϕ]:=Eˆ[ϕ(X)] for each ϕ∈C (Rd).
X b.Lip
It is easy to verify that (Rd,C (Rd),Fˆ ) forms a nonlinear (resp. sublinear) expectation space. Fˆ is
b.Lip X X
calledthedistributionofX. Twod-dimensionalrandomvectorsX andX definedrespectivelyonnonlinear
1 2
expectation spaces (Ω ,H ,Eˆ ) and (Ω ,H ,Eˆ ) are called identically distributed, denoted by X =d X , if
1 1 1 2 2 2 1 2
Fˆ =Fˆ , i.e.,
X1 X2
Eˆ [ϕ(X )]=Eˆ [ϕ(X )] for each ϕ∈C (Rd).
1 1 2 2 b.Lip
2
Similar to the classical case, Peng [13] gave the following definition of convergence in distribution.
Definition 2.2 Let X , n ≥ 1, be a sequence of d-dimensional random vectors defined respectively on
n
nonlinear (resp. sublinear) expectation spaces (Ω ,H ,Eˆ ). {X :n≥1} is said to converge in distribution
n n n n
if, for each fixed ϕ∈C (Rd), {Fˆ [ϕ]:n≥1} is a Cauchy sequence. Define
b.Lip Xn
Fˆ[ϕ]= lim Fˆ [ϕ],
n→∞ Xn
the triple (Rd,C (Rd),Fˆ) forms a nonlinear (resp. sublinear) expectation space.
b.Lip
If X , n ≥ 1, is a sequence of d-dimensional random vectors defined on the same sublinear expectation
n
space (Ω,H,Eˆ) satisfying
lim Eˆ[|X −X |]=0,
n m
n,m→∞
then we can deduce that {X :n≥1} converges in distribution by
n
|Fˆ [ϕ]−Fˆ [ϕ]|=|Eˆ[ϕ(X )]−Eˆ[ϕ(X )]|≤C Eˆ[|X −X |],
Xn Xm n m ϕ n m
where C is the Lipschitz constant of ϕ.
ϕ
The following definition of tightness is important for obtaining a subsequence which converges in distri-
bution.
Definition 2.3 Let X be a d-dimensional random vector defined on a sublinear expectation space (Ω,H,Eˆ).
The distribution of X is called tight if, for each ε > 0, there exists an N > 0 and ϕ ∈ C (Rd) with
b.Lip
I ≤ϕ such that Fˆ [ϕ]=Eˆ[ϕ(X)]<ε.
{x:|x|≥N} X
Definition 2.4 Let {Eˆ : λ ∈ I} be a family of nonlinear expectations and Eˆ be a sublinear expectation
λ
defined on (Ω,H). {Eˆ :λ∈I} is said to be dominated by Eˆ if, for each λ∈I,
λ
Eˆ [X]−Eˆ [Y]≤Eˆ[X −Y] for each X,Y ∈H.
λ λ
Definition 2.5 LetX ,λ∈I,beafamilyofd-dimensionalrandomvectorsdefinedrespectivelyonnonlinear
λ
expectation spaces (Ω ,H ,Eˆ ). {Fˆ : λ ∈ I} is called tight if there exists a tight sublinear expectation Fˆ
λ λ λ Xλ
on (Rd,C (Rd)) which dominates {Fˆ :λ∈I}.
b.Lip Xλ
Remark 2.6 A family of sublinear expectations {Fˆ : λ ∈ I} on (Rd,C (Rd)) is tight if and only if
Xλ b.Lip
Fˆ[ϕ]=sup Fˆ [ϕ] for each ϕ∈C (Rd) is a tight sublinear expectation.
λ∈I Xλ b.Lip
Theorem 2.7 ([13]) Let X , n ≥ 1, be a sequence of d-dimensional random vectors defined respectively
n
on nonlinear expectation spaces (Ω ,H ,Eˆ ). If {Fˆ : n ≥ 1} is tight, then there exists a subsequence
n n n Xn
{X :i≥1} which converges in distribution.
ni
The following definition of independence is fundamental in nonlinear expectation theory.
