Optimal Martingale Transport problem in higher dimensions

Transcription

1 University of British Columbia Based on joint work with Young-Heon Kim & Tongseok Lim (UBC) September, 2015 Varese

2 Main Points of the talk: Comparison between optimal transport problem and optimal martingale transport problem. The one-dimensional case. Optimal martingale transport problem and Skorohod embeddings in Brownian motion. Recent progress on the optimal martingale transport problem in higher dimension. Conjectures / open problems.

3 Optimal Transport Problem Given two Borel probability measures µ, ν on R d, one considers the set Π(µ, ν) of probability measures π on R d R d whose marginals are µ, ν. They are called a transport plans from µ to ν. Given a cost function c : R d R d R, one needs to characterize the optimal transfer plans, that is those π Π(µ, ν) that minimize or maximize Motivation: min π Π(µ,ν) X Y c(x, y)dπ(x, y). [Economics] find a most cost-effective matching between resources/agents: [Physics] density functional theory: describe the state of many correlated particles. [Mathematics] Appropriate distances between measures? Notions of barycenters of a family of measures?

4 Equivalent probabilistic statement A coupling of µ and ν is a pair of random variables X : Ω R d, Y : Ω R d on a probability space (Ω, F, P) such that Law(X) = µ, Law(Y ) = ν. An equivalent problem is to study the couplings that minimize the expected cost min X µ,y ν E Pc(X, Y ). Existence of optimal π usually follows easily from standard compactness. When is an optimal measure π unique? When is an optimal measure π concentrated on a graph {(x, T (x))}, where T : X Y is a map? Originated by Monge (19th century) and Kantorovitch (1940 s). Brenier 87, Gangbo 95, Caffarelli 96, Gangbo-McCann 96, Levin 96, Ma-Trudinger-Wang 05,... : Assume µ << dx and that c satisfies the twist condition, i.e,: y D xc(x, y) is injective (e.g. c(x, y) = x y 2.) Then π is concentrated on a graph and is unique.

5 A dual problem There is an important dual problem (Monge-Kantorovich) { } c(x, y)dπ = sup β(y)dν α(x)dµ; (β, α) D(c, µ, ν) min π Π(µ,ν) X Y D(c, µ, ν) is the set of (α, β) L 1 (µ) L 1 (ν) so that Y β(y) α(x) c(x, y) for all (x, y) X Y. If c(x, y) = x, y, then α can be taken to be any convex function and β = α, the Legendre transform of α (Fenchel-Young inequality).. X

6 Optimal Martingale Transport Problem Consider now Borel probability measures µ, ν on R d in convex order; µ C ν, that is φ dµ φ dν for all φ : R d R convex. R d R d Consider the set MT (µ, ν) of probability measures π on R d R d with marginals µ, ν, but also whose disintegration (π x) x R d satisfy that the barycenter of each π x is x (martingale constraint). These are the martingale transport plans. Recall that the disintegration (π x) x R d of π with respect to the first variable is: dπ(x, y) = dπ x(y)dµ(x). The basic problem is to characterize the optimal solutions of the maximization or minimization problem min c(x, y)dπ(x, y). π MT (µ,ν) R d R d Motivation: [Finance] find the maximum / minimum price of the option whose value depends on the stock π, given the information that can be observed from the market (the marginals).

7 Equivalent probabilistic statement and duality Consider all couplings X : Ω R d, Y : Ω R d on a probability space (Ω, F, P) that form a 1-step martingale, that is Law(X) = µ, Law(Y ) = ν, E(Y X) = X. Note that for B R d, π x(b) = P(Y B X = x). Study the (1-step) martingales (stocks) (X, Y ) with prescribed marginals, which maximize / minimize the expected cost (option price) We also have a duality problem { } min c(x, y) dπ; π MT (µ, ν) R 2d min E P c(x, Y ). X µ,y ν,e(y X)=X { } = sup βdν αdµ; (α, β) D m(c) R d R d D m(c) is the set of (α, β) L 1 (µ) L 1 (ν) so that for some γ C b (R d ), β(y) α(x) γ(x)(y x) c(x, y) for all (x, y) R d R d.

