English

An injective martingale coupling

Probability 2025-03-17 v2

Abstract

We give an injective martingale coupling; in particular, given measures μ\mu and ν\nu in convex order on R\mathbb R such that ν\nu is continuous, we construct a martingale transport such that for each yy in the support of the target law ν\nu there is a {\em unique} xx in {a support of} the initial law μ\mu such that (some of) the mass at xx is transported to yy. Then π\pi has disintegration π(dx,dy)=ν(dy)δθ(y)(dx)\pi(dx,dy) = \nu(dy) \delta_{\theta(y)}(dx) for some function θ\theta. More precisely we construct a martingale coupling π\pi of the measures μ\mu and ν\nu such that there is a set Γμ\Gamma_\mu such that μ(Γμ)=1\mu(\Gamma_\mu)=1 and a disintegration (πx)xΓμ(\pi_x)_{x \in \Gamma_\mu} of π\pi of the form π(dx,dy)=πx(dy)μ(dx)\pi(dx,dy) = \pi_x(dy) \mu(dx) such that, with Γπx\Gamma_{\pi_x} a support of πx\pi_x, we have #{xΓμ:yΓπx}{0,1}\# \{ x \in \Gamma_\mu : y \in \Gamma_{\pi_x} \} \in \{ 0,1 \} for all yy and {y:#{xΓμ:yΓπx}=1}=supp(ν)\{ y : \# \{ x \in \Gamma_\mu : y \in \Gamma_{\pi_x} \} = 1 \} = supp(\nu). Moreover, if μ\mu is continuous we may take Γπx=supp(πx)\Gamma_{\pi_x} = supp(\pi_x) for each xx. However, we cannot also insist that Γμ=supp(μ)\Gamma_\mu = supp (\mu).

Keywords

Cite

@article{arxiv.2303.01578,
  title  = {An injective martingale coupling},
  author = {David Hobson and Dominykas Norgilas},
  journal= {arXiv preprint arXiv:2303.01578},
  year   = {2025}
}

Comments

55 pages, 6 figures

R2 v1 2026-06-28T08:58:16.825Z