Dynamic Programming Principles for Optimal Stopping with ...

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Dynamic Programming Principles for Optimal Stopping with Expectation Constraint

arXiv:1708.02192v1 [math.OC] 7 Aug 2017

Erhan Bayraktar∗† , Song Yao‡§

Abstract We analyze an optimal stopping problem with a constraint on the expected cost. When the reward function and cost function are Lipschitz continuous in state variable, we show that the value of such an optimal stopping problem is a continuous function in current state and in budget level. Then we derive a dynamic programming principle (DPP) for the value function in which the conditional expected cost acts as an additional state process. As the optimal stopping problem with expectation constraint can be transformed to a stochastic optimization problem with supermartingale controls, we explore a second DPP of the value function and thus resolve an open question recently raised in [S. Ankirchner, M. Klein, and T. Kruse, A verification theorem for optimal stopping problems with expectation constraints, Appl. Math. Optim., 2017, pp. 1-33]. Based on these two DPPs, we characterize the value function as a viscosity solution to the related fully non-linear parabolic Hamilton-JacobiBellman equation. MSC: 60G40, 49L20, 93E20, 49L25. Keywords: Optimal stopping with expectation constraint, dynamic programming principle, shifted processes, shifted stochastic differential equations, flow property, stochastic optimization with supermartingale controls, fully non-linear parabolic Hamilton-Jacobi-Bellman equation, viscosity solution, Monge-Amp`ere type equation.

1

Introduction

In this article, we analyze a continuous-time optimal stopping problem with expectation constraint on the accumulated cost. Suppose that the game begins at time t over the canonical space Ωt of continuous paths. Under the t  t Wiener measure Pt , the coordinator process W t = {Wst }s∈[t,∞) of Ωt is a Brownian motion. Let F = F s s∈[t,∞) be

the Pt −augmentation of the filtration generated by W t , and let the Rl −valued state flow X t,x evolve from position x ∈ Rl according to a stochastic differential equation Z s Z s Xs = x+ b(r, Xr )dr+ σ(r, Xr ) dWrt , s ∈ [t, ∞). (1.1) t

t



f (r, Xrt,x )dr and a terminal reward π(τ, Xτt,x ) by Rτ choosing an F −stopping time τ , which, however, has to satisfy a budget constraint Et [ t g(r, Xrt,x )dr] ≤ y. So the value of such a optimal stopping problem with expectation constraint is in form of   V(t, x, y) := sup Et R(t, x, τ ) , (1.2) We aim to maximize the sum R(t, x, τ ) of a running reward t

t

τ ∈Txt (y)

 Rτ with Txt (y) := τ : Et [ t g(r, Xrt,x )dr] ≤ y and Et [·] = EPt [·]. In particular, when the cost rate g(r, x) is a power function of r, the budget constraint specifies as a moment constraint on stopping times. Kennedy [37] initiated the study of optimal stopping problem with expectation constraint. The author used a Lagrange multiplier method to reduce a discrete-time optimal stopping problem with first-moment constraint to an ∗ Department

of Mathematics, University of Michigan, Ann Arbor, MI 48109; email: [email protected]. Bayraktar is supported in part by the National Science Foundation under DMS-1613170, and in part by the Susan M. Smith Professorship. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of the National Science Foundation. ‡ Department of Mathematics, University of Pittsburgh, Pittsburgh, PA 15260; email: [email protected]. § S. Yao is supported by the National Science Foundation under DMS-1613208. † E.

DPPs for Optimal Stopping with Expectation Constraint

2

unconstrained optimal stopping problem and showed that the optimal value of the dual problem is equal to that of the primal problem. Since then, the Lagrangian technique has been prevailing in research of optimal stopping problems with expectation constraints. In the present paper, we develop a new approach to analyze the optimal stopping problem with expectation constraint (1.2). Our main contributions are obtaining the continuity of the value function V and establishing two dynamic programming principles (DPPs) for V. When reward/cost functions f, π, g are Lipschitz continuous in state variable x and the cost function g is nondegenerate in sense of (g3), we first demonstrate over a general probability setting that the value function is continuous in (t, x, y) by utilizing a priori estimates of the state process X t,x and delicately constructing approximate stopping strategies (see Theorem 2.1). This continuity result together with the properties of shifted processes then allow us to derive in Theorem 4.1 a DPP for the value function V over the canonical space:  Z  t,x t,x,τ  , Y + V(t, x, y) = sup Et 1{τ ≤ζ(τ )}R(t, x, τ )+1{τ >ζ(τ )} V ζ(τ ), Xζ(τ ) ζ(τ ) τ ∈Txt (y)

ζ(τ )

t

Here the conditional expected cost Yst,x,τ := Et t

Rτ

τ ∧s

f (r, Xrt,x )dr

 .

(1.3)

 t g r, Xrt,x dr F s acts as an additional state process and the

intermediate horizon ζ can be a general F −stopping time depending on the stopping rule τ we select. For the “≤” part of (1.3), we exploit the flow property of shifted stochastic differential equations (Proposition 3.6) as well as the regular conditional probability distribution due to [57]; while in the “≥” part, we carefully paste together local ε−optimal stopping strategies and utilize the continuity of value function V. Also, we can transform the optimal stopping problem with expectation constraint to an unconstrained stochastic optimization problem whose controls are supermartingales starting from budget level y: Let At (y) denote all uniformly integrable continuous supermartingales α = {αs }s∈[t,∞) with αt = y. As shown in Proposition 4.2, for each nontrivial τ ∈ Txt (y) there exists α ∈ At (y) such that τ coincides with the first hitting time τ (t, x, α) of the process Yst,x,α :=  Rs Rτ αs − t g(r, Xrt,x )dr, s ∈ [t, ∞) to 0 If Et [ t g(r, Xrt,x )dr] = y, α is indeed a true martingale . So the value function   V can be alternatively expressed as V(t, x, y) = sup Et R t, x, τ (t, x, α) . Correspondingly, we establish a second α∈At (y)

DPP for the value function V over the canonical space (Theorem 4.2)   V(t, x, y) = sup Et 1{τ (t,x,α)≤ζ(α)}R t, x, τ (t, x, α) α∈At (y)

Z  t,x t,x,α  +1{τ (t,x,α)>ζ(α)} V ζ(α), Xζ(α) , Yζ(α) +

t

ζ(α)

f (r, Xrt,x )dr



,

(1.4)

and thus justify a postulate recently made by [2] (see Remark 3.3 therein). Although the “≤” part of (1.4) can be easily deduced from (1.3), the “≥” part entails an intricate pasting of approximately optimal supermartingale controls. In light of these two DPPs, we then show that the value function V of the optimal stopping problem with expectation constraint is a viscosity solution to a related fully non-linear parabolic Hamilton-Jacobi-Bellman (HJB) equation    −∂t u(t, x, y)− 12 trace σ(t, x)·σ T (t, x)·Dx2 u(t, x, y) −bT (t, x)·Dx u(t, x, y)    (1.5) +g(t, x)∂y u(t, x, y)−Hu(t, x, y) = f (t, x), ∀ (t, x, y) ∈ (0, ∞)×Rl ×(0, ∞),    u(t, x, 0) = π(t, x), ∀ (t, x) ∈ [0, ∞)×Rl , with the Hamiltonian Hu(t, x, y) := sup

a∈Rd

1

2 |a|

2 2 ∂y u(t, x, y)+(Dx (∂y u(t, x, y)))T·σ(t, x)·a



. As pointed out in [44], the

non-linear HJB equation (1.5) is a Monge-Amp`ere type equation. Relevant Literature. Since Arrow et al. [3] and Snell [56], the general theory of (unconstrained) optimal stopping has been plentifully developed over decades. Expositions of this theory are presented in the monographs [21, 46, 55, 27, 33, 51], which contain extensive bibliographies and references to the literature. For the recent development of the optimal stopping under model uncertainty/non-linear expectations and the closely related controller-stopper-games, see [34, 35, 28, 20, 22, 36, 54, 8, 9, 6, 19, 5, 26, 10, 47, 12, 11] among others.

1. Introduction

3

As to the optimal stopping with expectation constraint, the Lagrange multiplier method introduced in [37] was later developed by many researches (see e.g. [52, 45, 41, 25, 4, 58, 42]), and has been applied to various economic and financial problems such as Markov decision processes with constrained stopping times [30, 29], non-exponential discounting and mean-variance portfolio optimization [48, 49] and quickest detection problem [50]. Our stochastic control approach in deriving the second DPP resembles those of two recent papers [2], [44]. By applying the martingale representation to the conditional expected cost, Ankirchner et al. [2] transformed the optimal stopping problem with expectation constraint to a stochastic optimization problem in which the stochastic integral of locally square-integrable controls is regarded as an additional state process. Miller [44] independently employed the same method to address the optimal stopping problem with first-moment constraint that is embedded in a time-inconsistent optimal stopping problem. The idea of expanding the state space by the conditional probability/expectation process has also appeared in the literature dealing with stochastic target problems, see e.g. [15, 17, 18, 16, 14]. Our paper is distinct from [2], [44] in four aspects: First, we first obtain the continuity of the value function V, and using this establish the two DPPs (1.3) and (1.4), which were not addressed by them. Second, our value function V takes the starting moment t of the game as an input, so the related non-linear HJB equation (1.5) is of Rτ parabolic type rather than elliptic type. Third, we need the constraint Et [ t g(r, Xrt,x )dr] ≤ y for the continuity and the DPPs of the value function, although the auxiliary optimal stopping problem considered in [44] is subject to constraint E[τ ] = y and the dynamic programming equation studied by [2] is for the value function U of the optimal Rτ stopping with constraint E[ 0 g(Xrx )dr] = y. See Remark 4.1 for a comparison of these two types of constraints. Fourth, our discussion of related non-linear HJB equations seems different from theirs. Our Theorem 5.1 obtains that the value function V is a viscosity supersolution of (1.5), and is only a viscosity subsolution of (1.5) with the upper semi-continuous envelope Hu of Hu. By assuming that the value U is a smooth function satisfying the DPP, Proposition 3.4 of [2] showed that U is a supersolution to a similar non-linear HJB equation to (1.5), and is further a subsolution if the Hamiltonian is continuous (see Subsection 6.1 of [2] for an example of discontinuous Hamiltonian). However, possible discontinuity of the Hamiltonian was not discussed in [44]. Lately, the optimal stopping with constraint on the distribution of stopping time has attracted a lot of research interests. Bayraktar and Miller [7] studied the optimal stopping of a Brownian motion with the restriction that the distribution of the stopping time must equal to a given measure consisting of finitely-many atoms. The applications of such a distribution-constrained optimal stopping problem in mathematical finance include model-free superhedging with an outlook on volatility and inverse first-passage-time problem. Within a weak formulation on the canonical path space, Kallblad [31] extended the distribution-constrained optimal stopping problem for a general target measure and for path-dependent cost functions. From the perspective of mass transport, Beiglboeck et al. [13] obtained a monotonicity principle for the optimal stopping of a Brownian motion under distribution constraint, and thus characterized the constrained optimal stopping rule as the first hitting time of a barrier in a suitable phase space. Very recently, Ankirchner et al. [1] showed that for optimally stopping a one-dimensional Markov process with first-moment constraint on stopping times, one only needs to consider those stopping times at which the law of the Markov process is a weighted sum of three Dirac measures. There are also some other types of optimal stopping problems with constraints: see [24] for an optimal stopping problem with a reward constraint; see [38, 39, 43, 40] for optimal stopping with information constraint. The rest of the paper is organized as follows: In Subsection 1.1, we introduce notations and make standing assumptions on drift/diffusion coefficients and reward/cost functions. In Section 2, we set up the optimal stopping problem with expectation constraint over a general probability space and show the continuity of its value function in current state and budget constraint level. Section 3 explores the measurability/integrability properties of shifted processes and the flow property of shifted stochastic differential equations as technical preparation for proving our main result, two types of DPPs. Then in Subsection 4.1, we derive over the canonical space a DPP for the value function V of the optimal stopping with expectation constraint in which the conditional expected cost acts as an additional state process. In subsection 4.2, we transform the the optimal stopping problem with expectation constraint to a stochastic optimization problem with supermartingale controls and establish a second DPP for V. Based on two DPPs, we characterize V as the viscosity solution to the related fully nonlinear parabolic HJB equation in Section 5. Section 6 contains proofs of our results while the demonstration of some auxiliary statements with starred labels in these proofs are relegated to the Appendix. We also include some technical lemmata in the appendix.

DPPs for Optimal Stopping with Expectation Constraint

1.1

4

Notation and Preliminaries

For a generic Euclidian space E, we denote its Borel sigma−field by B(E). For any x ∈ E and δ ∈ (0, ∞), Oδ (x) := {x′ ∈ E : |x−x′ | < δ} denotes the open ball centered at x with radius δ and its closure is O δ (x) := {x′ ∈ E : |x−x′ | ≤ δ}. R∞ Fix l ∈ N and p ∈ [1, ∞). Let c(t) : [0, ∞) → (0, ∞) be a continuous function with 0 c(t)dt < ∞, and let C be a R∞ constant with C ≥ 1+ 0 c(t)dt. As lim c(t) = 0, the continuity of c(·) implies that kc(·)k := sup c(t) < ∞. Also, let t→∞

t∈[0,∞)

ρ be a modulus of continuity function and denote its inverse function by ρ−1 . We shall consider the following drift/diffusion coefficients and reward/cost functions throughout the paper.  • Let b : (0, ∞) × Rl → Rl be a B(0, ∞) ⊗ B(Rl ) B(Rl )−measurable function and let σ : (0, ∞) × Rl → Rl×d be a  B(0, ∞)⊗B(Rl ) B(Rl×d )−measurable function such that for any t ∈ (0, ∞) and x1 , x2 ∈ Rl b(t, x1 )−b(t, x2 ) ≤ c(t)|x1 −x2 |, |b(t, 0)| ≤ c(t), (1.6) p p and σ(t, x1 )−σ(t, x2 ) ≤ c(t)|x1 −x2 |, |σ(t, 0)| ≤ c(t) . (1.7)

 • The running reward function f : (0, ∞)×Rl → R is a B(0, ∞)⊗B(Rl ) B(R)−measurable function such that for any t ∈ (0, ∞) and x1 , x2 ∈ Rl  f (t, x1 )−f (t, x2 ) ≤ c(t) |x1 −x2 |∨|x1 −x2 |p and f (t, 0) ≤ c(t). (1.8)

• The terminal reward function π : [0, ∞)×Rl → R is a continuous function such that for any t, t′ ∈ [0, ∞) and x, x′ ∈ Rl   and π(t, 0) ≤ C. (1.9) |π(t, x)−π(t′ , x′ )| ≤ ρ |t−t′ | +C |x−x′ |∨|x−x′ |p  l • The cost rate function a B(0, ∞)⊗B(Rl ) B(0, ∞)−measurable function satisfying g : (0, ∞)×R → (0, ∞)pis  (g1) g(t, x1 )−g(t, x2 ) ≤ c(t) |x1 −x2 |∨|x1 −x2 | , ∀ t ∈ (0, ∞), ∀ x1 , x2 ∈ Rl ; Rt (g2) 0 g(t, 0)dr < ∞, ∀ t ∈ (1, ∞); (g3) For any R ∈ (0, ∞), there exists κR ∈ (0, ∞) such that g(t, x) ≥ κR , ∀ t ∈ (0, ∞), ∀ x ∈ Rl with |x| ≤ R. The constant κR can be regarded as the basic cost rate when the long-term state radius is R. Moreover, we will use the convention inf ∅ := ∞ as well as the inequality X q n n n X X q q−1 ai ≤ (1 ∨ nq−1 ) aqi (1.10) (1 ∧ n ) ai ≤ i=1

i=1

i=1

for any q ∈ (0, ∞) and any finite subset {a1 , · · · , an } of (0, ∞).

2

Continuity of Value Functions for General Optimal Stopping with Expectation Constraint

For an optimal stopping problem with expectation constraint, we first discuss the continuity of its value function over a general complete probability space (Ω, F , P ). Let B be a d−dimensional standard Brownian motion on (Ω, F , P ). The P −augmentation of its natural filtration    F = Ft := σ σ(Bs ; s ∈ [0, t])∪N t∈[0,∞) satisfies the usual hypothesis, where N := N ⊂ Ω : N ⊂ A for some A ∈ F with P (A) = 0} collects all P −null sets in F . Let T stand for all F−stopping times τ with τ < ∞, P −a.s. For any F−adapted continuous process X, we set X∗ := sup |Xs |. s∈[0,∞)

2.1

Reward Processes

Let (t, x) ∈ [0, ∞) × Rl . It is well-known that under (1.6) and (1.7), the following stochastic differential equation (SDE) on Ω Z s Z s Xs = x+ b(t+r, Xr )dr+ σ(t+r, Xr )dBr , s ∈ [0, ∞) (2.1) 0

0

admits a unique solution X t,x = {Xst,x}s∈[0,∞) , which is an Rl −valued, F−adapted continuous process satisfying

2.2

Expectation Constraints

5

Lemma 2.1. Let q ∈ [1, ∞) and (t, x) ∈ [0, ∞)×Rl . R∞ (1 ) For some constant Cq ≥ 1 depending on q and 0 c(s)ds, we have     q q  ′ E sup Xst,x ≤ Cq 1+|x|q ; E sup Xst,x −Xst,x ≤ Cq |x′ −x|q , ∀ x′ ∈ Rl ; E



and

(2.2)

s∈[0,∞)

s∈[0,∞)

 t,x  q q t,x q ≤ Cq 1+|x|q kc(·)kq δ q +kc(·)k 2 δ 2 , ∀ δ ∈ (0, ∞), ∀ τ ∈ T . sup Xτ +λ −Xτ

(2.3)

λ∈(0,δ]

(2 ) Given ̟ ∈ [1, ∞), assume functions b and σ additionally satisfy that for any 0 ≤ t1 < t2 < ∞ and x′ ∈ Rl p   b(t2 , x′ )−b(t1 , x′ ) ≤ c(t1 )ρ(t2 −t1 ) 1+|x′ |̟ and σ(t2 , x′ )−σ(t1 , x′ ) ≤ c(t1 )ρ(t2 −t1 ) 1+|x′ |̟ . Then it holds for any t′ ∈ (t, ∞) that   t′,x  q t,x q ≤ Cq,̟ 1+|x|q̟ ρ(t′−t) , E sup Xs −Xs

(2.4)

(2.5)

s∈[0,∞)

where Cq,̟ ≥ 1 is some constant depending on q, ̟ and

R∞ 0

c(s)ds.

Given t ∈ [0, ∞), let the state process evolve from position x ∈ Rl according to SDE (2.1). If the player chooses   Rτ to exercise at time τ ∈ T , she will receive a running reward 0 f t+s, Xst,x ds and a terminal reward π t+τ, Xτt,x , whose totality is Z τ   (2.6) R(t, x, τ ) := f t+s, Xst,x ds+π t+τ, Xτt,x . 0

One can deduce from (1.8), (1.9) and the first inequality in (2.2) that    E |R(t, x, τ )| ≤ 2C 2+Cp (1+|x|p ) := Ψ(x).

(2.7)

Given another initial position x′ ∈ Rl , (1.8), (1.9), H¨ older’s inequality and the second inequality in (2.2) imply that Z τ    f (t+r, Xrt,x)−f (t+r, Xrt,x′ ) dr+ π(t+τ, Xτt,x )−π(t+τ, Xτt,x′ ) E R(t, x, τ )−R(t, x′ , τ ) ≤ E 0 Z ∞  h i  1 ′ p ′ ≤ c(t+r)dr+C E X t,x−X t,x ∗ + X t,x−X t,x ∗ ≤ 2C (Cp ) p |x−x′ |+Cp |x−x′ |p . (2.8) 0

2.2

Expectation Constraints

Let (t, x) ∈ [0, ∞)×Rl . As the first inequality in (2.2) shows that (X∗t,x )p < ∞, P −a.s., (g1)−(g3) imply that P −a.s. Z ∞ Z s Z s    t,x t,x t,x p g t+r, Xrt,x dr = ∞. g t+r, Xr dr ≤ g(t+r, 0)dr+C X∗ +(X∗ ) < ∞, ∀ s ∈ (0, ∞) and (2.9) 0

0

0

Given y ∈ [0, ∞), we try to maximize the player’s expected total wealth R(t, x, τ ) when her expected cost is subject to the following constraint: Z τ  E g(t+r, Xrt,x)dr ≤ y. (2.10) 0

Like reward processes

  t,x dr s∈[0,∞) and π(t + s, Xst,x) s∈[0,∞) , this expectation constraint is also 0 f t + r, Xr

Rs

state-related. Hence, starting from the initial state x ∈ Rl , the value of the general optimal stopping problem with expectation constraint y is   V (t, x, y) := sup E R(t, x, τ ) , (2.11) τ ∈Tt,x (y)

 where Tt,x (y) := F−stopping time τ : E 0 g(t+r, Xrt,x)dr ≤ y .  Rτ Rτ For any τ ∈ Tt,x (y), as E 0 g(t+r, Xrt,x )dr ≤ y < ∞, one has 0 g(t+r, Xrt,x )dr < ∞, P −a.s. The second part of   Rτ (2.9) then implies that τ < ∞, P −a.s. So Tt,x (y) = τ ∈ T : E 0 g(t+r, Xrt,x)dr ≤ y . 

Rτ

DPPs for Optimal Stopping with Expectation Constraint

6

Example 2.1. (Moment Constraints) For q ∈ (1, ∞), a ∈ [0, ∞) and b ∈ (0, ∞), take g(t, x) := aqtq−1 + b, (t, x) ∈ (0, ∞)×Rl . Then the constraint (2.10) for t = 0 specify as the moment constraint E[aτ q +bτ ] ≤ y. Let (t, x) ∈ [0, ∞)×Rl . It is clear that

V (t, x, y) is increasing in y. As Tt,x (0) = {0}, we see from (2.7) that

  Ψ(x) ≥ V (t, x, y) ≥ V (t, x, 0) = E π(t, X0t,x ) = π(t, x),

(2.12)

∀ (t, x, y) ∈ [0, ∞)×Rl ×[0, ∞).

(2.13)

When y ∈ (0, ∞), we even have the following update of (2.11).

  Lemma 2.2. It holds for any (t, x, y) ∈ [0, ∞)×Rl ×(0, ∞) that V (t, x, y) = sup E R(t, x, τ ) , where Tbt,x (y) := τ ∈Tbt,x (y)  τ ∈ Tt,x (y) : τ > 0, P −a.s. .

The value function V (t, x, y) of the general optimal stopping problem with expectation constraint is continuous in the following way:

Theorem 2.1. (1 ) Given t ∈ [0, ∞), V (t, x, y) is continuous in (x, y) ∈ Rl ×[0, ∞) in the sense that for any (x, ε) ∈ Rl ×(0, 1), there exists δ = δ(t, x, ε) ∈ (0, 1) such that for any y ∈ [0, ∞)   V (t, x, y)−V (t, x, y) ≤ ε, ∀ (x, y) ∈ O δ (x)× (y −δ)+ , y +δ .