8 1-dimensional results Theorem (Beiglböck-Juillet 13) Suppose µ C ν on R and µ << L 1. Let c(x, y) = x y. Then, there exists a unique minimizing martingale transport plan π that is concentrated on a set Γ R R such that Γ x 3 for every x R. More precisely, π can be decomposed into π stay + π go, where π stay = (Id Id) # (µ ν) (this measure is concentrated on the diagonal of R 2 ) and π go is concentrated on graph(t 1 ) graph(t 2 ) where T 1, T 2 are two decreasing real functions. Theorem (Hobson-Neuberger 13) Suppose µ C ν on R and µ << L 1. Let c(x, y) = x y. Then, there exists a unique maximizing martingale transport plan π that is concentrated on a set Γ R R such that π is concentrated on Γ and Γ x 2 for every x R. More precisely, π is concentrated on graph(t 1 ) graph(t 2 ) where T 1, T 2 are two increasing real functions. Question What is a right extension of these results to higher dimensions?

9 Equivalent problem; a.k.a. the Skorokhod embedding problem Let (Ω, F, F t, P) be a probability space with Brownian motion B x t valued in R. If µ and ν are in convex order on R, then there exists a (randomized) stopping time τ such that Law(B 0 ) = µ and Law(B τ ) = ν. So, given a cost function c : R d R d R, it suffices to study the optimal stopping time τ which "embeds" µ to ν, that is: min E Pc(B 0, B τ ). B 0 µ,b τ ν Problem What can be said on the optimal time? Is it unique? Is it non-random? Is it a hitting time of a "barrier"?

10 The higher-dimensional case is much richer Suppose now that µ and ν are such that R d φ(y) dµ(y) R d φ(y) dν(y) for all φ subharmonic on R d, then we say that µ SH ν in the subharmonic order. If µ = δ x, then x is the subharmonic barycenter for ν. The corresponding problem is then min π SHMT (µ,ν) R d R d c(x, y)dπ(x, y) where SHMT (µ, ν) is set of probability measures π on R d R d with marginals µ, ν and such that their disintegration (π x) x R d have subharmonic barycenter at x. In R 1 every convex function is subharmonic and vice versa. In higher dimension, genuine differences between MT and SHMT problem emerge; in particular, not every measure in P(R d ) has a subharmonic barycenter. This stringent constraint on π x makes it difficult to vary a given disintegration (π x) x R d to another, in order to use "calculus of variation". A similar set of problems if one consider plurisubharmonic functions.

11 However, we have the Brownian approach If B x t is an R d -valued Brownian motion starting at x R d, and if f is a subharmonic function on R d, then f (B x t ) t 0 is a submartingale. If τ is a stopping time, and ν is the distribution ν of B τ, where B is Brownian motion starting at µ, then µ SH ν in the subharmonic order. Vice-versa, if µ SH ν, then there exists a Brownian motion starting at µ and a randomized stopping time τ such that ν B τ. Theorem (Ghoussoub, Kim & Lim, to appear) Let µ and ν be radially symmetric probability measures on R d which are in subharmonic order. Then, there exists a stopping time τ that minimizes min E P B 0 B τ. B 0 µ,b τ ν Moreover, it is of barrier-type, and in particular, it is unique. Same for the case of a plurisubharmonic order. Deep problem: Can one do without assuming radial symmetry?

12 Back to the convex order in higher-dimensions The first result in higher dimension is due to Lim. He considers the case when the marginals are radially symmetric. Theorem (Lim 14) Assume that µ and ν are radially symmetric probability measures in convex order on R d, and that µ << L d. Consider the minimization problem. Then: Under any optimal solution π, the common mass µ ν does not move. WLOG suppose µ ν = 0. Then for µ a.e. x, the disintegration (conditional probability) π x is concentrated on two points; say {T 1 (x), T 2 (x)}, where T 1 (x), T 2 (x) lie on the 1-dimensional subspace spanned by x. In particular, the optimal solution π is unique. Idea By using radial symmetry, one can use variational methods and show that the disintegrations (π x) x R d must be supported on the one-dimensional subspace spanned by x. Then one can reduce the problem to one-dimensional situation. Remark Optimal solutions in the maximization problem behave very differently even in the case of radially symmetric marginals. Main problem What if µ and ν are non-radial?