Definition 2.8 Let (Ω,H,Eˆ) be a nonlinear expectation space. A d-dimensional random vector Y is said
to be independent from another m-dimensional random vector X under Eˆ[·] if, for each test function ϕ ∈
C (Rm+d), we have
b.Lip
Eˆ[ϕ(X,Y)]=Eˆ[Eˆ[ϕ(x,Y)] ].
x=X
3
Remark 2.9 It is important to notethat “Y is independent from X”does not imply that “X is independent
from Y” (see [4]).
A family of d-dimensional random vectors (X ) on the same nonlinear expectation space (Ω,H,Eˆ) is
t t≥0
called a d-dimensional stochastic process.
Definition 2.10 Two d-dimensional processes (X ) and (Y ) defined respectively on nonlinear expec-
t t≥0 t t≥0
tation spaces (Ω ,H ,Eˆ ) and (Ω ,H ,Eˆ ) are called identically distributed, denoted by (X ) =d (Y ) ,
1 1 1 2 2 2 t t≥0 t t≥0
if for each n∈N, 0≤t <···<t , (X ,...,X )=d (Y ,...,Y ), i.e.,
1 n t1 tn t1 tn
Eˆ [ϕ(X ,...,X )]=Eˆ [ϕ(Y ,...,Y )] for each ϕ∈C (Rn×d).
1 t1 tn 2 t1 tn b.Lip
Definition 2.11 A d-dimensional process (X ) with X = 0 on a nonlinear expectation space (Ω,H,Eˆ)
t t≥0 0
is said to have independent increments if, for each 0 ≤ t < ··· < t , X −X is independent from
1 n tn tn−1
(X ,...,X ). A d-dimensional process (X ) with X =0 is said to have stationary increments if, for
t1 tn−1 t t≥0 0
d
each t, s≥0, X −X =X .
t+s s t
We give a typical example of process with stationary and independent increments.
Example 2.12 Let Γ be a given bounded subset in Rd×d, where Rd×d denotes the set of all d×d matrices.
Define G:S(d)→R by
1
G(A)= suptr[AQQT] for each A∈S(d),
2
Q∈Γ
where S(d) denotes the set of all d×d symmetric matrices. A d-dimensional process (B ) on a sublinear
t t≥0
expectation space (Ω,H,Eˆ) is called a G-Brownian motion if the following properties are satisfied:
(1) B =0;
0
(2) It is a process with independent increments;
(3) For each t, s≥0, Eˆ[ϕ(B −B )]=uϕ(t,0) for each ϕ∈C (Rd), where uϕ is the viscosity solution
t+s s b.Lip
of the following G-heat equation:
 ∂ u(t,x)−G(D2u(t,x))=0,
t x

u(0,x)=ϕ(x).

Obviously, (3) implies that the process (B ) has stationary increments.
t t≥0
3 Main result
Let (M ) and (N ) be two d-dimensional processes with independent increments defined respectively
t t≥0 t t≥0
on nonlinear (resp. sublinear) expectation spaces (Ω ,H ,Eˆ ) and (Ω ,H ,Eˆ ). We need the following
1 1 1 2 2 2
assumption:
(A) There exist two sublinear expectations E˜ :H →R and E˜ :H →R satisfying:
1 1 2 2
4
(1) E˜ and E˜ dominate Eˆ and Eˆ respectively;
1 2 1 2
(2) For each t≥0,
lim(E˜ [|M −M |]+E˜ [|N −N |])=0.
1 s t 2 s t
s→t
Remark 3.1 If Eˆ and Eˆ are sublinear expectations, then we can get
1 2
lim(Eˆ [|M −M |]+Eˆ [|N −N |])=0
1 s t 2 s t
s→t
by Eˆ [·]≤E˜ [·] and Eˆ [·]≤E˜ [·]. So we can replace E˜ and E˜ by Eˆ and Eˆ .