(2 ) Given ̟ ∈ [1, ∞), assume b, σ additionally satisfy (2.4) and f, g additionally satisfy that for any 0 ≤ t1 < t2 < ∞ and x′ ∈ Rl  f (t2 , x′ )−f (t1 , x′ ) ∨ g(t2 , x′ )−g(t1 , x′ ) ≤ c(t1 )ρ(t2 −t1 ) 1+|x′ |̟ , (2.14)

then V (t, x, y) is continuous in (t, x, y) ∈ [0, ∞)× Rl × [0, ∞) in the sense that for any (t, x, ε) ∈ [0, ∞)× Rl × (0, 1), there exists δ ′ = δ ′ (t, x, ε) ∈ (0, 1) such that for any y ∈ [0, ∞)     V (t, x, y)−V (t, x, y) ≤ ε, ∀ (t, x, y) ∈ (t−δ ′ )+ , t+δ ′ ×Oδ′ (x)× (y −δ ′ )+ , y +δ ′ .

3

Shifted Processes

Let us review the properties of shifted processes on the canonical space so that we can study two types of dynamic programming principles of the optimal stopping problem with expectation constraint over the canonical space.   Fix d ∈ N and let t ∈ [0, ∞). From now on, we consider the canonical space Ωt := ω ∈ C [t, ∞); Rd : ω(t) = 0 of continuous paths over period [t, ∞), which is a separable complete metric space under the uniform norm kωkt := sup |ω(s)|. Let F t := B(Ωt ) be the Borel sigma field of Ωt under k · kt . The canonical process W t = {Wst }s∈[t,∞) s∈[t,∞)  of Ωt is a d−dimensional standard Brownian motion on Ωt , F t under the Wiener measure Pt . Let N t collect all  t Pt −null sets, i.e., N t := N ⊂ Ωt : N ⊂ A for some A ∈ F t with Pt (A) = 0}, and set F := σ(F t∪N t ). The completion   t of Ωt , F t , Pt is the probability space Ωt , F , P t with P t = Pt . For simplicity, we still write Pt for P t and Ft

t

denote the expectation under P t by Et [·]. For any sub sigma−field G of F , let L1 (G) be the space of all real-valued,   G−measurable random variables ξ with Et |ξ| < ∞.   t We denote the natural filtration of W t by Ft = Fst := σ Wrt ; r ∈ [t, s] s∈[t,∞) . Its Pt −augmentation F consists  t t of F s := σ Fst ∪N t , s ∈ [t, ∞) and satisfies the usual hypothesis. Let T stand for all stopping times τ with respect  t t t to the filtration F such that τ < ∞, Pt −a.s., and set T ♯ := τ ∈ T : τ takes countably many values in [t, ∞) . For t

t

t easy reference, we set F∞ := F t and F ∞ := F . The following spaces will be used in the sequel.  t • For any q ∈ [1, ∞), let Cqt (E) = Cq t [t, ∞), E be the space of all E−valued, F −adapted processes {Xs }s∈[t,∞) with F  q Pt −a.s. continuous paths such that Et X∗ < ∞ with X∗ := sup |Xs |. s∈[t,∞)

|Xr |2 dr < ∞, ∀ s ∈ [t, ∞) = 1.  t • Let Mt denote all real-valued, uniformly integrable continuous martingales with respect to F , Pt .  • Set Kt := K ∈ C1t (R) : for Pt −a.s. ω ∈ Ωt , K· (ω) is an continuous increasing path starting from 0 . • Let

Ht2,loc

d

t

denote all R −valued, F −predictable processes {Xs }s∈[t,∞) with Pt

Rs t

3.1

Concatenation of Sample Paths

3.1

7

Concatenation of Sample Paths

Let 0 ≤ t ≤ s < ∞. We define a translation operator Πts from Ωt to Ωs by  Πts (ω) (r) := ω(r)−ω(s), ∀ (r, ω) ∈ [s, ∞)×Ωt .

On the other hand, one can concatenate ω ∈ Ωt and ω e ∈ Ωs at time s by:   ω ⊗s ω e (r) := ω(r) 1{r∈[t,s)} + ω(s)+ ω e (r) 1{r∈[s,∞)} ,

∀ r ∈ [t, ∞),

which is still of Ωt .  e := ω⊗s ω e for any A e ⊂ Ωs . e: ω e ∈A Given ω ∈ Ωt , we set As,ω := {e ω ∈ Ωs : ω⊗s ω e ∈ A} for any A ⊂ Ωt ; and set ω⊗s A In particular, ∅s,ω := ∅ and ω ⊗s ∅ := ∅. The next result shows that each A ∈ Fst consists of all branches ω ⊗s Ωs with ω ∈ A. Lemma 3.1. Let 0 ≤ t ≤ s < ∞ and A ∈ Fst . It holds for any ω ∈ A that ω ⊗s Ωs ⊂ A or As,ω = Ωs . Let ω ∈ Ωt . For any Fst −measurable random variable ξ, since the set {ω ′ ∈ Ωt : ξ(ω ′ ) = ξ(ω)} = ξ −1 {ξ(ω)} belongs to Fst , Lemma 3.1 implies that ω ⊗s Ωs ⊂ {ω ′ ∈ Ωt : ξ(ω ′ ) = ξ(ω)}

i.e., ξ(ω ⊗s ω e ) = ξ(ω),

∀ω e ∈ Ωs .



(3.1)

To wit, the value ξ(ω) depends only on ω|[t,s] . For any r ∈ [s, ∞], the operation ()s,ω projects an Frt −measurable set to an Frs −measurable set while the operation ω ⊗s · transforms an Frs −measurable set into an Frt −measurable set. e ∈ F t for any Lemma 3.2. Let 0 ≤ t ≤ s < ∞, ω ∈ Ωt and r ∈ [s, ∞]. We have As,ω ∈ Frs for any A ∈ Frt and ω ⊗s A r s e ∈ Fr . A

3.2

Measurability and Integrability of Shifted Processes

Let 0 ≤ t ≤ s < ∞, let ξ be an E−valued random variable on Ωt and let X = {Xr }r∈[t,∞) be an E−valued process on Ωt . For any ω ∈ Ωt , we define the shifted random variable ξ s,ω and the shifted process X s,ω by ξ s,ω (e ω ) := ξ(ω ⊗s ω e ) and X s,ω (r, ω e ) := X(r, ω ⊗s ω e ),

∀ (r, ω e ) ∈ [s, ∞)×Ωs .

By Lemma 3.2, shifted random variables and shifted processes inherit the measurability of original ones. Proposition 3.1. Let 0 ≤ t ≤ s < ∞ and let ω ∈ Ωt . (1 ) Let ξ be an E−valued random variable on Ωt . If ξ is Frt −measurable for some r ∈ [s, ∞], the shifted random variable ξ s,ω is Frs −measurable.  (2 ) Let X = {Xr }r∈[t,∞) be an E−valued process on Ωt . If X is Ft −adapted, the shifted process X s,ω = Xrs,ω r∈[s,∞) is Fs −adapted. In virtue of regular conditional probability distribution by [57], the shifted random variables carry on the integrability as follows: Proposition 3.2. Let 0 ≤ t ≤ s < ∞. If ξ ∈ L1 (F t ), then it holds for Pt −a.s. ω ∈ Ωt that ξ s,ω ∈ L1 (F s ) and     Et ξ Fst (ω) = Es ξ s,ω ∈ R.

(3.2)

Consequently, the shift of a Pt −null set still has zero Ps −probability.

Proposition 3.3. Let 0 ≤ t ≤ s < ∞. (1 ) For any Pt −null set N ∈ N t , it holds for Pt −a.s. ω ∈ Ωt that N s,ω ∈ N s . Then for any two real-valued random variables ξ1 and ξ2 on Ωt with ξ1 ≤ ξ2 , Pt −a.s., it holds for Pt −a.s. ω ∈ Ωt that ξ1s,ω ≤ ξ2s,ω , Ps −a.s. t s (2 ) For any τ ∈ T with τ ≥ s, it holds for Pt −a.s. ω ∈ Ωt that τ s,ω ∈ T . Based on Proposition 3.3 (1) and Lemma A.4, we can extend Proposition 3.2 from raw filtration Ft to augmented t filtration F , and can show that the shifted processes inherit the integrability of original ones.

DPPs for Optimal Stopping with Expectation Constraint

8

Proposition 3.4. Let 0 ≤ t ≤ s < ∞. t (1 ) For any F s −measurable random variable ξ, it holds for Pt −a.s. ω ∈ Ωt that ξ s,ω = ξ(ω), Ps −a.s. t s (2 ) For any r ∈ [s, ∞] and F r −measurable random variable ξ, it holds for Pt −a.s. ω ∈ Ωt that ξ s,ω is F r −measurable.     t If ξ is integrable, then it holds for Pt −a.s. ω ∈ Ωt that ξ s,ω is integrable and Et ξ F s (ω) = Es ξ s,ω ∈ R. t

(3 ) Let X = {Xr }r∈[t,∞) be an F −adapted process with Pt −a.s. continuous paths. It holds for Pt −a.s. ω ∈ Ωt that the  s shifted process X s,ω = Xrs,ω r∈[s,∞) is F −adapted with Ps −a.s. continuous paths. If X ∈ Cqt (E) for some q ∈ [1, ∞), then X s,ω ∈ Cqs (E) for Pt −a.s. ω ∈ Ωt . Moreover, the shift of a uniformly integrable martingale are still uniformly integrable martingales under the augmented filtrations. Proposition 3.5. Let 0 ≤ t ≤ s < ∞. For any M = {Mr }r∈[t,∞) ∈ Mt , it holds for Pt −a.s. ω ∈ Ωt that M s,ω = {Mrs,ω }r∈[s,∞) is of Ms .

3.3

Shifted Stochastic Differential Equations t

Let (t, x) ∈ [0, ∞)×Rl . The SDE (1.1) has a unique solution X t,x = {Xst,x }s∈[t,∞) , which is an Rl −valued, F −adapted continuous process. As it holds Pt −a.s. that Z t+s Z t+s Z s Z t σ(t+r, Xt+r ) dWt+r , s ∈ [0, ∞), Xt+s = x+ b(r, Xr )dr+ σ(r, Xr ) dWrt = x+ b(t+r, Xt+r )dr+ t



t

t,x

we see that Xt+s

s∈[0,∞)

0

r∈[0,s]

is exactly the unique solution of (2.1) with the probabilistic specification

    t t Ω, F , P, N , {Bs }s∈[0,∞) , {Fs }s∈[0,∞) = Ωt , F t , Pt , N t , {Wt+s }s∈[0,∞) , F t+s s∈[0,∞) .

(3.3)

 t t Clearly, τ is an F −stopping time if and only if τe := τ−t is a stopping time with respect to the filtration F t+s s∈[0,∞) .  t So the corresponding T under setting (3.3) is T = τe = τ −t : τ ∈ T . It then follows from Lemma 2.1 that

Corollary 3.1. Let q ∈ [1, ∞) and (t, x) ∈ [0, ∞)×Rl . For the same constant Cq as in Lemma 2.1,     q q  ′ Et sup Xst,x ≤ Cq 1+|x|q ; Et sup Xst,x −Xst,x ≤ Cq |x′ −x|q , ∀ x′ ∈ Rl ; and Et



(3.4)

s∈[t,∞)

s∈[t,∞)

 t,x  q q t t,x q sup Xτ +λ −Xτ ≤ Cq 1+|x|q kc(·)kq δ q +kc(·)k 2 δ 2 , ∀ δ ∈ (0, ∞), ∀ τ ∈ T .

(3.5)

λ∈(0,δ]

The shift of X t,x given path ω|[t,s] turns out to be the solution of the shifted stochastic differential equation (2.1) over period [s, ∞) with initial state Xst,x (ω): Proposition 3.6. (Flow Property) Let 0 ≤ t ≤ s < ∞, x ∈ Rl and set X := X t,x . It holds for Pt −a.s. ω ∈ Ωt that  s,X (ω) e ∈ Ωs : Xr (ω ⊗s ω e ) = Xr s (e ω ), ∀ r ∈ [s, ∞) = 1. Ps ω

The proof of Proposition 3.6 depends on the following result about the convergence of shifted random variables in probability.  t Lemma 3.3. For any {ξi }i∈N ⊂ L1 F that converges to 0 in probability Pt , we can find a subsequence ξbi i∈N of  it such that for Pt −a.s. ω ∈ Ωt , ξbs,ω converges to 0 in probability Ps . i

4

i∈N

Two Dynamic Programming Principle of Optimal Stopping with Expectation Constraint

In this section, we exploit the flow property of shifted stochastic differential equations to establish two types of dynamic programming principles (DPPs) of the optimal stopping problem with expectation constraint over the canonical space.

4.1

4.1

The First Dynamic Programming Principle for V

9

The First Dynamic Programming Principle for V

Let the state process now evolve from time t ∈ [0, ∞) and position x ∈ Rl according to SDE (1.1). If the player selects  Rτ t to exercise at time τ ∈ T , she will receive a running reward t f (r, Xrt,x )dr and a terminal reward π τ, Xτt,x . So the player’s total wealth is R(t, x, τ ) :=

Z

τ t

f (s, Xst,x )ds+π

τ, Xτt,x



=

Z

τe

0

t,x t,x  f (t+s, Xt+s )ds+π t+e τ , Xt+e τ ,

 which is the payment R t, x, τe in (2.6) under the specification (3.3). By (2.7) and (2.8), one has

   1 and Et R(t, x, τ )−R(t, x′ , τ ) ≤ 2C (Cp ) p |x−x′ |+Cp |x−x′ |p , ∀ x′ ∈ Rl .

  Et |R(t, x, τ )| ≤ Ψ(x)

(4.1)

   R τe Rτ  Rτ t t,x )dr , Given y ∈ [0, ∞), set Txt (y) := τ ∈ T : Et [ t g(r, Xrt,x )dr] ≤ y . As Et t g(r, Xrt,x )dr = Et 0 g(t+r, Xt+r  we see that τe = τ −t : τ ∈ Txt (y) is the corresponding Tt,x (y) under setting (3.3). Then the maximum of the player’s  Rτ expected wealth subject to the budget constraint Et t g(r, Xrt,x )dr ≤ y, i.e.,   V(t, x, y) := sup Et R(t, x, τ ) = τ ∈Txt (y)

sup

τe∈Tt,x (y)

  Et R t, x, τe

(4.2)

is exactly the value function (2.11) of the constrained optimal stopping problem under the specification (3.3). Then (2.13) and Lemma 2.2 show that Ψ(x) ≥ V(t, x, y) ≥ V(t, x, 0) = π(t, x), ∀ (t, x, y) ∈ [0, ∞)×Rl ×[0, ∞)   V(t, x, y) = sup E R(t, x, τ ) , ∀ (t, x, y) ∈ [0, ∞)×Rl ×(0, ∞),

and

(4.3) (4.4)

τ ∈Tbt,x (y)

 where Tbxt (y) := τ ∈ Txt (y) : τ > t, Pt −a.s. . Also,

Theorem 2.1 still holds for the value function V. t

Now, let (t, x) ∈ [0, ∞)× Rl and let τ ∈ T with Et continuous process: Yst,x,τ

:= Et

Z

τ

g

t

r, Xrt,x

Since it holds for any s ∈ [t, ∞) that Yst,x,τ = Et

Z

τ τ ∧s

Rτ t

t

s

the continuity of Y t,x,τ implies that

 t g(r, Xrt,x )dr < ∞. We define a real-valued, F −adapted

 Z  t dr F s −

Z   t g r, Xrt,x dr F s = Et

(4.5)

τ ∨s

τ ∧s

 g r, Xrt,x dr,

s ∈ [t, ∞).

  t g r, Xrt,x dr F s ∈ [0, ∞),

 Nt,x,τ := Yst,x,τ ∈ / [0, ∞) for some s ∈ [t, ∞) ∈ N t .

Pt − a.s.,

(4.6)

(4.7)

Then we have the first dynamic programming principle for the value function V in which the conditional expected cost Y t,x,τ acts as an additional state process. Theorem 4.1. Let t ∈ [0, ∞). t (1 ) For any (x, y) ∈ Rl ×[0, ∞), let {ζ(τ )}τ ∈Txt (y) be a family of T ♯ −stopping times. Then we have the DPP (1.3), where sup Et [·] can be replaced by sup Et [·] if y > 0. τ ∈Txt (y)

τ ∈Tbxt (y)

(2 ) If V(s, x, y) is continuous in (s, x, y) ∈ [t, ∞)×Rl ×(0, ∞), then (1.3) holds for any (x, y) ∈ Rl ×[0, ∞) and any t family {ζ(τ )}τ ∈Txt (y) of T −stopping times.

DPPs for Optimal Stopping with Expectation Constraint

4.2

10

An Alternative Stochastic Control Problem and the Second Dynamic Programming Principle for V

Fix t ∈ [0, ∞) and set At := {α = M −K : (M, K) ∈ Mt ×Kt }. Clearly, each α ∈ At is a uniformly integrable continuous  t supermartingales with respect to F , Pt .  t Let x ∈ Rl and α ∈ At . We define a continuous supermartingale with respect to F , Pt Yst,x,α := αs −

t

and define an F −stopping time

Z

t

s

 g r, Xrt,x dr,

s ∈ [t, ∞),

 τ (t, x, α) := inf s ∈ [t, ∞) : Yst,x,α = 0 .

(4.8)

The uniform integrability of α implies that the limit lim αs exists in R, Pt −a.s. Since s→∞ t,x  dr = ∞, Pt −a.s. by (2.9), one can deduce that r, Xt+r τ (t, x, α) < ∞,

R∞

Pt −a.s.

t

 R∞ g r, Xrt,x dr = 0 g t+ (4.9)

t

Namely, τ (t, x, α) ∈ T .   Given α ∈ At , the expected wealth Et R t, x′ , τ (t, x′ , α) is continuous in x ∈ Rl , which will play an important role in the demonstration of the second DPP for V (Theorem 4.2). Proposition 4.1. Let (t, x) ∈ [0, ∞)×Rl and let α ∈ At . For any ε ∈ (0, 1), there exists δ = δ(t, x, ε) ∈ (0, 1) such that h   i Et R t, x′ , τ (t, x′ , α) −R t, x, τ (t, x, α) ≤ ε, ∀ x′ ∈ O δ (x).  For any y ∈ (0, ∞), we set At (y) := α ∈ At : αt = y, Pt −a.s. .

Proposition 4.2. Given (t, x, y) ∈ [0, ∞)×Rl ×(0, ∞), α → τ (t, x, α) is a surjective mapping from At (y) to Tbxt (y).

Remark 4.1. Let (t, x, y) ∈ [0, ∞)×Rl ×(0, ∞). 1 ) Let τ ∈ Tbxt (y). Proposition 4.2 shows that τ = τ (t, x, α) for some α ∈ At (y). In particular, we see from (6.87) of its Rτ  proof that α is a martingale (resp. supermartingale) if Et t g(r, Xrt,x )dr −y = 0 (resp. ≤ 0). To wit, the constraint Rτ  Et t g(r, Xrt,x )dr = y (resp. ≤ y) corresponds to martingale (resp. supermartingale) controls in the alternative stochastic optimization problem. Rs In case that α is a martingale, we know from the martingale representation theorem that αs = y + t qr dWrt , R s qr dWrt , s ∈ [t, ∞) could be a strict local s ∈ [t, ∞) for some q ∈ H2,loc . However reversely, for a e q ∈ H2,loc ,α es := y + t e t t  R τ (t,x,e  α) g(r, Xrt,x )dr < y, see Example A.1 in the appendix. This is the reason why [44] requires martingale with Et t E[τ 2 ] < ∞ (see line -4 in page 3 therein) for the one-to-one correspondence between constrained stopping rules and squarely-integrable controls. Rτ  2 ) Define the value of the optimal stopping under the constraint Et t g(r, Xrt,x )dr = y by  hZ   t U(t, x, y) := sup Et R(t, x, τ ) : τ ∈ T with Et

t

τ

g(r, Xrt,x )dr

i

 =y .

Clearly, U(t, x, y) ≤ V(t, x, y). However, we do not know whether they are equal since U(t, x, y) may not be increasing in y (cf. line 5 of Lemma 1.1 of [2] ).  Rτ 3 ) The constraint Et t g(r, Xrt,x )dr ≤ y is necessary for proving the continuity and the first DPP of the value Rτ  function V: Even if τ1 in (6.18) has E 0 1 g(t+r, Xrt,x )dr = y, the approximately optimal stopping time τb1 constructed  R τb1   R τb  in the case (6.20) may satisfy E 0 g(t+r, Xrt,x )dr < (y −δ)+ rather than E 0 1 g(t+r, Xrt,x )dr = (y −δ)+ . Even Rτ  if the τ ∈ Txt (y) given in Lemma 6.1 reaches Et t g(r, Xrt,x )dr = y, the pasting τ of τ with the locally ε−optimal Rτ    R τ stopping times τni ’s in (6.72) satisfies Et t g(r, Xrt,x )dr < y + ε but Et t g(r, Xrt,x )dr = y + ε after a series of estimations in (A.2).

5. Related Fully Non-linear Parabolic HJB Equations

11

By Proposition 4.2 and (4.4), we can alternatively express the optimal stopping problem with expectation constraints (2.10) as a stochastic control problem:   (4.10) V(t, x, y) = sup Et R t, x, τ (t, x, α) , ∀ (t, x, y) ∈ [0, ∞)×Rl ×(0, ∞). α∈At (y)

Moreover, we have the second dynamic programming principle for the value function V in which the controlled supermartingale Y t,x,α serves as an additional state process.

Theorem 4.2. Let t ∈ [0, ∞). t (1 ) For any (x, y) ∈ Rl ×[0, ∞), let {ζ(α)}α∈At (y) be a family of T ♯ −stopping times. Then we have the DPP (1.4). (2 ) If V(s, x, y) is continuous in (s, x, y) ∈ [t, ∞)×Rl ×(0, ∞), then (1.4) holds for any (x, y) ∈ Rl ×[0, ∞) and any t family {ζ(α)}α∈At (y) of T −stopping times.

5

Related Fully Non-linear Parabolic HJB Equations

In this section, we show that the value function of the optimal stopping problem with expectation constraint is the viscosity solution to a related fully non-linear parabolic Hamilton-Jacobi-Bellman (HJB) equation.  For any φ(t, x, y) ∈ C 1,2,2 [0, ∞)×Rl ×[0, ∞) , we set  Dφ(t, x, y) := Dx φ, Dx2 φ, ∂y φ, ∂y2 φ, Dx (∂y φ) (t, x, y) ∈ Rl ×Sl ×R×R×Rl, ∀ (t, x, y) ∈ [0, ∞)×Rl ×[0, ∞),

where Sl denotes the set of all Rl×l −valued symmetric matrices. Recall the definition of viscosity solutions to a parabolic equation with a general (non-linear) Hamiltonian H : [0, ∞)×Rl ×R×Rl ×Sl ×R×R×Rl → [−∞, ∞]. Definition 5.1. An upper (resp. lower ) semi-continuous function u : [0, ∞)× Rl × [0, ∞) → R is called a viscosity subsolution (resp. supersolution) of   −∂t u(t, x, y) − H t, x, u(t, x, y), Du(t, x, y) = 0, ∀ (t, x, y) ∈ (0, ∞)×Rl ×(0, ∞), u(t, x, 0) = π(t, x),

∀ (t, x) ∈ [0, ∞)×Rl

if u(t, x, 0) ≤ (resp. ≥) π(t, x), ∀ (t, x) ∈ [0, ∞)×Rl , and if for any (to , xo , yo ) ∈ (0, ∞)×Rl×(0, ∞) and φ ∈ C 1,2,2 [0, ∞)×  Rl ×[0, ∞) such that u−φ attains a strict local maximum 0 (resp. strict local minimum 0) at (to , xo , yo ), one has  −∂t φ(to , xo , yo )−H to , xo , φ(to , xo , yo ), Dφ(to , xo , yo ) ≤ (resp. ≥) 0.  For any φ ∈ C 1,2,2 [0, ∞)×Rl ×[0, ∞) , we also define  1 trace σ(t, x)·σ T (t, x)·Dx2 φ(t, x, y) +bT (t, x)·Dx φ(t, x, y), 2 o n1 T |a|2 ∂y2 φ(t, x, y)+ Dx (∂y φ(t, x, y)) ·σ(t, x)·a ≥ 0, (t, x, y) ∈ [0, ∞)×Rl ×[0, ∞), Hφ(t, x, y) := sup a∈Rd 2

Lx φ(t, x, y) :=

as well as the upper semi-continuous envelope of Hφ (the smallest upper semi-continuous function above Hφ) Hφ(t, x, y) :=

lim

Hφ(t′ , x′ , y ′ ) = lim ↓

(t′ ,x′ ,y ′ )→(t,x,y)

δ→0

sup

Hφ(t′ , x′ , y ′ ), (t, x, y) ∈ [0, ∞)×Rl ×[0, ∞),

(5.1)

(t′ ,x′ ,y ′ )∈Oδ (t,x,y)

    where Oδ (t, x, y) := (t−δ)+ , t+δ ×Oδ (x)× (y −δ)+ , y +δ .

