A new family of one dimensional martingale couplings

In this paper, we exhibit a new family of martingale couplings between two one-dimensional probability measures $\mu$ and $\nu$ in the convex order. This family is parametrised by two dimensional probability measures on the unit square with respective marginal densities proportional to the positive and negative parts of the difference between the quantile functions of $\mu$ and $\nu$. It contains the inverse transform martingale coupling which is explicit in terms of the associated cumulative distribution functions. The integral of $\vert x-y\vert$ with respect to each of these couplings is smaller than twice the $W_1$ distance between $\mu$ and $\nu$. When $\mu$ and $\nu$ are in the decreasing (resp. increasing) convex order, the construction is generalised to exhibit super (resp. sub) martingale couplings.

We develop in Section 1 an abstract construction of a new family of martingale couplings between two probability measures µ and ν on the real line with finite first moments and comparable in the convex order. This family is parametrised by two dimensional probability measures on the unit square with respective marginal densities proportional to the positive and the negative parts of the difference F −1 µ − F −1 ν between the quantile functions of µ and ν. It contains the law of the above coupling (F −1 µ (U ), Z) when µ and ν are symmetric. Moreover, each coupling in the family is obtained as the image by (u, y) which implies (0.1) as soon as the set of probability measures on the unit square in non empty.
In Section 2, we give an explicit example of such a probability measure on the unit square, which yields the inverse transform martingale coupling. This coupling is explicit in terms of the cumulative distribution functions of the above-mentionned densities and their left-continuous generalised inverses. It is therefore more explicit than the left-curtain and right-curtain couplings introduced by Beiglböck and Juillet. According to Henry-Labordère and Touzi [4], under the condition that ν has no atoms and the set of local maximal values of F ν − F µ is finite, the left-curtain coupling can be explicited by solving two coupled ordinary differential equations starting from each right-most local maximiser. We also check that the inverse transform martingale coupling is stable with respect to its marginals µ and ν for the weak convergence topology. The building brick of the inverse transform martingale coupling is a martingale coupling between pδ F −1 ) to help ensuring that the second marginal is equal to ν when the first is equal to µ. Then p is given by the equality of the means which is equivalent to the convex order: . The above symmetric coupling corresponds to the choice v = 1 − u for 0 < u < 1/2, which, by symmetry, leads to p = 1/2. Of course, in general, this choice is no longer possible. We instead rely on the necessary condition of Theorem 3.A.5 Chapter 3 [9]: ) and x − := max(−x, 0) respectively denote the positive and negative parts of a real number x. We now choose explains why the construction succeeds. More details are given in Section 2.

Definition
Let µ and ν be two probability measures on R with finite first moment and such that µ < cx ν. Let U − and U + be defined by The probability measures µ and ν have finite first moments, so their mean are well defined and equal since they are in the convex order.
. We note Q the set of probability measures Q on (0, 1) 2 such that Proposition 1.1. Let µ, ν ∈ P 1 (R) such that µ < cx ν. The set Q is non-empty.
This is direct consequence of Proposition 2.1 below. Let Q be an element of Q. Let π − and π + be two Markov kernels on (0, 1) such that Let ( m(u, dy)) u∈(0,1) be the Markov kernel defined by For all x ∈ R such that F µ (x) > 0 and F µ (x − ) < 1, m(x, dy) can be rewritten as The proof of Proposition 1.2 relies on the two following lemmas.
With a symmetric reasoning, we obtain that for du-almost all u ∈ U + , Proof. Let h : R 2 → R be a measurable and nonnegative function. By Lemma 4.
By Lemma 4.2 and the inverse transform sampling, Thanks to Lemma 1.3, we get for du-almost all u ∈ (0, 1), R h(y) m(u, dy) hence M is a martingale coupling between µ and ν. In particular with h : y → |y|, using the inverse transform sampling, we have In the same way, for du-almost all u ∈ U + , Since by Lemma 4.1, So µ(dx) m(x, dy) is a martingale coupling between µ and ν.
Let us complete this section with simple examples. Let µ, ν ∈ P 1 (R) be such that µ < cx ν. Suppose that the difference of the quantile functions change sign only once, that is there exists p ∈ (0, 1) such that and nonincreasing on [p, 1]. Then one can easily see that any probability measure Q defined on (0, 1) satisfying properties (i) and (ii) of the definition of Q is concentrated on (0, p) × (p, 1) and therefore satisfies (iii). In particular, the probability measure Q 1 defined on (0, 1) 2 by is an element of Q. Suppose now that µ and ν are symmetric with common mean α ∈ R and that their respective quantile on (1/2, 1). We saw in the introduction an explicit coupling between µ and ν in the case α = 0. Let us show here that this coupling is in fact associated to a particular element of Q. According to Lemma 4.3, for du-almost all u ∈ (0, 1), which is helpful in order to see that the probability measure Q 2 defined on (0, 1) 2 by is an element of Q. For that element Q 2 , using Lemma 1.3 and Lemma 4.3, we have for du-almost all u ∈ (0, 1), Let U and V be two independent random variables uniformly distributed on [0, 1] and let Z the random variable defined as in the introduction with the mean α taken into account, that is Let f : R → R be a bounded and measurable function. Then ) and using the inverse transform sampling, we have In the same way, we have , where we used (1.3) for the last equality. So the law of (

