Martingale optimal transport with Monge s cost function

Transcription

1 Martingale optimal transport with Monge s cost function Martin Klimmek, joint work with David Hobson (Warwick) klimmek@maths.ox.ac.uk Crete July, 2013

2 ca c(x, y) = x y No splitting, transport along straight lines, some paths are forbidden

3 The Monge transport problem Monge s earthwork problem in one dimension: Transport earth from a given area to a target site in a way that minimises the cost of carriage. inf s:s µ=ν R x s(x) µ(dx). The s µ are push-forward maps of µ through s. s µ(u) = µ(s 1 (U)) for Borel sets U R.

4 The Kantorovich relaxation and duality Let C(µ, ν) = {(X, Y ) : X µ, Y ν}. The Kantorovich relaxation (primal problem): The dual problem: { sup Ψ,Φ Ψ(x)µ(dx) + R inf E[c(X, Y )]. (X,Y ) C(µ,ν) R } Φ(y)ν(dy) ; Ψ(x) + Φ(y) c(x, y).

5 Monge s bakeries tavernas Ψ(x) : Cost of calamari from fisherman x Φ(y) : Price of calamari sold at taverna y minimise costs: or maximise profits: { sup Ψ,Φ Ψ(x)µ(dx) + R Ψ(x) + Φ(y) c(x, y) x, y. inf E[c(X, Y )]. (X,Y ) C(µ,ν) R } Φ(y)ν(dy) ; Ψ(x) + Φ(y) c(x, y).

6 Monotone structure Define the c-convex dual of f to be f c (y) = sup x {c(x, y) f (x)}. f is c-convex if f = (f c ) c. Gangbo and McCann: If c(x, y) = f ( y x ) (f strictly convex), the optimal coupling is given by (X, Y ) = (X, s(x )) where s is the generalized subdifferential of a c-convex function Ψ satisfying. Ψ (x) = c x (x, s(x)).

7 Martingale coupling M(µ, ν) = {(X, Y ) : X µ, Y ν, E[Y X ] = X }. The primal problem: inf E[c(X, Y )]. (X,Y ) M(µ,ν)

8 Dual (Lagrangian) formulation { } min y x ρ(dx, dy) subject to ρ ρ(dx, dy) = ν(dy), ρ(dx, dy) = µ(dx), x y (y x)ρ(dx, dy) = 0. y Lagrangian objective function: L(x, y) = y x α(y) β(x) θ(x)(x y).

9 Economic motivation Let S be set of triples (α, β, γ) such that L(x, y) 0 where L(x, y) = y x α(y) β(x) θ(x)(x y). For such triples in S, E[ Y X ] β(x)µ(dx) + α(y)ν(dy) (F t ) T0 t T 1 forward price process (a martingale). Suppose marginal laws F T0 µ and F T1 ν are known. Purchase β(f T0 ) and α(f T1 ) Sell forward θ(f T0 ) of stock over [T 0, T 1 ] Payoff is α(f T1 ) + β(f T0 ) θ(f T0 )(F T1 F T0 ) F T1 F T0 The strategy is a sub-hedge.

10 Economic motivation Let S be set of triples (α, β, γ) such that L(x, y) 0 where L(x, y) = y x α(y) β(x) θ(x)(x y). For such triples in S, E[ Y X ] β(x)µ(dx) + α(y)ν(dy) (F t ) T0 t T 1 forward price process (a martingale). Suppose marginal laws F T0 µ and F T1 ν are known. Purchase β(f T0 ) and α(f T1 ) Sell forward θ(f T0 ) of stock over [T 0, T 1 ] Payoff is α(f T1 ) + β(f T0 ) θ(f T0 )(F T1 F T0 ) F T1 F T0 The strategy is a sub-hedge.

11 Guess and check L(x, y) = y x α(y) β(x) θ(x)(x y) L(x, x) = L(x, p(x)) = L(x, q(x)) = 0 L(x, y) = y x α(y) + α(x) θ(x)(x y)

12 Map η = (µ ν) + to γ = (ν µ) + z a z a f η (u)du = uf η (u)du = q(z) q(z) a f γ (u)du + uf γ (u)du + p(z) a p(z) f γ (u)du uf γ (u)du

13 Differential equations Map η = (µ ν) + to γ = (ν µ) + with constraints z a z a f η (u)du = uf η (u)du = q(z) q(z) a f γ (u)du + uf γ (u)du + p(z) a p(z) f γ (u)du, uf γ (u)du, where p : [a, b] (, a], q : [a, b] [b, ). Differentiate to obtain a pair of differential equations: p (x) = q (x) = q(x) x f µ (x) f ν (x) q(x) p(x) f µ (p(x)) f ν (p(x)), x p(x) f µ (x) f ν (x) q(x) p(x) f µ (q(x)) f ν (q(x)).