Theorem 5.1. Assume that b, σ additionally satisfy (2.4) and f, g additionally satisfy (2.14). Then the value function V in (4.2) is a viscosity supersolution of  −∂t u(t, x, y)−Lx u(t, x, y)+g(t, x)∂y u(t, x, y) − Hu(t, x, y)−f (t, x) = 0, ∀ (t, x, y) ∈ (0, ∞)×Rl ×(0, ∞), (5.2) u(t, x, 0) = π(t, x), ∀ (t, x) ∈ [0, ∞)×Rl , and is a viscosity subsolution of  −∂t u(t, x, y)−Lx u(t, x, y)+g(t, x)∂y u(t, x, y) − Hu(t, x, y)−f (t, x) = 0, u(t, x, 0) = π(t, x),

∀ (t, x) ∈ [0, ∞)×Rl .

∀ (t, x, y) ∈ (0, ∞)×Rl ×(0, ∞),

(5.3)

DPPs for Optimal Stopping with Expectation Constraint

12

Remark 5.1. See Section 5.2 of [44] for the connection between the fully non-linear parabolic HJB equation (5.2) and generalized Monge-Amp`ere equations.

6

Proofs

6.1

Proofs of Section 2

R∞ Proof of Lemma 2.1: In this proof, we set c := 0 c(s)ds and let cq q, whose form may vary from line to line. 1) Let T ∈ (0, ∞) and set qe:= q∨2. Given s ∈ [0, T ], we set Φs := sup

denote a generic constant depending only on t,x Xr , (2.1) and (1.6) show that

r∈[0,s]

Φs ≤ |x|+

≤ |x|+

Z

s

0

Z

s

Z   |b(t+r, 0)|+ b t+r, Xrt,x −b(t+r, 0) dr+ sup

c(t+r)dr+

0

Z

s

0

Z c(t+r)|Xrt,x |dr + sup s′ ∈[0,s]

s′

0

s′ ∈[0,s]

s′

0

σ(t+r, Xrt,x)dBr

σ(t+r, Xrt,x )dBr ,

P −a.s.

(6.1)

Taking qe−th power of (6.1), we can deduce from H¨ older’s inequality, the Burkholder-Davis-Gundy inequality, (1.7) and Fubini’s Theorem that    Z s qe−1  Z s  Z s qe   2  2qe Xrt,x qedr +cq E e (t+r)dr E Φqse ≤ 4qe−1 |x|qe+4qe−1 c qe+4qe−1 E c q−1 c(t+r) 1+|Xrt,x| dr Z

0

T

qe−1 Z

0

s

Z

0

T

 q2e −1 Z

s   (t+r)dr (t+r)dr c E E (1+Φr )qe dr 0 0 0 0   Z T qe  q2e −1   qe−1  q2e −1 Z s    Z T qe  Z T qe qe qe−1 qe q−2 e q−1 e q−2 e (t+r)dr (t+r)dr (t+r)dr +cq c ≤ cq |x| +c +T + 4 c E Φqre dr. c

≤4

qe−1

qe

|x| +4

qe−1 qe

c +4

qe−1

c

q e q−1 e



Φqre

 dr+cq

q e q−2 e

0

0

0

0

An application of Gronwall’s inequality then gives that   q2e −1   Z T qe     e (t+r)dr c q−2 E Φqs ≤ 1+E Φqse ≤ 1+cq |x|qe+c qe+T  Z qe−1 × exp 4

0

T

c

q e q−1 e

(t+r)dr

0

qe−1

s+cq

Z

T

c

q e q−2 e

0

 2qe −1  s < ∞, (t+r)dr

∀ s ∈ [0, T ].

(6.2)

Let s ∈ [0, T ]. Since the Burkholder-Davis-Gundy inequality and (1.6) also show that   Z s   Z s  Z s′ q   2q   2q t,x 2 t,x t,x 2 ≤ cq E E sup c(t+r) 1+|Xr | dr σ(t+r, Xr )dBr ≤ cq E σ(t+r, Xr ) dr s′ ∈[0,s]



0

0

0

   Z s  q  q 1 ≤ cq E (1+Φs ) 2 c(t+r) 1+|Xrt,x| dr 2 ≤ E 81−q (1+Φs )q +cq c(t+r) 1+|Xrt,x| dr 2 0 0  Z s Z s   q  q   1 1−q q ≤ 4 c(t+r)dr +cq E 1+E Φs +cq c(t+r)Φr dr , 2 0 0 Z q

s

taking q−th power of (6.1) and using Fubini’s Theorem yield that  Z s  Z q   q q q 1−q c(t+r)Φr dr +E sup 4 E Φs ≤ |x| +c +E 0

  1 ≤ |x|q +cq cq + 41−q 1+E Φqs +cq 2

Z

s′ ∈[0,s] s

s′ 0

c(t+r)dr

0

q 

σ(t+r, Xrt,x )dBr

 q−1  Z s E c(t+r)Φqr dr .

(6.3)

0

Rs  q−1   1 1 R s Rs q−1 q Here, we applied H¨ older’s inequality 0 ar br dr ≤ 0 |ar |q dr q 0 |br | q−1 dr q with ar , br = c q (t+r)Φr , c q (t+    q r) . As E sup Xrt,x < ∞ by (6.2), it follows from (6.3) that for any s ∈ [0, T ] r∈[0,s]

  E Φqs ≤ 1+2×4q−1|x|q +cq cq +cq cq−1

Z

0

s

  c(t+r)E Φqr dr.

(6.4)

6.1

Proofs of Section 2

13

    Rs Applying Gronwall’s inequality again yields that E Φqs ≤ 1+2×4q−1|x|q +cq cq exp cq cq−1 0 c(t+r)dr , ∀ s ∈ [0, T ]. In particular, taking s = T and then letting T → ∞, one can deduce from the monotone convergence theorem that   q   E sup Xrt,x ≤ 1+2×4q−1|x|q +cq cq exp cq cq . (6.5) r∈[0,∞)

′ e s := sup Xr . Since an analogy to (6.1) shows that 2) Let Xs := Xst,x−Xst,x , ∀ s ∈ [0, ∞). Given s ∈ [0, ∞), we set Φ r∈[0,s]

e s ≤ |x′ −x|+ Φ

Z

0

s

Z c(t+r)|Xr |dr+ sup s′ ∈[0,s]

s′

0

′  σ(t+r, Xrt,x )−σ(t+r, Xrt,x ) dBr ,

P −a.s.,

the Burkholder-Davis-Gundy inequality and (1.7) imply that Z s  Z s q   2q   q 2 e ≤ |x′ −x|q +E 31−q E Φ c(t+r)|X |dr +c E c(t+r)|X | dr r q r s 0



q

≤ |x −x| +E

Z

s

0

c(t+r)|Xr |dr

q 



0

e q/2 +cq E Φ s

Z

s

0

c(t+r)|Xr |dr

 Z s q  1 1−q  e q  ≤ |x −x| + 3 E Φs +cq E c(t+r)|Xr |dr . 2 0 i h  q  ′ e s ≤ 2q−1 E X∗t,x q + X∗t,x q < ∞ by Part 1, an analogy to (6.4) shows that Since E Φ ′

q

 e q ≤ 2×3q−1 |x′ −x|q +cq cq−1 E Φ s 

Z

s

0

 q e dr, c(t+r)E Φ r

 2q #

∀ s ∈ [0, ∞).

  q R e s ≤ 2×3q−1 |x′−x|q exp cq cq−1 s c(t+r)dr , ∀ s ∈ [0, ∞). As s → ∞, Then we see from Gronwall’s inequality that E Φ 0   q  ′ the monotone convergence theorem implies that E sup Xrt,x −Xrt,x ≤ 2×3q−1|x′ −x|q exp cq cq . r∈[0,∞)

3) Let δ ∈ (0, ∞) and τ ∈ T . For any λ ∈ (0, δ], since it holds P −a.s. that

t,x Xτt,x +λ −Xτ =

Z

τ

τ +λ

b(t+r, Xrt,x)dr+

Z

τ +λ

τ

σ(t+r, Xrt,x)dBr =

Z

τ +λ

τ

Z

b(t+r, Xrt,x)dr+

τ +λ

0

1{τ 0 (In this case, one must have y > 0). Since both E 0 1 g(t+   R s g(t+r, Xrt,x)dr s∈[0,∞) are F−adapted continuous processes, r, Xrt,x )dr Fs s∈[0,∞) and 0  hZ τb1 = τb1 (t, x, x, y, ε) := inf s ∈ [0, ∞) : E

0

defines an F−stopping time which satisfies E E[·] yields that E

hZ

0

τb1

g(t+r, Xrt,x)dr

i

=E

hZ

τ1 0

 R τ1 0

τ1



g(t+r, Xrt,x)dr Fs

 i Z s t,x − g(t+r, Xr )dr ≤ a

(6.20)

0

 R τb g(t+ r, Xrt,x)dr Fτb1 − 0 1 g(t+ r, Xrt,x)dr = a. Taking expectation

i g(t+r, Xrt,x)dr −a = y −δy = (y −δ)+ ,

 so τb1 ∈ Tt,x (y −δ)+ .

(6.21)

DPPs for Optimal Stopping with Expectation Constraint

16

Rτ  Rτ As E 0 1 g(t+r, Xrt,x)dr Fτ1 − 0 1 g(t+r, Xrt,x)dr = 0 < a, we also see that τb1 ≤ τ1 . The condition (g1), H¨ older’s inequality, the second inequality in (2.2) and (6.17) show that Z h ∞ i g(t+r, Xrt,x)−g(t+r, Xrt,x) dr E 0   Z  ∞ 1 ≤ E (X t,x−X t,x)∗ +(X t,x−X t,x)p∗ c(t+r)dr ≤ C(Cp ) p |x−x|+CCp |x−x|p ≤ λ.

(6.22)

0

Since E

 R τ1 0

 1 g(t+r, Xrt,x)dr ≤ y and since λ ≥ λ∧εo > C(Cp ) p δ ≥ δ ≥ δy by (6.17), one has a=E

hZ

0

≤E

hZ

0

τ1

i hZ g(t+r, Xrt,x)dr −y +δy < E

0

τ1

i



g(t+r, Xrt,x)−g(t+r, Xrt,x) dr

 i g(t+r, Xrt,x)−g(t+r, Xrt,x) dr +λ

+λ ≤ 2λ.

Using (6.22) again, we can deduce from (6.21) that h Z τ1 i h Z τ1 i h Z ∞ i g(t+r, Xrt,x)−g(t+r, Xrt,x) dr 2λ > a = E g(t+r, Xrt,x)dr ≥ E g(t+r, Xrt,x)dr −E τ b1 τb1 0   Z τ1 √ √  g(t+r, Xrt,x)dr −λ ≥ κR λ P AcR ∩{τ1 > τb1 + λ } −λ. ≥ E 1Ac ∩{τ1 >bτ1 +√λ} R

(6.23)

τb1

√ √  It follows from (6.15) that P AcR ∩{τ1 > τb1 + λ } < 3κ λ ≤ λo /2, proving the claim (6.19). R √ √ τ1 ≤ τ1 ≤ τb1 + λ}. Since (6.19) shows that Set A := {τ1 ≤ τb1 + λ} = {b √ √   P (Ac ) = P {τ1 > τb1 + λ} ≤ P (AR )+P AcR ∩ τ1 > τb1 + λ < λo < εo ,

(6.9)−(6.15) imply that  Z E

τ b1

0

f (t+r, Xrt,x)dr−

Z

τ1



f (t+r, Xrt,x)dr

≤E



2+(X∗t,x)p



Z



 c(t+r)dr+1A kc(·)k(τ1 −b τ1 )

1 0 √  < C 2P (Ac )+εo + λ(2+M)kc(·)k < (1+3C)εo ,        ≤ 2CE 1Ac 2+(X t,x)p < 2C 2P (Ac )+εo < 6Cεo . and E 1Ac π τb1 , Xτbt,x −π τ1 , Xτt,x ∗ 1 1 0

Ac

Also, we can deduce from (1.9), (6.15), H¨ older’s inequality, (2.3) and (6.16) that h  i       t,x X t,x−Xτt,x + X t,x−Xτt,x p E 1A π τb1 , Xτbt,x −π τ , X +CE 1 ≤ E 1 ρ τ −b τ 1 A A 1 1 τ τ b τ b 1 1 1 1 1 1  p1     √  t,x t,x t,x p t,x p +CE 1A sup Xτb1 +r−Xτb1 Xτb1 +r−Xτb1 ≤ ρ λ +C E 1A sup √ √ r∈(0, λ ]

≤ εo +C(Cp )

1 p



1 2

1 2

1+|x| kc(·)kλ +kc(·)k λ

Combining (6.24), (6.25) and (6.26) yields that

1 4



(6.24) (6.25)

r∈(0, λ ]

 p p p +CCp 1+|x|p kc(·)kp λ 2 +kc(·)k 2 λ 4 ≤ (1+C)εo .

  E R(t, x, τb1 )−R(t, x, τ1 ) < (2+10C)εo,

(6.26)

(6.27)

which together with (2.8) and (6.17) show that

      E R(t, x, τb1 )−R(t, x, τ1 ) ≤ E R(t, x, τb1 )−R(t, x, τb1 ) +E R(t, x, τb1 )−R(t, x, τ1 ) < (4+10C)εo = ε−εo.

Then it follows from (2.12) and (6.18) that for any (x, y) ∈ O δ (x)×[(y −δ)+ , ∞),

     V (t, x, y) ≥ V t, x, (y −δ)+ ≥ E R(t, x, τb1 ) > E R(t, x, τ1 ) −ε+εo ≥ V (t, x, y)−ε.

1b) To show V (t, x, y) ≤ V (t, x, y)+ε, ∀ (x, y) ∈ O δ (x)×[0, y +δ], we let x ∈ O δ (x).

(6.28)

6.1

Proofs of Section 2

17

There exists τ2 = τ2 (t, x, y, ε) ∈ Tt,x (y +δ) such that

   E R(t, x, τ2 ) ≥ V t, x, y +δ −εo.

(6.29)

We claim that we can also construct a stopping time τb2 = τb2 (t, x, x, y, ε) ∈ Tt,x (y) satisfying τb2 ≤ τ2

√  and P AcR ∩{τ2 > τb2 + λ } < λo /2.

(6.30)

Rτ   Rτ  If E 0 2 g(t+r, Xrt,x)dr ≤ y i.e. τ2 ∈ Tt,x (y) , we directly set τb2 := τ2 . Otherwise, set b := E 0 2 g(t+r, Xrt,x)dr −y > 0.  R   R τ2 s Similar to (6.20), τb2 = τb2 (t, x, x, y, ε) := inf s ∈ [0, ∞) : E 0 g(t + r, Xrt,x)dr Fs − 0 g(t + r, Xrt,x)dr ≤ b is an  R τb Rτ F−stopping time satisfying E 0 2 g(t+r, Xrt,x)dr Fτb2 − 0 2 g(t+r, Xrt,x)dr = b. Taking expectation E[·] yields that E

hZ

0

τ b2

i hZ g(t+r, Xrt,x)dr = E

0

τ2

i g(t+r, Xrt,x)dr −b = y,

so τb2 ∈ Tt,x (y).

(6.31)

Rτ  Rτ As E 0 2 g(t+r, Xrt,x)dr Fτ2 − 0 2 g(t+r, Xrt,x)dr = 0 < b, we also see that τb2 ≤ τ2 . Rτ  Since E 0 2 g(t+r, Xrt,x)dr ≤ y +δ < y +λ, we can deduce from (6.22) and (6.31) that hZ

Z

 i g(t+r, Xrt,x)−g(t+r, Xrt,x) dr +λ 0 0   Z τ2 h Z τ2 i h Z τ2 i t,x t,x >E g(t+r, Xrt,x)dr g(t+r, Xr )dr −y = b = E g(t+r, Xr )dr ≥ E 1Ac ∩{τ2 >bτ2 +√λ} R 0 τb2 τ b2 √ √  c ≥ κR λ P AR ∩{τ2 > τb2 + λ } .

2λ ≥ E

h i g(t+r, Xrt,x)−g(t+r, Xrt,x) dr +λ ≥ E



τ2

√ √  By (6.15), P AcR ∩{τ2 > τb2 + λ } < 2κ λ < λo /2, proving the claim (6.30). R   An analogy to (6.24)−(6.26) yields that E R(t, x, τb2 )−R(t, x, τ2 ) < (2+10C)εo, so we see from (2.8) and (6.17)

      E R(t, x, τb2 )−R(t, x, τ2 ) ≤ E R(t, x, τb2 )−R(t, x, τ2 ) +E R(t, x, τ2 )−R(t, x, τ2 ) < (4+10C)εo = ε−εo.

It then follows from (2.12) and (6.29) that for any (x, y) ∈ O δ (x)×[0, y +δ],

     V (t, x, y) ≤ V t, x, y +δ ≤ E R(t, x, τ2 ) +εo < E R(t, x, τb2 ) +ε ≤ V (t, x, y)+ε,

  which together with (6.28) leads to that V (t, x, y)−V (t, x, y) ≤ ε, ∀ (x, y) ∈ O δ (x)× (y −δ)+ , y +δ . 2) Next, let ̟ ∈ [1, ∞), we further assume that b, σ additionally satisfy (2.4) and f, g additionally satisfy (2.14). Fix (t, x, ε) ∈ [0, ∞)×Rl ×(0, 1). Given t ∈ [0, ∞) and ζ ∈ T , (1.8), (1.9), (2.14), H¨ older’s inequality, (2.5), (1.10) and the first inequality in (2.2) imply that   E |R(t, x, ζ)−R(t, x, ζ)|  Z ζ     f (t+r, X t,x)−f (t+r, X t,x) + f (t+r, X t,x)−f (t+r, X t,x) dr+ π t+ζ, X t,x −π t+ζ, X t,x ≤E r r r r ζ ζ   0 Z Z h i   ∞  ∞ c(t∧t+r)dr ≤ E (X t,x−X t,x )∗ +(X t,x−X t,x)p∗ c(t+r)dr+C +ρ(|t−t|)+ρ(|t−t|)E 1+|X∗t,x|̟ 0 0  p 1/p (6.32) ≤ 2CCp,̟ (1+|x|̟ )ρ(|t−t|)+2CCp,̟ (1+|x|p̟ ) ρ(|t−t|) +ρ(|t−t|)+Cρ(|t−t|) 1+C̟ (1+|x|̟ ) .

Let us still set εo , M and take λo = λo (t, x, ε), R = R(t, x, ε), λ = λ(t, x, ε) as in Part 1. We now choose δ = δ ′ (t, x, ε) ∈ (0, 1) such that ′

 p  λ∧εo  1 1/p , (Cp ) p δ ′ +Cp (δ ′ )p +Cp,̟ 1+|x|̟ ρ(δ ′ )+Cp,̟ 1+|x|p̟ ρ(δ ′ ) +ρ(δ ′ )+ρ(δ ′ ) 1+C̟ (1+|x|̟ ) ≤ C

(6.33)

and fix y ∈ [0, ∞). 2a) To show that V (t, x, y) ≥ V (t, x, y)−ε, ∀ (t, x, y) ∈ [(t−δ ′ )+ , t+δ ′ ]×O δ′ (x)×[(y−δ ′ )+ , ∞), we let (t, x) ∈ [(t−δ ′ )+ , t+ δ ′ ]×Oδ′ (x).

DPPs for Optimal Stopping with Expectation Constraint

18

The condition (g1), (2.14), H¨ older’s inequality, (2.2), (2.5), (1.10) and (6.33) show that Z h ∞ i g(t+r, X t,x)−g(t+r, X t,x) dr E r r 0   Z ∞  g(t+r, Xrt,x)−g(t+r, Xrt,x) + g(t+r, Xrt,x)−g(t+r, Xrt,x) + g(t+r, Xrt,x)−g(t+r, Xrt,x) dr ≤E  0  Z  ∞ ≤ E (X t,x −X t,x )∗ +(X t,x −X t,x )p∗ +(X t,x−X t,x)∗ +(X t,x−X t,x)p∗ c(t+r)dr 0   Z ∞  +ρ(|t−t|)E 1+(X∗t,x)̟ c(t∧t+r)dr 0

   p 1/p 1+|x|̟ ρ(|t−t|)+CCp,̟ 1+|x|p̟ ρ(|t−t|) +Cρ(|t−t|) 1+C̟ (1+|x|̟ ) ≤ C(Cp ) |x−x|+CCp |x−x| +CCp,̟   p  1 1/p ≤ C(Cp ) p δ ′ +CCp (δ ′ )p +CCp,̟ 1+|x|̟ ρ(δ ′ )+CCp,̟ 1+|x|p̟ ρ(δ ′ ) +Cρ(δ ′ ) 1+C̟ (1+|x|̟ ) ≤ λ. (6.34) 1 p

p

Let τ3 = τ3 (t, x, y, ε) ∈ Tt,x (y) such that

  E R(t, x, τ3 ) ≥ V (t, x, y)−εo .

