# Jensen's Inequality for Generalized Peng'sg-Expectations and Its Applications

Authors: Zhaojun Zong
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study Jensen's inequality for generalized Peng'sg-expectations and give four equivalent conditions on Jensen's inequality for generalized Peng's g-expectations without the assumption that the generator g is continuous with respect to t. This result includes and extends some existing results. Furthermore, we give some applications of Jensen's inequality for generalized Peng's g-expectations.


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## Body

## 1. Introduction

By Pardoux and Peng [1], we know that there exists a unique adapted and square integrable solution to a backward stochastic differential equation (BSDE for short) of the type
(1)yt=ξ+∫tT‍g(s,ys,zs)ds-∫tT‍zs·dWs,t∈[0,T],
provided that the function g is Lipschitz in both variables y and z, and ξ and (g(t,0,0))t∈[0,T] are square integrable. g is said to be the generator of BSDE (1). We denote the unique adapted and square integrable solution of BSDE (1) by (yt(T,g,ξ),zt(T,g,ξ))t∈[0,T].Based on such a BSDE, Peng [2] introduced the notion of g-expectation. He proved that the g-expectation preserves many of properties of the classical mathematical expectation, but not the linearity property, and thus the g-expectation is a type of nonlinear mathematical expectation. Indeed, g-expectation is a kind of nonlinear expectation, which can be considered as a nonlinear extension of the well-known Girsanov transformations. The original motivation for studying g-expectation comes from the theory of expected utility. Since the notion of g-expectation was introduced, many properties of g-expectation have been investigated by many researchers. In 1997, Peng [3] introduced the notions of conditional g-expectation and g-martingale. Later, Briand et al. [4] studied Jensen's inequality for g-expectations and gave a counter example and a proposition to indicate that even for a linear function, Jensen's inequality might fail for some g-expectations. This yields a natural question: under which conditions on g in the g-expectation does Jensen's inequality hold for any convex function? Under the assumptions that g does not depend on y and is convex, Chen et al. [5, 6] studied Jensen's inequality for g-expectations and gave a necessary and sufficient condition on g under which Jensen's inequality holds for convex functions. Provided that g only does not depend on y, Jiang [7] gave another necessary and sufficient condition on g under which Jensen's inequality holds for convex functions. It was an improved result in comparison with the result that Chen et al. yielded. Later, this result was improved by Hu [8] and Jiang [9] showing that, in fact, g must be independent of y. But these results need the assumption that the generator g is continuous with respect to t.In this paper, without the assumption that the generatorg is continuous with respect to t, we study Jensen's inequality for generalized Peng's g-expectations and give four equivalent conditions on Jensen's inequality for generalized Peng's g-expectations, which generalize the known results on Jensen's inequality for g-expectations in Chen et al. [5, 6], Jiang [7, 9], and Hu [8]. Furthermore, we give some applications of Jensen's inequality for generalized Peng's g-expectations.This paper is organized as follows: in Section2, we introduce some notations, assumptions, notions, and lemmas which will be useful in this paper; in Section 3, we give our main results including the proofs and applications.


## 2. Preliminaries

Firstly, let us list some notations, assumptions, notions, lemmas, and propositions that are used in this paper. Let(Ω,ℱ,P) be a probability space and let (Wt)t≥0 be a d-dimensional standard Brownian motion with respect to filtration (ℱt)t≥0 generated by Brownian motion and all P-null subsets, that is,
(2)ℱt=σ{Ws;s≤t}∨𝒩,
where 𝒩 is the set of all P-null subsets. Fix a real number T>0. For any positive integer n and z∈Rn, |z| denotes its Euclidean norm.We define the following usual spaces of processes (random variables):(i)
ConsiderLp(Ω,ℱT,P)={ξ:ξ is ℱT-measurable random variable such that E[|ξ|p]<∞,p≥1};(ii)
Considerℒ(Ω,ℱT,P)=⋃p>1Lp(Ω,ℱT,P);(iii)
Consider𝒮ℱp(0,T;P;R)={V:V is a continuous process with E[sup0≤t≤T|Vt|p]<∞,p≥1};(iv)
Consider𝒮ℱ(0,T;P;R)=⋃p>1𝒮ℱp(0,T;P;R);(v)
Considerℒℱp(0,T;P;Rn)={V:Vis  a  progressively measurable process with E[(∫0T‍|Vs|2ds)p/2]<∞,p≥1};(vi)
Considerℒℱ(0,T;P;Rn)=⋃p>1Lℱp(0,T;P;Rn).Suppose the generatorg(ω,t,y,z):Ω×[0,T]×R×Rd↦R satisfies the following assumptions:(A.1)
there exists a constantμ>0, such that P-a.s., we have: ∀t∈[0,T], ∀y1,y2∈R, z1,z2∈Rd, |g(t,y1,z1)-g(t,y2,z2)|≤μ(|y1-y2|+|z1-z2|);(A.2)
P-a.s., ∀(t,y)∈[0,T]×R, g(t,y,0)≡0.The following lemma is a special case of Theorem 4.2 in Briand et al. [10].Lemma 1.