1 1 2 2 1 2 1 2
Now we give our main theorem.
Theorem 3.2 Let(M ) and(N ) betwod-dimensional processes withindependent incrementsdefined
t t≥0 t t≥0
respectively on nonlinear (resp. sublinear) expectation spaces (Ω ,H ,Eˆ ) and (Ω ,H ,Eˆ ) satisfying the
1 1 1 2 2 2
assumption(A).Then thereexistsa2d-dimensionalprocess (M˜ ,N˜ ) withindependent incrementsdefined
t t t≥0
on a nonlinear (resp. sublinear) expectation space (Ω,H,Eˆ) such that (M˜ ) =d (M ) and (N˜ ) =d
t t≥0 t t≥0 t t≥0
(N ) . Furthermore, (M˜ ,N˜ ) is a process with stationary and independent increments if (M ) and
t t≥0 t t t≥0 t t≥0
(N ) are two processes with stationary and independent increments.
t t≥0
In the following, we only prove the sublinear expectation case. The nonlinear expectation case can be
proved by the same method. Moreover,the following lemma shows that we only need to prove the theorem
for t∈[0,1].
Lemma 3.3 Let (Xi,Yi) , i ≥ 0, be a sequence of 2d-dimensional processes with independent in-
t t t∈[0,1]
crements defined respectively on sublinear expectation spaces (Ω¯ ,H¯ ,E¯ ) such that (Xi) =d (M −
i i i t t∈[0,1] i+t
M ) and (Yi) =d (N −N ) . Then there exists a 2d-dimensional process (M˜ ,N˜ ) with
i t∈[0,1] t t∈[0,1] i+t i t∈[0,1] t t t≥0
independent increments defined on a sublinear expectation space (Ω,H,Eˆ) such that (M˜ ) =d (M ) and
t t≥0 t t≥0
(N˜ ) =d (N ) .
t t≥0 t t≥0
Proof. Set (Ω,H,Eˆ) = (Π∞ Ω¯ ,⊗∞ H¯ ,⊗∞ E¯ ) which is the product space of {(Ω¯ ,H¯ ,E¯ ) : i ≥ 0} (see
i=0 i i=0 i i=0 i i i i
[12]). For each ω =(ω )∞ , define
i i=0
[t]−1 [t]−1
M˜ (ω)= Xi(ω )+X[t] (ω ), N˜ (ω)= Yi(ω )+Y[t] (ω ).
t X 1 i t−[t] [t] t X 1 i t−[t] [t]
i=0 i=0
By Proposition3.15 in Chapter I in [12], we can easily obtain that (M˜ ,N˜ ) has independent increments
t t t≥0
property, (M˜ ) =d (M ) and (N˜ ) =d (N ) . (cid:3)
t t≥0 t t≥0 t t≥0 t t≥0
Set Ω=Ω ×Ω ={ω =(ω ,ω ):ω ∈Ω ,ω ∈Ω }. For each ω =(ω ,ω )∈Ω, define
1 2 1 2 1 1 2 2 1 2
M˜ (ω)=M (ω ), N˜ (ω)=N (ω ) for t∈[0,1].
t t 1 t t 2
For notation simplicity, we denote X =(M˜ ,N˜ ). Define the space of random variables as follows:
t t t
H={ϕ(X ,X −X ,...,X −X ):∀n≥1,∀0≤t <t <···<t ≤1,
t1 t2 t1 tn tn−1 1 2 n
∀ϕ∈C (Rn×2d)}.