(6.35)  If E 0 g(t + r, Xrt,x)dr ≤ (y − δ ′ )+ , we directly set τb3 := τ3 . Otherwise, we define τb3 = τb3 (t, t, x, x, y, ε) := inf s ∈    R τ3  R R s τ [0, ∞) : E 0 g(t+r, Xrt,x)dr Fs − 0 g(t+r, Xrt,x)dr ≤ a′ with a′ := E 0 3 g(t+r, Xrt,x)dr −(y −δ ′ )+ > 0. Similar to  ′ + (6.19), one can deduce from (6.34) that τb3 is a Tt,x (y −δ ) −stopping time satisfying √   τb3 ≤ τ3 and P AcR ∩ τ3 > τb3 + λ < λo /2.  Using similar arguments to those that lead to (6.27), one can deduce from (6.14)−(6.16) that E R(t, x, τb3 ) −   R(t, x, τ3 ) < (2+10C)εo. Then applying (2.8) with (t, x, x′ , τ ) = t, x, x, τb3 and applying (6.32) with ζ = τb3 , we see from (6.33) that         E R(t, x, τb3 )−R(t, x, τ3 ) ≤ E R(t, x, τb3 )−R(t, x, τb3 ) +E R(t, x, τb3 )−R(t, x, τb3 ) +E R(t, x, τb3 )−R(t, x, τ3 ) p 1 1/p (1+|x|̟ )ρ(δ ′ )+2CCp,̟ (1+|x|p̟ ) ρ(δ ′ ) ≤ 2C(Cp ) p δ ′ +2CCp (δ ′ )p +2CCp,̟  +ρ(δ ′ )+Cρ(δ ′ ) 1+C̟ (1+|x|̟ ) +(2+10C)εo < (4+10C)εo = ε−εo.  R τ3



It follows from (2.12) and (6.35) that for any (t, x, y) ∈ [(t−δ ′ )+ , t+δ ′ ]×Oδ′ (x)×[(y −δ ′ )+ , ∞),      V (t, x, y) ≥ V t, x, (y −δ ′ )+ ≥ E R(t, x, τb3 ) > E R(t, x, τ3 ) −ε+εo ≥ V (t, x, y)−ε.

(6.36)

2b) We next show that V (t, x, y) ≤ V (t, x, y)+ε, ∀ (t, x, y) ∈ [(t−δ ′ )+ , t+δ ′ ]×Oδ′ (x)×[0, y +δ ′]. Let (t, x) ∈ [(t−δ ′ )+ , t+δ ′ ]×Oδ′ (x). There exists τ4 = τ4 (t, t, x, x, y, ε) ∈ Tt,x (y +δ ′ ) such that    E R(t, x, τ4 ) ≥ V t, x, y +δ ′ −εo . (6.37)  R τ4   If E 0 g(t+ r, Xrt,x )dr we directly set τb4 := τ4 . Otherwise, we define τb4 = τb4 (t, t, x, x, y, ε) := inf s ∈ [0, ∞) : ≤ y, Rs Rτ Rτ  E 0 4 g(t+r, Xrt,x )dr Fs − 0 g(t+r, Xrt,x )dr ≤ b′ with b′ := E 0 4 g(t+r, Xrt,x )dr −y > 0. Analogous to (6.30), we can deduce from (6.34) that τb4 is a Tt,x (y)−stopping time satisfying √  τb4 ≤ τ4 and P AcR ∩{τ4 > τb4 + λ } < λo /2.   Since an analogy to (6.24)−(6.26) gives that E R(t, x, τb4 )−R(t, x, τ4 ) < (2+10C)εo, applying (6.32) with ζ = τ4  and applying (2.8) with (t, x, x′ , τ ) = t, x, x, τ4 , we see from (6.33) that         E R(t, x, τb4 )−R(t, x, τ4 ) ≤ E R(t, x, τb4 )−R(t, x, τ4 ) +E R(t, x, τ4 )−R(t, x, τ4 ) +E R(t, x, τ4 )−R(t, x, τ4 ) p 1/p ≤ (2+10C)εo +2CCp,̟ (1+|x|̟ )ρ(δ ′ )+2CCp,̟ (1+|x|p̟ ) ρ(δ ′ ) +ρ(δ ′ )  1 +Cρ(δ ′ ) 1+C̟ (1+|x|̟ ) +2C(Cp ) p δ ′ +2CCp (δ ′ )p < (4+10C)εo = ε−εo. It then follows from (2.12) and (6.37) that for any (t, x, y) ∈ [(t−δ ′ )+ , t+δ ′ ]×Oδ′ (x)×[0, y +δ ′]      V (t, x, y) ≤ V t, x, y +δ ′ ≤ E R(t, x, τ4 ) +εo < E R(t, x, τb4 ) +ε ≤ V (t, x, y)+ε,   which together with (6.36) yields V (t, x, y)−V (t, x, y) ≤ ε, ∀ (t, x, y) ∈ [(t−δ ′ )+ , t+δ ′ ]×O δ′ (x)× (y−δ ′ )+ , y+δ ′ . 

6.2

Proofs of Section 3

6.2

19

Proofs of Section 3

n o Proof of Lemma 3.1: Set Λ := A ⊂ Ωt : A = ∪ ω ⊗s Ωs . Clearly, ∅, Ωt ∈ Λ. For any A ∈ Λ, we claim that ω∈A

ω ⊗s Ωs ⊂ Ac for any ω ∈ Ac .

(6.38)

 Assume not, there exist an ω ∈ Ac and an ω e ∈ Ωs such that ω ⊗s ω e ∈ A. Then ω ⊗s ω e ⊗s Ωs ⊂ A and it follows that  ω ∈ ω ⊗ s Ωs = ω ⊗ s ω e ⊗s Ωs ⊂ A. A contradiction appear. So (6.38) holds, which shows that Ac ∈ Λ.    ω⊗s Ωs , namely, ∪ An ∈ Λ. ∪ ω⊗s Ωs = ∪ For any {An }n∈N ⊂ Λ, one can deduce that ∪ An = ∪ n∈N ω∈ ∪ An n∈N n∈N ω∈An n∈N  d t −1 s Given r ∈ [t, s] and E ∈ B(R ), if ω ∈ (Wr ) (E), it holds for any ω e ∈ Ω that ω ⊗s ω e (r) = ω(r) ∈ E or ω ⊗s ω e∈ (Wrt )−1 (E), which implies that (Wrt )−1 (E) ∈ Λ. Hence, Λ is a sigma−field of Ωt containing all generating sets of Fst . It follows that Fst ⊂ Λ, proving the lemma.  Proof of Lemma 3.2: Let us regard ω ⊗s · as a mapping Γ from Ωs to Ωt , i.e., Γ(e ω ) := ω ⊗s ω e, ∀ ω e ∈ Ωs . So s,ω −1 t A = Γ (A) for any A ⊂ Ω . 1) Assume first that r ∈ [s, ∞). Given t′ ∈ [t, r] and E ∈ B(Rd ), we can deduce that Γ−1

 (Wtt′ )−1 (E) = {e ω ∈ Ωs : Wtt′ (ω ⊗s ω e ) ∈ E} =

 s   Ω ,

if t′ ∈ [t, s) and ω(t′ ) ∈ E;

∅, if t′ ∈ [t, s) and ω(t′ ) ∈ / E;     ω s ′ s −1 ′ s ′ e ∈ Ω : ω(s)+ ω e (t ) ∈ E = (W ′ ) (E ) ∈ F , if t ∈ [s, r]; t

r

 where E ′ := E −ω(s) = {x−ω(s) : x ∈ E} ∈ B(Rd ). So all generating sets of Frt belong to Λr := A ⊂ Ωt : Γ−1 (A) ∈ Frs , which is clearly a sigma−field of Ωt . It follows that Frt ⊂ Λr , or As,ω = Γ−1 (A) ∈ Frs for any A ∈ Frt .  e ∈ Frs . We know from Lemma A.2 (1) that (Πts )−1 A e ∈ Frt . Since the continuity of On the other hand, let A   paths in Ωt shows that ω ⊗s Ωs = ω ′ ∈ Ωt : ω ′ (t′ ) = ω(t′ ), ∀ t′ ∈ (t, s)∩Q = ′ ∩ (Wtt′ )−1 {ω(t′ )} ∈ Fst ⊂ Frt , one t ∈(t,s)∩Q   s t t −1 e e A ∩ ω ⊗ s Ω ∈ Fr . can deduce that ω ⊗s A = (Πs ) 2) Next, we consider the case of r = ∞. Given r′ ∈ [s, ∞), since Γ−1 (A) ∈ Frs′ ⊂ F s for any A ∈ Frt′ , we see that  −1 Frt′ ⊂  Λ := A ⊂ Ωt : Γ (A) ∈ F s , which is clearly a sigma−field of Ωt . It follows from Lemma A.1 (1) that

Frt′ ⊂ Λ. So As,ω = Γ−1 (A) ∈ F s for any A ∈ F t .   e ∈ F t′ ⊂ F t for any A e ∈ F s′ , one has F s′ ⊂ Λ e = ω ⊗s A e := A e ⊂ Ωs : On the other hand, let r′ ∈ [s, ∞). Since Γ A r r r    c s s t t t e is a disjoint union of Γ(Ω ) = ω ⊗s Ω ∈ Fs ⊂ F . It follows e ∪Γ A e ∈ F . Given A e ∈ Λ, e it is clear that Γ A Γ A      c t s e = (ω ⊗s Ω )\Γ A e ∈ F . Also, it holds for any {A en }n∈N ⊂ Λ e that Γ ∪ A en = ∪ Γ A e is a en ∈ F t . So Λ that Γ A n∈N n∈N   e or sigma−field of Ωs that contains all Frs′ , r′ ∈ [s, ∞). Then Lemma A.1 (1) implies that F s = σ ′ ∪ Frs′ ⊂ Λ, r ∈[s,∞)  e= Γ A e ∈ F t for any A e∈ F s.  ω ⊗s A Ft =σ



r ′ ∈[t,∞)

Frt′ = σ



r ′ ∈[s,∞)

Proof of Proposition 3.1: 1) Let ξ be an E−valued random variable on Ωt that is Frt −measurable for some −1   r ∈ [s, ∞]. For any E ∈ B(E), since ξ −1 (E) ∈ Frt , Lemma 3.2 shows that ξ s,ω (E) = ω e ∈ Ωs : ξ(ω ⊗s ω e) ∈ E = ω e∈  s,ω Ωs : ω ⊗ s ω e ∈ ξ −1 (E) = ξ −1 (E) ∈ Frs . So ξ s,ω is Frs −measurable. 2) Let {Xr }r∈[t,∞) be an E−valued, Ft −adapted process. For any r ∈ [s, ∞) and E ∈ B(E), since Xr ∈ Frt , one can −1    s,ω (E) = ω e ∈ Ωs : X r, ω ⊗s ω e ∈E = ω e ∈ Ωs : ω ⊗ s ω e ∈ Xr−1 (E) = Xs−1 (E) ∈ deduce from Lemma 3.2 that Xrs,ω   Frs , which shows that Xrs,ω r∈[s,∞) is Fs −adapted.

Proof of Proposition 3.2: In virtue of Theorem 1.3.4 and (1.3.15) of [57], there exists a family {Psω }ω∈Ωt of probabilities on (Ωt , F t ), called the regular conditional probability distribution of Pt with respect to the sigma-field Fst , such that ( i) For any A ∈ F t , the mapping ω → Psω (A) is Fst −measurable;   ( ii) For any ξ ∈ L1 (F t ), EPsω [ξ] = Et ξ Fst (ω) for Pt −a.s. ω ∈ Ωt ;  (iii) For any ω ∈ Ωt , Psω ω ⊗s Ωs = 1.

(6.39) (6.40)

DPPs for Optimal Stopping with Expectation Constraint

20

e ∈ F t for any A e ∈ F s . Then one can deduce from (6.40) that 1) Given ω ∈ Ωt , Lemma 3.2 shows that ω ⊗s A   e , ∀A e ∈ F s defines a probability measure on (Ωs , F s ). We claim that for Pt −a.s. ω ∈ Ωt e := P ω ω ⊗s A P s,ω A s   e , e = Ps A P s,ω A

e∈ F s. ∀A

(6.41)

 e ∈ F s . Since (Πt )−1 A e ∈ F t by Lemma A.2 (1), (6.40) and (6.39) imply that for Pt −a.s. ω ∈ Ωt To see this, we let A s h i      e = Et 1 t −1 e F t (ω). e ∩ (ω ⊗s Ωs ) = P ω (Πt )−1 A e = P ω (Πt )−1 A e = P ω ω ⊗s A (6.42) P s,ω A s s s s s s (Π ) (A) s

We can deduce from Lemma A.1 (1) that

    (Πts )−1 (F s ) = (Πts )−1 σ (Wrs )−1 (E) : r ∈ [s, ∞), E ∈ B(Rd ) = σ (Πts )−1 (Wrs )−1 (E) : r ∈ [s, ∞), E ∈ B(Rd )   = σ (Wrt −Wst )−1 (E) : r ∈ [s, ∞), E ∈ B(Rd ) = σ Wrt −Wst ; r ∈ [s, ∞) ,

which is independent of Fst under Pt . Then (6.42) and Lemma A.2 (2) show that for Pt −a.s. ω ∈ Ωt , i h h i    e . e = Ps A e = Et 1 t −1 e F t (ω) = Et 1 t −1 e = Pt (Πt )−1 A P s,ω A s s (Π ) (A) (Π ) (A) s

s

o  ∩ (Wssi )−1 Oδi (xi ) : m ∈ N, si ∈ Q∪{s} with s ≤ s1 ≤ · · · ≤ sm , xi ∈ Qd , δi ∈ Q+ is a countable set, i=1    e holds for each A e ∈ C s . To wit, C s ⊂ Λ := A e∈ e = Ps A we can find a N ∈ N s such that for any ω ∈ N c , P s,ω A ∞ ∞   c s s,ω e s e , ∀ ω ∈ N . It is easy to see that Λ is a Dynkin system. As C∞ is closed under intersection, Ω :P A = Ps A s Lemma A.1 (2) and Dynkin System Theorem show that F s = σ(C∞ ) ⊂ Λ. Namely, it holds for any ω ∈ N c that   s s,ω e e e P A = Ps A , ∀ A ∈ F , proving (6.41). 2) Now, let ξ ∈ L1 (F t ). Proposition 3.1 (1) shows that ξ s,ω is F s −measurable for any ω ∈ Ωt . Also, we can deduce from (6.39)−(6.41) that for Pt −a.s. ω ∈ Ωt Z Z Z s,ω     ξ(ω ′ ) dP ω (ω ′ ) ξ ω ⊗ s ω ξ (e e = e dPsω ω ⊗s ω ω ) dP s,ω (e ω) = Es |ξ s,ω | = s s ω ′ ∈ω⊗s Ωs ω e ∈Ωs Zωe ∈Ω     ξ(ω ′ ) dPsω (ω ′ ) = EP ω |ξ| = Et |ξ| Fst (ω) < ∞, = s s As C∞ :=

n

m

ω ′ ∈Ωt

     thus ξ s,ω ∈ L1 F s . Similarly, it holds for Pt −a.s. ω ∈ Ωt that Es ξ s,ω = Et ξ Fst (ω) ∈ R.



Proof of Proposition 3.3: 1) Let N be a Pt −null set, so there exists an A ∈ F t with Pt (A) = 0 such that N ⊂ A. For any ω ∈ Ωt , Lemma 3.2 shows that N s,ω = {e ω ∈ Ωs : ω⊗s ω e ∈ N } ⊂ {e ω ∈ Ωs : ω⊗s ω e ∈ A} = As,ω ∈ F , and we see that s s,ω ω ), ∀ ω e ∈ Ω . Then (3.2) implies that for Pt −a.s. ω ∈ Ωt (1A ) (e ω) = 1{ω⊗s ωe ∈A} = 1{eω∈As,ω } = 1As,ω (e        (6.43) Ps As,ω = Es 1As,ω = Es (1A )s,ω = Et 1A Fst (ω) = 0, and thus N s,ω ∈ N s .

Next, let ξ1 and ξ2 be two real-valued random variables with ξ1 ≤ ξ2 , Pt −a.s. Since N := {ω ∈ Ωt : ξ1 (ω) > ξ2 (ω)} ∈ N t , (6.43) leads to that for Pt −a.s. ω ∈ Ωt ,      ω) . ω ) > ξ2s,ω (e e ∈ Ωs : ξ1s,ω (e e = Ps ω e > ξ2 ω ⊗s ω e ∈ Ωs : ξ1 ω ⊗s ω 0 = Ps N s,ω = Ps ω

t t er ∈ F t such that Nr := Ar ∆ A er ∈ N t 2) Let τ ∈ T with τ ≥ s and let r ∈ [s, ∞). As Ar := {τ ≤ r} ∈ F r , there exists an A r t s,ω b (see e.g. Problem 2.7.3 of [32]). By Part (1), it holds for all ω ∈ Ω except on a Pt −null set Nr that Nr ∈ N s .  s brc , since As,ω es,ω e s,ω = Nrs,ω ∈ N s and since A es,ω Given ω ∈ N r ∆ Ar = Ar ∆ Ar r ∈ Fr by Lemma 3.2, we can deduce s that As,ω ∈ F r and it follows that r s

Let ω ∈

{τ s,ω ≤ r} = {e ω ∈ Ωs : τ s,ω (e ω ) ≤ r} = {e ω ∈ Ωs : τ (ω ⊗s ω e ) ≤ r} = {e ω ∈ Ωs : ω ⊗ s ω e ∈ Ar } = As,ω r ∈ F r. ∩

(6.44)

brc . For any r ∈ [s, ∞), there exists a sequence {rn }n∈N in (s, ∞)∩Q such that lim ↓ rn = r. Then N

r∈(s,∞)∩Q

n→∞

s

s

s

(6.44) and the right-continuity of Brownian filtration F (under Ps ) imply that {τ s,ω ≤ r} = ∩ {τ s,ω ≤ rn } ∈ F r+ = F r . s

Hence τ s,ω ∈ T .

n∈N



6.2

Proofs of Section 3

21 t

Proof of Proposition 3.4: 1) Let r ∈ [s, ∞] and ξ be an F r −measurable random variable. By Lemma A.4 (2), there exists an Frt −measurable random variable ξe that equals to ξ except on a N ∈ N t . Proposition 3.1 (1) shows that ξes,ω is Frs −measurable for any ω ∈ Ωt . Also, we see from Proposition 3.3 (1) that for Pt −a.s. ω ∈ Ωt ,   ω e ∈ Ωs : ξes,ω (e ω ) 6= ξ s,ω (e ω) = ω e ∈ Ωs : ω ⊗ s ω e ∈ N = N s,ω ∈ N s (6.45)

s t and thus ξ s,ω ∈ F r . In particular, if ξ is an F s −measurable and ξe is Fst −measurable, then (6.45) and (3.1) imply e = ξ(ω), Ps −a.s. that Pt −a.s. ω ∈ Ωt , ξ s,ω = ξes,ω = ξ(ω)  Suppose next that ξ is integrable so is ξe . Proposition 3.2 and Lemma A.4 (1) show that for Pt −a.s. ω ∈ Ωt , ξes,ω          t is integrable (so is ξ s,ω ) and Et ξ F s (ω) = Et ξ Fst (ω) = Et ξe Fst (ω) = Es ξes,ω = Es ξ s,ω ∈ R.  t 2a) Let X = {Xr }r∈[t,∞) be an F −adapted process with Pt −a.s. continuous paths and set N1 := ω ∈ Ωt : the path X· (ω) is not continuous ∈ N t . In light of Lemma A.4 (3), we can find an E−valued, Ft −predictable  er (ω) 6= Xr (ω) for some r ∈ [t, ∞)} ∈ N t . In particular, X e is an e= X er such that N2 := {ω ∈ Ωt : X process X r∈[t,∞) t F −adapted process. e s,ω is Fs −adapted for any ω ∈ Ωt , and Proposition 3.3 (1) Proposition 3.1 (2) shows that the shifted process X s,ω ∈ N s . Let ω ∈ N3c . Since implies that for any ω ∈ Ωt except on a Pt −null set N3 that N1 ∪N2

s,ω   e s,ω (e ∈ N s, ω e ∈ Ωs : X·s,ω (e ω ) is not continuous ∪ ω e ∈ Ωs : X ω ) 6= Xrs,ω (e ω ) for some r ∈ [s, ∞) ⊂ N1 ∪N2 r s

one can deduce that X s,ω is an F −adapted process with Ps −a.s. continuous paths. 2b) Next, let us further assume that X ∈ Cqt (E) for some q ∈ [1, ∞). Define ξ := sup X∗q

t X∗q ∈ F and s s,ω

 Et [ξ] = Et X∗q < ∞. 

r∈[t,∞)∩Q

q X er ∈ F t . As ξ equals to

According to Part (1), it holds for all ω ∈ Ωt except thus on (N1 ∪N2 ) , one has on a Pt −null set N4 that ξ is F −measurable and Ps −integrable. c s,ω c Let ω ∈ (N3 ∪ N4 ) . For any ω e ∈ (N1 ∪ N2 )s,ω = (N1 ∪ N2 )c , the continuity of the path X·s,ω (e ω) = q q s,ω q er (ω⊗s ω e ). It follows that e ) ≤ ξ(ω⊗s ω e ) = sup X ω ) = sup Xr (ω⊗s ω X· (ω⊗s ω e ) implies that sup Xr (e r∈[s,∞)∩Q r∈[s,∞)∩Q r∈[s,∞) i h  s,ω  s,ω q s,ω q < ∞. Hence, X ∈ Cs (E) for any ω ∈ (N3 ∪N4 )c .  Es sup |Xr | ≤ Es ξ c

r∈[s,∞)

Proof of Proposition 3.5: Let M = {Mr }r∈[t,∞) ∈ Mt . By Proposition 3.4 (3), it holds for Pt −a.s. ω ∈ Ωt that s M s,ω is an F −adapted process with Ps −a.s. continuous paths. So we only need to show that M s,ω is a uniformly  s integrable martingale with respect to F , Ps for Pt −a.s. ω ∈ Ωt . t By the uniform integrability of M , there exists ξ ∈ L1 F such that for any r ∈ [s, ∞),  t Mr = Et ξ F r ,

Pt −a.s.

(6.46)

Set N := {ω ∈ Ωt : the path M· (ω) is not continuous} ∈ N t . Proposition 3.3 (1) and Proposition 3.4 (2) imply that s for all ω ∈ Ωt except on a No ∈ N t , one has N s,ω ∈ N s and ξ s,ω ∈ L1 F . t Fix r ∈ [s, ∞). As Mr ∈ L1 F r , Proposition 3.4 (2) shows that for all ω ∈ Ωt except on a Pt −null set Nr1 ,  s Mrs,ω ∈ L1 F r .    e belongs to F tr , so 1A Mr ∈ L1 F tr and 1A ξ ∈ L1 F t . e ∈ F sr . By Lemma A.3 (2), the set A := (Πts )−1 A Let A

Since it holds for any ω ∈ Ωt and ω e ∈ Ωs that (1A )s,ω (e ω ) = 1{ω⊗s ωe∈A} = 1{Πt (ω⊗s ωe )∈A} ω ), Proposition e(e e = 1{e e = 1A ω ∈A} s t 3.4 (2) and (6.46) yield that for Pt −a.s. ω ∈ Ω i i h  h        t t  t t  t t Es 1AeMrs,ω = Et 1A Mr |F s (ω) = Et 1A Et ξ|F r F s (ω) = Et Et 1A ξ|F r F s (ω) = Et 1A ξ|F s (ω) = Es 1Aeξ s,ω . o  ∩ (Wssi )−1 Oδi (xi ) : m ∈ N, si ∈ Q+ ∪{s} with s ≤ s1 ≤ · · · ≤ sm ≤ r, xi ∈ Qd , δi ∈ Q+ is a countable i=1     e ∈ C s . To wit, set, there exists a Nr2 ∈ N s such that for any ω ∈ (Nr2 )c , Es 1AeMrs,ω = Es 1Aeξ s,ω holds for each A r      e ⊂ Ωs : Es 1 eM s,ω = Es 1 eξ s,ω , ∀ ω ∈ (N 2 )c . It is easy to see that C s is closed under intersection Crs ⊂ Λr := A r r A A r and Λ is a Dynkin system. Then Lemma A.1 (2) and Dynkin System Theorem show that Frs = σ(Crs ) ⊂ Λr . Clearly, N s also belongs to Λr , so As Crs :=

n

m

s

F r = σ(Frs ∪N s ) ⊂ Λr .