Supposeg satisfies (A.1) and (A.2). Then for each given ξ∈Lp(Ω,ℱT,P), where 1<p<2, the BSDE (1) has a unique pair of adapted processes (yt(T,g,ξ),zt(T,g,ξ))t∈[0,T]∈𝒮ℱp(0,T;P;R)×lℱp(0,T;P;Rd).From Lemma1, we have the following.Remark 2.
Supposeg satisfies (A.1) and (A.2). Then for each given ξ∈ℒ(Ω,ℱT,P), the BSDE (1) has a unique pair of adapted processes (yt(T,g,ξ),zt(T,g,ξ))t∈[0,T]∈𝒮ℱ(0,T;P;R)×ℒℱ(0,T;P;Rd).Now, we introduce the notions of generalized Peng'sg-expectation and generalized conditional Peng's g-expectation.Definition 3 (generalized Peng'sg-expectation [11]).
Supposeg satisfies (A.1) and (A.2). For any ξ∈ℒ(Ω,ℱT,P), let (yt(T,g,ξ),zt(T,g,ξ))t∈[0,T] be the solution of BSDE (1). Consider the mapping ℰg[·]:ℒ(Ω,ℱT,P)↦R, denoted by ℰg[ξ]=y0(T,g,ξ). One calls ℰg[ξ] the generalized Peng's g-expectation of ξ.Definition 4 (generalized Peng's conditionalg-expectation [11]).
Supposeg satisfies (A.1) and (A.2). The generalized Peng's conditional g-expectation of ξ with respect to ℱt is defined by
(3)ℰg[ξ∣ℱt]=yt(T,g,ξ),t∈[0,T].Then, let us list some basic properties of generalized Peng'sg-expectation.Proposition 5 (see [11]).
Considerℰg[ξ∣ℱt] is the unique random variable η in ℒ(Ω,ℱt,P) such that
(4)ℰg[1Aξ]=ℰg[1Aη],∀A∈ℱt.Proposition 6 (see [11]).
Supposeg satisfies (A.1) and (A.2). If g does not depend on y, that is, g(ω,t,z):Ω×[0,T]×Rd↦R, then
(5)ℰg[ξ+η∣ℱt]=ℰg[ξ∣ℱt]+η,∀ξ∈ℒ(Ω,ℱT,P),∀η∈ℒ(Ω,ℱt,P).Proposition 7 (see [11]).
Supposeg satisfies (A.1) and (A.2). For ξ,ηn∈Lp(Ω,ℱT,P), where n=1,2,… and p>1, if E[|ξ-ηn|p∣ℱt]→0,a.s.,t∈[0,T], then
(6)limn→∞ℰg[ηn∣ℱt]=ℰg[ξ∣ℱt],a.s.,t∈[0,T].
Applying Proposition 7, one can immediately obtain the following.Remark 8.
(i) For anyξ∈ℒ(Ω,ℱT,P), let ξn=(ξ∧n)∨(-n), n=1,2,…, then limn→∞ℰg[ξn∣ℱt]=ℰg[ξ∣ℱt], a.s., ∀t∈[0,T].
(ii) For anyξn∈ℒ(Ω,ℱT,P), if limn→∞ξn=ξ a.s. and |ξn|≤η a.s. with η∈ℒ(Ω,ℱT,P), then limn→∞ℰg[ξn∣ℱt]=ℰg[ξ∣ℱt], a.s., ∀t∈[0,T].Lemma 9.