b.Lip
5
In the following, we will construct a sublinear expectation Eˆ : H → R such that (M˜ ) =d (M ) ,
t t∈[0,1] t t∈[0,1]
(N˜ ) =d (N ) and (M˜ ,N˜ ) possessing independent increments. In order to construct Eˆ, we
t t∈[0,1] t t∈[0,1] t t t∈[0,1]
set, for each fixed n≥1,
Hn ={ϕ(X ,X −X ,...,X −X ):∀ϕ∈C (R2n×2d)},
δn 2δn δn 2nδn (2n−1)δn b.Lip
where δ =2−n. Define Eˆn :Hn →R as follows:
n
Step 1. For each given φ(X −X ) = φ(M˜ −M˜ ,N˜ −N˜ ) ∈ Hn with k ≤ 2n
kδn (k−1)δn kδn (k−1)δn kδn (k−1)δn
and φ∈C (R2d), define
b.Lip
Eˆn[φ(X −X )]=Eˆ [ψ(M −M )],
kδn (k−1)δn 1 kδn (k−1)δn
where
ψ(x)=Eˆ [φ(x,N −N )] for each x∈Rd.
2 kδn (k−1)δn
Step 2. For eachgivenϕ(Xδn,X2δn−Xδn,...,X2nδn−X(2n−1)δn)∈Hn withϕ∈Cb.Lip(R2n×2d),define
Eˆn[ϕ(X ,X −X ,...,X −X )]=ϕ ,
δn 2δn δn 2nδn (2n−1)δn 0
where ϕ is obtained backwardly by Step 1 in the following sense:
0
ϕ (x ,x ,...,x )=Eˆn[ϕ(x ,x ,...,x ,X −X )],
2n−1 1 2 2n−1 1 2 2n−1 2nδn (2n−1)δn
ϕ (x ,x ,...,x )=Eˆn[ϕ (x ,x ,...,x ,X −X )],
2n−2 1 2 2n−2 2n−1 1 2 2n−2 (2n−1)δn (2n−2)δn
.
.
.
ϕ (x )=Eˆn[ϕ (x ,X −X )],
1 1 2 1 2δn δn
ϕ =Eˆn[ϕ (X )].
0 1 δn
Lemma 3.4 Let (Ω,Hn,Eˆn) be defined as above. Then
(1) (Ω,Hn,Eˆn) forms a sublinear expectation space;
(2) For each 2≤k ≤2n, X −X is independent from (X ,...,X −X );
kδn (k−1)δn δn (k−1)δn (k−2)δn
(3) (M˜ ,M˜ −M˜ ,...,M˜ −M˜ )=d (M ,M −M ,...,M −M )and(N˜ ,N˜ −
δn 2δn δn 2nδn (2n−1)δn δn 2δn δn 2nδn (2n−1)δn δn 2δn
N˜δn,...,N˜2nδn −N˜(2n−1)δn)=d (Nδn,N2δn −Nδn,...,N2nδn −N(2n−1)δn).
Proof. (1)ItiseasytocheckthatEˆn :Hn →Riswell-defined. WeonlyproveEˆn satisfiesmonotonicity,the
otherpropertiescanbesimilarlyobtained. ForeachgivenY =ϕ1(Xδn,X2δn−Xδn,...,X2nδn−X(2n−1)δn),
Z = ϕ (X ,X −X ,...,X −X ) ∈ Hn with Y ≥ Z, it is easy to verify that Y = (ϕ ∨
2 δn 2δn δn 2nδn (2n−1)δn 1
ϕ2)(Xδn,X2δn−Xδn,...,X2nδn−X(2n−1)δn). Then by the definition ofEˆn andthe monotonicity ofEˆ1 and
Eˆ , we can get Eˆn[Y]≥Eˆn[Z]. (2) and (3) can be easily obtained by the definition of Eˆn. (cid:3)
2
Obviously, Hn ⊂Hn+1 for each n≥1. We set
L= Hn.
[
n≥1
6
It is easily seen that L is a subspace of H such that if Y ,...,Y ∈ L, then ϕ(Y ,...,Y ) ∈ L for each
1 m 1 m
ϕ ∈ C (Rm). In the following, we want to define a sublinear expectation Eˆ : L → R. Unfortunately,
b.Lip
Eˆn+1[·]6=Eˆn[·]onHn,becausetheorderofindependenceundersublinearexpectationspaceisunchangeable.
But the following lemma will allow us to construct Eˆ.