Stability inequality
where the constant 2 is sharp. Moreover, provided that µ = ν, for all Q ∈ Q, the martingale coupling M is such that The proof of Theorem 1.5 relies on Proposition 1.6 below. Note that since Π M (µ, ν) ⊂ Π(µ, ν), we always have The optimal martingale coupling P ∈ Π M (µ, ν) which minimises R×R |x − y| P (dx, dy) was actually exposed by Hobson and Klimmek [5] under the dispersion assumption that there exists an interval E of positive length such that (µ − ν) + (E ∁ ) = (ν − µ) + (E) = 0. They show that the optimal coupling Π HK is unique. Moreover, in the simpler case where µ ∧ ν = 0, if a < b denote the endpoints of E, then there exist two nonincreasing functions R : (0, 1) → (−∞, a] and S : (0, 1) → [b, +∞) such that for all u ∈ (0, 1), denoting π HK (u, dy) = π HK (F −1 µ (u), dy), one has If µ and ν are reduced to two atoms each and are such that µ ≤ cx ν, then there exists a unique martingale coupling between µ and ν, so Π HK derives of course from Q. However, for any Q ∈ Q, we can see thanks to Lemma . Since R and S are nonincreasing and F −1 ν is nondecreasing, we necessarily have π HK = m as long as F −1 ν takes at least three different values on (0, 1), so Π HK does not derive from an element of Q in that case. For example, if µ(dx) Note that the maximisation problem sup P ∈Π M (µ,ν) R×R |x − y| P (dx, dy) is discussed by Hobson and Neuberger [6].
Let us show now that the constant 2 in sharp, that is Let a, b ∈ R such that 0 < a < b. Let µ = 1 2 δ a + 1 2 δ −a and ν = 1 2 δ b + 1 2 δ −b . Since µ and ν are two probability measures with equal means such that µ is concentrated on [−a, a] and ν on R\[−a, a], then µ < cx ν. Any coupling H between µ and ν is of the form where p, p ′ , r, r ′ ≥ 0 and p + p ′ = r + r ′ = p + r ′ = p ′ + r = 1/2. One can easily see that H is a martingale coupling iff b(p − p ′ ) = a/2 and b(r ′ − r) = −a/2, that is Since there is only one martingale coupling, we trivially have which tends to 2 as b tends to a.
Also, the stability inequality (1.8) does not generalise for a cost function of the form (x, y) → |x − y| ρ with W ρ ρ (µ, ν) replacing W 1 (µ, ν) for ρ > 1, as shown in the next proposition in general dimension. Proposition 1.8. Let d ≥ 1 and ρ > 1. Then The proof of Proposition 1.8 will need the following lemma for the case 1 < ρ < 2.
Lemma 1.9. Let d ≥ 1 and ρ ∈ (1, 2). There exists C ρ > 0 such that where, by convention, for all y ∈ R d and for x = 0 we choose |x| ρ−2 x, y R d equal to its limit 0 as x → 0. Note that when ρ = 2, both sides of the inequality are equal with C 2 = 1.
Proof of Proposition 1.8. The case ρ > 2 was done in the introduction in the one dimensional case. Its extension to dimension d is immediate. We now consider the case 1 < ρ < 2. Let µ, ν ∈ P ρ (R d ) be such that µ < cx ν, and let P be a martingale coupling between µ and ν. Thanks to Lemma 1.9, there exists C ρ > 0 such that Since P is a martingale coupling, we have for µ(dx)-almost all x ∈ R d , R d |x| ρ−2 x, y R d π(x, dy) = |x| ρ , where both sides are equal to 0 when x = 0, so Let us now specify a choice of µ and ν. Let X be a non-constant d-dimensional random variable such that E[|X| ρ ] < +∞. Then the mapping which associates to t ∈ R + the law of E[X] + t(X − E[X]) is increasing in the convex order. Indeed, for all 0 ≤ s < t and for all convex functions ϕ :

For all n ∈ N, let µ n and µ n+1 be respectively the laws of E[X]+n(X −E[X]) and E[X]+(n+1)(X −E[X]).
To set this idea on a simple example, one could choose X ∼ N d (0, I d ) so that µ n = N d (0, n 2 I d ) and µ n+1 = N d (0, (n + 1) 2 )I d . So, if X is non-constant, centered and such that E[|X| ρ ] < +∞, then for all n ∈ N,