Supposeg satisfies (A.1) and (A.2). Let {Ai}i=1m be a ℱt-measurable partition of Ω (i.e., Ai∈ℱt, Ai⋂‍Aj=∅ if i≠j and ⋃i=1mAi=Ω), where t≤T. Then for each Xi∈ℒ(Ω,ℱT,P), i=1,…,m, one has
(7)∑i=1m‍1Aiℰg[Xi∣ℱt]=ℰg[∑i=1m‍1AiXi∣ℱt]a.s.Proof.
We consider the following BSDEs:(8)ℰg[Xi∣ℱt]=Xi+∫tT‍g(s,ℰg[Xi∣ℱs],zs(T,g,Xi))ds-∫tT‍zs(T,g,Xi)·dWs,i=1,…,m,(9)ℰg[∑i=1m‍1AiXi∣ℱt]=∑i=1m‍1AiXi+∫tT‍g(s,ℰg[∑i=1m‍1AiXi∣ℱs],∑i=1m‍1AiXi+∫tT‍gzs(T,g,∑i=1m1AiXi){[∑i=1m‍1AiXi∣ℱs]})ds-∫tT‍zs(T,g,∑i=1m1AiXi)·dWs.
By the fact that ∑i=1m1Aig(s,ℰg[Xi∣ℱs],zs(T,g,Xi)) = g(s,∑i=1m1Aiℰg[Xi∣ℱs],∑i=1m1Aizs(T,g,Xi)),t≤s≤T and from (8), we have
(10)∑i=1m‍1Aiℰg[Xi∣ℱt]=∑i=1m‍1AiXi+∫tT‍g(s,∑i=1m‍1Aiℰg[Xi∣ℱs],∑i=1m‍1Aizs(T,g,Xi))ds-∫tT‍∑i=1m‍1Aizs(T,g,Xi)·dWs.
Comparing this with (9), it follows that ∑i=1m1Aiℰg[Xi∣ℱt]=ℰg[∑i=1m1AiXi∣ℱt] a.s. The proof of Lemma 9 is complete.Proposition 10.
Supposeg satisfies (A.1) and (A.2). Then the following two statements are equivalent:(i)
consider∀(X,k)∈ℒ(Ω,ℱT,P)×R, ℰg[X+k∣ℱt]=ℰg[X∣ℱt]+ka.s.,(ii)
consider∀(X,η)∈ℒ(Ω,ℱT,P)×ℒ(Ω,ℱt,P), ℰg[X+η∣ℱt]=ℰg[X∣ℱt]+ηa.s.Proof.
It is obvious that (ii) implies (i). We only need to prove that (i) implies (ii). Suppose (i) holds. Let{Ai}i=1m be a ℱt-measurable partition of Ω and let λi∈R (i=1,2,…,m). From Lemma 9 and (i), we deduce that for each X∈ℒ(Ω,ℱT,P),
(11)ℰg[X+∑i=1m‍λi1Ai∣ℱt]=ℰg[∑i=1m‍1Ai(X+λi)∣ℱt]=∑i=1m‍1Aiℰg[X+λi∣ℱt]=∑i=1m‍1Ai(ℰg[X∣ℱt]+λi)=ℰg[X∣ℱt]+∑i=1m‍λi1Aia.s.
In other words, for any X∈ℒ(Ω,ℱT,P) and any simple function η∈ℒ(Ω,ℱt,P),
(12)ℰg[X+η∣ℱt]=ℰg[X∣ℱt]+ηa.s.
Let
(13)ηn:=∑i=0n2n-1‍i2n1{(i/2n)≤η<((i+1)/2n)}+n1{η≥n}+∑i=0n2n-1‍-i2n1{-((i+1)/2n)≤η<-(i/2n)}+(-n)1{η<-n},n=1,2,….
Obviously, for each n, ηn is a simple function in ℒ(Ω,ℱt,P). From (12), we have
(14)ℰg[X+ηn∣ℱt]=ℰg[X∣ℱt]+ηna.s.