(6.47)

DPPs for Optimal Stopping with Expectation Constraint Now, let ω ∈ Noc ∩



22

c     (Nr1 ∪ Nr2 ) . For any r ∈ [s, ∞), (6.47) shows that Es 1AeMrs,ω = Es 1Aeξ s,ω ,



r∈[s,∞)∩Q

e ∈ F s and thus Es [ξ s,ω |F s ] = M s,ω , Ps −a.s. Since {e ∀A ω ∈ Ωs : path M·s,ω (e ω ) is not continuous} ⊂ N s,ω ∈ N s , r r r   s,ω  s s s,ω we can deduce from the continuity of process Es [ξ |F r ] r∈[s,∞) that Ps Mr = Es ξ s,ω |F r , ∀ r ∈ [s, ∞) = 1.  s  Therefore, M s,ω is a uniformly integrable continuous martingale with respect to F , Ps .  t Proof of Lemma 3.3: Let {ξi }i∈N be a sequence of L1 F that converges to 0 in probability Pt , i.e.    lim ↓ Et 1{|ξi |>1/n} = lim ↓ Pt |ξi | > 1/n = 0, ∀ n ∈ N. (6.48) i→∞

i→∞

 1 ξi i∈N from {ξi }i∈N such that   lim 1{|ξi1 |>1} = 0, Pt −a.s. Clearly, S1 also satisfies (6.48). Then by lim ↓ Et 1{|ξi1 |>1/2} = 0, we can find a i→∞ i→∞  subsequence S2 = ξi2 i∈N of S1 such that lim 1{|ξi2 |>1/2} = 0, Pt −a.s. Inductively, for each n ∈ N we can select a i→∞ = 0, Pt −a.s. subsequence Sn+1 = {ξ n+1 }i∈N of Sn = {ξ n }i∈N such that lim 1 n+1 



In particular, lim ↓ Et 1{|ξi |>1} = 0 allows us to extract a subsequence S1 = i→∞

i

i

i→∞

|ξi

1 |> n+1

For any i ∈ N, we set ξei := ξii , which belongs to Sn for n = 1, · · · , i. Given n ∈ N, since {ξei }∞ i=n ⊂ Sn , it holds Pt −a.s. that lim 1 e 1 = 0. Then a conditional-expectation version of the bound convergence theorem and |ξi |> n

i→∞

s Proposition 3.4 (2) imply that for all ω ∈ Ωt except on a Pt −null set Nn , ξei is F −measurable and h h ti s,ω i . 0 = lim Et 1{|ξei |>1/n} F s (ω) = lim Es 1{|ξei |>1/n} i→∞

i→∞

Let ω ∈



∪ Nn

n∈N

1{|ξei |>1/n}

c

(6.49)

. For any n ∈ N, one can deduce that

s,ω

(e ω ) = 1 e

= 1 s,ω e

|ξi (ω⊗s ω e )|>1/n

ξi

  (e = 1 s,ω ω ), e (e ω ) >1/n |ξ i |>1/n

∀ω e ∈ Ωs ,

h   s,ω i = 0. which together with (6.49) leads to that lim Ps |ξeis,ω | > 1/n = lim Es 1{|ξei |>1/n} i→∞

i→∞



Proof of Proposition 3.6: As X ∈ C2t (Rl ) by Corollary 3.1, we know from Proposition 3.4 (3) that for Pt −a.s.  ∈ C2s (Rl ). ω ∈ Ωt , Xs,ω r r∈[s,∞) To show that for Pt −a.s. ω ∈ Ωt , Xs,ω solves (1.1) over [s, ∞) with initial state Xs (ω), we let N1 be the Pt −null R s′ set such that X satisfies (1.1) on N1c . Define Ms′ := t 1{r>s} σ(r, Xr )dWrt , s′ ∈ [t, ∞). 1) By Proposition 3.4 (1), there exists a Pt −null set N2 such that for any ω ∈ N2c , Xs (ω ⊗s ω e ) = Xs (ω) holds for all ω e ∈ Ωs except on a Nω ∈ N s . Let ω ∈ N1c ∩N2c and ω e ∈ Nωc . Implementing (1.1) on the path ω ⊗s ω e over period [s, ∞) yields that Xs,ω ω) s′ (e

e ) = Xs (ω ⊗ s ω e )+ = Xs′ (ω ⊗ s ω

= Xs (ω)+

Z

s′

s

Z

s

s′

  b r, Xr (ω ⊗ s ω e ) dr+

 b r, Xs,ω ω) dr + Mss,ω ω ), ′ (e r (e

Z

s′ ∈ [s, ∞).

s

s′

 e) σ(r, Xr )dWrt (ω ⊗ s ω

(6.50)

So it remains to show that for Pt −a.s. ω ∈ Ωt , it holds Ps −a.s. that Mss,ω = ′

Z

s

s′

 dWrs , σ r, Xs,ω r

s′ ∈ [s, ∞).

(6.51)

 t 2) Since {Ms′ }s′ ∈[t,∞) is a square-integrable martingale with respect to F , Pt by (1.7) and Corollary 3.1, n we t l×d know that (see e.g. Problem 3.2.27 of [32]) there is a sequence of R −valued, F −simple processes Φnr = o  P t n  , r ∈ [t, ∞) η 1 where {tni }i∈N is an increasing sequence in [t, ∞) and ηin ∈ F tni for i ∈ N i n n i∈N r∈(ti ,ti+1 ]

n∈N

such that

Pt − lim

n→∞

Z

t



trace

n

 T o dr = 0 Φnr −σ(r, Xr ) Φnr −σ(r, Xr )

and Pt − lim

sup Msn′ − Ms′ = 0,

n→∞ s′ ∈[t,∞)

6.3

Proof of Section 4

where Msn′ := Pt − lim

R s′ t

n→∞

Z

23 P

Φnr dWrt = ∞

trace

s

n

i∈N

 ηin Wst′ ∧tn −Wst′ ∧tn . Then it directly follows that i+1

i

 T o Φnr −σ(r, Xr ) Φnr −σ(r, Xr ) dr = 0

n→∞ s′ ∈[s,∞)

n bn = P By Lemma 3.3, {Φn }n∈N has a subsequence Φ b in 1 r i∈N η

bn r∈(b tn i ,t i+1 ]

except on a Pt −null set N4

0 = Ps − lim

Z

n→∞

Z



s ∞

trace

n

sup Msn′ − Ms′ = 0.

and Pt − lim

o , r ∈ [t, ∞)

 n  o b n −σ(r, Xr ) Φ b −σ(r, Xr ) T dr Φ r r

n

n∈N

such that for any ω ∈ Ωt

s,ω

 T o  s,ω s,ω n s,ω b −σ(r, X ) −σ(r, X ) dr Φ r r r r n→∞ s   n s,ω cs′ − M csn −Ms′ sup M and 0 = Ps − lim = Ps − lim

n→∞

= Ps − lim

trace

bn Φ

s,ω

(6.52)

s′ ∈[s,∞)

 cn s,ω cn s,ω − M s,ω M − sup M ′ ′ s , s s

(6.53)

n→∞ s′ ∈[s,∞)

cn′ := where M s

R s′ t

P

b nr dWrt = Φ

i∈N

  ηb in Wst′ ∧bt n −Wst′ ∧bt n . i+1

i

Given n ∈ N, let ℓn be the largest integer such that b t nℓn < s. For any i = ℓn, ℓn +1, · · · , we set sni := b t ni ∨s. Since  t t s s,ω ηb in ∈ F bt ni ⊂ F sni . Proposition 3.4 (2) shows that ηb in ∈ F sni holds for any ω ∈ Ωt except on a Pt −null set Nin . Let    ∞ b := N4c ∩ ∩ ∩ (N n )c . As sn = s, one has ηb n s,ω ∈ F ss . For any s′ ∈ [s, ∞) and ω e ∈ Ωs , ω ∈Ω i ℓn ℓn n∈N i=ℓn

s,ω

bn Φ

s′

b ns′ (ω ⊗s ω (e ω) = Φ e) =

X i∈N

ηb in (ω ⊗s ω e ) 1 ′ bn bn s ∈(t ,t i

i+1 ]

= ηb n ℓn

s,ω

(e ω )1

{s′ ∈[s,sn ℓn +1 ]}

+

∞ X

i=ℓn +1

ηb in

s,ω

(e ω ) 1{s′ ∈(sni ,sni+1 ]} .

 n s,ω s b ′ is an Rl×d −valued, F −simple process. Applying Proposition 3.2.26 of [32] and using (6.52) So Φ s′ ∈[s,∞) s yield that Z ′ Z s′ s  n s,ω s,ω s s b 0 = Ps − lim sup Φ r dWr − σ(r, Xr )dWr . (6.54) n→∞ s′ ∈[s,∞) s s For any ω e ∈ Ωs , one can deduce that

cn M

s,ω s′

cn (e ω )− M

s,ω s

(e ω) =

∞ X i=ℓ

=

∞ X i=ℓ

∞     s,ω  X  e s′∧sni (e ω) ω e s′∧sni+1 − ω ηb in e ) s′∧sni = ηb in (ω ⊗s ω e ) (ω ⊗s ω e ) s′∧sni+1 −(ω ⊗s ω

 n s,ω

ηb i

Z   (e ω ) Wss′ ∧sni+1 −Wss′ ∧sni (e ω) =

i=ℓ

s′

b Φ

s

 n s,ω r

 dWrs (e ω ),

s′ ∈ [s, ∞),

b Eventually, we see from (6.50) which together with (6.53) and (6.54) shows that (6.51) holds Ps −a.s. for any ω ∈ Ω.  s,Xs (ω) s s,ω b e ∈ Ω : Xr (ω ⊗s ω e ) = Xr (e ω ) = Xr (e ω ), ∀ r ∈ [s, ∞) = 1 for any ω ∈ Ω.  that Ps ω

6.3

Proof of Section 4

The proof of the first DPP (Theorem 4.1) is based on the following auxiliary result. t

Lemma 6.1. Given (t, x, y) ∈ [0, ∞)×Rl ×[0, ∞), let τ ∈ Txt (y) and ζ ∈ T ♯ . Then  Z ζ     t,x t,x,τ t,x f (r, Xr )dr ≤ V(t, x, y). Et R(t, x, τ ) ≤ Et 1{τ ≤ζ} R(t, x, τ )+1{τ >ζ} V(ζ, Xζ , Yζ )+ t

(6.55)

DPPs for Optimal Stopping with Expectation Constraint

24

Proof: 1) Let us start with some basic settings. Denote (X, Y) := (X t,x , Y t,x,τ ) and let ζ take values in a countable subset {ti }i∈N of [t, ∞). In light of Lemma  e r (ω) 6= e= X er such that N := {ω ∈ Ωt : X A.4 (3), there exists an Rl −valued, Ft −predictable process X r∈[t,∞) Xr (ω) for some r ∈ [t, ∞)} ∈ N t . Let i ∈ N. By Proposition 3.3 (1), we can find a Pt −null set Ni such that for any ω ∈ Nic , N ti ,ω is a Pti −null set. e r ∈ F t ⊂ F t , (3.1) implies that For any r ∈ [t, ti ], since X r ti e e r (ω ⊗t ω e) = X Xr (ω ⊗ti ω i e ) = Xr (ω) = Xr (ω),

∀ ω ∈ N c ∩Nic , ∀ ω e ∈ (N c )ti ,ω .

ei , Also Proposition 3.6 shows that for all ω ∈ Ωt except on a Pt −null set N n o ti ,Xti (ω) ω ) 6= Xr Nωi := ω e ∈ Ωti : Xtri ,ω (e (e ω ), for some r ∈ [ti , ∞) ∈ N

ti

(6.56)

.

(6.57)

t

Let τi be a T −stopping time with τi ≥ ti . According to Proposition 3.3 (2) and Proposition 3.4 (2), it holds for bi that τωi := τ ti ,ω ∈ T ti , all ω ∈ Ωt except on a Pt −null set N i  Z τi  Z τi  ti ,ω  h i  ti ,ω t  t Et R(t, x, τi ) F ti (ω) = Eti R(t, x, τi ) g(r, Xr )dr F ti (ω) = Eti g(r, Xr )dr . (6.58) and Et t

t

 i c

 i c

e c ∩N b c . Given ω Let ω ∈ N c ∩Nic ∩N e ∈ N ti ,ω ∪Nω i i

ti ,Xt (ω)

i e ) = Xr (e ω ) for = (N c )ti ,ω ∩ Nω , (6.57) shows Xr (ω⊗ti ω   t i ,Xti (ω) i i e = Xτ i e = X τω (e ω ), ω ⊗ti ω e ), ω ⊗ti ω any r ∈ [ti , ∞). In particular, taking r = τω (e ω ) yields that X τi (ω ⊗ti ω (e ω ), ω which together with (6.56) leads to that Z τi (ω⊗t ωe) i  ti ,ω  e) e ), Xτi (ω ⊗ti ω R(t, x, τi ) (e ω) = e ) dr+π τi (ω ⊗ti ω f r, Xr (ω ⊗ti ω

t

=

Z

ti

t

=

Z

t

R τi

and similarly,

t

g(r, Xr )dr

ti ,ω

ti

 f r, Xr (ω) dr+

Z

i τω (e ω)

    ti ,Xti (ω) ti ,Xt (ω) f r, Xr (e ω ) dr+π τωi (e ω ), Xτ i i (e ω) ω

ti

  ω ), f r, Xr (ω) dr+ R(ti , Xti (ω), τωi ) (e

(e ω) =

R

i τω g ti

   Rt  ti ,Xti (ω) r, Xr dr (e ω )+ t i g r, Xr (ω) dr. Taking expectation Et [·], we ti

see from (6.58) that for Pt −a.s. ω ∈ Ωt , τωi is a T −stopping time satisfying Z ti t     i Et R(t, x, τi ) F ti (ω) = Eti R(ti , Xti (ω), τω ) + f r, Xr (ω) dr, and Et

Z

τi

t

Z  t g(r, Xr )dr F ti (ω) = Eti

(6.59)

t

i τω

ti

   Z ti ,Xti (ω) g r, Xr dr +

t

ti

 g r, Xr (ω) dr.

(6.60)

2) We next show the first inequality in (6.55). t Let i ∈ N and set τi := τ ∨ti ∈ T . We can deduce from (6.60), (4.6), (4.7) and (6.59) that for Pt −a.s. ω ∈ Ωt , t i t ,ω τωi := τi i is a T −stopping time satisfying Eti

Z

i τω

ti

and

Z    ti ,Xti (ω) g r, Xr dr = Et

τ ∨ti

t

t    Et R(t, x, τi ) F ti (ω) = Eti R ti , Xti (ω), τωi + t t > ζ} ∈ F τ ∧ζ ⊂ F ζ

Z  t g(r, Xr )dr F ti (ω)−

t

Z

t

ti

ti

 g r, Xr (ω) dr = Yti (ω) ∈ [0, ∞),

  f r, Xr (ω) dr ≤ V ti , Xti (ω), Yti (ω) +

Z

t

ti

 f r, Xr (ω) dr. (6.61) t

(see e.g. Lemma 1.2.16 of [32]), one has {τ > ζ = ti } = {τ > ζ}∩{ζ = ti } ∈ F ti . Then (6.61) As {τ shows that h t i      Et 1{τ >ζ=ti } R(t, x, τ ) = Et 1{τ >ζ=ti } R(t, x, τi ) = Et 1{τ >ζ=ti } Et R(t, x, τi ) F ti     Z ζ Z ti         (6.62) f r, Xr dr . f r, Xr dr = Et 1{τ >ζ=ti } V ζ, Xζ , Yζ + ≤ Et 1{τ >ζ=ti } V ti , Xti , Yti + t

t

6.3

Proof of Section 4

25

Since (4.3), (6.9), (6.10) and the first inequality in (3.4) imply that     Z ∞ Z ζ   p p c(r) 2+|Xζ | dr Et V(ζ, Xζ , Yζ ) + f (r, Xr ) dr ≤ Et 2C 2+Cp (1+|Xζ | ) + t  tp  ≤ 2C(3+Cp)+C(1+2Cp)Et X∗ < ∞,

(6.63)

taking summation over i ∈ N in (6.62), we can deduce from the first inequality in (4.1) and the dominated convergence theorem that " # X X     Et 1{τ >ζ} R(t, x, τ ) = Et 1{τ >ζ=ti } R(t, x, τ ) = Et 1{τ >ζ=ti } R(t, x, τ ) i∈N

i∈N

" #  Z ζ Z ζ   X X       f r, Xr dr = Et f r, Xr dr 1{τ >ζ=ti } V ζ, Xζ , Yζ + ≤ Et 1{τ >ζ=ti } V ζ, Xζ , Yζ + i∈N



t

t

i∈N

  Z ζ     f r, Xr dr . = Et 1{τ >ζ} V ζ, Xζ , Yζ +

(6.64)

t

h Rζ i   It follows that Et R(t, x, τ ) ≤ Et 1{τ ≤ζ} R(t, x, τ )+1{τ >ζ} V(ζ, Xζ , Yζ )+ t f (r, Xr )dr .

3) Now, we demonstrate the second inequality in (6.55). Fix ε ∈ (0, 1) and let i ∈ N, x ∈ Rl . In light of (4.5) and Theorem 2.1 (1), there exists δi (x) ∈ (0, ε/2) such that p 1 C(Cp ) p δi (x)+CCp δi (x) < ε/4, (6.65)

and that for any y ∈ [0, ∞), V(ti , x′ , y′ )−V(ti , x, y) ≤ ε/4,

  ∀ (x′ , y′ ) ∈ O δi (x) (x)× (y−δi (x))+ , y+δi (x) .

(6.66)

Then (g1), H¨ older’s inequality and the second inequality in (3.4) imply that Z ∞  Z ς  t ,x t ,x′ t ,x t ,x′ p    ti ,x′ ti ,x i i i i dr ≤ Eti Eti −g r, Xr c(r) Xr −Xr + Xr −Xr g r, Xr dr ti ti Z ∞ i  h 1 ′ p ′ ≤ c(r)dr Eti X ti ,x−X ti ,x ∗ + X ti ,x−X ti ,x ∗ ≤ C(Cp ) p |x−x′ |+CCp |x−x′ |p 0

p 1 ti (6.67) ≤ C(Cp ) p δi (x)+CCp δi (x) < ε/4, ∀ ς ∈ T , ∀ x′ ∈ O δi (x) (x) .  i i We can find a sequence (xn , yn ) n∈N in Rl ×[0, ∞) such that Rl ×[0, ∞) = ∪ Oni ×Dni with Oni := Oδi (xin ) (xin ) n∈N ( + i  i i i i y −δi (xn ) , yn +δi (xn ) , if yn > 0, and Dni :=  n  i 0, δi (xn ) , if yni = 0.  /  t t i c ∈ F ti . There ∪ A Let n ∈ N. We set Ain := {τ > ζ = ti }∩ (Xti , Yti ) ∈ Oni ×Dni ∩Nt,x,τ ∈ F ti and Ain := Ain ′ n ′ ζ = ti }∩ (Xti , Yti ) ∈ ∪ Oni ×Dni n∈N i,n∈N i∈N   c c = ∪ {τ > ζ = ti } ∩Nt,x,τ = {τ > ζ}∩Nt,x,τ .

∪ Ain =

i,n∈N

∪ Ain satisfies that

   Z ti ,Xti (ω) g r, Xr dr +

∪ Ain = i∈N



(6.71)

DPPs for Optimal Stopping with Expectation Constraint

26

We claim that τ := 1{τ ≤ζ} τ +

X

i,n∈N

1Ain τni (Πtti )+1{τ >ζ}∩Nt,x,τ t belongs to Txt (y +ε).

(6.72*)

c Let i, n ∈ N and ω ∈ Ain ∩ N i,n . As Xti (ω) ∈ Oni = Oδi (xin ) (xin ), (6.65) and the second inequality in (4.1) imply  h i p  1 Eti R(ti , Xti (ω), τni )−R(ti , xin , τni ) ≤ 2C (Cp ) p δi (xin )+Cp δi (xin ) < ε/2.

(6.73)

 Since Xti (ω)−xin ∨ Yti (ω)−yni < δi (xin ), applying (6.66) with (x, y) = (xin , yni ) and (x′ , y′ ) = Xti (ω), Yti (ω) , we can deduce from (6.69), (6.73) and (6.68) that h

Et R

Z    t i F t (ω) > Eti R(ti , xin , τni ) + i

t, x, τni (Πtti )

ti

  f r, Xr (ω) dr−ε/2 ≥ V ti , xin , yni + t Z ti  f (r, Xr (ω))dr−ε. ≥ V ti , Xti (ω), Yti (ω) +

Z

ti

t

 3 f r, Xr (ω) dr− ε 4

t

Taking expectation Et [·] over Ain yields that

  h h  t i i   Et 1Ain R(t, x, τ ) = Et 1Ain R t, x, τni (Πtti ) = Et 1Ain Et R t, x, τni (Πtti ) F ti 

h



≥ Et 1Ain V ζ, Xζ , Yζ +

Z

ζ

f (r, Xr )dr−ε

t



.

Similar to (6.64), taking summation up over i, n ∈ N, we can deduce from (6.71), (6.63), the first inequality in (4.1) and the dominated convergence theorem that  Z     Et 1{τ >ζ} R(t, x, τ ) ≥ Et 1{τ >ζ} V ζ, Xζ , Yζ +

ζ

f (r, Xr )dr−ε

t



.

h Rζ i   It thus follows that V(t, x, y +ε) ≥ Et R(t, x, τ ) ≥ Et 1{τ ≤ζ} R(t, x, τ )+1{τ >ζ} V(ζ, Xζ , Yζ )+ t f (r, Xr )dr −ε. As  ε → ∞, the second inequality in (6.55) follows from the continuity of V in y i.e. (4.5) and Theorem 2.1 (1) . 

Proof of Theorem 4.1: Fix t ∈ [0, ∞). t 1) Let (x, y) ∈ Rl ×[0, ∞) and let {ζ(τ )}τ ∈Txt (y) be a family of T ♯ −stopping times. For any τ ∈ Txt (y), taking ζ = ζ(τ ) in (6.55) yields that  Z   t,x t,x,τ Et R(t, x, τ ) ≤ Et 1{τ ≤ζ(τ )}R(t, x, τ )+1{τ >ζ(τ )} V(ζ(τ ), Xζ(τ ) , Yζ(τ ) )+ 

t

ζ(τ )

f (r, Xrt,x )dr



≤ V(t, x, y).

 Taking supremum over τ ∈ Txt (y) or taking supremum over τb ∈ Txt (y) if y > 0 , we can deduce (1.3) from (4.4). 2) Next, assume that V(s, x, y) is continuous in (s, x, y) ∈ [t, ∞)×Rl ×(0, ∞). t We fix (x, y) ∈ Rl ×[0, ∞) and a family {ζ(τ )}τ ∈Txt (y) of T −stopping times. Let τ ∈ Txt (y), n ∈ N and define ζn = ζn (τ ) := 1{ζ(τ )=t} t+

X i∈N

t

1{ζ(τ )∈(t+(i−1)2−n,t+i2−n ]} (t+i2−n ) ∈ T .

