ROOT S BARRIER, VISCOSITY SOLUTIONS OF OBSTACLE PROBLEMS AND REFLECTED FBSDES

Transcription

1 OOT S BAIE, VISCOSITY SOLUTIONS OF OBSTACLE POBLEMS AND EFLECTED FBSDES HAALD OBEHAUSE AND GONÇALO DOS EIS Abstract. Following work of Dupire 005, Carr Lee 010 and Cox Wang 011 on connections between oot s solution of the Skorokhod embedding problem, free boundary PDEs and model-independent bounds on options on variance we propose an approach with viscosity solutions. Besides extending the previous results, it gives a link with reflected FBSDEs as introduced by El Karoui et. al and allows for easy convergence proofs of numerical schemes via the Barles Souganidis 1991 method. 1. Introduction Let B be standard, real-valued Brownian motion started at 0 and µ a centered integrable probability measure on, B. In 1961 Skorokhod [36 asked whether it is possible to find a stopping time τ such that B τ µ and B τ = B τ t t is uniformly integrable; a stopping time τ with these properties is now usually referred to as a solution of the Skorokhod embedding problem or Skorokhod stopping problem. Further, he provided an affirmative answer by constructing a stopping time which depends on an additional random variable independent of B. Skorokhods motivation came from invariance properties of random walks and subsequently this question attracted the interested of many researchers and a large number of different approaches and generalizations were developed, each approach giving rise to a different stopping time which solves the same Skorokhod embedding problem we do not even attempt to give a brief summary but refer to the surveys [7, 8, 1 which list many different approaches and discusses several applications. One of the earliest approaches after Skorokhod was initiated by oot in 1969 [30 who showed that if µ is centered and has a second moment, then it is possible to find a a subset of [0, [,, the so-called oot barrier, such that its first hitting by the time space process t, B t, τ = inf {t 0 : t, B t }, solves the Skorokhod embedding problem. In addition to this intuitive solution, Loynes [6 introduced extra conditions under which oot s barrier is uniquely defined as a subset of [0, [, and ost [34 proved that among all stopping times τ which solve the Skorokhod embedding problem, oot s solution minimizes E [ τ ; further he gave important generalizations and connections with potential theory, characterized the possible stopping distributions of Markov processes by introducing a filling scheme in continuous time and proved the existence of another barrier which leads to a hitting time which maximizes E [ τ. Despite all these nice properties of oot s and ost s approach they had until very recently the significant disadvantage that it was not known how to calculate the barriers in fact, only for a handful of very simple target measures µ the explicit form of the barriers was known. One of the more recent motivations to study the Skorokhod embedding problem comes from mathematical finance where Skorokhod embeddings naturally appear if one is interested in model-independent prices of exotic derivatives, i.e. extremal solutions of the Skorokhod problem lead to robust lower and upper bounds for arbitrage-free prices of exotics we refer to the survey s of Hobson [1 and Obłój [8. Motivated by this, Dupire [14 gave in 005 a presentation where he firstly pointed out that in the case of options on variance these extremal solutions are the ones of oot [30 resp. ost [34 and, secondly, that these barriers can be calculated by solving nonlinear parabolic PDEs, namely solutions to free boundary problems. This was further developed by Carr Lee [8 and in 011 Cox Wang [11 made this connection precise using the variational approach as developed in the 1970 s by A. Bensoussan J.L. Lions et. al, cf. [6 and do much more we refer the reader to [11. This article is inspired by all these beautiful results and our goal is to study the connection of oot s barrier with the parabolic obstacle problem from the perspective of viscosity theory; more precisely, given two probability measures µ, ν on, B and a function σ : [0, we want to find under appropriate conditions on σ, µ, ν the oot barrier such that its first hitting time τ solves the following Skorokhod embedding problem 1 { dxt = σ t, X t db t, X 0 µ, X τ ν and X τ = X t τ t is uniformly integrable. Key words and phrases. Skorokhod embedding problem, oot and ost barrier, viscosity solutions of obstacle problems, reflected BSDEs, bounds on variance options. 1

2 OOT S BAIE, VISCOSITY SOLUTIONS OF OBSTACLE POBLEMS AND EFLECTED FBSDES by identifying with a continuous function u C [0,, which is a viscosity solution of { min u + u. x ν dx, t σ u = 0 on 0,, u 0,. =. x µ dx on. and with the quadruple of process X, Y, Z, K, solution to the reflected Forward Backward SDE for h := y ν dy, T > 0, t, x [0, T and s [t, T Xs t,x = x + s Ys t,x = u 0, X t,x T Ys t,x h Xs t,x t σ T r, X r dw r, + K t,x T Kt,x s, t < s T and T T s t Zt,x Y t,x s r dw r, h Xs t,x dks t,x = 0. What we consider among the benefits of the viscosity approach are that it covers the case when σ is time dependent and only elliptic, i.e. no time-homogeneity or uniform ellipticity assumptions as in [11 are needed especially the important case for model-independent bounds when σ t, x = x is immediately in our setting without an extra transformation, gives a direct link to reflected FBSDEs and their Feynman Kac representation via classic work of El Karoui, Kapoudjian, Pardoux, Peng and Quenez[15, uses continuous functions instead of functions with values in weighted Sobolev spaces and allows for short proofs of convergence of numerical schemes via the Barles Souganidis method. Further, we introduce a generalization of Loynes notion of regular barriers, so-called µ, ν-regular barriers which allows to establish a one to one correspondence between the problems 1 and. In Section we recall several results from the literature and discuss the general oot embedding problem, in Section 3 we recall several results in viscosity theory and the connection with reflected FBSDES. Section 4 contains our main result: the link between the embedding problem, PDE in viscosity sense and reflected FBSDE. In Section 5 apply the Barles Souganidis [3 method in this context and we finish in Section 6 by recalling the relevance of oot s barrier for the derivation of bounds on variance options. Acknowledgement 1. Gd is also affiliated with CMA/FCT/UNL, Caparica, Portugal. Gd acknowledges partial support by the Fundação para a Ciência e a Tecnologia Portuguese Foundation for Science and Technology through PEst-OE/MAT/UI097/011 CMA. HO is grateful for partial support from the European esearch Council under the European Union s Seventh Framework Programme FP7/ /EC grant agreement nr and DFG Grant SPP-134. Acknowledgement. The authors would like to thank Alexander Cox, Paul Gassiat, Martin Keller essel and Jan Obłój for helpful remarks.. Convex order, potential functions, root barriers and root embedding Already for the case of Brownian motion one cannot hope to find Skorokhod embeddings for two arbitrary probability measures µ and ν, i.e. B 0 µ, B τ ν. Intuitively, a necessary condition is that the target distribution ν should have a higher dispersion than the initial distribution µ. This can be expressed by a classic order relation on the set of probability measures, namely the so called convex order. In the context of oot solution this order relation appears also naturally when one thinks in terms of the potential theoretic picture as developed in this context by ost [31, 3, 33, 34, Chacon [10 and others. Definition 1. Let k N, m and denote with M k m the set of probability measures on, B with a finite k-th moment and mean m i.e. µ M k m iff xµ dx = m and x k µ dx < and set M k = m Mk m, M = k N {0} Mk. Definition. Let m. A pair of probability measures µ, ν M 1 M 1 is in increasing convex 1 order, µ cx ν, if ˆ ˆ 3 f x µ dx f x ν dx for every convex function f : provided the integrals exist. We also say that two measures µ, ν are in convex order if either µ cx ν or ν cx µ. Two real-valued random variables X, Y are in increasing convex order, denoted X cx Y, if L X cx L Y. emark 1. Assume µ M 1 m, ν M 1 n and µ cx ν. Then m = n, this follows from choosing f x = x and f x = x. The intuition being that µ cx ν signifies that a µ distributed random variable is less likely to take very large values than a ν distributed random variable since extremal values of convex functions are obtained on intervals of the form, a b,. As it turns out, an especially important case is given when 3 is applied with f. =. m resp. f. =. m for fixed m. 1 Sometimes this is also called the Choquet order.