On the other hand, limn→∞(X+ηn)=X+η, |X+ηn|≤|X|+|η|. Thus, from Remark 8 (ii), it follows that (ii) is true. The proof of Proposition 10 is complete.


## 3. Main Results and Applications

Definition 11.
Letg: Ω×[0,T]×R×Rd↦R. The function g is said to be superhomogeneous if for each (y,z)∈R×Rd and any real number λ, then g(t,λy,λz)≥λg(t,y,z), dP×dt a.s. The function g is said to be positively homogeneous if for each (y,z)∈R×Rd and any real number λ≥0, then g(t,λy,λz)=λg(t,y,z), dP×dt a.s.Before we give our main results, let us see an example.Example 12.
FixT=1 and d=1. Let ξ=f(W1), where f(x)=exp((x2/2p1)-x)1(x≥p1), 1<p1<2.
Obviously,f is an increasing function. We can easily get
(15)E[|ξ|p1]=∫p1∞‍exp(x22-p1x)12πe-(1/2)x2dx=12πp1e-p12<∞,E[|ξ|p]=∞,∀p>p1.
Hence, ξ∈ℒ(Ω,ℱ1,P), but ξ∉L2(Ω,ℱ1,P).
Letξn=ξ∧n, n=1,2,…. Clearly, for each n, ξn∈L2(Ω,ℱ,P). For simplicity, we will write ℰμ[·]≡ℰg[·] for g=μ|z|. From Theorem 1 in Chen and Kulperger’s [12], we know that ℰμ[ξn]=EQ[ξn], where dQ/dP=e-(1/2)μ2+μW1.
By Remark8(i), we have ℰμ[ξn]→ℰμ[ξ], as n→∞. On the other hand, applying Hölder's inequality and noting that E[e-(1/2)μ2+μW1]=1 and E[e-(1/2)μ2q2+μqW1]=1, we obtain
(16)EQ[ξ]≤(E[|ξ|p1])1/p1(E[(dQdP)q])1/q≤e(1/2)(q-1)μ2(E[|ξ|p1])1/p1<∞,
where (1/p1)+(1/q)=1. It then follows from the monotonic convergence theorem that
(17)EQ[ξn]⟶EQ[ξ],asn⟶∞.
Thus
(18)ℰμ[ξ]=EQ[ξ].
Letφ(x)=(x-k)+, where k∈R. Obviously, φ(x) is a convex and increasing function. From this, we know that φ∘f is an increasing function. In a similar manner of the above, we can deduce that
(19)ℰμ[φ(ξ)]=EQ[φ(ξ)].
From (18), (19), and the classical Jensen's inequality, we have
(20)φ(ℰμ[ξ])=φ(EQ[ξ])≤EQ[φ(ξ)]=ℰμ[φ(ξ)].
This problem yields a natural question: in general, under which conditions ong do generalized Peng's g-expectations satisfy Jensen's inequality for convex functions?The following theorem will answer this question.Theorem 13.
Letg satisfy (A.1) and (A.2). Then the following four statements are equivalent.(i)
Jensen's inequality for generalized Peng'sg-expectation ℰg[·∣ℱt] holds in general, that is, for each convex function φ(x):R↦R and each ξ∈ℒ(Ω,ℱT,P), if φ(ξ)∈ℒ(Ω,ℱT,P), then one has
(21)ℰg[φ(ξ)ℱt]≥φ(ℰg[ξ∣ℱt])a.s.;(ii)
consider∀(ξ,a,b)∈L2(Ω,ℱT,P)×R×R, ℰg[aξ+b]≥aℰg[ξ]+b;(iii)
consider∀(ξ,a,b)∈L2(Ω,ℱT,P)×R×R, ℰg[aξ+b∣ℱt]≥aℰg[ξ∣ℱt]+b a.s.;(iv)
considerg is independent of y, superhomogeneous, and positively homogeneous with respect to z.Proof.
(i)⇒(ii) is obvious.
(ii)⇒(iii): let η=ξ+b. By (ii), we have
(22)ℰg[η-b]≥ℰg[η]-b,
That is,
(23)ℰg[ξ]+b≥ℰg[ξ+b].
Thus, for each (ξ,b)∈L2(Ω,ℱT,P)×R,
(24)ℰg[ξ+b]=ℰg[ξ]+b.