Applying (6.55) with ζ = ζn yields that  Z    t,x,τ  + , Y Et R(t, x, τ ) ≤ Et 1{τ ≤ζn } R(t, x, τ )+1{τ >ζn } V ζn , Xζt,x ζn n

t

ζn

f (r, Xrt,x )dr



≤ V(t, x, y).

(6.74)

An analogy to (6.63) shows that

 V ζn , X t,x , Y t,x,τ + ζn ζn

Z

t

ζn

 f (r, Xrt,x ) dr ≤ 2C(3+Cp )+C(1+2Cp)Xp∗ ∈ L1 F t .

(6.75)

6.3

Proof of Section 4

27

 t,x,τ t t,x,τ We claim that Yζ(τ ) > 0, Pt −a.s. on {τ > ζ(τ )}. To see it, we set A := {τ > ζ(τ )}∩ Yζ(τ ) = 0 ∈ F ζ(τ ) and can deduce that   hZ τ i Z ζ(τ )  h i t t,x,τ t,x t,x − g(r, X )dr 0 = Et 1A Yζ(τ = E 1 E g(r, X )dr F t A t ζ(τ ) r r ) t t  h Z τ  Z τ h i i t g(r, Xrt,x )dr F ζ(τ ) = Et 1A = Et Et 1A g(r, Xrt,x )dr , ζ(τ )

ζ(τ )



which implies that 1A ζ(τ ) g(r, Xrt,x )dr = 0, Pt −a.s. It follows from the strict positivity of function g that Pt (A) = 0, proving the claim. As lim ↓ ζn = ζ(τ ), one has lim ↓ 1{τ ≤ζn } = 1{τ ≤ζ(τ )}. The continuity of function V in (s, x, y) ∈ n→N n→N   , Yζt,x,τ [t, ∞) × Rl × (0, ∞) and the continuity of processes X t,x , Y t,x,τ then show that lim 1{τ ≤ζn } V ζn , Xζt,x = n n n→N  t,x t,x,τ 1{τ ≤ζ(τ )}V ζ(τ ), Xζ(τ ) , Yζ(τ ) , Pt −a.s. Letting n → ∞ in (6.74), we can deduce from (6.75), the first inequality in (4.1) and the dominated convergence theorem that  Z ζ(τ )     t,x t,x,τ  t,x ≤ V(t, x, y). (6.76) , Y + f (r, X )dr Et R(t, x, τ ) ≤ Et 1{τ ≤ζ(τ )}R(t, x, τ )+1{τ >ζ(τ )} V ζ(τ ), Xζ(τ r ) ζ(τ ) t

 Taking supremum over τ ∈ Txt (y) or taking supremum over τb ∈ Txt (y) if y > 0 , we obtain (1.3) again from (4.4). 

Proof of Proposition 4.1: Let us simply denote τ (t, x, α) by τo . For n ∈ N, an analogy to (4.9) shows that   τ n := inf s ∈ [t, ∞) : Yst,x,α = 1/n and τ n := inf s ∈ [t, ∞) : Yst,x,α = −1/n t

define two T −stopping times. By definition, α = M −K for some (M, K) ∈ Mt ×Kt . It holds for all ω ∈ Ωt except on a Pt −null set N that M· (ω) is a continuous path, that K· (ω) is an continuous increasing path and that τ n (ω) < ∞ for any n ∈ N. 1) We first show that lim ↑ τ n = lim ↓ τ n = τo

n→∞

n→∞

Pt −a.s.

(6.77)

 Let ω ∈ N c and set τ (ω) := lim ↑ τ n (ω) ≤ τo (ω). The continuity of path Y·t,x,α (ω) implies that Y t,x,α τ n (ω), ω = n→∞   1/n, ∀ n ∈ N and thus Y t,x,α τ (ω), ω = lim Y t,x,α τ n (ω), ω = 0. It follows that τo (ω) = τ (ω) = lim ↑ τ n (ω). n→∞ t

n→∞

On the other hand, we define a T −stopping time τ := lim ↓ τ n ≥ τo and let ω ∈ N c . For any n ∈ N, as n→∞  τ n (ω) < ∞, the continuity of path Y·t,x,α (ω) again gives that Y t,x,α τ n (ω), ω = −1/n. Letting n → ∞ yields that   Y t,x,α τ (ω), ω = lim Y t,x,α τ n (ω), ω = 0. n→∞ Since M is a uniformly integrable martingale, we know from the optional sampling theorem that   Z τ h i t,x g(r, Xr )dr = Et Mτ −Mτo −Yτt,x,α +Yτt,x,α = 0, Et Kτ −Kτo + o τo



which implies Kτ −Kτo + τo g(r, Xrt,x )dr = 0, Pt −a.s. Then one can deduce from the strict positivity of function g that τo = τ = lim ↓ τ n , Pt −a.s., proving (6.77). n→∞   2) Next, let ε ∈ (0, 1) and set εo := (4+10C)−1 ε. As M := Et (X∗t,x )p < ∞ by the first inequality in (3.4), we can find  λo = λo (t, x, ε) ∈ 0, εo such that   t Et 1A (X∗t,x )p < εo for any A ∈ F with Pt (A) < λo .

(6.78)

 t There exists R = R(t, x, ε) ∈ (0, ∞) such that the set AR := X∗t,x > R ∈ F satisfies Pt (AR ) < λo /2. Let λ = λ(t, x, ε) ∈ (0, 1) satisfy that εo ∧ ρ−1 (εo ) and (2+M)kc(·)k   p p 1 1 1+|x| kc(·)kλ+kc(·)k 2 λ 2 +Cp 1+|x|p kc(·)kp λp +kc(·)k 2 λ 2 ≤ εo . λ≤

1

(Cp ) p

(6.79) (6.80)

DPPs for Optimal Stopping with Expectation Constraint  1 We pick up δ = δ(t, x, ε) ∈ 0, Cn

λo 2Cp

 p1 

28

such that 1

C(Cp ) p δ+CCp δ p ≤ λ∧εo ,

(6.81)

t

Set Ωn := {τo −λ ≤ τ n ≤ τ n ≤ τo +λ} ∈ F for any n ∈ N. As (6.77) implies that Pt

 ∪ Ωn = 1, there exists n ∈ N such

n∈N

that Pt (Ωn ) > 1−λo /2.  ′ t Now, fix x′ ∈ O δ (x) and simply denote τ (t, x′ , α) by τ ′ . We define A′ := (X t,x − X t,x )∗ ≤ (Cn)−1 ∈ F . The second inequality in (3.4) shows that    ′ Pt (A′ )c = Cp np Et (X t,x −X t,x )p∗ ≤ Cp Cp np |x′ −x|p ≤ Cp Cp np δ p < λo /2.

   t So the set A := A′ ∩Ωn ∈ F satisfies that Pt (Ac ) = Pt (A′ )c ∪Ωcn ≤ Pt (A′ )c +Pt Ωcn < λo < εo . Let ω ∈ A. Since it holds for any s ∈ [t, ∞) that Z s Z s  p  t,x′ ,α   ′ ′ Ys g r, Xrt,x′ (ω) −g r, Xrt,x (ω) dr ≤ (ω)−Yst,x,α (ω) ≤ c(r) Xrt,x −Xrt,x ∨ Xrt,x −Xrt,x (ω)dr t t Z −1 ∞ ≤ Cn c(r)dr ≤ 1/n, t

we see that

    ′ ′ Yst,x ,α (ω) ≥ Yst,x,α (ω)−1/n > 0, ∀ s ∈ t, τ n (ω) and Yst,x,α (ω) ≥ Yst,x ,α (ω)−1/n > −1/n, ∀ s ∈ t, τ ′ (ω) .

The former implies that τ ′ (ω) ≥ τ n (ω) while the latter means that τ n (ω) ≥ τ ′ (ω). In summary, τo −λ ≤ τ n ≤ τ ′ ≤ τ n ≤ τo +λ on A.

(6.82)

By an analogy to (6.24) and (6.25), we can deduce from (6.9), (6.10), (6.78), (6.82) and (6.79) that   Z ∞   f (r, Xrt,x ) dr ≤ Et 2+(X∗t,x )p 1Ac c(r)dr+1A kc(·)k|τ ′ −τo | t τo ∧τ ′  c < C 2Pt (A )+εo +λ(2+M)kc(·)k < (1+3C)εo ,       t,x  ′ t,x and Et 1Ac π τ , Xτ ′ −π τo , Xτo ≤ 2CEt 1Ac 2+(X∗t,x)p < 2C 2Pt (Ac )+εo < 6Cεo . Et

Z

τo ∨τ ′

And similar to (6.26), H¨ older’s inequality, (1.9), (6.79), (6.82), (3.5) and (6.80) imply that h  i       t,x p t,x t,x ≤ Et 1A ρ |τ ′ −τo | +CEt 1A X t,x −π τo , Xτt,x Et 1A π τ ′ , Xτt,x + X ′ ′ −Xτ ′ −Xτ τ τ o o o  p1     t,x t,x t,x p t,x p ≤ ρ(λ)+C Et 1A sup Xτ ′ ∧τo +r−Xτ ′ ∧τo +CEt 1A sup Xτ ′ ∧τo +r−Xτ ′ ∧τo r∈(0,λ]

≤ εo +C(Cp )

1 p



1 2

1+|x| kc(·)kλ+kc(·)k λ

Combining (6.83), (6.84) and (6.85) yields that   Et R(t, x, τ ′ )−R(t, x, τo ) ≤ Et

Z

τo ∨τ ′ τo ∧τ ′

1 2



+CCp 1+|x|

p



(6.83) (6.84)

r∈(0,λ]

p p kc(·)kp λp +kc(·)k 2 λ 2 ≤ (1+C)εo .

(6.85)

  t,x  ′ |f (r, Xrt,x )|dr+ π τo , Xτt,x −π τ , X τ ′ < (2+10C)εo , o

which together with (2.8) and (6.81) leads to that       Et R(t, x′ , τ ′ )−R(t, x, τo ) ≤ Et R(t, x′ , τ ′ )−R(t, x, τ ′ ) +Et R(t, x, τ ′ )−R(t, x, τo ) < (4+10C)εo = ε .



Proof of Proposition 4.2: Let (t, x, y) ∈ [0, ∞)×Rl ×(0, ∞). 1) Let α ∈ At (y). Since τ (t, x, α) < ∞, Pt −a.s. by (4.9), the continuity of process Y t,x,α implies that ατ (t,x,α) =

Z

t

τ (t,x,α)

g(r, Xrt,x )dr,

Pt −a.s.

(6.86)

6.3

Proof of Section 4

29

 t , P −supermartingale α and the optional sampling One can then deduce from the uniform integrability of the F t hR i   τ (t,x,α) t,x g(r, Xr )dr = Et ατ (t,x,α) ≤ Et [αt ] = y, namely, τ (t, x, α) ∈ Txt (y). As Ytt,x,α = αt = y > 0, theorem that Et t

Pt −a.s., we also derive from the continuity of process Y t,x,α that τ (t, x, α) > t, Pt −a.s. Thus α → τ (t, x, α) is a mapping from At (y) to Tbxt (y). i hR  Rτ τ t 2) Next, let τ ∈ Tbxt (y) and set δ := y−Et t g(r, Xrt,x )dr ≥ 0. Clearly, Ms := δ+Et t g(r, Xrt,x )dr F s ≥ 0, s ∈ [t, ∞)  t is a uniformly integrable continuous martingale with respect to F , Pt , i.e., M ∈ Mt .   t Define Js := ′ inf Et τ − t|F s′ , s ∈ [t, ∞) and let N be the P −null set such that for any ω ∈ N c , the path s ∈[t,s]   t t Et τ − t|F · (ω) is continuous and Et τ − t|F s (ω) ≥ 0, ∀ s ∈ [t, s] ∩ Q. For any ω ∈ N c , we can deduce that  t  J· (ω) is a nonnegative, continuous decreasing process. Given s ∈ [t, ∞), set ξs := ′ inf Et τ − t|F s′ , which is s ∈[t,s]∩Q  t t F s −measurable random variable. The continuity of process Et τ −t|F s , s ∈ [t, ∞) shows that Js = ξs on N c , so Js t

is also F s −measurable. It follows that

   s−t + Ks := δ 1∧ ∈ [0, δ], Js

s ∈ [t, ∞)

(6.87)

 t t is an F −adapted continuous increasing process. Since τ > t, Pt −a.s., one has Jt = Et τ − t|F t = Et [τ − t] > 0, Pt −a.s. and thus Kt = 0, Pt −a.s. To wit, K ∈ Kt . Set α := M −K. It is clear that Z τ Z τ   t t t,x t,x αs = Ms −Ks ≥ δ+Et g(r, Xr )dr F s −δ = Et g(r, Xr )dr F s , ∀ s ∈ [t, ∞). t

t

 t As αt = Mt −Kt = δ +Et t g(r, Xrt,x )dr +0 = y, Pt −a.s., we see that α ∈ At (y). Since Jτ ≤ Et (τ −t)|F τ = τ −t, Pt −a.s., one has Kτ = δ, Pt −a.s. and thus Z τ g(r, Xrt,x )dr, Pt −a.s. (6.88) ατ = Mτ − δ = Rτ



t

This shows τ (t, x, α) ≤ τ , Pt −a.s. On the other hand, subtracting (6.86) from (6.88) and applying the optional   Rτ  sampling theorem to α again yield that 0 ≤ Et τ (t,x,α) g(r, Xrt,x )dr = Et ατ −ατ (t,x,α) ≤ 0. The strict positivity of function g then implies that τ (t, x, α) = τ , Pt −a.s.  Similar to Lemma 6.1, the following auxiliary result is crucial for proving the second DPP of V (Theorem 4.2). t

Lemma 6.2. Given (t, x, y) ∈ [0, ∞)×Rl ×(0, ∞), let α ∈ At (y) and let ζ ∈ T ♯ . Then  Z ζ      t,x t,x,α  t,x + f (r, Xr )dr Et R t, x, τ (t, x, α) ≤ Et 1{τ (t,x,α)≤ζ}R t, x, τ (t, x, α) +1{τ (t,x,α)>ζ} V ζ, Xζ , Yζ t

≤ V(t, x, y).

(6.89)  Proof: Suppose that α = M −K for some (M, K) ∈ Mt ×Kt . We denote (X, Y, τb) := X t,x , Y t,x,α , τ (t, x, α) and let ζ take values in a countable subset {ti }i∈N of [t, ∞). 1) Let us start with the first inequality in (6.89).  R τb t Since α is a uniformly integrable continuous supermartingales with respect to F , Pt , one has ατb = t (r, Xr )dr and the optional sampling theorem implies that Z τb∧ζ Z τb∧ζ  t  Yτb∧ζ = ατb∧ζ − g(r, Xr )dr ≥ Et ατb F τb∧ζ − g(r, Xr )dr = Et

hZ

t

t

τb

i Z t g(r, Xr )dr F τb∧ζ −

t

t

τb∧ζ

τ g(r, Xr )dr = Yτbt,x,b ∧ζ ,

As τb ∈ Tbxt (y) by Proposition 4.2, we see from (6.55) that  Z ζ    Et 1{bτ ≤ζ} R t, x, τb +1{bτ >ζ} V(ζ, Xζ , Yζ )+ f (r, Xr )dr 



t

 ≥ Et 1{bτ ≤ζ} R t, x, τb +1{bτ >ζ} V(ζ, Xζ , Yζt,x,bτ )+

Z

t

ζ

f (r, Xr )dr

Pt −a.s.



  ≥ Et R(t, x, τb) ,

(6.90)

DPPs for Optimal Stopping with Expectation Constraint

30

proving the first inequality in (6.89). 2) The proof of the second inequality in (6.89) is relatively lengthy, we split it into several steps. By an analogy to (6.12), we must have either Pt {ζ = t} = 1 or Pt {ζ > t} = 1. If Pt {ζ = t} = 1, as Yt = αt = y > 0, Pt −a.s., one has τb = τ (t, x, α) > t = ζ, Pt −a.s. Then  Z ζ         f (r, Xr )dr = Et V(t, Xt , Yt ) = Et V(t, x, y) = V(t, x, y). Et 1{bτ ≤ζ} R t, x, τb +1{bτ >ζ} V ζ, Xζ , Yζ + t

So let us suppose that t1 > t in the rest of this proof. There exists a Pt −null set N such that for any ω ∈ N c , M· (ω) is a continuous path and K· (ω) is an continuous increasing path. By the uniform integrability of M , there t exists ξ ∈ L1 F such that Pt −a.s.  t Ms = Et ξ F s ,

∀ s ∈ [t, ∞).

(6.91)

For any i ∈ N, similar to (6.56) and (6.57), it holds for all ω ∈ Ωt except on a Pt −null set Ni that o n ti ,Xti (ω) e ) 6= Xr e ) 6= Xs (ω) for some s ∈ [t, ti ] or Xr (ω ⊗ti ω (e ω ) for some r ∈ [ti , ∞) ∈ N Nωi := ω e ∈ Ωti : Xs (ω ⊗ti ω

ti

. (6.92)

2a) Fix ε ∈ (0, 1). The first inequality in (4.1) and an analogy to (6.63) show that

Z ζ h i   f (r, Xr ) dr ≤ Ψ(x)+2C(3+Cp)+C(1+2Cp)Et Xp∗ < ∞. Et R(t, x, τb) + V(ζ, Xζ , Yζ ) + t

So there exists λ = λ(t, x, α, ε) ∈ (0, 1) such that    Z ζ  t f (r, Xr ) dr < ε/5 for any A ∈ F with Pt (A) < λ . Et 1A R(t, x, τb) + V(ζ, Xζ , Yζ ) +

(6.93)

t

We can find Io ∈ N such that Pt {ζ > tIo } < λ/2.  Let i = 1, · · · , Io and (x, y) ∈ Rl ×(0, ∞). In light of (4.5) and Theorem 2.1 (1), there exists δi (x, y) ∈ 0, 1∧y∧ε such that   V(ti , x′ , y′ )−V(ti , x, y) ≤ ε/5, ∀ (x′ , y′ ) ∈ O δ (x,y) (x)× y−δi (x, y), y+δi (x, y) . (6.94) i  By (4.10), there exists α(ti , x, y) ∈ Ati y−δi (x, y) such that h h i  i (6.95) V ti , x, y−δi (x, y) = sup e) ≤ Eti R ti , x, τ (ti , x, α(ti , x, y)) +ε/5, Eti R ti , x, τ (ti , x, α α e∈Ati (y−δi (x,y))

 and Proposition 4.1 shows that for some δbi (x, y) ∈ 0, δi (x, y) h   i Eti R ti , x′ , τ (ti , x′ , α(ti , x, y)) −R ti , x, τ (ti , x, α(ti , x, y)) ≤ ε/5,

∀ x′ ∈ O δbi (x,y) (x).

(6.96)

 Let us simply write Oi (x, y) for the open set Oδbi (x,y) (x)× y− δbi (x, y), y+ δbi (x, y) .

τ > ti }, one has Since (4.8) implies that Yti (ω) > 0 for any ω ∈ {b     τ > ti }∩ (Xti , Yti ) ∈ O R (0)×[R−1 , R] . τ > ti }∩ (Xti , Yti ) ∈ Rl ×(0, ∞) = lim ↑ Pt {b Pt {b τ > ti } = Pt {b R→∞

So there exists Ri ∈ (0, ∞) such that

  λ / O Ri (0)×[Ri−1 , Ri ] ≤ i+1 , τ > ti }∩ (Xti , Yti ) ∈ Pt {b (6.97) 2  ni ni and we can find a finite subset (xin , yni ) n=1 of O Ri (0)×[Ri−1 , Ri ] such that ∪ Oi (xin , yni ) ⊃ O Ri (0)×[Ri−1 , Ri ]. n=1  t Let n = 1, · · · , ni and define Ain := {b τ > ζ = ti }∩ (Xti , Yti ) ∈ Oi (xin , yni ) ∈ F ti . Clearly,   Yti (ω)−yni ∈ − δbi (xin , yni ), δbi (xin , yni ) ⊂ − δi (xin , yni ), δi (xin , yni ) ,

∀ ω ∈ Ain .

6.3

Proof of Section 4 31  /  t t i ∈ F ti and define a F ti −measurable random variable ηni := 1Ain Yti−yni +δi (xin , yni ) ∈ ∪ A We also set Ain := Ain ′ n ′ 0. ω ) = Mti,n αi,n ti (e i

(6.99)

As (Πtti )−1 (N i,n ) is a Pt −null set by Lemma A.3 (1), one can deduce from Lemma A.3 (2) that t

Msi,n (Πtti ), s ∈ [ti , ∞) is an F −adapted continuous process with Mti,n (Πtti ) = yni −δi (xin , yni ), Pt −a.s. and i

(6.100)

t

(Πtti ) = 0, Pt −a.s. Ksi,n (Πtti ), s ∈ [ti , ∞) is an F −adapted, continuous increasing process with Kti,n i

Rs  t t An analogy to (4.8) and (4.9) shows that νni := inf s ∈ [ti , ∞) : αi,n s (Πti )− ti g(r, Xr )dr = 0 defines a T −stopping   i i t t time. Since αi,n ti (Πti ) > 0, Pt −a.s. by (6.99), we see that νn > ti , Pt −a.s. and thus Et νn −ti F ti > 0, Pt −a.s.   t t Similar to the proof of Proposition 4.2, Jsi,n := ′ inf Et νni −ti F s′ , s ∈ [ti , ∞) is an F −adapted, non-negative, s ∈[ti ,s] h t i t  i i,n i continuous decreasing process such that Jti = Et νn −ti F ti > 0, Pt −a.s. and that Jνi,n = νni −ti , i ≤ Et νn −ti F ν i n n i h  t + i ≥ 0, s ∈ [ti , ∞) defines an F −adapted, continuous increasing process over period Pt −a.s. Then Ksi,n := ηni 1∧ s−t J i,n s

[ti , ∞) such that

=0 Kti,n i Io

Set A♯ := ∪

ni

i,n i and Kνi,n i = ηn holds except on a Pt −null set NK .

(6.101)

n

t

∪ Ain ∈ F tIo and N♯ := N ∪

i=1 n=1



Io



 ∪ (Πtti )−1 (N i,n ) ∈ N t . We claim that

ni

i=1 n=1

 M s := Ms + i=1 n=1 1{s≥ti }∩Ain Msi,n (Πtti )−Ms +Mti −yni +δi (xin , yni ) , s ∈ [t, ∞) is of Mt ,  PIo Pni i,n i,n t and K s := Ks + i=1 , s ∈ [t, ∞) is of Kt . n=1 1{s≥ti }∩Ain Ks (Πti )−Ks +Kti +Ks PIo Pni

(6.102*)

As t1 > t by assumption, it holds Pt −a.s. that M t = Mt = y. So α := M −K ∈ At (y).  2b) Setting τ := τ (t, x, α), we next show that τ = τb, Pt −a.s. on {b τ ≤ ζ}∪ {b τ > ζ}∩Ac♯ . Since (6.102) shows that    M s (ω), K s (ω) = Ms (ω), Ks (ω) , ∀ (s, ω) ∈ [t, ∞)×Ac♯ ∪[[t, ζ[[ ,

(6.103)

we obtain that

   (6.104) ∀ (s, ω) ∈ [t, ∞)×Ac♯ ∪ t, ζ . n o n  Rs  So for any ω ∈ Ac♯ , one has τb(ω) = τ (t, x, α) (ω) = inf s ∈ [t, ∞) : αs (ω)− t g r, Xr (ω) dr = 0 = inf s ∈ [t, ∞) : o   Rs αs (ω)− t g r, Xr (ω) dr = 0 = τ (t, x, α) (ω) = τ (ω). Let ω ∈ {b τ ≤ ζ ≤ Io }∩N♯c . By (6.104), αs (ω) = αs (ω),

α(s, ω) = α(s, ω),

∀ s ∈ [t, ζ(ω)).

