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| Main Authors: | , |
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| Format: | Preprint |
| Published: |
2023
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| Subjects: | |
| Online Access: | https://arxiv.org/abs/2303.01578 |
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| _version_ | 1866912273880055808 |
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| author | Hobson, David Norgilas, Dominykas |
| author_facet | Hobson, David Norgilas, Dominykas |
| contents | We give an injective martingale coupling; in particular, given measures $μ$ and $ν$ in convex order on $\mathbb R$ such that $ν$ is continuous, we construct a martingale transport such that for each $y$ in the support of the target law $ν$ there is a {\em unique} $x$ in {a support of} the initial law $μ$ such that (some of) the mass at $x$ is transported to $y$. Then $π$ has disintegration $π(dx,dy) = ν(dy) δ_{θ(y)}(dx)$ for some function $θ$.
More precisely we construct a martingale coupling $π$ of the measures $μ$ and $ν$ such that there is a set $Γ_μ$ such that $μ(Γ_μ)=1$ and a disintegration $(π_x)_{x \in Γ_μ}$ of $π$ of the form $π(dx,dy) = π_x(dy) μ(dx)$ such that, with $Γ_{π_x}$ a support of $π_x$, we have $\# \{ x \in Γ_μ: y \in Γ_{π_x} \} \in \{ 0,1 \}$ for all $y$ and $\{ y : \# \{ x \in Γ_μ: y \in Γ_{π_x} \} = 1 \} = supp(ν)$. Moreover, if $μ$ is continuous we may take $Γ_{π_x} = supp(π_x)$ for each $x$. However, we cannot also insist that $Γ_μ= supp (μ)$. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2303_01578 |
| institution | arXiv |
| publishDate | 2023 |
| record_format | arxiv |
| spellingShingle | An injective martingale coupling Hobson, David Norgilas, Dominykas Probability 60G42 We give an injective martingale coupling; in particular, given measures $μ$ and $ν$ in convex order on $\mathbb R$ such that $ν$ is continuous, we construct a martingale transport such that for each $y$ in the support of the target law $ν$ there is a {\em unique} $x$ in {a support of} the initial law $μ$ such that (some of) the mass at $x$ is transported to $y$. Then $π$ has disintegration $π(dx,dy) = ν(dy) δ_{θ(y)}(dx)$ for some function $θ$. More precisely we construct a martingale coupling $π$ of the measures $μ$ and $ν$ such that there is a set $Γ_μ$ such that $μ(Γ_μ)=1$ and a disintegration $(π_x)_{x \in Γ_μ}$ of $π$ of the form $π(dx,dy) = π_x(dy) μ(dx)$ such that, with $Γ_{π_x}$ a support of $π_x$, we have $\# \{ x \in Γ_μ: y \in Γ_{π_x} \} \in \{ 0,1 \}$ for all $y$ and $\{ y : \# \{ x \in Γ_μ: y \in Γ_{π_x} \} = 1 \} = supp(ν)$. Moreover, if $μ$ is continuous we may take $Γ_{π_x} = supp(π_x)$ for each $x$. However, we cannot also insist that $Γ_μ= supp (μ)$. |
| title | An injective martingale coupling |
| topic | Probability 60G42 |
| url | https://arxiv.org/abs/2303.01578 |