For each (X,t,k)∈L2(Ω,ℱT,P)×[0,T]×R, by (24), we know that for each A∈ℱt,
(25)ℰg[1A(X+k)]=ℰg[1AX+1Ak-k]+k=ℰg[1AX+1AC(-k)]+k=ℰg[ℰg[1AX+1AC(-k)∣ℱt]]+k=ℰg[1Aℰg[X∣ℱt]+1AC(-k)]+k=ℰg[1Aℰg[X∣ℱt]+1AC(-k)+k]=ℰg[1A(ℰg[X∣ℱt]+k)].
Thus,
(26)ℰg[X+k∣ℱt]=ℰg[X∣ℱt]+ka.s.,∀t∈[0,T].
On the other hand, for each λ≠0, define
(27)ℰλ[·∣ℱt]=ℰg[λ·∣ℱt]λ,∀t∈[0,T].
It is easy to check that ℰg[·∣ℱt] and ℰλ[·∣ℱt] are two ℱ-expectations on L2(Ω,ℱT,P) (the notion of ℱ-expectation can be seen in [13]). From (ii), we have if λ>0, for each ξ∈L2(Ω,ℱT,P)(28)ℰλ[ξ]≥ℰg[ξ].
Hence, by Lemma 4.5 in [13], we have
(29)ℰλ[ξ∣ℱt]≥ℰg[ξ∣ℱt]a.s.,∀t∈[0,T].
Similarly, if λ<0, for each ξ∈L2(Ω,ℱT,P)(30)ℰλ[ξ]≤ℰg[ξ].
Hence, by Lemma 4.5 in [13] again, we have
(31)ℰλ[ξ∣ℱt]≤ℰg[ξ∣ℱt]a.s.,∀t∈[0,T].
Thus from (29) and (31), we have ∀(ξ,λ)∈L2(Ω,ℱT,P)×R,
(32)ℰg[λξ∣ℱt]≥λℰg[ξ∣ℱt]a.s.,∀t∈[0,T].
From (26) and (32), we have
(33)∀(ξ,a,b)∈L2(Ω,ℱT,P)×R×R,ℰg[aξ+b∣ℱt]≥aℰg[ξ∣ℱt]+ba.s.,∀t∈[0,T].
(iii)⇒(iv): Firstly, we prove that g is independent of y. From (iii), we can obtain that for each (ξ,y)∈L2(Ω,ℱT,P)×R,
(34)ℰg[ξ-y∣ℱt]=ℰg[ξ∣ℱt]-y,a.s.,∀t∈[0,T].
For each (t,y,z)∈[0,T]×R×Rd, let Y·t,y,z be the solution of the following SDE defined on [t,T]:
(35)Yst,y,z=y-∫ts‍g(r,Yrt,y,z,z)dr+z·(Ws-Wt).
From (34), we have
(36)Yrt,y,z-y=ℰg[Yst,y,z∣ℱr]-y=ℰg[Yst,y,z-y∣ℱr],t≤r≤s≤T.
Let Ys=Yst,y,z-y, s∈[t,T] and Z be the corresponding part of Itô's integrand. It then follows that
(37)Ys=-∫ts‍g(r,Yrt,y,z,z)dr+∫ts‍z·dWr=-∫ts‍g(r,Yr,Zr)dr+∫ts‍Zr·dWr.
Thus, Zr≡z and
(38)g(r,Yr,z)=g(r,Yrt,y,z-y,z)=g(r,Yrt,y,z,z).
Then, we can apply Lemma 4.4 in Peng [14] to obtain that for each (y,z)∈R×Rd,
(39)g(t,y,z)=g(t,0,z),dP×dta.s.
Namely, g is independent of y.
Now we prove thatg is superhomogeneous with respect to z. From (iii), we can obtain that for each (ξ,λ)∈L2(Ω,ℱT,P)×R,
(40)λℰg[ξ∣ℱt]≤ℰg[λξ∣ℱt],a.s.,∀t∈[0,T].
For each (t,z)∈[0,T]×Rd, let Y·t,z be the solution of the following SDE defined on [t,T]:
(41)Yst,z=-∫ts‍g(r,z)dr+z·(Ws-Wt).