If τb(ω) < ζ(ω), one can deduce from (6.105) that Z s Z s n o n o   τb(ω) = inf s ∈ [t, ∞) : αs (ω)− g r, Xr (ω) dr = 0 = inf s ∈ [t, ζ(ω)) : αs (ω)− g r, Xr (ω) dr = 0 t t Z s n o  = inf s ∈ [t, ζ(ω)) : αs (ω)− g r, Xr (ω) dr = 0 , t

(6.105)

DPPs for Optimal Stopping with Expectation Constraint

32

 Rs  which implies that τ (ω) = inf s ∈ [t, ∞) : αs (ω)− t g r, Xr (ω) dr = 0 = τb(ω). Otherwise, suppose that τb(ω) = ζ(ω). The definition of τ (t, x, α) and (6.105) show that α(s, ω) = α(s, ω) >

Z

s

t

 g r, Xr (ω) dr, ∀ s ∈ [t, ζ(ω))

 and α ζ(ω), ω =

Z

ζ(ω)

t

 g r, Xr (ω) dr.

(6.106)

As M · (ω), M· (ω), K · (ω), K· (ω) are all continuous paths by the proof of (6.102), we see from (6.103) and (6.106) that      α ζ(ω), ω = M −K ζ(ω), ω = (M −K) ζ(ω), ω = α ζ(ω), ω =

Z

ζ(ω)

t

 which means that τ (ω) = τ (t, x, α) (ω) = ζ(ω) = τb(ω). Hence, we have verified that

 g r, Xr (ω) dr,

 τ = τb, Pt −a.s. on Ac♯ ∪{b τ ≤ ζ ≤ Io } = {b τ ≤ ζ}∪ {b τ > ζ}∩Ac♯ .

(6.107)

2c) Let i = 1, · · · , Io and n ∈ 1, · · · , ni . In this step, we demonstrate that h     Et 1Ain R(t, x, τ ) ≥ Et 1Ain V ti , Xti , Yti + b i,n := {ω ∈ Ωt : νni (ω) = ∞} ∈ N Set N definition of νni shows that t αi,n s (Πti (ω)) >

Z

s ti

 g r, Xr (ω) dr,

t

Z

ti

f (r, Xr )dr−4ε/5

t

i .

 b i,n ∪N i,n ∪(Πtt )−1 (N i,n ) c ∈ F tt . Let ω ∈ Gin . The and Gin := Ain ∩ Ni ∪ N K i i

   ∀ s ∈ ti , νni (ω) and αi,n νni (ω), Πtti (ω) =

Z

i νn (ω)

ti

 g r, Xr (ω) dr.

(6.108)

Since ω ∈ Ain ⊂ {b τ > ti } and since

Z   t (ω) + Π αs (ω) = 1{s Πtti (ω) +

αs (ω) = αs (ω) >

t



νni (ω), Πtti (ω)

+

Z

t

t

ti

 g r, Xr (ω) dr =

 τ (ω) = τ (t, x, α) (ω) = νni (ω),

Z

t

∀ ω ∈ Gin .

i νn (ω)

  ∀ s ∈ ti , νni (ω) ,

 g r, Xr (ω) dr,

(6.109)

e in ∈ N t . By Proposition 3.3 (1), it ein ∈ Ftt such that N i,n := Gin ∆G Similar to Problem 2.7.3 of [32], there exists G G i ti ,ω i,n i,n t t b holds for all ω ∈ Ω except on a Pt −null set N ∈N i. G that NG c c ti ,ω c i,n i,n i i t ,ω i ei and ω e ∩ N b e ∈ (NGi,n )ti ,ω = (NGi,n )c , Lemma Now, let ω ∈ Gn ∩G and ω e ∈ (NG ) i ∪ Nω . As ω ∈ G n n G i,n c i i i i i e e e e , we see = Gn ∩ Gn ⊂ Gn . Applying (6.109) with ω = ω⊗ti ω e ∈ Gn ∩ N e ∈ Gn and thus ω⊗ti ω 3.1 shows that ω⊗ti ω G

from (6.92) that

Z s   e ) dr = 0 g r, Xr (ω ⊗ti ω e ) = inf s ∈ [ti , ∞) : αi,n (e ω )− e ) = νni (ω ⊗ti ω τ (ω ⊗ti ω s ti Z s   o n  ti ,Xti (ω) i,n ω) =: τωi,n (e ω ). (e ω ) dr = 0 = τ ti , Xti (ω), αi,n (e = inf s ∈ [ti , ∞) : αs (e ω )− g r, Xr ti

6.3

Proof of Section 4

33

Then (6.92) again shows that ti ,ω  e) = (e ω ) = R(t, x, τ ) (ω ⊗ti ω R(t, x, τ ) =

Z

ti

t

=

Z

ti

t

 f r, Xr (ω) dr+

Z

Z

τ (ω⊗ti ω e)

t i,n τω (e ω)

ti

   e ) dr+π τ (ω ⊗ti ω f r, Xr (ω ⊗ti ω e e ), X τ (ω ⊗ti ω e ), ω ⊗ti ω

    ti ,Xti (ω) f r, Xr (e ω ) dr+π τωi,n (e ω ), X ti ,Xti (ω) τωi,n (e ω ), ω e

  ω ). f r, Xr (ω) dr+ R(ti , Xti (ω), τωi,n ) (e

e ∈ Ωti except the Pti −null set (NGi,n )ti ,ω ∪ Nωi yields that Taking expectation Eti [·] over ω Eti

h

  ti ,ω i R(t, x, τ ) = Eti R ti , Xti (ω), τ (ti , Xti (ω), αi,n ) +

Z

ti

t

 f r, Xr (ω) dr.

   Since Xti (ω), Yti (ω) ∈ Oδbi (xi ,yi ) (xin )× yni −δbi (xin , yni ), yni +δbi (xin , yni ) , using (6.96) with (x, y, x′ ) = xin , yni , Xti (ω) and n n   applying (6.94) with (x, y, x′ , y′ ) = xin , yni , xin , yni − δi (xin , yni ) and (x, y, x′ , y′ ) = xin , yni , Xti (ω), Yti (ω) respectively, we can deduce from (6.95) that Z ti h h  i ti ,ω i Eti R(t, x, τ ) f (r, Xr (ω))dr ≥ Eti R ti , xin , τ (ti , xin , αi,n ) −ε/5 ≥ V ti , xin , yni −δi (xin , yni ) −2ε/5 − t    ein ∩ N b i,n c . ≥ V ti , xin , yni −3ε/5 ≥ V ti , Xti (ω), Yti (ω) −4ε/5, ∀ ω ∈ Gin ∩ G (6.110) G h t ti ,ω i  for The first inequality in (4.1) and Proposition 3.4 (2) imply that Et R(t, x, τ ) F ti (ω) = Eti R(t, x, τ )

Pt −a.s. ω ∈ Ωt . As 1Gi ∩Gei = 1Gin 1Gei = 1Gin = 1Ain , Pt −a.s., we can derive from (6.110) that n

n

n

h h t i   i  Et 1Ain R(t, x, τ ) = Et 1Ain Et R(t, x, τ ) F ti = Et 1Gi ∩Gei Eti (R(t, x, τ ))ti ,ω n n   Z Z ti      f (r, Xr )dr−4ε/5 = Et 1Ain V ζ, Xζ , Yζ + ≥ Et 1Gi ∩Gei V ti , Xti , Yti + 

n

n

ζ

f (r, Xr )dr−4ε/5

t

t

Taking summation over n ∈ 1, · · · , ni and i = 1, · · · , Io and using the conclusion of Part 2 yield that  Z ζ      c Et R(t, x, τ ) ≥ Et 1{bτ ≤ζ}∪({bτ >ζ}∩A♯ ) R(t, x, τb)+1A♯ V ζ, Xζ , Yζ + f (r, Xr )dr −4ε/5.



.

(6.111)

t

o n ni ni ni τ > ζ = ti }∩ (Xti , Yti ) ∈ ∪ Oi (xin , yni ) for i = 1, · · · , Io , one can deduce that 2d) Since ∪ Ain = ∪ Ain = {b n=1

n=1

n=1

{b τ > ζ}∩Ac♯ = {b τ > ζ > Io }∪



Io



i=1



o n ni i i , / ∪ Oi (xn , yn ) {b τ > ζ = ti }∩ (Xti , Yti ) ∈ n=1

   PIo and (6.97) implies that Pt {b τ > ζ}∩Ac♯ ≤ Pt {ζ > Io }+ i=1 / O Ri (0)×[Ri−1 , Ri ] < λ. It Pt {b τ > ti }∩ (Xti , Yti ) ∈ then follows from (6.93) that  Z ζ    Et 1{bτ >ζ}∩Ac R(t, x, τb)−V ζ, Xζ , Yζ − f (r, Xr )dr ♯ t

 Z ζ   ≤ Et 1{bτ >ζ}∩Ac♯ R(t, x, τb) + V(ζ, Xζ , Yζ ) + f (r, Xr ) dr < ε/5, 

t

which together with (6.111) and (4.10) leads to that

 Z     V(t, x, y) ≥ Et R(t, x, τ ) ≥ Et 1{bτ ≤ζ} R(t, x, τb)+1{bτ >ζ} V ζ, Xζ , Yζ +

t

Letting ε → ∞ yields the second inequality in (6.89). Proof of Theorem 4.2: Fix t ∈ [0, ∞).

ζ

f (r, Xr )dr



−ε. 

DPPs for Optimal Stopping with Expectation Constraint

34 t

1) Let (x, y) ∈ Rl ×(0, ∞) and {ζ(α)}α∈At (y) be a family of T ♯ −stopping times. For any α ∈ At (y), taking ζ = ζ(α) in (6.89) yields that     Et R t, x, τ (t, x, α) ≤ Et 1{τ (t,x,α)≤ζ(α)}R t, x, τ (t, x, α) Z  t,x t,x,α  +1{τ (t,x,α)>ζ(α)} V ζ(α), Xζ(α) , Yζ(α) +

ζ(α)

t

f (r, Xrt,x )dr



≤ V(t, x, y).

Taking supremum over α ∈ At (y), we obtain (1.4) from (4.10). 2) Next, suppose that V(s, x, y) is continuous in (s, x, y) ∈ [t, ∞)×Rl ×(0, ∞). t We fix (x, y) ∈ Rl ×(0, ∞) and a family {ζ(α)}α∈At (y) of T −stopping times. Let α ∈ At (y), n ∈ N and define ζn = ζn (α) := 1{ζ(α)=t} t+

X i∈N

t

1{ζ(α)∈(t+(i−1)2−n ,t+i2−n ]} (t+i2−n) ∈ T .

Applying (6.89) with ζ = ζn yields that     Et R t, x, τ (t, x, α) ≤ Et 1{τ (t,x,α)≤ζn} R t, x, τ (t, x, α)

Z  t,x,α  , Y +1{τ (t,x,α)>ζn} V ζn , Xζt,x + ζn n

ζn

t

f (r, Xrt,x )dr



≤ V(t, x, y).

As n → ∞, using similar arguments to those that lead to (6.76) we can deduce from the continuity of function V in  (s, x, y) ∈ [t, ∞)×Rl ×(0, ∞), the continuity of processes X t,x , Y t,x,α , and the dominated convergence theorem that     Et R t, x, τ (t, x, α) ≤ Et 1{τ (t,x,α)≤ζ(α)}R t, x, τ (t, x, α)

Z  t,x,α  t,x +1{τ (t,x,α)>ζ(α)} V ζ(α), Xζ(α) , Yζ(α) +

t

ζ(α)

f (r, Xrt,x )dr



≤ V(t, x, y).

Taking supremum over α ∈ At (y) and using (4.10) yield (1.4) again.

6.4



Proof of Section 5

Proof of Theorem 5.1: Under (2.4) and (2.14), Theorem 2.1 (2) and (4.5) show that V is continuous in (t, x, y) ∈ [0, ∞)×Rl ×[0, ∞). By (4.3), V(t, x, 0) = π(t, x) for any (t, x) ∈ [0, ∞)×Rl . 1) We first show that V is a viscosity supersolution of (5.2).  Let (to , xo , yo ) ∈ (0, ∞)×Rl ×(0, ∞) and let φ ∈ C 1,2,2 [0, ∞)×Rl ×[0, ∞) such that V −φ attains a strict local   minimum 0 at (to , xo , yo ). So there exists a δo ∈ 0, to ∧yo such that for any (t, x, y) ∈ Oδo (to , xo , yo ) (to , xo , yo )

(V −φ)(t, x, y) > (V −φ)(to , xo , yo ) = 0 and Dx φ(t, x, y)−Dx φ(to , xo , yo ) ∨ ∂y φ(t, x, y)−∂y φ(to , xo , yo ) < 1. (6.112) According to (2.4) and (2.14), the functions b, σ, f, g are continuous in (t, x). Then φb (t, x, y) := −∂t φ(t, x, y)−Lx φ(t, x, y)+g(t, x)∂y φ(t, x, y)−f (t, x),

∀ (t, x, y) ∈ [0, ∞)×Rl ×[0, ∞)

is also a continuous function. To show φb (to , xo , yo )−Hφ(to , xo , yo ) ≥ 0, it suffices to verify that for any a ∈ Rd

T φb (to , xo , yo )− 21 |a|2 ∂y2 φ(to , xo , yo )− Dx (∂y φ(to , xo , yo )) ·σ(to , xo )·a ≥ 0.

Assume not, i.e. there exists an a ∈ Rd such that

T 1 ε := |a|2 ∂y2 φ(to , xo , yo )+ Dx (∂y φ(to , xo , yo )) ·σ(to , xo )·a− φb (to , xo , yo ) > 0. 2

(6.113)

6.4

Proof of Section 5

35

b we can find some δ ∈ (0, δo ) such that Using the continuity of σ, φ and φ,

T 1 1 2 2 |a| ∂y φ(t, x, y)+ Dx (∂y φ(t, x, y)) ·σ(t, x)·a− φb (t, x, y) ≥ ε > 0, 2 2

∀ (t, x, y) ∈ O δ (to , xo , yo ).

(6.114)

 to Clearly, Ms := yo +aT · Wsto , s ∈ [to , ∞) is a uniformly integrable continuous martingale with respect to F , Pto .  o to By taking K ≡ 0, we have αo := M ∈ Ato (yo ). As Θs := s, Xsto ,xo , Ysto ,xo ,α , s ∈ [to , ∞) are F −adapted continuous  to / O δ (to , xo , yo ) defines an F -stopping time with processes with Θto = (to , xo , yo ), Pt −a.s., ζ := inf s ∈ [to , ∞) : Θs ∈ to < ζ ≤ to +δ, Pto −a.s. Since Θs ∈ O δ (to , xo , yo ) on the stochastic interval [[to , ζ[[ ,

(6.115)

(6.114), (6.112), (1.7) and (2.4) imply that T 1 2 2 1 |a| ∂y φ(Θr )+ Dx (∂y φ(Θr )) ·σ(r, Xrto ,xo )·a− φb (Θr ) ≥ ε > 0 2 2 p p   to ,xo ) + ∂y φ(Θr ) |a| ≤ 1+|Dxφ(to , xo , yo )| |σ(to , xo )|+ kc(·)k δ+ kc(·)kρ(δ)(1+|xo |̟ ) and Dx φ(Θr ) σ(r, Xr  + 1+|∂y φ(to , xo , yo )| |a| < ∞ (6.116)

 holds on [[to , ζ[[ . Applying Itˆ o’s formula to process φ(Θs ) s∈[to ,∞) then yields that φ(Θζ )−φ(to , xo , yo ) =

Z ζ  T 1 ∂t φ(Θr )−g(r, Xrto ,xo )∂y φ(Θr )+Lx φ(Θr )+ |a|2 ∂y2 φ(Θr )+ Dx (∂y φ(Θr )) ·σ(r, Xrto ,xo )·a dr 2 to Z ζ  (Dx φ(Θr ))T ·σ(r, Xrto ,xo )+∂y φ(Θr )·aT dWrto , + to ζ

≥− Set m1 :=

Z

to

f (r, Xrto ,xo )dr+

min

(t,x,y)∈∂Oδ (to ,xo ,yo )

Z

ζ

to

 (Dx φ(Θr ))T ·σ(r, Xrto ,xo )+∂y φ(Θr )·aT dWrto ,

Pto − a.s.

(6.117)

(V −φ)(t, x, y) > 0 by (6.112). The continuity of process Θ and (6.115) show that

  o Pto Θζ ∈ ∂Oδ (to , xo , yo ) = Pto Ysto ,xo ,α ≥ yo −δ > 0, ∀ s ∈ [to , ζ] = 1,

(6.118)

the latter of which implies that

τ (to , xo , αo ) > ζ > to ,

Pto − a.s.

(6.119)

Taking expectation Eto [·] in (6.117) and applying Theorem 4.2 (2) with ζ(α) ≡ ζ, we can derive from (6.116), (6.118) that     Z ζ Z ζ f (r, Xrto ,xo )dr f (r, Xrto ,xo )dr +m1 ≤ Eto V(Θζ )+ φ(to , xo , yo )+m1 ≤ Eto φ(Θζ )+ t

t

 Z ζ    = Eto 1{τ (to ,xo ,αo )≤ζ} R to , xo , τ (to , xo , αo ) +1{τ (to,xo ,αo )>ζ} V(Θζ )+ f (r, Xrto ,xo )dr ≤

sup α∈Ato (yo )

Eto



t

   1{τ (to ,xo ,α)≤ζ} R to , xo , τ (to , xo , α) +1{τ (to ,xo ,α)>ζ} V ζ, Xζto ,xo , Yζto ,xo ,α +

= V(to , xo , yo ) = φ(to , xo , yo ).

(6.120) Z

t

ζ

f (r, Xrto ,xo )dr



A contradiction appears. We can also employ the first DPP (Theorem 4.1) to induce the incongruity: Denote τo := τ (to , xo , αo ). By the o continuity of process Y to ,xo ,α , Z τo  (6.121) g r, Xrto ,xo dr, Pto − a.s. yo +aT ·Wτtoo = to

DPPs for Optimal Stopping with Expectation Constraint

36

 R τo

  g r, Xrto ,xo dr = yo , which together with (6.119) shows τo ∈ Tbxtoo (yo ). On the other hand, taking conditional  to  expectation Eto · F ζ in (6.121), one can deduce from (6.119) and the optional sampling theorem that So Eto

to

o

Yζto ,xo ,α = yo +aT ·Wζto − = Eto

hZ

τo

Z

ζ

to

to    g r, Xrto ,xo dr = Eto yo +aT ·Wτtoo F ζ −

Z

ζ

to

Z ζ   to i to ,xo dr F ζ − g r, Xrto ,xo dr = Yζto ,xo ,τo , g r, Xr

 g r, Xrto ,xo dr

to

to

Pto − a.s.

Then we can apply Theorem 4.1 (2) with ζ(α) ≡ ζ to continue the deduction in (6.120)

 Z ζ    φ(to , xo , yo )+m1 ≤ Eto 1{τo ≤ζ} R to , xo , τo +1{τo >ζ} V(ζ, Xζto ,xo , Yζto ,xo ,τo )+ f (r, Xrto ,xo )dr ≤

sup τ ∈Tbxtoo (yo )

Eto



t

Z ζ   to ,xo ,τ  to ,xo to ,xo )dr + f (r, Xr , Yζ 1{τ ≤ζ} R to , xo , τo +1{τ >ζ} V ζ, Xζ

= V(to , xo , yo ) = φ(to , xo , yo ).



t

The contradiction recurs. Therefore, V is a viscosity supersolution of (5.2). 2) Next, we demonstrate that V is also a viscosity subsolution of (5.3).  Let (to , xo , yo ) ∈ (0, ∞)×Rl ×(0, ∞) and let ϕ ∈ C 1,2,2 [0, ∞)×Rl ×[0, ∞) such that V −ϕ attains a strict local   maximum 0 at (to , xo , yo ). So there exists a λo ∈ 0, to ∧yo such that for any (t, x, y) ∈ Oλo (to , xo , yo ) (to , xo , yo )

(V −ϕ)(t, x, y) < (V −ϕ)(to , xo , yo ) = 0 and Dx ϕ(t, x, y)−Dx ϕ(to , xo , yo ) ∨ ∂y ϕ(t, x, y)−∂y ϕ(to , xo , yo ) < 1. (6.122)

Similar to (6.113), ϕ b (t, x, y) := −∂t ϕ(t, x, y)−Lx ϕ(t, x, y)+g(t, x)∂y ϕ(t, x, y)−f (t, x), ∀ (t, x, y) ∈ [0, ∞)×Rl×[0, ∞) defines a continuous function. If Hϕ(to , xo , yo ) = ∞, then ϕ b (to , xo , yo )−Hϕ(to , xo , yo ) ≤ 0 holds automatically. eo ∈ (0, λo ) such that Hϕ(t, x, y) ≤ So let us just consider the case Hϕ(to , xo , yo ) < ∞. By (5.1), there exists λ 2 Hϕ(to , xo , yo )+1 < ∞ and thus ∂y ϕ(t, x, y) ≤ 0, ∀ (t, x, y) ∈ Oλeo (to , xo , yo ). If one had ∂y ϕ(to , xo , yo ) ≤ 0, (2.12) and (6.122) would imply that Z yZ s Z y ϕy (to , xo , s)ds = ϕ(to , xo , yo )+(y −yo)·∂y ϕ(to , xo , yo )+ ϕ2y (to , xo , r)drds ϕ(to , xo , y) = ϕ(to , xo , yo )+ yo

≤ ϕ(to , xo , yo ) = V(to , xo , yo ) ≤ V(to , xo , y),

 eo . ∀ y ∈ yo , yo + λ

yo

yo

which contradicts with the strict local maximum of V −ϕ at (to , xo , yo ). Hence we must have ∂y ϕ(to , xo , yo ) > 0.

(6.123)

To draw a contradiction, we assume that ǫ := ϕ b (to , xo , yo )−Hϕ(to , xo , yo ) > 0.

According to (6.123) and the continuity of ϕ, b there exists λ ∈ (0, λo ) such that for any (t, x, y) ∈ O λ (to , xo , yo ) ∂y ϕ(t, x, y) ≥ 0

b (to , xo , yo )−ǫ/2 ≤ ϕ(t, b x, y). and Hϕ(t, x, y) ≤ Hϕ(to , xo , yo )+ǫ/2 = ϕ

(6.124)

Fix α ∈ Ato (yo ), so α = M α−K α for some (M α , K α ) ∈ Mto ×Kto . In light of the martingale representation theorem, such that one can find qα ∈ H2,loc to Pto

nZ

s

to

2 |qα r | dr < ∞,

Z s o n o T to α (qα ∀ s ∈ [to , ∞) = Pto Ms = r ) dWr , ∀ s ∈ [to , ∞) = 1.