13 Two conjectures Let Γ R d R d be a support of an optimal martingale measure π. Γ x is the transported image of x; Γ x := {y R d ; (x, y) Γ}, that is a support of π x for each x. What can be said about the "geometry" of the fiber Γ x, for each x? Assume that c(x, y) = x y, µ << L d. Conjecture 1: If π is an optimal martingale solution for either maximization or minimization problem. Then, for µ almost every x, Γ x is supported on the set of extreme points) of the closed convex hull of Γ x, i.e., ( ) Γ x Ext conv(γ x). Conjecture 2: If µ ν = 0 and π is an optimal martingale solution for the minimization problem. Then for µ almost every x, the set Γ x consists of k + 1 points that form the vertices of a k-dimensional polytope, where k := k(x) is the dimension of the linear span of Γ x and therefore, the minimizing solution is unique.

14 Back to optimal transport Let X, Y be arbitrary sets, and c : X Y R be a function. A subset Γ X Y is said to be c-cyclically monotone if, for any N N, and any family (x 1, y 1 ),..., (x N, y N ) of points in Γ, holds the inequality N c(x i, y i ) i=1 N c(x i, y i+1 ) i=1 (with the convention y N+1 = y 1 ). A transport plan is said to be c-cyclically monotone if it is concentrated on a c-cyclically monotone set. Roughly, it is a plan that cannot be improved: it is impossible to perturb it (in the sense considered before, by rerouting mass along some cycle) and get something more economical. Theorem A transfer plan π is optimal if and only if its support Γ is c-cyclically monotone.

15 Dual statement to the c-cyclical monotonicity If Γ X Y, we let X Γ be the projection of Γ onto the first coordinate space R d and Y Γ to the second. Theorem Γ X Y is c-cyclically monotone if and only if for each finite subset H Γ, there exist a pair of functions α H : X H R, β H : Y H R, such that β H (y) α H (x) c(x, y) x X H, y Y H, β H (y) α H (x) = c(x, y) (x, y) H. Remark This follows from the duality theorem in finite-dimensional linear programming. Question Can one find a dual pair (α, β) that works for the whole set Γ? Answer Yes; for example for the canonical cost c(x, y) = 1 2 x y 2, this follows from Rockafellar s theorem. Theorem (Rockafellar 69) Suppose that Γ R d R d is a cyclically monotone set. Then there exists a convex function α on R d such that Γ α.

16 Implication of Rockafellar s theorem Let c(x, y) = 1 2 x y 2 and let Γ R d R d be c-cyclically monotone set. Rockafellar s theorem yield convex functions (α, β) on R d such that β(y) α(x) 1 2 x y 2 x R d, y R d, β(y) α(x) = 1 2 x y 2 (x, y) Γ. In particular, whenever (x, y) Γ, we have α(x) x y 2 α(x ) x y 2 x R d. Now if α is differentiable at x, then x(α(x) x y 2 ) = 0 y = x + α(x). The presence of regular solution for the dual problem implies good properties on the optimal transport plans.

17 Monotonicity principle for martingale transport problem Definition: Say that 2 measures π 1, π 2 on R d R d are competitors, if they have the same marginals and if the barycenters of their respective disintegrations are the same: R d y dπ 1 x (y) = R d y dπ 2 x (y), x R d. Definition: Say that Γ R d R d is martingale-monotone if for any finite subset H Γ and any measure π H supported on H, we have for any competitor π of π H, c(x, y)dπ H (x, y) c(x, y)dπ (x, y) Theorem(Monotonicity principle) (Beiglböck-Juillet 13) Let π be an optimal solution for martingale transport problem. Then there exists a martingale-monotone support Γ of π. An Interpretation of martingale-monotonicity: spt(π H ) Γ means that π H is a subplan" of the full transport plan π A competitor means that if we change the subplan π H to π, then the martingale structure of π H will not be disrupted. Now if c dφ > c dψ, then we may modify π to have ψ as its subplan, achieving less cost, meaning that the current plan π is not a minimizer.

18 Dual statement of the Monotonicity principle Let c : R d R d R be a cost function. Say that a subset G of R d R d admits a c-dual, or that G is c-dualizable, if there exists a triple {α, β, γ}, α : X G R, β : Y G R, γ : X G R d, such that the following duality relation (for the minimization problem) holds: β(y) α(x) γ(x) (y x) c(x, y) x X G, y Y G, (0.1) β(y) α(x) γ(x) (y x) = c(x, y) (x, y) G. (0.2) In the maximization problem, the inequality in (0.1) is reversed. We will call the triple {α, β, γ} a c-dual for G. Theorem Suppose Γ R d R d is martingale-monotone. Then any finite subset H Γ admits a c-dual. Question Can one then find a dual triplet (α, β, γ) for the whole martingale monotone set Γ? Is there a Rockafellar-type theorem? Answer No. At least in the maximization case.