14 Deterministic transport Optimal martingale transport The optimal coupling The optimal sub-hedge picture R Better: A potential R Let D(x) = (y x)+ ν(dy ) (y x)+ µ(dy ) Manipulating the two equations a bit, find that (p, q) is given by D 0 (x) = D 0 (p(x)) + D 0 (q(x)) D(x) xd 0 (x) = D(q(x)) q(x)d 0 (q(x)) + D(p(x)) p(x)d 0 (p(x)) Example

15 Deterministic transport Optimal martingale transport The optimal coupling The optimal sub-hedge Example

16 The optimal coupling Theorem There exists a unique pair of decreasing functions (p, q) such that if If X µ and Y {p(x ), X, q(x )} with E[Y X ] = X then Y ν and (X, Y ) minimises E[ Y X ] over all martingale couplings. f η (x) q(x) x q(x) p(x) I {y=p(x)} y < x, ρ(x, y) = f µ (x) f η (x) y = x, f η (x) x p(x) q(x) p(x) I {y=q(x)} y > x.

17 The dual picture L(x, y) = y x α(y) + α(x) θ(x)(x y) L(x, x) = L(x, p(x)) = L y (x, p(x)) = L(x, q(x)) = L y (x, q(x)) = 0

18 Expanding, for x [a, b], x p(x) + α(x) α(p(x)) θ(x)(x p(x)) = 0 (1) 1 α (p(x)) + θ(x) = 0 (2) q(x) x + α(x) α(q(x)) θ(x)(x q(x)) = 0 (3) Differentiating (1) and using (2), 1 α (q(x)) + θ(x) = 0 (4) 1 + α (x) θ(x) θ (x)(x p(x)) = 0. (5) Similarly, 1 + α (x) θ(x) θ (x)(x q(x)) = 0 (6). Subtracting (6) from (5) gives Adding (5) and (6), θ (x) = 2 q(x) p(x). α (x) = θ(x) + θ (x) (2x p(x) q(x)). 2

19 Properties L(x, y) = y x + α(x) α(y) θ(x)(x y). q(x) is a local minimum in y for L(x,y), expect 0 L yy (x, q(x)) = α (q(x)). So α is concave at y = q(x). Also α (q(x)) = 1 + θ(x), so α (q(x))q (x) = θ (x) > 0. So q (x) < 0.

20 The optimal sub-hedge Fix x 0 [a, b], define θ = [a, b] R and α = [a, b] R via θ(x) = α(x) = x x 0 x x 0 2dz q(z) p(z), 2x q(z) p(z) dz = xθ(x) q(z) p(z) x x 0 q(z) + p(z) q(z) p(z) dz. Extend to R by defining δ : R R and ψ : [a, b] R via θ(p 1 (x)) x < a, δ(x) = θ(x) x [a, b], θ(q 1 (x)) x > b. α(p 1 (x)) + (p 1 (x) x)(1 θ(p 1 (x))) x < a, ψ(x) = α(x) x [a, b], α(q 1 (x)) + (q 1 (x) x)( 1 θ(q 1 (x))) x > b.

21 The optimal sub-hedge Theorem L(x, y) = y x + ψ(x) ψ(y) δ(x)(x y) 0 for all x, y R, with equality for y {p(x), x, q(x)}.

22 Uniform example Suppose µ U[ 1, 1] and ν U[ 2, 2]. D(x) = For 1 < x < 1 we have 1 8 x x x 1, x 2 1 < x < 1, 1 8 x x x 2. D (x) = D (p(x)) + D (q(x)) x = q(x) + p(x) D(x) xd (x) = D(q(x)) q(x)d (q(x)) + D(p(x)) p(x)d (p(x)) = D(q(x)) q(x)d (q(x)) + D( x q(x)) + (x + q(x))d ( x q(x)). After some simplification we find q(x) 2 + q(x)x + x 2 3 = 0, so q(x) = x+ 12 3x 2, and p(x) = x 12 3x

23 Uniform example: Dual Setting x 0 = 0, we have θ(x) = and x 0 α(x) = xθ(x) u 2 du = 1 3 x x 0 0 udu 12 3u 2 = 2x du 1 u 2 /4 = 2 ( sin 1 x 3 2 ( sin 1 x ) 3 2 ) x 2 3.