The inverse transform martingale coupling 2.1 Definition of the inverse transform martingale coupling
Let µ, ν ∈ P 1 (R) such that µ < cx ν.
which are well defined thanks to the equality Ψ − (1) = Ψ + (1), consequence of the equality of the means.
Let Q be the probability measure defined on (0, 1) 2 by Proposition 2.1. The probability measure Q is an element of Q as defined in Section 1. Moreover, For this particular Q ∈ Q, Definition (1.1) of the Markov kernel ( m(u, dy)) u∈(0,1) becomes Note that if ϕ(u) = 0, then Ψ + (u) = 0 so for all v ∈ (0, u), ) makes sense. We define the Markov kernel (m(x, dy)) x∈R as in Definition (1.2). Then by Proposition 1.2, M (dx, dy) = µ(dx) m(x, dy) is a martingale coupling, called the inverse transform martingale coupling.

Remark 2.2. Let us mention that the inverse transform martingale coupling permits to give the following short constructive proof of Strassen's theorem. Let
Hence the respective Fenchel-Legendre transforms ϕ * µ and ϕ * ν of ϕ µ and ϕ ν satisfy ϕ * µ ≥ ϕ * ν . For all u ∈ (0, 1) and for all t ∈ R, 1], we get the well known fact (see for instance Lemma A.22 [3]) that for all q ∈ R, ϕ * µ (q) = ( 1] which is an equality for q = 1 implies the existence of the inverse transform martingale coupling M between µ and ν. Proof of Proposition 2.1. Let h : (0, 1) 2 → R be a measurable and bounded function.
Since Ψ − is continuous, one has Ψ − (Ψ −1 − (u)) = u for all u ∈ (0, 1). By Lemma 4.1, we deduce that ϕ(ϕ(u)) = u, (F −1 µ (u) − F −1 ν (u)) + du-almost everywhere on (0, 1). Therefore, on the one hand, by Lemma 4.5, On the other hand, still by Lemma 4.5, With the same reasoning, one can show that on (0, 1). So, by Lemma 4.5, We saw in Section 1.1 a concrete example of an element Q 1 ∈ Q (see (1.5)) when µ, ν ∈ P 1 (R) are such that µ < cx ν and there exists p ) dv is nondecreasing on [0, p] and nonincreasing on [p, 1]. Any probability measure Q defined on (0, 1) satisfying properties (i) and (ii) of the definition of Q is concentrated on (0, p) × (p, 1) and therefore satisfies (iii). The probability measure Q 1 is the simplest example that comes to mind, but it is associated with a not very straightforward martingale coupling between µ and ν. Of course, the inverse transform martingale coupling presented in this section is a valid example, but we show here that it inspires another coupling which is sort of the nonincreasing twin of the inverse transform martingale coupling. Let which is well defined since χ + (1) = χ + (p) = χ − (1 − p) = γ, consequence of the equality of the means. Let Q be the probability measure defined on (0, 1) 2 by In the symmetric case, that is when µ and ν are symmetric and p = 1/2, we have Γ(u) = u, so we get Q 2 (see (1.6)), hence Q is a generalisation of the symmetric coupling. Note that Γ(1) = 1 − p, hence Γ(u) < 1 for all u ∈ (0, 1). It is clear that Q satisfies property (i) of the definition of Q. By Lemma 4.5 applied with the functions f 1 : u ∈ (0, 1) for any measurable and bounded function h : (0, 1) → R. So Q satisfies (ii) as well, and therefore (iii). We saw with Proposition 1.6 that for all Q ∈ Q, µ, ν). The next proposition shows that the inverse transform martingale coupling and its nonincreasing twin, when it exists, play particular roles among the martingale couplings which derive from Q when 1 is replaced by ρ > 1. ( m IT (u, dy)) u∈(0,1) be defined as in (2.2), and ( m N IT (u, dy)) u∈(0,1) its nonincreasing twin deriving from (2.4) when it exists. Then, for all ρ > 2, ( m IT (u, dy)) u∈(0,1) and ( m N IT (u, dy)) u∈(0,1) respectively minimises and maximises
If µ, ν ∈ P 1 (R) are such that µ ≤ icx ν, denoting by µ and ν the respective images of µ and ν by x → −x, one can easily see that µ ≤ icx ν iff µ ≤ dcx ν. So if M d (dx, dy) denotes a supermartingale coupling between µ and ν, then the image of M d (dx, dy) by (x, y) → (−x, −y) is a submartingale coupling between µ and ν. In particular, the image of the inverse transform supermartingale coupling by (x, y) → (−x, −y) is a submartingale coupling between µ and ν.
Proof. Let f : R → R be a measurable and bounded function. Then where we used for the last but one equality the inverse transform sampling, and for the last equality the fact that F µ (F −1 µ (u)) = u if F µ is continuous in F −1 µ (u). One can easily see that for all x ∈ R and u ∈ (0, 1), h(Ψ −1 2 (Ψ 2 (v))) dΨ 2 (v).