Lemma 3.5 For each fixed n ≥ 1, let Fˆn, k ≥ n, be the distribution of (X ,X −X ,...,X −
k δn 2δn δn 2nδn
X ) under Eˆk. Then {Fˆn :k ≥n} is tight.
(2n−1)δn k
Proof. For each given N >1 and ϕ∈C (R2n×2d) with I ≤ϕ≤I , we have
b.Lip {x:|x|≥N} {x:|x|≥N−1}
Fˆn[ϕ]=Eˆk[ϕ(X ,X −X ,...,X −X )]
k δn 2δn δn 2nδn (2n−1)δn
1
≤ Eˆk[|(X ,X −X ,...,X −X )|]
N −1 δn 2δn δn 2nδn (2n−1)δn
2n
1
≤ Eˆk[ (|M˜ −M˜ |+|N˜ −N˜ |)]
N −1 X iδn (i−1)δn iδn (i−1)δn
i=1
2n 2n
1
≤ (Eˆk[ |M˜ −M˜ |]+Eˆk[ |N˜ −N˜ |])
N −1 X iδn (i−1)δn X iδn (i−1)δn
i=1 i=1
2n 2n
1
= (Eˆ [ |M −M |]+Eˆ [ |N −N |]),
N −1 1 X iδn (i−1)δn 2 X iδn (i−1)δn
i=1 i=1
where the last equality in the above formula is due to (3) in Lemma 3.4. Then for each ε >0, we can take
an N >1 and ϕ ∈C (R2n×2d) with I ≤ϕ ≤I such that
0 0 b.Lip {x:|x|≥N0} 0 {x:|x|≥N0−1}
2n 2n
1
supFˆn[ϕ ]≤ (Eˆ [ |M −M |]+Eˆ [ |N −N |])<ε.
k 0 N −1 1 X iδn (i−1)δn 2 X iδn (i−1)δn
k≥n 0
i=1 i=1
Thus {Fˆn :k ≥n} is tight. (cid:3)
k
Now we will use this lemma to construct a sublinear expectation Eˆ :L→R.
Lemma 3.6 Set D = {i2−n : n ≥ 1,0 ≤ i ≤ 2n}. Then there exists a sublinear expectation Eˆ : L → R
satisfying the following properties:
(1) For each 0≤t <···<t with t ∈D, i≤n, X −X is independent from (X ,...,X );
1 n i tn tn−1 t1 tn−1
(2) For each 0 ≤ t < ··· < t with t ∈ D, i ≤ n, (M˜ ,...,M˜ ) =d (M ,...,M ) and (N˜ ,...,N˜ ) =d
1 n i t1 tn t1 tn t1 tn
(N ,...,N ).
t1 tn
Proof. For n = 1, by Lemma 3.5, we know {Fˆ1 : k ≥ 1} is tight. Then, by Theorem 2.7, there exists a
k
subsequence {Fˆ1 :j ≥1} which converges in distribution, i.e., for each ϕ∈C (R2×2d), {Fˆ1 [ϕ]:j ≥1}
k1 b.Lip k1
j j
is a Cauchy sequence. Note that Fˆ1k1[ϕ]=Eˆkj1[ϕ(X2−1,X1−X2−1)], then for each Y ∈H1, {Eˆkj1[Y]:j ≥1}
j
is a Cauchy sequence.
For n=2, by Lemma 3.5 and Theorem 2.7, we can find a subsequence {k2 :j ≥1}⊂{k1 :j ≥1} such
j j
that for each Y ∈H2, {Eˆkj2[Y]:j ≥1} is a Cauchy sequence.
7
Repeat this process, for each n ≥ 2, we can find a subsequence {kn : j ≥1}⊂ {kn−1 : j ≥1} such that
j j
for each Y ∈ Hn, {Eˆkjn[Y] : j ≥ 1} is a Cauchy sequence. Taking the diagonal sequence {kj : j ≥ 1}, then
j
for each Y ∈L, {Eˆkjj[Y]:j ≥1} is a Cauchy sequence, where Eˆkjj[Y]=∞ if Y 6∈Hkjj. Define
Eˆ[Y]= lim Eˆkjj[Y] for each Y ∈L.
j→∞
For each Y, Z ∈L, there exists a j0 such that Y, Z ∈Hkjj for j ≥j0. From this we can easily deduce that
Eˆ is a sublinear expectation.