19 Dual problem is NOT attained in general Example Let µ = ν be two identical probability measures on the interval [0, 1], then the only martingale (say π) from µ to itself is the identity transport, hence it is obviously the solution of the maximization problem with respect to the distance cost, and its support is Γ = {(x, x) : x [0, 1]}. If now {α, β, γ} is a solution to the dual problem, then β(y) x y + γ(x) (y x) + α(x) x [0, 1], y [0, 1]; β(y) = x y + γ(x) (y x) + α(x) (x, y) Γ. The above relations easily yield that for any 0 < a < b < 1, we have γ(a) + 2 γ(b), which means that it is impossible to define a suitable real-valued function γ for a.e. x in [0, 1]. Remark In the case of the minimization problem, the triplet (0, 0, 0) is a c-dual. Conjecture 3: In the minimization problem, the dual problem is always attained.

20 Good things happen if duality is attained Lemma: Suppose Γ R d R d admits a dual triplet (α, β, γ). If α and γ are differentiable at x, then Γ x is supported on the set of extreme points of the closed convex hull of Γ x. Proof: If (x, y) Γ, then x y + γ(x) (y x) + α(x) x y + γ(x ) (y x ) + α(x ) x Hence x( x y + γ(x) (y x) + α(x)) = x y + γ(x) (y x) γ(x) + α(x) = 0. x y Now suppose that we can find {y, y 0,..., y s} Γ x with y = Σ s i=0p i y i, Σ s i=0p i = 1, p i > 0. Then we get x y s x y = i=0 p i x y i x y i. But this can hold only if all y i lie on the ray emanating from x.

21 Once there is a dual, it can be extended & improved Lemma: Suppose Γ admits a c-dual triplet {α, β, γ}. Set Ω := int(conv(y Γ )). Then, there exists α : Ω R, β : R d R, and γ : Ω R d such that {α, β, γ} coincides with {α, β, γ} on X Γ, Y Γ, X Γ, respectively, (thus, is a c-dual triplet for Γ), and α is locally Lipschitz. Furthermore, γ is differentiable at points where α is differentiable. Proof. We define the extensions as α(x) := inf{c R : a R d such that β(y) x y a (y x) + c y Y Γ } γ(x) := {a R d : β(y) x y a (y x) + α(x) y Y Γ β(y) x y = a (y x) + α(x) y Γ x} Then... β(y) := inf { x y + γ(x) (y x) + α(x)}. x Ω

22 The conclusion in the presence of a dual Theorem Let π be an optimal solution for martingale transport problem and let Γ be its support. Suppose that µ << L d and Γ admits a dual. Then for µ a.e. x, ( ) Γ x Ext conv(γ x). We can find a triplet (α, β, γ) such that α is locally Lipshitz, hence differentiable for a.e x. Since µ << L d, it is differentiable for µ a.e x, hence Conjecture 1 holds. As in the optimal transport case, exploiting the dual side of the problem can be powerful. However, the dual is not always attained in the martingale transport case, at least in the maximization problem.

23 Splitting the problem We then try to partition the support Γ of a martingale transport into subsets so that the restricted Martingale transport on each component attains its dual. Theorem: Decomposition into irreducible components: Let Γ be a support of a martingale transport. Then, there exists an equivalence relation on X Γ such that or each x X Γ, there exists an open convex subset C(x) of IC(Y Γ ) such that 1. x x if and only if C(x) = C(x ); 2. If x x, then C(x) C(x ) =. 3. If Γ is martingale monotone (hence its finite subsets admit c-duals), then for each x X, Γ (C(x) R d ) admits a c-dual. Γ (C(x) R d ) is the largest subset of Γ for which one can get a dual. But in the minimization problem with distance cost, we suspect that the dual problem is always attained for the whole optimal support Γ.