From (40), we have
(42)ℰg[λYst,z∣ℱr]≥λℰg[Yst,z∣ℱr]=λYrt,z,t≤r≤s≤T.
Thus, (λYst,z)s∈[t,T] is an ℰg-submartingale. From the decomposition theorem of ℰg-supermartingale (see [15]), it follows that there exists an increasing process (As)s∈[t,T] such that
(43)λYst,z=-∫ts‍g(r,Zr)dr+As-At+∫ts‍Zr·dWr,s∈[t,T].
This with λYst,z=-∫ts‍λg(r,z)dr+∫ts‍λz·dWr yields Zr≡λz and
(44)λg(r,z)≤g(r,λz),dP×dta.s.
At last, we prove thatg is positively homogeneous with respect to z. From (iii), we can obtain that for each fixed λ>0 and ξ∈L2(Ω,ℱT,P),
(45)1λℰg[λξ∣ℱt]≤ℰg[ξ∣ℱt],a.s.,∀t∈[0,T],
that is,
(46)ℰg[λξ∣ℱt]≤λℰg[ξ∣ℱt],a.s.,∀t∈[0,T].
Thus, we have
(47)ℰg[λξ∣ℱt]=λℰg[ξ∣ℱt],a.s.,∀t∈[0,T].
Obviously, if λ=0, (47) still holds. Thus, for each λ≥0,
(48)ℰg[λξ∣ℱt]=λℰg[ξ∣ℱt],a.s.,∀t∈[0,T].
For each (t,z)∈[0,T]×Rd, let Y·t,z be the solution of SDE (34). From (48), for each λ≥0, we have
(49)ℰg[λYst,z∣ℱr]=λℰg[Yst,z∣ℱr]=λYrt,z,t≤r≤s≤T.
This implies that there exists a process Z·t,z,λ such that
(50)λYst,z=-∫ts‍g(r,Zrt,z,λ)dr+∫ts‍Zrt,z,λ·dWr,s∈[t,T].
Comparing this with λYst,z=-∫ts‍λg(r,z)dr+∫ts‍λz·dWr, it follows that Zrt,z,λ≡λz and
(51)λg(r,z)=g(r,λz),dP×dta.s.
(iv)⇒(iii): By comparison theorem (for example, we can see [3]), it is easy to obtain (iii).
(iii)⇒(i): Suppose (iii) holds. From (iii) and by Remark 8 (i), we have
(52)∀(X,k)∈ℒ(Ω,ℱT,P)×R,ℰg[X+k∣ℱt]=ℰg[X∣ℱt]+ka.s.,(53)∀(X,λ)∈ℒ(Ω,ℱT,P)×R,ℰg[λX∣ℱt]≥λℰg[X∣ℱt]a.s.
From (53), we can deduce that for each bounded variable ζ∈ℱt,
(54)∀X∈ℒ(Ω,ℱT,P),ℰg[ζX∣ℱt]≥ζℰg[X∣ℱt]a.s.
In fact, let {Ai}i=1m be a ℱt-measurable partition of Ω and let λi∈R (i=1,2,…,m). By (53), we have
(55)ℰg[∑i=1m‍λi1AiX∣ℱt]=∑i=1m‍1Aiℰg[λiX∣ℱt]≥∑i=1m‍1Aiλiℰg[X∣ℱt]a.s.
In other words, for each X∈ℒ(Ω,ℱT,P) and each simple function ζ∈ℒ(Ω,ℱt,P),
(56)ℰg[ζX∣ℱt]≥ζℰg[X∣ℱt]a.s.
Thus, thanks to Remark 8(ii), it follows that (54) is true.
The main idea of the following proof is derived from [7]. Given ξ∈ℒ(Ω,ℱT,P) and convex function φ such that φ(ξ)∈ℒ(Ω,ℱT,P), we set ηt=φ-′(ℰg[ξ∣ℱt]). Then ηt is ℱt-measurable. Since φ is convex, we have
(57)φ(x)-φ(y)≥φ-′(y)(x-y),∀x,y∈R.
Take x=ξ, y=ℰg[ξ∣ℱt]. Then we have
(58)φ(ξ)-φ(ℰg[ξ∣ℱt])≥ηt(ξ-ℰg[ξ∣ℱt])a.s.