Nowwe provethatthis Eˆ satisfies(1)and(2). Foreach0≤t <···<t witht ∈D, i≤n,thereexists
1 n i
a j0 such that ϕ(Xt1,...,Xtn−1,Xtn−Xtn−1)∈Hkjj for eachj ≥j0 and ϕ∈Cb.Lip(Rn×2d). Thus, from (2)
in Lemma 3.4, we can get
Eˆ[ϕ(Xt1,...,Xtn−1,Xtn −Xtn−1)]=jl→im∞Eˆkjj[ϕ(Xt1,...,Xtn−1,Xtn −Xtn−1)]
=jl→im∞Eˆkjj[ψj(Xt1,...,Xtn−1)],
whereψj(x1,...,xn−1)=Eˆkjj[ϕ(x1,...,xn−1,Xtn−Xtn−1)]foreachxi ∈R2d,i≤n−1. Defineψ(x1,...,xn−1)=
Eˆ[ϕ(x ,...,x ,X −X )] for each x ∈ R2d, i ≤ n−1. In order to prove (1), we only need to show
1 n−1 tn tn−1 i
that
jl→im∞Eˆkjj[ψj(Xt1,...,Xtn−1)]=Eˆ[ψ(Xt1,...,Xtn−1)]. (3.1)
It is clear that ψ , j ≥ j , and ψ are bounded Lipschitz functions with the common bound K and the
j 0 ϕ
commonLipschitzconstantL ,whereK andL arerespectivethe boundandLipschitzconstantofϕ. On
ϕ ϕ ϕ
the other hand, for each x = (x ,...,x ) ∈ R(n−1)×2d, by the definition of Eˆ, we can get ψ (x) → ψ(x).
1 n−1 j
Thus, from the common Lipschitz constant L and pointwise convergence, we can easily obtain that {ψ :
ϕ j
j ≥j } converges uniformly to ψ on any compact set in R(n−1)×2d. For each given N >0, we have
0
2K
|ψ (x)−ψ(x)|≤a +2K I ≤a + ϕ|x|,
j j ϕ {x:|x|>N} j N
where a =sup |ψ (x)−ψ(x)|. From the uniform convergenceon any compact set, we know a →0 as
j |x|≤N j j
j →∞. Thus
|Eˆkjj[ψj(Xt1,...,Xtn−1)]−Eˆkjj[ψ(Xt1,...,Xtn−1)]|
≤Eˆkjj[|ψj(Xt1,...,Xtn−1)−ψ(Xt1,...,Xtn−1)|]
≤Eˆkjj[aj + 2NKϕ|(Xt1,...,Xtn−1)|]
≤aj + 2NKϕ(Eˆkjj[|(M˜t1,...,M˜tn−1)|]+Eˆkjj[|(N˜t1,...N˜tn−1)|])
2K
=a + ϕ(Eˆ [|(M ,...,M )|]+Eˆ [|(N ,...N )|]),
j N 1 t1 tn−1 2 t1 tn−1
wherethelastequalityintheaboveformulaisdueto(3)inLemma3.4. NotethatEˆkjj[ψ(Xt1,...,Xtn−1)]→
Eˆ[ψ(X ,...,X )], then
t1 tn−1
limsup|Eˆkjj[ψj(Xt1,...,Xtn−1)]−Eˆ[ψ(Xt1,...,Xtn−1)]|
j→∞
2K
≤ ϕ(Eˆ [|(M ,...,M )|]+Eˆ [|(N ,...N )|]).