3 OOT S BAIE, VISCOSITY SOLUTIONS OF OBSTACLE POBLEMS AND EFLECTED FBSDES 3 Definition 3 Potential function. Associate with every µ M 1 a continuous non-positive function u µ C,, 0 given as ˆ u µ x := x y µ dy. We say that u µ is the potential function of the probability measure µ. Example 1. Denote with δ m the Dirac measure at m. Then u δm x = x m and it is easy to verify that u µ u δm for every µ M 1 m, resp. µ cx δ m. Lemma 1. Let m and µ, ν M 1 m. Then the following are equivalent 1 µ cx ν, y x + µ dy y x + ν dy, x, 3 u µ x u ν x, x. Moreover, µ cx ν implies that xµ dx = xν dx and if µ, ν M m then x µ dx x ν dx. Proof. Follows from [35, Section 3.A. Before we justify the name potential function let us note that the classic connection between positive [non-positive adon measures and second derivatives of concave [convex functions reads in this setting as Proposition 1. There exists a one to one correspondence between potential functions and probability measures. More precisely: 1 Let µ M 1 m for some m. Then u µ x = x y µ dy is a concave function satisfying u µ x u δm x = x m and Let u : be a concave function satisfying u x x m and lim u µ x u δm x = 0 x ± lim u x + x m = 0 x ± for some m. Then there exists a probability measure µ M 1 m such that u is the potential function of µ, i.e. u = u µ. Moreover, the connection is given by the relation µ = 1 u in the sense of distributions. Proof. See [0, Proposition.1 and [5, Proposition 4.1. emark. To explain the name potential function we briefly recall the classic potential theory for Markov processes cf. [7, 7, 8 for details: consider a real-valued Brownian motion B and denote its semigroup operator with Pt B. The potential kernel is defined as U B = P B 0 t dt, i.e. U B can be seen as a linear operator on the space of measures on, B by setting µu B = µp B 0 t dt which is of course nothing else than the occupation measure along Brownian trajectories started with B 0 µ. If µ is a signed measure with µ = 0 and finite first moment, then the adon Nikodym density with respect to the Lebesgue measure is given as dµu B ˆ = x y µ dy. dx Since in dimension one Brownian motion is recurrent, µu B is infinite if µ is positive. However, the right hand side x y µ dy is still well defined for every µ M1 and Chacon [10 demonstrated that this is indeed a very useful quantity in the study of Martingales and Markov processes with respect to hitting times. We summarize some properties of the potential functions of probability measures which we use throughout. The proofs are standard and can be found in [7, Proposition.3 and [8, Section 3.. Proposition. Let µ, ν M 1. Then 1 µ cx ν iff µ cv ν iff u µ. u ν., u µ is concave and Lipschitz continuous with Lipschitz constant equal to one, 3 if µ = m then u µ x u δm x = x m, 4 if ν M 1 then b we have u µ [b, = u ν [b, iff µ b, ν b,, 5 if µ n M 1, µ n n µ weakly and u µn converges for some point x 0 then u µn converges for all x and c 0 s.th lim n u µn = u µ + c. Further, if ν M 1 such that u µn u ν then µ M 1 and c = 0. Conversely, if u µn x n u µ x x for some µ n, µ M 1 then µ n µ weakly, 6 if µ cx ν then lim x u µ x u ν x = 0. 7 u µ is almost everywhere differentiable on and its right derivative at x equals 1 µ, x and its left derivative equals 1 µ, x. The symbol denotes weak convergence of probability measures.

4 OOT S BAIE, VISCOSITY SOLUTIONS OF OBSTACLE POBLEMS AND EFLECTED FBSDES 4 Since we want to embed a measure via a Markov process, the potential function which is given by the image measure of this Markov process is a fundamental quantity. Definition 4. If X is a real-valued, continuous Markov process carried on a probability space Ω, F, F t, P with P Xt 1 M 1 for every t 0 then we call the potential function of X at time t. u X t, x := u P X 1 t As shown by Chacon [10, Lemma 3.1a, potential functions give an intuitive characterization of Martingales. Proposition 3. Let X be a real-valued, adapted process. Then X is a martingale iff for every pair τ 1,τ of bounded stopping times with τ 1 τ, u X τ 1,. u X τ,...1. oot s solution and classic results on barriers. Our aim is to provide explicit solutions of the Skorokhod embedding problem not only for Brownian motion but for diffusion martingales. Assumption 1. Ω, F, F t, P is a filtered probability space that satisfies the usual conditions and carries a standard Brownian motion B. Assume that σ C [0,, is Lipschitz in space and of linear growth, both uniformly in time, that is 4 σ Lip := sup sup t [0, x y σ t, x σ t, y x y x < and σ LG := sup sup t [0, x σ t, x 1 + x Further, let m, µ, ν M m, µ cx ν, and X 0 µ, X 0 independent of the σ-algebra generated by the Brownian motion B and assume that with probability one σ t, X t > 0 t 0 where X = X t denotes the unique strong solution of the SDE 6, i.e. the real-valued, local martingale X such that 5 X. = X 0 + X 0 µ. ˆ. 0 σ t, X t db t P a.s Further, assume that for every t > 0, X t has a continuous but not necessarilty smooth density with respect to Lebesgue measure. Definition 5 Skorokhod embedding problem. Let σ, µ, ν and Ω, F, F t, X t, P be as in Assumption 1. If there exists a stopping time τ wrt the filtration F t such that X τ ν, X τ = X t τ t 0 is uniformly integrable u.i. then we say that τ solves the Skorokhod embedding problem given by σ, µ, ν. We denote with S σ, µ, ν the possibly empty set of stopping times τ which solve the Skorokhod embedding problem given by σ, µ, ν. emark 3. Unless stated otherwise, all stopping times are stopping times in the filtration F t. As pointed out in the introduction there are many different solutions of the embedding problem, that is in general the set S σ, µ, ν can contain much more than one stopping time in the survey article of Obłój [8 at least 0 different solutions are given for embedding laws in Brownian motion, i.e. σ, µ, ν = 1, δ 0, ν. In terms of applications a natural requirement is to consider stopping times which are minimal. Definition 6. A stopping time τ is called minimal for the process X if for every other stopping time ρ such that ρ τ and X τ X ρ it follows that ρ = τ. This article is focused on the approach initiated by oot in 1968 [30 and further developed by ost, Loynes, Kiefer et. al [34, 3, 31, 5, 6 of solving Skorokhod embeddings by stopping times which are the first entrance times of the time-space process t, X t t 0 into a closed subset of [0, [,. Definition 7. A closed subset of [0, [, is a oot barrier if 1 t, x implies t + r, x r 0, +, x x [,, 3 t, ± t [0, +. We denote by the set of all oot barriers. Given, its barrier function f : [, [0, is defined as f x := inf {t 0 : t, x }, x [,. Barrier functions have several nice properties such as being lower semi-continuous and that f x, x for any x see [6, Proposition 3 for more properties. <.