24 Disintegration of martingale plans Let (µ, ν) be probability measures on R d in convex order, and π MT (µ, ν). Assume further that µ ν = 0 in the case of minimization problem. Then, there exists a support Γ of π and a probability measure π on K(R d ) such that: 1. For each Borel set S R d R d, we have π(s) = π C (S)d π(c), K(R d ) where for π-a.e. C, π C is a probability supported on Γ C := Γ (C R d ). 2. For π-a.e. C, there exist probability measures µ C, ν C such that (µ C, ν C ) is in convex order, µ C is supported on X C := X C, ν C on Y ΓC, and π C MT (µ C, ν C ); 3. If π is optimal on MT(µ, ν), then for π-a.e. C, π C is optimal for MT (µ C, ν C ), and its corresponding dual problem is also attained, that is Γ C := Γ (C R d ) has a c-dual. 4. If µ C is absolutely continuous with respect to Lebesgue measure on V (C), and c(x, y) = x y, then for µ C -almost all x, Γ x Ext(conv(Γ x)).

25 If µ could be disintegrated along the partitions in an a.c. way, we are done. But... We know that each piece Γ (C(x) R d ) admits a dual. So if the source measure µ were "restricted" (disintegrated) on each partition in an absolutely continuous way, then one would apply the previous theorem on each partition and conclude that, for µ a.e. x, Γ x Ext ( conv(γ x) ). Unfortunately, this is not always the case. A counterexample (Ambrosio, Kirchheim, and Pratelli) They constructed a Nikodym set in R 3 having full measure in the unit cube. Yet, it intersects each element of a family of pairwise disjoint open lines only in one point.

26 Nikodym set connected with MG transport; unhappy disintegration can indeed happen Consider the following obvious inequality, 1 2ɛ ( x y ɛ)2 0, hence 1 2ɛ y 2 x y + 1 ɛ x (y x) + 1 2ɛ x 2 ɛ. (0.3) Letting α ɛ(x) = 1 2ɛ x 2 ɛ, β ɛ(y) = 1 2ɛ y 2 and γ ɛ(x) = 1 x, this is a dual ɛ relation for the maximization problem. It shows that every martingale π ɛ := (X, Y ) with X Y = ɛ a.s. is optimal with its own marginals X µ and Y ν. Fix ɛ > 0 and let X be a random variable whose distribution µ has uniform density on [ 1, 1] 3. Define Y conditionally on X by evenly distributing the mass along the lines l x in the Nikodym set at distance ɛ. That is Y splits equally in two pieces from x X along l x with distance ɛ. Then the martingale (X, Y ) is optimal for the maximization problem. But each equivalence class [x] is the singleton {x}, so the disintegration of µ along the partitions C(x) is the Dirac mass δ x, which is obviously not absolutely continuous w.r.t. L 1. Hence, the decomposition is not useful in this case. Note that the martingale (X, Y ) defined here has codimension 2.

27 The codimension 1 case Theorem Let c(x, y) = x y, and let π MT (µ, ν) be an optimal solution with a support Γ. Assume µ << L d and suppose that for µ a.e. x, dim(v (C(x))) d 1. Then for µ almost every x R d, Γ x Ext(conv(Γ x)). Idea If each convex partition {C(x)} x X has codimension at most 1, then by mutual disjointness, their relative "direction" cannot be as wild as the Nikodym set case. In fact we can define a bi-lipschitz map by which the partitions are parallelized, so the disintegration of µ can be done in an a.c. way. Corollary Let π be a solution of the optimization problem with c(x, y) = x y and Γ be its support. Suppose µ << L d. Then for µ-almost every x, the Hausdorff dimension of Γ x is at most d-1.

28 Conjecture 2 in the case of a discrete target Theorem Let c(x, y) = x y, suppose µ << L d and that ν is discrete; i.e. ν is supported on a countable set. Then for µ a.e. x, Γ x consists of exactly d + 1 points which are vertices of a polytope in R d, and therefore the optimal solution is unique. Remark The result holds for both max/min problem. Note that not only the µ << L d, but also the "singular" nature of ν help here. This result along with the radial symmetric case.

29 What is unsatisfactory? The codimension 1 assumption is not "practically checkable", unless d = 2. We do not have an example of non-existence of a dual for minimization problem. Even though we have an example of non-dual existence for maximization problem, still we do not have an example of optimal solution which is NOT supported on the Choquet boundary; hence the conjecture may still be valid for the maximization problem as well. For conjecture 2, we have no idea. More precisely, we do not understand what really distinguishes the min / max problems.

30 Thank You!