N 1 t1 tn−1 2 t1 tn−1
8
Since N can be arbitrarilylarge,we getrelation (3.1). Thus (1) is obtained. (2) is obviousby the definition
of Eˆ and (3) in Lemma 3.4. (cid:3)
Proof of Theorem 3.2. We first extend the sublinear expectation Eˆ : L→ R to Eˆ : H →R. Here we
stilluse Eˆ fornotationsimplicity. Foreachϕ(X ,X −X ,...,X −X )∈Hwith ϕ∈C (Rn×2d),
t1 t2 t1 tn tn−1 b.Lip
we can choose ti ∈D, k ≤n, i≥1, such that ti <ti and ti ↓t as i→∞. By assumption (A), we have
k k k+1 k k
|Eˆ[ϕ(X ,X −X ,...,X −X )]−Eˆ[ϕ(X ,X −X ,...,X −X )]|
ti1 ti2 ti1 tin tin−1 tj1 tj2 tj1 tjn tjn−1
n
≤L Eˆ[ |X −X −X +X |]
ϕ X tik tjk tik−1 tjk−1
k=1
n
≤L Eˆ[ |(X −X )−(X −X )−(X −X )+(X −X )|]
ϕ X tik tk tjk tk tik−1 tk−1 tjk−1 tk−1
k=1
n n
≤2L {Eˆ[ (|M˜ −M˜ |+|M˜ −M˜ |)]+Eˆ[ (|N˜ −N˜ |+|N˜ −N˜ |)]}
ϕ X tik tk tjk tk X tik tk tjk tk
k=1 k=1
n n
=2L {Eˆ [ (|M −M |+|M −M |)]+Eˆ [ (|N −N |+|N −N |)]}
ϕ 1 X tik tk tjk tk 2 X tik tk tjk tk
k=1 k=1
→0 as i,j →∞,
where L is the Lipschitz constant of ϕ and ti =0 for i≥1. So we can define
ϕ 0
Eˆ[ϕ(X ,X −X ,...,X −X )]= lim Eˆ[ϕ(X ,X −X ,...,X −X )].
t1 t2 t1 tn tn−1 i→∞ ti1 ti2 ti1 tin tin−1
It is easy to check that the limit does not depend on the choice of ti by using the same estimate as above.
k
On the other hand, if ϕ(X ,X − X ,...,X − X ) = ϕ˜(X ,X − X ,...,X − X ) for ϕ,
t1 t2 t1 tn tn−1 t1 t2 t1 tn tn−1
ϕ˜∈C (Rn×2d), then
b.Lip
|ϕ(X ,X −X ,...,X −X )−ϕ˜(X ,X −X ,...,X −X )|
ti1 ti2 ti1 tin tin−1 ti1 ti2 ti1 tin tin−1
≤|ϕ(X ,X −X ,...,X −X )−ϕ(X ,X −X ,...,X −X )|
ti1 ti2 ti1 tin tin−1 t1 t2 t1 tn tn−1
+|ϕ˜(X ,X −X ,...,X −X )−ϕ˜(X ,X −X ,...,X −X )|
t1 t2 t1 tn tn−1 ti1 ti2 ti1 tin tin−1
n
≤2(L +L ) (|M −M |+|N −N |).
ϕ ϕ˜ X tik tk tik tk
k=1
Thus, by assumption (A), Eˆ : H → R is well-defined. It is easy to verify that Eˆ is a sublinear expectation.
For each 0 ≤ t < ··· < t ≤ 1, we can choose ti ∈ D as above. By the definition of Eˆ and (1) in Lemma
1 n k
3.6, we can get that for each ϕ∈C (Rn×2d),
b.Lip
Eˆ[ϕ(X ,...,X ,X −X )]= lim Eˆ[ϕ(X ,...,X ,X −X )]
t1 tn−1 tn tn−1 i→∞ ti1 tin−1 tin tin−1
= lim Eˆ[ψ (X ,...,X )],
i→∞ i ti1 tin−1
where ψ (x ,...,x )=Eˆ[ϕ(x ,...,x ,X −X )]. Define
i 1 n−1 1 n−1 tin tin−1
ψ(x ,...,x )=Eˆ[ϕ(x ,...,x ,X −X )].