5 OOT S BAIE, VISCOSITY SOLUTIONS OF OBSTACLE POBLEMS AND EFLECTED FBSDES 5 Definition 8. We denote with σ, µ, ν as the possibly empty subset of of all σ, µ, ν such that τ S σ, µ, ν where τ is the first hitting time of by t, X t, i.e. τ = inf {t > 0 : t, X t }. We also say that σ, µ, ν embeds the law µ into ν via X and call τ the oot stopping time. Before we discuss conditions on σ, µ, ν which imply the existence of a oot barrier, i.e. that σ, µ, ν, we note that simple examples show that the oot barrier is in general not unique, i.e. σ, µ, ν > 1 already for generic situations. Example. The set = [0, [1, [0, [, 1 is an element of 1, δ 0, 1 δ 1 + δ 1, i.e. embeds the sum of two Diracs into a standard Brownian motion but obviously one can define differently, in fact any other oot barrier with a barrier function which coincides with f on [ 1, 1 is also an element of 1, δ 0, 1 δ 1 + δ 1. This non-uniqueness problem was in the Brownian case resolved by Loynes [6 in 1970 who introduced the notion of regular oot barriers. Definition 9 Loynes [6 regularity. Let. We say that resp. its barrier function f is Loynes regular if f vanishes outside the interval [ x, x+, where x + and x are the first [ positive resp. first negative zeros of f. Given Q, we say that Q, are Loynes-equivalent if f Q = f on x Q, x Q + and [ x, x+. emark 4. If Q, are Loynes-equivalent then x + = x Q + and x = x Q. We advise the reader to be suspicious about the word regular since Loynes-regular oot barriers can have spikes see examples in Section.. For the case σ, µ, ν = 1, δ 0, ν existence and uniqueness of Loynes-regular oot barriers are standard. Theorem 1 [30, 6. For every ν M 0 the set 1, δ 0, ν is not empty and there exists exactly one element in 1, δ 0, ν which is Loynes-regular. The second moment of the oot stopping time, E [ τ for 1, δ0, ν, is minimal among all elements of S 1, δ 0, ν. An advantage of oot s solution besides its intuitive appeal the stopping time being a hitting time the property that τ minimizes E [ τ among all solutions of the Skorokhod embedding problem, i.e. in our notation E [ τ = inf E [ τ where the inf is taken over all τ S σ, µ, ν. This was proven in ost [34 as well as in [11 and is the very reason why oot s solution is important for financial applications, see Section 6... egularity beyond Loynes. We now extend the discussion on oot barriers to our more general setting of oot embeddings for Itô-martingales with random initial distribution, i.e. we discuss σ, µ, ν. To that end we need to adapt several results of Loynes [6. First, note that already in the Brownian case σ 1 we need a modification of the notion of Loynes-regularity, a notion that is tailor made for the Dirac as initial distribution as the example below shows. Example 3. Let µ = 1 δ + δ and ν = 1 4 δ 3 + δ 1 + δ 1 + δ 3. It is easy to see by symmetry properties of the Brownian motion that for a = b = 0 the barrier Q a,b = [0, [3, [a, {1} [b, { 1} [0, [, 3 {+ } [, + is an element of 1, µ, ν, as is = [0, [3, [0, [1, 1 [0, [, 3 {+ } [, +. However, neither is Loynes-regular and in fact there cannot exist a Loynes-regular barrier in 1, µ, ν. Assume Q a,b 1, µ, ν, then a, b > 0 otherwise it would not be Loynes regular; now note that Q 0,0 1, µ, ν, hence every other Q a,b puts under ν more mass on the point 3 than the required 1 4 since the geometry of Q a,b implies that only more trajectories can hit the line [0, {3} than in the case a = b = 0; further, every element of 1, µ, ν must coincide with Q 0,0 on [0, [1, 3 [0, [ 3, 1. Motivated by the above we introduce the notion of µ, ν-regular barriers. Definition 10. For µ cx ν define N µ,ν := {x, + : u µ x = u ν x} {± } and N µ,ν = [0, + N µ,ν. A oot barrier is µ, ν-regular if = N µ,ν [or equivalently if f x = 0 x N µ,ν. Denote with reg σ, µ, ν the subset of σ, µ, ν of µ, ν-regular oot barriers. Further, two oot barriers, Q are µ, ν- equivalent if 3 \ [0, N µ,ν o = Q \ [0, N µ,ν o [or equivalently if f x = f Q x x N µ,ν c. 3 We denote with A the closure and with A o the interior of a given set A.