24 Uniform example: Differential equations µ U[ 1, 1] and ν U[ 2, 2]. p (x) = q (x) = q(x) x f µ(x) f ν(x) q(x) p(x) f µ(p(x)) f = x q(x) ν(p(x)) q(x) p(x) p(x) x q(x) p(x). Symmetry p(x) = q( x). Consider i.) j(x) = q(x) + p(x) and ii.) h(x) = q(x) p(x) i.) j (x) = q (x) + q ( x) = p(x) x h(x) + x q(x) h(x) = 1 symmetry j(0) = 0 j(x) = x. ii.) q (x)h(x) = p(x) x (h (x)+j (x))h(x) = j(x) h(x) 2x. So h (x)h(x) = j(x) 2x = 3x h(x) 2 = A 3x 2. Conclude: q(x) = (h(x) + j(x))/2 = q(0) x 2 x/2. Boundary condition q( 1) = 2 q(0) = 3 so q(x) = 12 3x 2 x 2. Then also a second boundary q(1) = 1 is satisfied. Further, p(x) = 12 3x 2 x 2.

25 Uniform example: Differential equations µ U[ 1, 1] and ν U[ 2, 2]. p (x) = q (x) = q(x) x f µ(x) f ν(x) q(x) p(x) f µ(p(x)) f = x q(x) ν(p(x)) q(x) p(x) p(x) x q(x) p(x). Symmetry p(x) = q( x). Consider i.) j(x) = q(x) + p(x) and ii.) h(x) = q(x) p(x) i.) j (x) = q (x) + q ( x) = p(x) x h(x) + x q(x) h(x) = 1 symmetry j(0) = 0 j(x) = x. ii.) q (x)h(x) = p(x) x (h (x)+j (x))h(x) = j(x) h(x) 2x. So h (x)h(x) = j(x) 2x = 3x h(x) 2 = A 3x 2. Conclude: q(x) = (h(x) + j(x))/2 = q(0) x 2 x/2. Boundary condition q( 1) = 2 q(0) = 3 so q(x) = 12 3x 2 x 2. Then also a second boundary q(1) = 1 is satisfied. Further, p(x) = 12 3x 2 x 2.

26 Uniform example: Differential equations µ U[ 1, 1] and ν U[ 2, 2]. p (x) = q (x) = q(x) x f µ(x) f ν(x) q(x) p(x) f µ(p(x)) f = x q(x) ν(p(x)) q(x) p(x) p(x) x q(x) p(x). Symmetry p(x) = q( x). Consider i.) j(x) = q(x) + p(x) and ii.) h(x) = q(x) p(x) i.) j (x) = q (x) + q ( x) = p(x) x h(x) + x q(x) h(x) = 1 symmetry j(0) = 0 j(x) = x. ii.) q (x)h(x) = p(x) x (h (x)+j (x))h(x) = j(x) h(x) 2x. So h (x)h(x) = j(x) 2x = 3x h(x) 2 = A 3x 2. Conclude: q(x) = (h(x) + j(x))/2 = q(0) x 2 x/2. Boundary condition q( 1) = 2 q(0) = 3 so q(x) = 12 3x 2 x 2. Then also a second boundary q(1) = 1 is satisfied. Further, p(x) = 12 3x 2 x 2.

27 Uniform example: Differential equations µ U[ 1, 1] and ν U[ 2, 2]. p (x) = q (x) = q(x) x f µ(x) f ν(x) q(x) p(x) f µ(p(x)) f = x q(x) ν(p(x)) q(x) p(x) p(x) x q(x) p(x). Symmetry p(x) = q( x). Consider i.) j(x) = q(x) + p(x) and ii.) h(x) = q(x) p(x) i.) j (x) = q (x) + q ( x) = p(x) x h(x) + x q(x) h(x) = 1 symmetry j(0) = 0 j(x) = x. ii.) q (x)h(x) = p(x) x (h (x)+j (x))h(x) = j(x) h(x) 2x. So h (x)h(x) = j(x) 2x = 3x h(x) 2 = A 3x 2. Conclude: q(x) = (h(x) + j(x))/2 = q(0) x 2 x/2. Boundary condition q( 1) = 2 q(0) = 3 so q(x) = 12 3x 2 x 2. Then also a second boundary q(1) = 1 is satisfied. Further, p(x) = 12 3x 2 x 2.

28 sas ɛυχαριστω para poli