An injective martingale coupling

David Hobson; Dominykas Norgilas

paper

An injective martingale coupling

arXiv:2303.01578

Abstract

We give an injective martingale coupling; in particular, given measures $I^¼$ and $I^½$ in convex order on $R$ such that $I^½$ is continuous, we construct a martingale transport such that for each $y$ in the support of the target law $I^½$ there is a {\em unique} $x$ in {a support of} the initial law $I^¼$ such that (some of) the mass at $x$ is transported to $y$ . Then $I ¨ $ has disintegration $I ¨ (d x, d y) = I^½ (d y) I^´_{I^¸ (y)} (d x)$ for some function $I^¸$ . More precisely we construct a martingale coupling $I ¨ $ of the measures $I^¼$ and $I^½$ such that there is a set $I^_{I^} ¼$ such that $I^¼ (I^_{I^} ¼) = 1$ and a disintegration $(I ¨ _{x})_{x \in I^_{I^} ¼}$ of $I ¨ $ of the form $I ¨ (d x, d y) = I ¨ _{x} (d y) I^¼ (d x)$ such that, with $I^_{I ¨ _{x}}$ a support of $I ¨ _{x}$ , we have $# {x \in I^_{I^} ¼ : y \in I^_{I ¨ _{x}}} \in {0, 1}$ for all $y$ and ${y : # {x \in I^_{I^} ¼ : y \in I^_{I ¨ _{x}}} = 1} = s u pp (I^½)$ . Moreover, if $I^¼$ is continuous we may take $I^_{I ¨ _{x}} = s u pp (I ¨ _{x})$ for each $x$ . However, we cannot also insist that $I^_{I^} ¼ = s u pp (I^¼)$ .

55 pages, 6 figures