6 OOT S BAIE, VISCOSITY SOLUTIONS OF OBSTACLE POBLEMS AND EFLECTED FBSDES 6 If F µ and F ν are the cumulative distribution functions of µ and ν respectively, then the set N µ,ν coincides with the set {x : F µ x = F ν x} {± }. The next Lemma shows that in the case σ 1 and µ = δ 0 the µ, ν-regularity of a oot barrier is equivalent to Loynes regularity. Lemma. Let ν M 0 and 1, δ 0, ν. Then is Loynes-regular iff is δ 0, ν-regular. Proof. Assume to be Loynes-regular, then f x = 0 for x / x, x + where x, x + are as in Definition 9. Since the embedding is possible, i.e. X τ ν, then via the regularity of the barrier one has that ν [ [ x, x+ = 1, i.e. ν puts no mass outside x [, x+ by definition so does δ0. In other words outside x, x + both measures ν and δ0 are the same which by point 4 of Proposition implies that u δ0 x = u ν x for x / [ x, x+ and by continuity of the potential functions for any x / x, x+. We conclude that is δ 0, ν-regular. For the inverse direction, just remark that being a δ 0, ν-regular barrier we have due to δ 0 cx ν that N δ0,ν has the form N δ0,ν = \ a, b for some a < 0 < b, in other words f x = 0 for any x / a, b. Notice that the convex order relation with δ 0 implies that if u ν c = 0 for some c > 0 then u ν x = 0 for any x c and corresponding condition for the case c < 0, this implies that a and b are the first negative resp. positive zero of f. Hence is Loynes-regular. Definition 10 allows to extend Loynes [6, Theorem 1 to the present setting, i.e. if there exists an embedding with a oot barrier then there also exists an embedding with a µ, ν-regular oot barrier. Theorem. Let µ, ν M, µ cx ν and let X be a real-valued continuous square integrable local Martingale X with a quadratic variation that is strictly increasing and X 0 µ. Assume additionally that there exists Q s.th. X τq ν and X τ Q is u.i.. Then Q is µ, ν-equivalent to a unique µ, ν-regular barrier s.th. X τ ν and X τ is u.i.; moreover E [[X τ <. We prepare the proof with a lemma concerning the union and intersection of oot barriers. Lemma 3. Take ν M 1 and let Y be a process with continuous paths. Assume there exists Q, with first hitting times τ Q and τ respectively and assume that Y τq ν and Y τ ν. Then Q, Q and both generate the same law ν with respective stopping times min {τ Q, τ } and max {τ Q, τ }. Proof. This proof is a straightforward modification of [6, Proposition 4, we fill in some details and sketch the proof for the case of Q : let f and f Q be the barrier functions of and Q and define the set K := {x : f Q x < f x}. Since both barriers with their respective stopping times generate ν we have that P [ Y τq K = P [Y τ K iff P [ Y τq K, Y τ K + P [ Y τq K, Y τ K c = P [ Y τq K, Y τ K + P [ Y τq K c, Y τ K iff P [ Y τq K, Y τ K c = P [ Y τq K c, Y τ K. A quick analysis shows that for a trajectory Y ω s.t. Y τ ω ends in K it is impossible that Y τq ω K c, in other words, since the paths of Y are continuous and in the set [0, K the barrier Q is to the left of then any path that hits Q must hit before hitting Q and hence P [ Y τq K c, Y τ K = 0 = P [ Y τq K, Y τ K c. Finally, by decomposing the probability space in disjoint events based on K and K c and using the just obtained result one concludes P [{ Y τq K, Y τ K } { Y τq K c, Y τ K c} = 1. The rest of the proof follows as in [6. We can now give the proof of Theorem. Proof of Theorem. To see that Q is µ, ν-equivalent to a µ, ν-regular barrier note that since u ν u µ and we embed by assumption the continuous time-space process t τ Q, X t τq does not enter [0, N µ,ν, hence := Q N µ,ν is a also an element of σ, µ, ν, and moreover an element of reg σ, µ, ν. To see the uniqueness, i.e. reg σ, µ, ν = 1 suppose that there are two µ, ν-regular barriers B, C, each embedding ν via X with u.i. stopping times τ B and τ C respectively. Then Γ = B C also embeds ν with stopping time γ = min {τ B, τ C } see Lemma 3; γ is also u.i. Furthermore, since X is a local martingale, E [[X τb = E [ X τ B = E [ X γ = E [[X γ. Since the quadratic variation t [X t is strictly increasing and γ τ B this already implies [X γ = [X τb. The strict monotonicity of t [X t then implies γ = τ B. The same arguments applies to τ C, hence we conclude that τ B = γ = τ C. To see that B, C and B C are the same outside N µ,ν since in N µ,ν this must hold we argue as in the proof of Lemma in [6 by showing that if B Γ then also τ B τ γ. Note that B Γ implies that the existence of x 0 such that x 0 / N µ,ν and wlog f B x 0 < f Γ x 0 where f Γ x 0 0. Since the barriers are regular and lower semi-continuous functions attain their minimum on a compact set, ɛ > 0 s.th f Γ x > ɛ on some neighborhood of x 0. Moreover ɛ can be chosen in such a way that ɛ < f Γ x 0 f B x 0, implying that a δ > 0 exists s.th f Γ x > f Γ x 0 ɛ > f B x 0 + ɛ for any x in the set {x : x x 0 < δ}. Since the embedding via X is possible, there exists a positive probability that the time-space process t, X t t 0 hits the line segment

7 OOT S BAIE, VISCOSITY SOLUTIONS OF OBSTACLE POBLEMS AND EFLECTED FBSDES 7 [f B x 0, f B x 0 + ɛ {x 0 } while hitting Γ only after time f B x 0 + ɛ. This implies that P [τ B τ Γ > 0 and the result follows. Since Q σ, µ, ν, E [ [X τq < and hence by construction also E [[X τ <..3. The general oot embedding and optimality. We give a generalization of Theorem 1 showing that σ, µ, ν is not empty, i.e. embedding ν M via the solution process of the SDE 6 dx t = σ t, X t db t, X 0 µ. oot s original construction for Brownian motion, i.e. σ 1 and µ = δ 0 relies only on the continuity of the Brownian trajectories, the strong Markov property, that trajectories exit compact intervals in finite time and the fact that the process has no interval of constancy hence as remarked several times in the literature the same construction generalizes. Here we give only the results, complete proofs are postponed to Appendix 7. It turns out to be natural for the proof to distinguish between the cases [X = and when X is a geometric Brownian motion which implies only [ln X = Existence and uniqueness of a oot barrier if [X =. We now state a general result on the existence of the oot embedding via the solution of 6 the proof is given in the appendix. Theorem 3 oot Embedding. Let σ, µ, ν be such that Assumption 1 is fulfilled and that [X = a.s. Then reg σ, µ, ν = 1. Moreover, ˆ E [[X τ = x²ν dx < and E [τ < σ, µ, ν. emark 5. One can also apply the results of ost and Cox Wang [11, 34 to show that τ minimizes E [f τ among all τ S σ, µ, ν where f can be a convex, increasing function with f 0 = 0 which further fulfills some properties which depend in general on σ..3.. Existence and Uniqueness for geometric Brownian motion σ t, x = x. The case when X is a Geometric Brownian motion GBM henceforth, i.e. σ t, x = x, is not covered by Theorem 3 since although E [[X = we have [X < a.s. otherwise one could write X as a time-changed Brownian motion via Dambis/Dubins Schwarz. However, this case is especially important for the applications in finance see Section 6. Assumption. µ, ν M, µ cx ν and there exists an ɛ > 0 such that supp µ 0, + and supp ν [ɛ, +. The assumption that ɛ is strictly greater than 0 is needed for our arguments see emark 17 in the appendix and does not cover the case ν [0, = 1, ν {0} = 0. However, note that one cannot hope for oot embeddings where the target measure has an atom at 0. We now state the embedding result for this situation again, the proof is given in the appendix. Theorem 4 oot Embedding via Geometric Brownian motion. Let σ x = x and Assumption hold. Then reg σ, µ, ν = 1. Moreover ˆ E [[X τ = x µ dx <, [ln X τ = τ σ, µ, ν Further, for every other stopping time τ S σ, µ, ν, E [f τ E [f τ for every function f : [0, [0, which is convex, f 0 = 0 and f has a bounded right-derivative. 3. Background on viscosity solutions and the obstacle problem We briefly recall the definition and some basic stability properties of viscosity solutions in the parabolic setting for a detailed exposition we refer to the User s guide [1 with some additional lemmas which will become useful in the subsequent sections. We then discuss the connection to reflected FBSDEs Sub- and supersolutions, properness and semirelaxed limits. Definition 11. Let O be a locally compact subset of and denote O T = 0, T O for T 0,. Consider a function u : O T and define for s, z O T the parabolic superjet P,+ O u s, z as the set of triples a, p, m which fulfill u t, x u s, z + a t s + p, x z m x z, x z + o t s + x z as O T t, x s, z Similarly we define the parabolic subjet P, u s, z such that P, u = P,+ u. O O