For each n∈N, we define
(59)Ωt,n:={|ℰg[ξ∣ℱt]|+|ηt|+|φ(ℰg[ξ∣ℱt])|≤n},
so we have
(60)ℰg[1Ωt,nφ(ξ)∣ℱt]≥ℰg[1Ωt,nφ(ℰg[ξ∣ℱt])-1Ωt,nηtℰg[ξ∣ℱt]+1Ωt,nηtξ∣ℱt]a.s.
By the definition of 1Ωt,n, we know
(61)1Ωt,nφ(ℰg[ξ∣ℱt])-1Ωt,nηtℰg[ξ∣ℱt]∈ℒ(Ω,ℱt,P).
Thus, in view of (52) and from Proposition 10, we can get
(62)ℰg[1Ωt,nφ(ξ)∣ℱt]≥1Ωt,nφ(ℰg[ξ∣ℱt])-1Ωt,nηtℰg[ξ∣ℱt]+ℰg[1Ωt,nηtξ∣ℱt]a.s.
Moreover, from (54), considering that 1Ωt,nηt∈ℱt and is bounded by n, we can get(63)ℰg[1Ωt,nηtξ∣ℱt]≥1Ωt,nηtℰg[ξ∣ℱt]a.s.
Hence, we can deduce that for each n∈N,
(64)ℰg[1Ωt,nφ(ξ)∣ℱt]≥1Ωt,nφ(ℰg[ξ∣ℱt])a.s.
Finally, thanks to Remark 8 (ii) again, we can get
(65)ℰg[φ(ξ)∣ℱt]≥φ(ℰg[ξ∣ℱt])a.s.
Hence, Jensen's inequality for ℰg[·∣ℱt] holds in general. The proof of Theorem 13 is complete.Example 14.
SupposeH is a bounded, convex, and closed subset of Rd and D= the set of Rd-valued continuous processes (vt)t∈[0,T] such that for each t, vt∈H a.s.. Define the probability measure Qv by
(66)dQvdP=e-(1/2)∫0T‍|vs|2ds+∫0T‍vs·dWs.
Thus, for any convex function φ,
(67)φ(esssupv∈DEQv[ξ∣ℱt])≤esssupv∈DEQv[φ(ξ)∣ℱt],a.s.,∀t∈[0,T],
whenever ξ, φ(ξ)∈ℒ(Ω,ℱT,P).Proof.
Letg(t,z)= ess supv∈Dvt·z. Obviously, g(t,z) is superhomogeneous and positively homogeneous with respect to z. and satisfies (A.1) and (A.2).
From El Karoui and Quenez [16], we have esssupv∈DEQv[ξ∣ℱt]=ℰg[ξ∣ℱt],a.s.,∀ξ∈L2(Ω,ℱT,P). Now we prove esssupv∈DEQv[ξ∣ℱt]=ℰg[ξ∣ℱt], a.s., ∀ξ∈ℒ(Ω,ℱT,P). Indeed, for any ξ∈ℒ(Ω,ℱT,P), there exists 1<p<2 such that ξ∈Lp(Ω,ℱT,P). Let ξn=(ξ∧n)∨(-n), n=1,2,…. Clearly, for each n, ξn∈L2(Ω,ℱT,P), then
(68)esssupv∈DEQv[ξn∣ℱt]=ℰg[ξn∣ℱt],a.s.
Since(69)esssupv∈DEQv[ξn∣ℱt]=esssupv∈D(EQv[ξn-ξ∣ℱt]+EQv[ξ∣ℱt])≤esssupv∈DEQv[ξn-ξ∣ℱt]+esssupv∈DEQv[ξ∣ℱt],
we have
(70)ℰg[ξn∣ℱt]-esssupv∈DEQv[ξ∣ℱt]≤esssupv∈DEQv[ξn-ξ∣ℱt].
With an approach similar to the one above, we can get easily that
(71)ℰg[ξn∣ℱt]-esssupv∈DEQv[ξ∣ℱt]≥essinfv∈DEQv[ξn-ξ∣ℱt].