(6.125)

to

 to to ,xo As Θα , Ysto ,xo ,α s ∈ [to , ∞) are F −adapted continuous processes with Θα to = (to , xo , yo ), Pto −a.s., s := s, Xs  ζ α := inf s ∈ [to , ∞) : Θα / O λ (to , xo , yo ) s ∈

(6.126)

6.4

Proof of Section 5

37

to

defines an F −stopping time with to < ζ α ≤ to +λ, Pto −a.s. The continuity of processes Θα implies that Pto −a.s. Θα s ∈ O λ (to , xo , yo ),

∀ s ∈ [to , ζ α ].

(6.127)

Similar to (6.116), we can deduce from (6.127), (6.124) and (6.122) that for Pto −a.s.

T 1 α 2 2 α α to ,xo q ∂y ϕ(Θα )·qα b (Θα b (Θα (6.128) ∂y ϕ(Θα r −ϕ r ) ≤ Hϕ(Θr )− ϕ r ) ≤ 0 and r )+ Dx (∂y ϕ(Θr )) ·σ(r, Xr r ) ≥ 0, 2 r p p   Dx ϕ(Θα ) σ(r, X to ,xo ) + ∂y ϕ(Θα ) qα ≤ 1+|Dxϕ(to , xo , yo )| |σ(to , xo )|+ kc(·)k δ+ kc(·)kρ(δ)(1+|xo |̟ ) r r r r  + 1+|∂y ϕ(to , xo , yo )| qα ∀ s ∈ [to , ζ α ]. (6.129) r < ∞,

 Rs to |2 dr > n ∧ ζ α ∈ T . Applying Itˆ Let n ∈ N and define ζnα := inf s ∈ [to , ∞) : to |qα o’s formula to process r  α ϕ(Θs ) s∈[to ,∞) , and using (6.128) yield that  ϕ Θα α −ϕ(to , xo , yo ) ζn Z ζnα   T 1 α 2 2 α to ,xo α )∂y ϕ(Θα ∂t ϕ(Θα = qr ∂y ϕ(Θα ·σ(r, Xrto ,xo )·qα r )+Lx ϕ(Θr )+ r )−g(r, Xr r dr r )+ Dx (∂y ϕ(Θr )) 2 to Z ζnα Z ζnα  α T T to ,xo α dWrto , )+∂y ϕ(Θα (Dx ϕ(Θα ∂y ϕ(Θα − r )·(qr ) r )) ·σ(r, Xr r )dKr + to α ζn

≤−

Z

to

f (r, Xrto ,xo )dr+

to α ζn

Z

to

 α T T to ,xo dWrto , )+∂y ϕ(Θα (Dx ϕ(Θα r )·(qr ) r )) ·σ(r, Xr

Pto − a.s.

h i  R ζnα to ,xo Taking expectation Eto [·], we see from (6.129) that ϕ(to , xo , yo ) ≥ Eto ϕ Θα f (r, X + )dr . Since (6.127) α r ζn to α  R ζ + n f (r, Xrto ,xo ) dr ≤ and the continuity of f show that ϕ Θα max |ϕ(t, x, y)|+λ max |f (t, x)| α ζn to (t,x,y)∈Oλ (to ,xo ,yo )

(t,x)∈Oλ (to ,xo )

and since lim ↑ ζnα = ζ α , Pto −a.s. by (6.125), we can derive from the dominated convergence theorem that n→∞

 Z  + ϕ(to , xo , yo ) ≥ lim Eto ϕ Θα α ζn n→∞

Set m2 :=

α ζn

to

min

(t,x,y)∈∂Oλ (to ,xo ,yo )

  Z   + f r, Xrto ,xo dr = Eto ϕ Θα α ζ

ζα

to

  f r, Xrto ,xo dr .

(6.130)

(ϕ−V)(t, x, y) > 0 by (6.122). The continuity of Θα and (6.127) show that

  to ,xo ,α Pto Θα ≥ yo −λ > 0, ∀ s ∈ [to , ζ α ] = 1. ζ α ∈ ∂Oλ (to , xo , yo ) = Pto Ys

The latter of which implies that τ (to , xo , α) > ζ α > to , Pto −a.s., which together with (6.130) leads to that     Z ζα Z ζα α to ,xo to ,xo )dr V(Θ )+ f (r, X )dr ≥ E )−m + f (r, X ϕ(to , xo , yo )−m2 ≥ Eto ϕ(Θα α α to 2 r ζ r ζ 

≥ Eto

t

t

 Z   α 1{τ (to ,xo ,α)≤ζ α } R to , xo , τ (to , xo , α) +1{τ (to ,xo ,α)>ζ α } V Θζ α +

ζα

f (r, Xrto ,xo )dr

t

Taking supremum over α ∈ Ato (yo ), we can deduce from Theorem 4.2 that   Z   α ϕ(to , xo , yo )−m2 ≥ sup Eto 1{τ (to ,xo ,α)≤ζ α } R to , xo , τ (to , xo , α) +1{τ (to ,xo ,α)>ζ α } V Θζ α + α∈Ato (yo )

t

ζα



. (6.131)

f (r, Xrto ,xo )dr



= V(to , xo , yo ) = ϕ(to , xo , yo ).

A contradiction appears. We can also use the first DPP (Theorem 4.1) to get the incongruity: Let τ ∈ Tbxtoo (yo ). By Proposition 4.2, to

τ = τ (to , xo , α) for some α ∈ Ato (yo ). Let ζ α be the F −stopping time defined in (6.126). Similar to (6.90), one has ,xo ,τ o ,xo ,α Yτt∧ζ ≥ Yτto∧ζ α α , Po −a.s. It follows from (6.131) that    Z ζα  f (r, Xrto ,xo )dr ϕ(to , xo , yo )−m2 ≥ Eto 1{τ ≤ζ α } R(to , xo , τ )+1{τ >ζ α} V ζ α , Xζtαo ,xo , Yζtαo ,xo ,α + 

≥ Eto

 Z to ,xo ,τ  to ,xo α + 1{τ ≤ζ α } R(to , xo , τ )+1{τ >ζ α} V ζ , Xζ α , Yζ α

t

t

ζα

f (r, Xrto ,xo )dr



.

DPPs for Optimal Stopping with Expectation Constraint

38

Taking supremum over τ ∈ Tbxtoo (yo ) and applying Theorem 4.1 (2) yield that ϕ(to , xo , yo )−m2 ≥

  Z  sup Eto 1{τ ≤ζ α } R(to , xo , τ )+1{τ >ζ α } V ζ α , Xζtαo ,xo , Yζtoα,xo ,τ +

τ∈Tbxtoo (yo )

t

ζα

f (r, Xrto ,xo )dr



= V(to , xo , yo ) = ϕ(to , xo , yo ). The contradiction appears again.

A



Appendix

   Lemma A.1. Let t ∈ [0, ∞). (1 ) The sigma−field F t satisfies B(Ωt ) = σ Wst ; s ∈ [t, ∞) = σ ∪ Fst . s∈[t,∞) n m  t t t −1 (2 ) For any s ∈ [t, ∞], the sigma−field Fs can be countably generated by Cs := ∩ (Wti ) Oδi (xi ) : m ∈ N, ti ∈ i=1 o Q+ ∪{t} with t ≤ t1 ≤ · · · ≤ tm ≤ s, xi ∈ Qd , δi ∈ Q+ . Proof: 1a) Let ω ∈ Ωt and δ ∈ (0, ∞). For any n ∈ N with n > 1/δ, since all paths in Ωt are continuous, we can deduce that Oδ−1/n (ω) = {ω ′ ∈ Ωt : |ω ′ (s)−ω(s)| ≤ δ−1/n, ∀ s ∈ [t, ∞)} = {ω ′ ∈ Ωt : |ω ′ (s)−ω(s)| ≤ δ−1/n, ∀ s ∈ [t, ∞)∩Q}   = ∩ {ω ′ ∈ Ωt : Wst (ω ′ ) ∈ O δ−1/n (ω(s))} = ∩ (Wst )−1 O δ−1/n (ω(s)) ∈ σ Wst ; s ∈ [t, ∞) . s∈[t,∞)∩Q

s∈[t,∞)∩Q

 It follows that Oδ (ω) = ∪ O δ−1/n (ω) ∈ σ Wst ; s ∈ [t, ∞) . As B(Ωt ) is generated by open sets {Oδ (ω) : ω ∈ Ωt , δ ∈ n∈N  (0, ∞)}, one thus has B(Ωt ) ⊂ σ Wst ; s ∈ [t, ∞) . Next, let s ∈ [t, ∞), x ∈ Rd and δ ∈ (0, ∞). Given ω ∈ (Wst )−1 (Oδ (x)), set λ = λ(s, x, ω) := δ−|Wst (ω)−x| > 0. Since |Wst (ω ′ )−x| ≤ |ω ′ (s)−ω(s)|+|ω(s)−x| ≤ kω ′ −ωkt +|ω(s)−x| < λ+|ω(s)−x| = δ, ∀ ω ′ ∈ Oλ (ω),   we see that Oλ (ω) ⊂ (Wst )−1 Oδ (x) and thus (Wst )−1 Oδ (x) is an open set under the uniform norm k · kt . Then Oδ (x) ∈ Λs := {E ⊂ Rd : (Wst )−1 (E) ∈ B(Ωt )}, which is a sigma−field of Rd . As B(Rd ) is generated by open sets   {Oδ (x) : x ∈ Rd , δ ∈ (0, ∞)}, one has B(Rd ) ⊂ Λs , which implies that σ Wst ; s ∈ [t, ∞) = σ (Wst )−1 (E) : s ∈ [t, ∞), E ∈ B(Rd ) ⊂ B(Ωt ).      1b) Clearly, Fst = σ Wrt ; r ∈ [t, s] ⊂ σ Wrt ; r ∈ [t, ∞) , ∀ s ∈ [t, ∞). It follows that σ ∪ Fst ⊂ σ Wst ; s ∈ [t, ∞) . s∈[t,∞)   t −1 t t On the other hand, since (Ws ) (E) ∈ Fs ⊂ σ ∪ Fr for any s ∈ [t, ∞) and E ∈ B(Rd ), we have σ Wst ; s ∈ r∈[t,∞)     [t, ∞) = σ (Wst )−1 (E) : s ∈ [t, ∞), E ∈ B(Rd ) ⊂ σ ∪ Fst . s∈[t,∞)

2) Fix s ∈ [t, ∞]. Define [t, si := [t, s] if s < ∞ and [t, si := [t, ∞) if s = ∞. Let r ∈ [t, si and let {ri }i∈N ⊂  {t} ∪ (t, r) ∩ Q+ with lim ↑ ri = r. For any x ∈ Qd and δ ∈ Q+ , the continuity of paths in Ωt implies that i→∞     ∞ b r := E ⊂ Rd : (Wrt )−1 (E) ∈ σ(Cst ) , (Wrt )−1 Oδ (x) = ∪ ∪ ∩ (Wrti )−1 O δ− n1 (x) ∈ σ(Cst ). Thus Oδ (x) ∈ Λ n=⌈2/δ⌉ m∈N i>m

br. which is a sigma−field of Rd . Since B(Rd ) can also be generated by {Oδ (x) : x ∈ Qd , δ ∈ Q+ }, we see that B(Rd ) ⊂ Λ   t −1 t t d t It follows that Fs = σ Wr ; r ∈ [t, si = σ (Wr ) (E) : r ∈ [t, si, E ∈ B(R ) ⊂ σ(Cs ). On the other hand, it is clear   that σ(Cst ) ⊂ σ (Wrt )−1 (E) : r ∈ [t, si, E ∈ B(Rd ) = σ Wrt ; r ∈ [t, si = Fst . 

Lemma A.2. Let 0 ≤ t ≤ s < ∞.  (1 ) The mapping Πts is Frt Frs −measurable for any r ∈ [s, ∞]. Then for each Fs −stopping time τ , τ (Πts ) is a Ft −stopping time with values in [s, ∞].   e , ∀A e∈ F s. e = Ps A (2 ) The law of Πts under Pt is Ps : i.e., Pt ◦ (Πts )−1 A  e ∈ Frt ⊂ F t for any A e ∈ Frs . Set Proof: 1) For any r ∈ [s, ∞), an analogy to Lemma A.1 of [10] shows that (Πts )−1 A   e := A e ⊂ Ωs : (Πt )−1 A e ∈ F t , which is a sigma−field of Ωs . As F s ⊂ Λ e for any r ∈ [s, ∞), we see from Lemma A.1 Λ r  s  e i.e., (Πt )−1 A e ∈ F t for any A e∈ F s. (1) that F s = σ ∪ Frs ⊂ Λ, s r∈[s,∞)

A. Appendix

39

 Let τ be an Fs −stopping time. For any r ∈ [s, ∞), since {τ ≤ r} ∈ Frs , we see that τ (Πts ) ≤ r = (Πts )−1 {τ ≤ r} ∈ Frt . Thus τ (Πts ) is a Ft −stopping time with values in [s, ∞]. 2) Next, let us demonstrate that the induced probability Pe := Pt ◦ (Πts )−1 equals to Ps on F s . Since the Wiener measure Ps on (Ωs , F s ) is unique (see e.g. Proposition I.3.3 of [53]), it suffices to show that the canonical process W s is a Brownian motion on Ωs under Pe : Let s ≤ r < r′ < ∞. For any E ∈ B(Rd ), one can deduce that 

(Πts )−1 Wrs′ −Wrs

−1

    (E) = ω ∈ Ωt : Wrs′ (Πts )(ω) −Wrs (Πts )(ω) ∈ E   = ω ∈ Ωt : ω(r′ )−ω(s)− ω(r)−ω(s) ∈ E = (Wrt′ −Wrt )−1 (E).

(A.1)

 −1  (E) = Pt (Wrt′ −Wrt )−1 (E) , which shows that the distribution of Wrs′ −Wrs under Pe is the same So Pe Wrs′ −Wrs as that of Wrt′ −Wrt under Pt (a d−dimensional normal distribution with mean 0 and variance matrix (r′ −r)Id×d ).  e ∈ Frt is independent of W t′ −Wrt under Pt , (A.1) implies that e ∈ Frs , since (Πts )−1 A On the other hand, for any A r e ∩ W s′ −W s Pe A r r

     e ∩ (Πt )−1 W s′ −W s −1 (E) (E) = Pt (Πts )−1 A r r s         e · Pe Wrs′ −Wrs −1 (E) , e · Pt (Πts )−1 Wrs′ −Wrs −1 (E) = Pe A A

−1

= Pt (Πts )−1

∀ E ∈ B(Rd ),

which shows that Wrs′ −Wrs is independent of Frs under Pe. Hence, W s is a Brownian motion on Ωs under Pe .



We have the following extension of Lemma A.2.

Lemma A.3. Let 0 ≤ t ≤ s < ∞.  e is a Pt −null set. e , (Πt )−1 N (1 ) For any Ps −null set N s t s s t (2 ) For any r ∈ [s, ∞], the mapping Πts is F r F r −measurable. Then for each τ ∈ T , τ (Πts ) is a T −stopping time with values in [s, ∞].   e holds for any A e∈ F s. e = Ps A (3 ) Pt ◦ (Πts )−1 A

 e = 0. Lemma A.2 implies that e ∈ N s , so there exists an A e ∈ F s such that N e ⊂A e and Ps A Proof: 1) Let N       e = 0. As (Πts )−1 N e ⊂ (Πts )−1 A e , we see that (Πts )−1 N e ∈ N t. e = Ps A e ∈ F t and that Pt (Πts )−1 A (Πts )−1 A     e ⊂ Ωs : (Πt )−1 A e ∈ e ⊂ Ωs : (Πt )−1 A e ∈ F t ⊂ Λr := A 2) Given r ∈ [s, ∞], Lemma A.2 (1) shows that Frs ⊂ Λr := A s s r t s F r , which is clearly a sigma−field of Ωs . Since N s ⊂ Λr by Part (1), it follows that F r = σ(Frs ∪N s ) ⊂ Λt , i.e.  e ∈ F t for any A e∈ F s. (Πts )−1 A r r   s s t Let τ ∈ T . For any r ∈ [s, ∞), since {τ ≤ r} ∈ F r , we see that τ (Πts ) ≤ r = (Πts )−1 {τ ≤ r} ∈ F r . Thus τ (Πts ) is t

a T −stopping time with values in [s, ∞]. e ∈ F s . Similar to Problem 2.7.3 of [32], there exists an A ∈ F s such that A∆A e 3) Let A ∈ N s . Since

  c c e ec ) = (Πt )−1 (A∩A e ec ) e )∪(A∩ A )∪(Πts )−1 (A∩ A = (Πts )−1 (A∩A (Πts )−1 A∆A s          e ∆(Πt )−1 A , e c = (Πt )−1 A e ∩ (Πt )−1 (A) c ∪ (Πt )−1 (A)∩ (Πt )−1 A = (Πts )−1 A s s s s s

  e ∆(Πts )−1 A is a Pt −null set. So by Part (2) and Lemma A.2 (1), the we know from Part (1) that (Πts )−1 A   t e equals to the F t −measurable random variable (Πts )−1 A , Pt −a.s. Then F −measurable random variable (Πts )−1 A   e e = Pt (Πt )−1 (A) = Ps (A) = Ps (A).  Lemma A.2 (2) yields that Pt (Πt )−1 A s

s

Lemma A.4. Let t ∈ [0, ∞).  t  t (1 ) For any ξ ∈ L1 F , E and s ∈ [t, ∞], Et ξ F s = Et [ξ|Fst ], Pt −a.s. Consequently, an E−valued martingale (resp. local martingale or semi-martingale) with respect to (Ft , Pt ) is also a martingale (resp. local martingale or  t semi-martingale) with respect to F , Pt . t

(2 ) For any s ∈ [t, ∞] and any E−valued, F s −measurable random variable ξ, there exists an E−valued, Fst −measurable random variable ξe such that ξe= ξ, Pt −a.s. t (3 ) For any E−valued, F −adapted process X = {Xs }s∈[t,∞) with Pt −a.s. left-continuous paths, there exists an   t e es e= X such that ω ∈ Ω : X (ω) = 6 X (ω) for some s ∈ [t, ∞) ∈ N t. E−valued, Ft −predictable process X s s s∈[t,∞)

DPPs for Optimal Stopping with Expectation Constraint

40

  t t Proof: 1) Let ξ ∈ L1 F , E and s ∈ [t, ∞]. For any A ∈ F s = σ Fst ∪N t , similar to Problem 2.7.3 of [32], there exists     R R R R e ∈ Fst such that A∆A e ∈ N t . Then we can deduce that ξdPt = Ae ξdPt = Ae Et ξ Fst dPt = A Et ξ Fst dPt , an A A  t   which implies that Et ξ F = Et ξ F t , Pt −a.s. s

s

t

2) Let s ∈ [t, ∞] and let ξ be an E−valued, F s −measurable random variable. We first assume E = R. For any n ∈ N,  t  t t e we set ξn := (ξ ∧n)∨(−n) ∈ Fs and see from Part (1) that ξn := Et ξn Fs = Et ξn F s = ξn , Pt −a.s. Clearly, the  is F t −measurable and satisfies e random variable ξ := lim ξen 1 s

lim ξen 0, s ∈ [t, ∞). As it holds Pt −a.s. that Z s g(r, Xrt,x )dr ≤ g(s−t) < y −1+Υs = αos , ∀ s ∈ [t, s], t

o

we see that s < τo := τ (t, x, α ), Pt −a.s. Rs (R). Then αs := y + t qr dWrt = αoτo ∧s , Next, let us define qs := 1{s≤τo } qos , s ∈ [t, ∞), which is clearly of L2,loc t ∀ s ∈ [t, ∞) and it follows that τ (t, x, α) = τo > s, Pt −a.s. Since αo = Υ+y−1 is a positive continuous supermartingale, we can deduce from the continuity of α and the optional sampling theorem that h Z τ (t,x,α) i       Et  g(r, Xrt,x )dr = Et ατ (t,x,α) = Et αoτo ≤ Et αos = E[Υs +y −1] < y. t

A.1

Proofs of Starred Statements in Section 6

A.1

41

Proofs of Starred Statements in Section 6

n ∈ N, Proof of (6.72): Given s ∈ [t, ∞), let io be the largest integer such that tio ≤ s. For any i = 1, · · · , io and    i t t t i since {τ ≤ ζ} ∈ F τ ∧ζ ⊂ F τ , one can deduce that {τ ≤ s} = {τ ≤ ζ}∩{τ ≤ s} ∪ ∪ ∪ An ∩ τn (Πti ) ≤ s ∪ {τ > i≤io n∈N  t t ζ}∩Nt,x,τ ∈ F s . So τ ∈ T . c  For i, n ∈ N and ω ∈ Ain ∩ N i,n ⊂ {τ > ζ = ti }, since Xti (ω) ∈ Oni = Oδi (xin ) (xin ) and Yti (ω) ∈ Dni ⊂ yni −ε/2, ∞ ,  applying (6.67) with (x, x′ , ς) = xin , Xti (ω), τni , we see from (6.70) that Et

Z

i τn (Πtt ) i

t

Z  t g(r, Xr )dr F ti (ω) < Eti

i τn

t ,xi  r, Xr i n dr

g

ti

< Yti (ω)+ε+

Z

t

ti



+ε/2+

Z

ti

t

 g r, Xr (ω) dr = Et

 g r, Xr (ω) dr ≤ yni +ε/2+

Z

τ

t

Z

 t g(r, Xr )dr F ti (ω)+ε.

t

ti

 g r, Xr (ω) dr

(A.2)

Taking summation over i, n ∈ N, one can deduce from (6.71) and the monotone convergence theorem that   X    X  Z τ Z τni (Πtt ) Z τni (Πtt ) i i Et 1{τ >ζ} 1Ain g(r, Xr )dr = Et g(r, Xr )dr = Et 1Ain g(r, Xr )dr t

X

=

i,n∈N

=

X

i,n∈N

It follows that Et

"

i,n∈N

Et 1Ain Et 

Et 1Ain

hR

τ t

Z

Z

i τn (Πtt ) i

t

τ

t

t g(r, Xr )dr F ti 

g(r, Xr )dr +ε = Et t

#

 X

i,n∈N



X

t

t

i,n∈N

Et 1Ain Et

i,n∈N Z τ

1Ain

"

Z

τ

t



#  t g(r, Xr )dr F ti +ε 

g(r, Xr )dr +ε = Et 1{τ >ζ}

i Rτ  g(r, Xr )dr ≤ Et t g(r, Xr )dr +ε ≤ y +ε. Thus τ ∈ Txt (y +ε).

Z

t

τ



g(r, Xr )dr +ε. 

t

t

Proof of (6.102): 1) Given i = 1, · · · , Io , if A ∈ F ti and if {Υs }s∈[ti ,∞) is an F −adapted continuous process over  t period [ti , ∞) with Υti = 0, Pt −a.s., one can easily deduce that 1{s≥ti }∩A Υs s∈[t,∞) is an F −adapted continuous t

process starting from 0. Then we see from (6.100) that M is an F −adapted continuous process and K is an t F −adapted continuous increasing process with K t = Kt = 0, Pt −a.s. For any ω ∈ Ac♯ ∩N c , K · (ω) = K· (ω) is an increasing path; for i = 1, · · · , Io and n = 1, · · · , ni , it holds for any   e i,n c that ω ∈ Ain ∩(Πtti )−1 N  K s (ω) = 1{s