8 OOT S BAIE, VISCOSITY SOLUTIONS OF OBSTACLE POBLEMS AND EFLECTED FBSDES 8 Example 4. If a function u C 1, O T, then { u P, u t, x = t t,x, u hence P,+ u t, x = x t,x, m { u t t,x, u x t,x, m P, u t, x P,+ u t, x = : m u } x t,x t,x : m u x t,x { u t, u x, u } t,x. x If u is only differentiable from the left or right, a similar statement holds and is important for the proof of our main result, see Lemma 5. emark 6. An equivalent definition is to consider the extrema of u ϕ when ϕ is taken from an appropriate set of smooth test functions, see [1. Definition 1. A function F : O T is proper if t, x, a, p O T F t, x, r, a, p, m F t, x, s, a, p, m m m, s r. Denote the real-valued, upper semicontinuous functions on O T with USC O T and the lower semicontinuous functions with LSC O T. A subsolution of the forward problem { F t, x, u, t u, Du, D 8 u = 0 u 0,. = u 0. is a function u USC O T such that F t, x, u, a, p, m 0 for t, x O T and a, p, m P,+ O u t, x u 0,. u 0. on O The definition of a supersolution follows by replacing USC O T by LSC O T, P,+ O } by P, O and by. If u is a supersolution of 8 then we also say that F t, x, u, t u, Du, D u 0, u 0, x u 0 x holds in viscosity sense similar for subsolutions. Similarly we call a function v a viscosity sub-,super- solution of the backward problem 9 if { G t, x, v, t v, Dv, D v = 0 v T,. = v T. G t, x, v, a, p, m 0 for t, x O T and a, p, m P,+ O v t, x v 0,. v T. on O. The next lemma shows that finding a viscosity solution to 9 is equivalent to finding one to 8. Lemma 4. Let T <. Then u is a viscosity solution of the forward problem 8 iff v t, x := u T t, x is a viscosity solution of the backward problem 9 with v T. = u 0. and G t, x, v, a, p, m = F T t, x, v, a, p, m. Proof. Assume u is a viscosity solution. Let ϕ C O T, and assume t, x v t, x ϕ t, x attains a local maximum at ˆt, ˆx O T. Then T ˆt, ˆx is a local maximum of t, x u t, x ϕ t, x with ϕ t, x = ϕ T t, x, hence F,, u, t ϕ, D ϕ, D ϕ T ˆt,ˆx = G,, v, t ϕ, Dϕ, D ϕ ˆt,ˆx which is by emark 6 sufficient to see that v is a viscosity solution of 9. similarly. The other implication follows A strong feature of viscosity solutions, which is also the key for the proof of our main theorem, is that they are very robust under perturbations. The type of reasoning presented below is classic and called the Barles Perthame method of semi-relaxed limits. Definition 13. Let A m and g n n a sequence of functions g n : A which is locally uniformly bounded. We define g, ḡ : A as g p = lim sup q p,n g n q. g p = lim inf n q. q p,n

9 OOT S BAIE, VISCOSITY SOLUTIONS OF OBSTACLE POBLEMS AND EFLECTED FBSDES 9 Proposition 4. Let u n n USC O T, O a locally compact subset of, F n n a sequence of functions such that each u n is a subsolution of F n : O T F n t, x, v, t v, D v = 0. Further, assume u n n and F n n are locally uniformly bounded. Then u is a subsolution of F t, x, v, t v, D v = 0 on O. The analogous statement holds for a sequence of LSC O T functions which are supersolutions. u = u then the convergence of u n to u = u is locally uniform. Further, if Proof. This is simply a restatement for parabolic PDEs of Proposition 4.3, Lemma 6.1 found in the User s guide [1 combined with their subsequent remarks: the definition of u as lim sup n,s,y t,x u n s, y guarantees the existence of sequences n j, t j, x j O T s.t. t j, x j t, x and u nj t j, x j u and it is clear that for every sequence O T t j, x j converging to t, x one has lim sup j u nj t j, x j u. This allows to apply [1, proposition 4.3 which implies that a, p, X P,+ u t, x there exists a sequence t j, x j O T, p j, m j P,+ O u n j t j, x j s.t. tj, x j, u nj x j, a j, m j t, x, u x, a, m P,+ O u t, x. However, each u nj is a subsolution so that for every j it holds that and hence from the definition of F it follows that F nj t j, x j, a j, p j, m j 0 F t, x, a, p, m 0. The proof for subsolutions follows the same argument. Finally, if u = u =: u then u must be continuous and the convergence u n u must be uniform as seen by [1, emark eflected FBSDEs and viscosity solutions of obstacle problems. Our goal is to show that knowing the oot barrier is equivalent to knowing the viscosity solution of a certain obstacle PDE, that is we are interested in the case F t, x, r, a, p, m = min r h x, a σ t,x where the initial condition u 0 and the barrier h are m given as the potential function of the initial and target measures and σ is continuous we discuss the motivation for looking for such a PDE solution at the beginning of Section 4. Definition 14. Let T 0,, u 0 : and h, σ [0, T. Denote with V T σ, u 0, h the possibly empty set of viscosity solutions { min u t, x h x, 10 t σ u t, x = 0, t, x 0, T u 0,. = u 0 x, x. ut,x which are of linear growth uniformly in t, i.e. t, x [0, T, sup t,x [0,T 1+ x <. We refer to this PDE as the obstacle problem on 0, T with barrier h. emark 7. F t, x, r, a, p, m = min r h x, a σ t,x m is proper and moreover the barrier h introduces an asymmetry in the definition of sub- and supersolutions: in this case Definition 1 says that u is a subsolution of 10 if u 0,. u 0. and for every t, x 0, T such that u t, x > h x we have 11 a σ t, x m 0 a, p, m P,+ u t, x. Whereas u is a supersolution if u 0,. u 0. and t, x 0, T 1 a σ t, x m 0 a, p, m P, u t, x and u t, x h x. We now discuss the abstract problem of existence and uniqueness of a viscosity solution to. First we recall the definition of a subclass of reflected forward-backward stochastic equation FBSDE [15, Section 8 applied with f = 0. Therefore fix T 0, and for each t [0, T denote with {F t s, t s T } the natural filtration of the Brownian motion {W s W t, t s T }, augmented by the P-null sets of F. Take t, x [0, T and define for s [t, T the forward-backward dynamics whose solution is the F-adapted quadruple X, Y, Z, K 13 Xs t,x = x + s σ T r, Xt,x t r dw r, Ys t,x = u 0 X t,x T + K t,x T Kt,x s T s Ys t,x h Xs t,x, t < s T and T t Zt,x Y t,x s r dw r, h X t,x s dks t,x = 0,