Combining (42) with (43), we have
(72)|ℰg[ξn∣ℱt]-esssupv∈DEQv[ξ∣ℱt]|≤({|esssupv∈DEQv[ξn-ξ∣ℱt]|}|essinfv∈DEQv[ξn-ξ∣ℱt]|∨|esssupv∈DEQv[ξn-ξ∣ℱt]|)≤esssupv∈DEQv[|ξn-ξ|∣ℱt].
By Hölder's inequality and noting that (e-(1/2)∫0t‍|vs|2ds+∫0t‍vs·dWs)t∈[0,T] and (e-(1/2)∫0t‍|qvs|2ds+∫0t‍qvs·dWs)t∈[0,T] are both martingales with respect to (ℱt)t∈[0,T], we can obtain
(73)EQv[|ξn-ξ|∣ℱt]=E[|ξn-ξ|(dQv/dP)∣ℱt]E[(dQv/dP)∣ℱt]≤(E[|ξn-ξ|p∣ℱt])1/p(E[(dQv/dP)q∣ℱt])1/qE[(dQv/dP)∣ℱt]≤e(1/2)(q-1)μ2T(E[|ξn-ξ|p∣ℱt])1/p,
where (1/p)+(1/q)=1. It then follows from Lebesgue's dominated convergence theorem that
(74)esssupv∈DEQv[|ξn-ξ|∣ℱt]⟶0,asn⟶∞.
Hence,
(75)|ℰg[ξn∣ℱt]-esssupv∈DEQv[ξ∣ℱt]|⟶0,asn⟶∞.
On the other hand, from Remark8 (i), we have
(76)ℰg[ξn∣ℱt]⟶ℰg[ξ∣ℱt],asn⟶∞.
Thus,
(77)ℰg[ξ∣ℱt]=esssupv∈DEQv[ξ∣ℱt],a.s.,∀ξ∈ℒ(Ω,ℱT,P).
Applying Theorem13, we have
(78)φ(esssupv∈DEQv[ξ∣ℱt])≤esssupv∈DEQv[φ(ξ)∣ℱt],a.s.Definition 15.
Supposeg satisfies (A.1) and (A.2). A process (Xt)t∈[0,T] satisfying that for each t, Xt∈ℒ(Ω,ℱt,P) is called a generalized Peng's g-martingale (resp., generalized Peng's g-supermartingale, generalized Peng's g-submartingale), if for any t, s satisfying t≤s≤T,
(79)ℰg[Xs∣ℱt]=Xt(resp.≤Xt,≥Xt),a.s.Applying Theorem13, immediately we have the following.Theorem 16.
Supposeg is independent of y, superhomogeneous and positively homogeneous with respect to z and satisfies (A.1) and (A.2). If (Xt)t∈[0,T] is a generalized Peng's g-martingale and φ is a convex function such that φ(Xt)∈ℒ(Ω,ℱt,P), then (φ(Xt))t∈[0,T] is a generalized Peng's g-submartingale.Remark 17.
Supposeg is independent of y, superhomogeneous and positively homogeneous with respect to z and satisfies (A.1) and (A.2). Similarly, we can get the following.(i)
If(Xt)t∈[0,T] is a generalized Peng's g-submartingale and φ is a convex and increasing function such that φ(Xt)∈ℒ(Ω,ℱt,P), then (φ(Xt))t∈[0,T] is a generalized Peng's g-submartingale.(ii)
If(Xt)t∈[0,T] is a generalized Peng's g-supermartingale and φ is a convex and decreasing function such that φ(Xt)∈ℒ(Ω,ℱt,P), then (φ(Xt))t∈[0,T] is a generalized Peng's g-submartingale.Example 18.
(i) Letg=μ|z| and φ(x)=(x-a)+ where a∈R. Obviously, g satisfies the assumptions of Remark 17 and φ is a convex and increasing function. By Remark 17 (i), we have the following: suppose (Xt)t∈[0,T] is a ℰμ-submartingale, then ((Xt-a)+)t∈[0,T] is a ℰμ-submartingale.
(ii) Letg=μ|z| and φ(x)=(x-b)- where b∈R. With the similar argument, we have the following: suppose (Yt)t∈[0,T] is a ℰμ-supermartingale, then ((Yt-b)-)t∈[0,T] is a ℰμ-submartingale.



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*Source: 683047-2013-03-12.xml*