1 n−1 1 n−1 tn tn−1
9
Then
|ψ (x ,...,x )−ψ(x ,...,x )|
i 1 n−1 1 n−1
≤L Eˆ[|X −X −X +X |]
ϕ tin tin−1 tn tn−1
≤L {Eˆ [|M −M |+|M −M |]+Eˆ [|N −N |+|N −N |]},
ϕ 1 tin tn tin−1 tn−1 2 tin tn tin−1 tn−1
which implies
|Eˆ[ψ (X ,...,X )]−Eˆ[ψ(X ,...,X )]|
i ti1 tin−1 ti1 tin−1
≤L {Eˆ [|M −M |+|M −M |]+Eˆ [|N −N |+|N −N |]}.
ϕ 1 tin tn tin−1 tn−1 2 tin tn tin−1 tn−1
From this we deduce
Eˆ[ϕ(X ,...,X ,X −X )]= lim Eˆ[ψ(X ,...,X )]
t1 tn−1 tn tn−1 i→∞ ti1 tin−1
=Eˆ[ψ(X ,...,X )].
t1 tn−1
Thus (X ) has independent increments. It follows from (2) in Lemma 3.6 and assumption (A) that
t t∈[0,1]
(M˜ ) =d (M ) and(N˜ ) =d (N ) . If(M ) and(N ) aretwoprocesseswithsta-
t t∈[0,1] t t∈[0,1] t t∈[0,1] t t∈[0,1] t t∈[0,1] t t∈[0,1]
tionary and independent increments, then from the construction of Eˆ, (X ) has stationary increments.
t t∈[0,1]
The proof is complete. (cid:3)
In the following, we give an example to calculate Eˆ.
Example 3.7 Let Γ , i=1,2, be two given bounded subset in Rd×d. Define G :S(d)→R by
i i
1
G (A)= sup tr[AQQT] for each A∈S(d).
i
2
Q∈Γi
Let (Bi) be two d-dimensional G -Brownian motion defined on sublinear expectation space (Ω ,H ,Eˆ ),
t t≥0 i i i i
i = 1, 2. In the above, we construct a sublinear expectation space (Ω,H,Eˆ) and a 2d-dimensional pro-
cess (B˜ ) = (B˜1,B˜2) with stationary and independent increments satisfying (B˜1) =d (B1) and
t t≥0 t t t≥0 t t≥0 t t≥0
(B˜2) =d (B2) . Since
t t≥0 t t≥0
Eˆ[|B˜t|3]≤4(Eˆ[|B˜t1|3]+Eˆ[|B˜t2|3])=4(Eˆ1[|Bt1|3]+Eˆ2[|Bt2|3])=4Ct23,
where C = Eˆ [|B1|3]+Eˆ [|B2|3], by Theorem 1.6 in Chapter III in [12], we can obtain that (B˜ ) is a
1 1 2 1 t t≥0
G-Brownian motion with
 A D 
1 1
G(A)= Eˆ[hAB˜ ,B˜ i] for each A= ∈S(2d).
1 1
2  
 DT A 
 2 
By our construction, it is easy to check that Eˆn[hDB˜2,B˜1i] = Eˆn[−hDB˜2,B˜1i] = 0 for each n ≥ 1.
1 1 1 1
Thus Eˆ[hDB˜2,B˜1i] = Eˆ[−hDB˜2,B˜1i] = 0. By subadditivity, we can get Eˆ[hAB˜ ,B˜ i] = Eˆ[hA B˜1,B˜1i+
1 1 1 1 1 1 1 1 1
hA B˜2,B˜2i]. Furthermore,
2 1 1
Eˆn[hA B˜1,B˜1i+hA B˜2,B˜2i]=Eˆ [hA B1,B1i]+Eˆ [hA B2,B2i]
1 1 1 2 1 1 1 1 1 1 2 2 1 1
for each n≥1 by our construction. Thus G(A)=G (A )+G (A ).
1 1 2 2
10