10 OOT S BAIE, VISCOSITY SOLUTIONS OF OBSTACLE POBLEMS AND EFLECTED FBSDES 10 where Ks t,x s [t,t is an increasing and continuous process verifying K t,x t = 0. In the above equation the meaning of the processes X and Y is clear and a rough interpretation of the processes Z and K is a follows: Z guides the evolution of Y via the Itô integral so that Y can hit the random variable u 0 X T at horizon time T. Note that both processes Y, Z are F t -adapted and so they do not see the future, nonetheless, u 0 X T is attained at time t = T. The process K ensures that Y does not go below the barrier h: K pushes Y upwards whenever Y touches and tries to go below the barrier h, else it remains inactive that is constant. Proposition 5 Stochastic representation for the obstacle PDE. Let u 0, h C,,σ C [0,,, h. u 0. and assume that for every t [0,, σ t,., h., u 0. are Lipschitz continuous σ uniformly in t. Moreover, assume that u0 x + σ t, x + h x sup sup <. 1 + x t [0, x Then there exists a v C [0,, which is a viscosity solution of { min v t, x h x, 14 t σ v t, x = 0, t, x 0, v 0, x = u 0 x, x. Moreover, for every T 0, 1 there exists t, x [0, T a unique quadruple Xs t,x, Ys t,x, Zs t,x, Ks t,x s [t,t of {Fs}-progressively t measurable processes which satisfies for some c T > 0 [ { sup E sup X t,x s + Y t,x } ˆ T s + Z t,x r dr + K t,x T c T 1 + x t [0,T s [t,t and this quadruple fulfills 13. Moreover, s, x [t, T the mapping s, x Xs t,x, Ys t,x is P-a.s. jointly continuous and in particular X t,x t, Y t,x t, Z t,x t, K t,x t 4 i.e. deterministic, t, x Y T t,x T t is the unique element of V T σ, u 0, h, 3 v [0,T t, x Y T t,x T t. Proof. Existence & uniqueness in [0, T, T < : Time reversion of see 9 and Lemma 4 shows that it is enough to deal with { min v t, x h x, t σ v t, x = 0, t, x O T = 0, T v T, x = u 0 x, x. where σ t, x = σ T t, x. Continuity, existence and uniqueness of the viscosity solution v follows from Lemma 8.4, Theorems 8.5 and 8.6 in [15 respectively. The linear growth of u in its spatial variable follows from standard manipulations for BSDEs. [15, Proposition 3.5 applied to the FBSDE setting above i.e. 13 with σ replaced by σ yields the existence of a constant k T > 0 such that t, x [0, T [ Y t,x t E Y t,x s k T E [ u0 X t,x [ T + E h + s, Xs t,x c T 1 + x sup s [t,t t sup s [t,t with h + := max {0, h} and where the last inequality follows from the linear growth assumptions on u 0 and h along with standard SDE estimates: sup t [0,T E [ X t,x T ĉ T 1 + x see e.g. equation 4.6 of [16. The solution to follows via our Lemma 4. Existence & uniqueness in [0, : Above gives a unique solution for every finite T > 0. First note that for T, T > 0, Y T t,x T t and Y T t,x T t coincide on [0, T T. By the comparison result in the appendix, Theorem 9, we can define v t, x as Y T t,x T t where T can be arbitrary chosen subject to T > t. Since the definition of a viscosity solution just relies on local properties, v is a viscosity solution of 14. emark 8. In the above link between the solutions of 13 and the time reversed version of 10, 14 one can in general not expect that the solutions of 14 exist in a classical sense. The following formal argument gives at least an intuition why FBSDE and obstacle PDEs are in a similar relation as SDEs and linear PDEs: suppose a sufficiently regular solution v of 14 exists. Via Itô s formula it follows that: Y t,x s = v s, Xs t,x, Z t,x s The last condition in 13 then reads as ˆ T t Y t,x r = σ x v s, Xs t,x and K t,x s = h Xr t,x dk t,x r = 0 iff ˆ T and the rhs explains the form of the PDE fulfilled by v. t ˆ s t t v σ v r, X t,x dr. [ v h t v σ r, v X t,x r dr = 0 r

11 OOT S BAIE, VISCOSITY SOLUTIONS OF OBSTACLE POBLEMS AND EFLECTED FBSDES 11 emark 9. Proposition 5 applies as well to moving barriers i.e. h could also depend on time however we only need barriers constant in time. Note also that the above shows { v [0,T } = VT σ, u µ, u ν but it does not show that {v } = V σ, u µ, u ν, viz. we only claim that v is a viscosity solution of the PDE 14 on the unbounded time domain but not that v is of linear growth uniformly in time however, we this will be a simple consequence of Theorem 5. The existence proof of 13 via penalization in [15 gives a lower bound for the first time the solution touches the barrier. This is intuitive clear but since it is useful for numerics we give the full proof. Corollary 1. Let v be the solution of 14 as given by Theorem 5 and let w 0 to be the unique viscosity solution 4 of { 15 t σ w 0 t, x = 0, on 0, T w 0 0, = u 0 Then w 0 t, x v t, x t, x [0, T. In other words, v hits the barrier h at time t only after w 0 has hit it at some time 0 s t. Furthermore, if σ is uniformly parabolic and bounded then w 0 C 1, 0, T, and can be written as a fundamental solution. Proof. The statement follows by noting that v is a viscosity solution of 14, hence a supersolution of the heat equation 15. Applying a standard comparison result immediately shows w 0 v. Another way to see this without resorting to comparison results and which gives another construction of a solutions to the obstacle problem is the following: the existence proof for 13 was done in [15 via penalization arguments. For n N consider the penalized obstacle problem t w n t, x 1 σ t, x w n t, x = n w n t, x h x, t, x O T, w n 0, x = u 0 x, x. where a := min {a, 0} for any a implying w n t, x h x := min {w n t, x h x, 0}. Intuitively, the solution of the penalized equation approximates that of as n since the term n w n h blows up for the points which do not satisfy the barrier condition and so in order for a well behaved solution to exist in the limit the barrier condition has to be verified. To shorten the discussion, along the proof it is shown the sequence w n n N converges and that w n u as n, i.e. the sequence is monotonic increasing see [15, Theorem 8.5. It follows that w 0 t, x u t, x for all t, x [0, T with w 0 the solution to the penalized PDE for n = 0, equation 15. Standard results see for example [18, Chapter 1, Section 7, Theorem 1 guarantee that under the stronger assumption of boundedness and uniform parabolicity of σ there exists a smooth solution and since every smooth solution is a viscosity solution the statement follows. 4. oot s barrier and the parabolic obstacle problem We are now in position to prove the main results: a one to one correspondence between regular oot barriers and viscosity solutions From the oot barrier to the obstacle problem. As pointed out in Section, potential functions are a powerful way to keep track of the evolution of the distribution of a local martingale. Let σ, µ, ν and consider the potential function of the stopped process X τ, namely u t, x = u X τ t, x E [ X τ t x. Tanaka s formula immediately gives u t, x = u µ x E [ L x t τ where L x t t,x denotes the local time of X t t, i.e. we see for every fixed x that t u t, x decreases until we have substracted enough such that it matches u ν x. While one cannot hope that u is a classical PDE solution due to the kinks that appear when u touches the barrier u ν we give a formal derivation for the PDE satisfied by u which then motivates our proof of Theorem 5. For brevity also assume σ 1 i.e. we embed into Brownian motion. Consider a sequence f n n C, such that f n. n. uniformly, hence f n. n sgn. and f n. n δ0.. Set 16 u n t, x := E [f n X τ t x. 4 This equation is a standard parabolic PDE whose unique solution is known to exist under the assumptions of Theorem 5, see e.g. the proof of Theorem 8.5 in [15 or think of the BSDE 13 when h = implying that K t = 0.

12 OOT S BAIE, VISCOSITY SOLUTIONS OF OBSTACLE POBLEMS AND EFLECTED FBSDES 1 and since f n is smooth, u n t, x = E [ f n X τ t x and from Itô s formula applied to the right hand side of 16 it follows that ˆ 17 u n t, x = f n y x µ dy 1 ˆ t E [ f n X r x 1 r τ dr. 0 Hence t u n t, x = 1 E [ f n X r x 1 t τr and putting the above together shows t u n 1 un = 1 E [ f n X τ t x f n X r x 1 t τr = 1 E [ f n X τ x 1 t>τ 0 on 0,. that is u is a supersolution of the heat equation. Further, equation 17 applied with t = and using that ν also gives X τ u n t, x u ν x = 1 ˆ t E [ f n X r x 1 r τ dr. Now for t, x, also s, x for s t and one expects that f n X τ s supp { f n x} {x}, hence the Itô correction vanishes and above reads in the limit as n as u = u ν on. x 0 a.s. since for n big enough Now if t, x c and t τ then the properties of the oot barrier imply that X τ x, hence f n X τ x 1 t>τ 0 for n big enough. We therefore arrive as n at t u = u for t, x c, u 0, x = u µ x. To conclude this very formal argument note that c = [0, and therefore putting all the above shows us that min u u ν, t u u = 0 on 0,, u 0, = u µ. The proof of Theorem 5 makes this type of reasoning precise via the method of semi-relaxed limits. Theorem 5. Let σ, µ, ν be either as in Theorem 3 or as in Theorem 4. Let σ, µ, ν and denote with u the potential function of the stopped process X τ = X t τ t, i.e. u t, x := u X τ t, x := E [ X τ t x. Then {u} = V σ, u µ, u ν, in other words, the potential function of X τ is the unique viscosity solution of linear growth of the obstacle problem 10 with initial data u µ, barrier u ν and heat coefficient σ, { min u u 18 ν, t u σ u = 0, on 0, u 0,. = u µ.. Moreover, 1 for every x t u t, x is non-increasing and u ν x u t, x u µ x t, x, u = u ν and t, x lim t u t, x = u ν x, 3 u is Lipschitz in space uniformly in time, sup sup t [0, x y u t, x u t, y x y Proof. We have to show that u is a viscosity solution of 18 and to show uniqueness of the solutions via the comparison theorem given in the appendix we also need to establish that u is of linear growth. First note, that to verify that u is the viscosity of 18 it is enough to show that in viscosity sense 19 <. u u ν 0 on [0,, t σ u 0 on [0,, u u ν = 0 on, t σ u = 0 on c,

13 OOT S BAIE, VISCOSITY SOLUTIONS OF OBSTACLE POBLEMS AND EFLECTED FBSDES 13 The first inequality follows immediately via conditional Jensen 0 u t, x = E [ X τ t x E [E [ X τ x F t τ = E [ X τ x = u ν x. For the other inequalities we approximate 18 and conclude by the Barles Perthame method of semi-relaxed limits. Step 1. u n t, x := E [ψ n X τ t x. Define the sequence ψ n C, as ψ n x = ˆ x ˆ y 0 0 nφ nz dzdy x where φ is the usual Gaussian scaled to the unit disc { exp 1 1 x for x < 1 φ x = 0 otherwise. Especially note that ψ n.. uniformly, ψ n. is continuous, 0 ψ n O n and supp ψ n [ 1 n, 1 n we could replace ψ n by any other sequence with this properties. Further introduce Since ψ n.. uniformly we have P-a.s. u n t, x = E [ψ n X τ t x, ˆ u n ν x = ψ n y x ν dy, ˆ u n µ x = ψ n y x µ dy. ψ n X τ.. X τ.. ;[0, = sup ψ n X τ t x X τ t x n 0 t,x [0, hence we get uniform convergence of u n, u n µ and u n ν, i.e. 1 Denote the complement of with u n u ;[0, n 0, u n ν u ν, n 0, u n µ u µ, n 0. c = [0, [, \ and note that since is closed, c is open. Further, by dominated convergence we have x, n N that lim t u t, x = u µ x and lim u n t, x = u n µ x. t Step. u u ν = 0 on. Fix x and apply the Itô-formula to ψ n. x and the local martingale X τ. After taking expectations and using Fubini s theorem we arrive at ˆ t [ σ u n t, x = u n r, X r 3 µ x E ψ n X r x 1 r<τ dr. Taking lim t on both sides in the above implies by that ˆ [ σ 4 u n ν x = u n r, X r µ x E ψ n X r x 1 r<τ dr. 0 0 Since σ ψn 0, 3 shows also that u n is non-increasing in time and by subtracting 4 from 3 we arrive at ˆ [ σ u n t, x u n r, X r ν x = E ψ n X r x 1 r<τb dr 0. t The uniform convergence given in 1 implies that t u t, x is non-increasing, i.e. t, x [0, we have that u t + r, x u t, x r 0. Fix a point t, x o o denoting the interior of. Because is a oot barrier this implies that r t also r, x o and since o is open we conclude that n 0 s.t. {r, y [0, [, : r t, x y < 1n0 } o. Since supp ψ n [ n 1, n 1 this shows ψ n X r x 1 r<τ = 0 for all r t and all n > n 0, hence ˆ [ σ lim n un u n r, X r ν t, x = E lim n ψn X r x 1 r<τ dr = 0 t, x. t

14 OOT S BAIE, VISCOSITY SOLUTIONS OF OBSTACLE POBLEMS AND EFLECTED FBSDES 14 By continuity of u and u ν we can conclude that u = u ν. Step 3. t σ u 0 on [0,. From 3 it follows that u n t, x has a right- and left- derivative t, x [0, ; to see this take t+ u n t, x = lim ɛ 1 u t + ɛ, x u t, x ɛ 0 [ ˆ t+ɛ = E lim ɛ 1 σ r, X r ψ n X r x 1 r<τ dr ɛ 0 t [ σ t, X t = E ψ n X t x 1 t<τ similarly it follows that the left derivative t u n is given as [ σ 5 t u n t, X t t, x = E ψ n X t x 1 t τ. Note that for every t [0, t u n t,., t+ u n t,. C, ; further, since σ t,x t ψ n X t x is nonnegative we conclude t u n t, x t+ u n t, x 0. Note that for n < we cannot expect the above inequality to be an equality; e.g. consider ν = δ 0, µ = N 0, 1, σ 1 which is solved by = {t, x : t 1}, hence τ 1. From the definition of u n it follows that we can exchange differentiation in space and expectation to arrive at 6 σ t, x u n t, x = σ t, x E [ ψ n X τ t x 0 on [0,, which is a continuous function in t, x. Lemma 5 shows that a, p, m P, O un t, x Hence, by 6 and 5 a t u n t, x and m u n t, x. a m t u n t, x σ t, x u n t, x Splitting the term inside the expectation gives = 1 E [ σ t, x ψ n X τ t x σ t, X t ψ n X t x 1 t τ. t u ɛ t, x m 1 E [ σ t, x ψ n X τ t x σ t, x ψ n X t x 1 t τ }{{} =:I nt,x + 1 E [ σ t, x ψ n X t x 1 t τ σ t, X t ψ n X t x 1 t τ }{{} =:II nt,x = 1 I n t, x + 1 II n t, x. We conclude that u n is a supersolution of t σ u 1 I n + II n = 0 on 0,. Further, I n t, x = E [ σ t, x ψ n X τ t x σ t, x ψ n X t x 1 t τ = σ t, x E [ ψ n X τ x 1 t>τ 0 hence u n is also a viscosity supersolution of { t u σ u 1 II n = 0 on 0, u 0,. = u n µ.. Using the Lipschitz property of σ, supp ψ n = [ n 1, n 1 and that ψ n c n we estimate II n t, x E [ σ t, x σ t, X t ψ n X t x 1 t τ σ Lip n σ LG 1 + x + n E [ ψn X t x 1 t τ 4σ Lip σ LG x + 1 n n E [ cn1 Xt x n 1 = 4cσ Lip σ LG x + n P X t x n 1

15 OOT S BAIE, VISCOSITY SOLUTIONS OF OBSTACLE POBLEMS AND EFLECTED FBSDES 15 σ Lip and σ LG as defined in 4 and for the second inequality we use the trivial estimate σ t, x σ t, X t σ t, x σ t, X t σ t, x + σ t, X t combined with x X t n, Lipschitzness and linear growth of σ. Now for every compact K [0, we have lim sup n t,x K P [ X t x n 1 = 0 since by Assumption 1 the process X has a continuous density with respect to the Lebesgue measure on, B and P [ X t x n 1 = ˆ x+ 1 n 1 n x 1 n ˆ y x f t, y dy f t, y dy 0 as n uniformly in t, x, therefore II n 0 locally uniformly on [0,. By step 1, u n u and u n 0,. u µ. locally uniformly as n. The stability of viscosity solutions, Proposition 4, implies that u is a viscosity supersolution of { 7 t σ u = 0 on [0, u 0,. = u µ. which proves the desired inequality. Step 4. t σ u = 0 on c. From Lemma 5 it follows that if t u n t, x < t+ u n t, x then P,+ u n t, x = in which case we are done and if t u n t, x = t u n t, x = t+ u n t, x then a, p, m P,+ u n t, x, a = t u n t, x and m u n t, x. Hence, in the latter case we have a, p, m P,+ u n t, x that a m t u n t, x σ t, x u n t, x. Proceeding as in step 3 we see that u n is a subsolution of { t u σ u 1 I n + II n = 0 on 0, u 0,. = u n µ.. We already know that II n 0 locally uniformly and now show that I n 0 locally uniformly on c : since is a oot barrier we have τ + r, X τ r 0, hence if t, x c and t τ then X τ x. Therefore lim sup n t,x K ψ n X τ x 1 t τ = 0 for every compact K c which is enough to conclude that I n converges locally uniformly on c to 0, i.e. for every compact K c lim sup I n t, x [ σ n,k lim E sup ψ n X τ x 1 t τ t,x K n t,x K = 0. The stability results, Proposition 4, implies that u is a subsolution of t σ u = 0 on c, u 0, = u µ. Together with the result from step 3 this shows that u is a solution in viscosity sense. Step 5. Putting the above together shows that u is a viscosity solution of the obstacle problem 18. To see that u is of linear growth, recall that by definition u 0, x = u µ x and in step we showed that x, t u t, x is non-increasing. Further, we know u t, x u ν x t, x from the estimate 0, hence u t, x u µ x + u ν x. This together, with the properties of potential functions, Proposition 1 and Lemma ut,x, implies sup t,x [0, 1+ x <. This shows that the assumptions of the comparison result, Theorem 9, are met and we conclude that {u} = V σ, u µ, u ν. { For a function v C 1, 0,, one has P, v t, x P,+ v v t, x = t t,x, v x t,x, v x t,x }. The lemma below is a generalization of this for functions which are only left- and right-differentiable which appear in the proof above.