There exists no ultimate solution to Skorokhod's problem

Abstract

Let (X,Y) be a mean zero martingale pair, i.e., X and Y possess mean zero and E(YIX)=X a.s.. It has been proved in various ways that (1) there exist stopping times τ on Brownian motion {B(t);t≥0} such that B(τ) is distributed like X and {B(tΛτ); t≥0} is uniformly integrable; and (2) for any such τ there exist stopping times τ′ such that τ≤τ′ a.s., (B(τ), B(τ′)) is distributed like (X,Y), and {B(tΛτ′); t≥0} is uniformly integrable. In other words (to explain the role of uniform integrability), a martingale pair can be embedded in a piece of Brownian motion that is itself a martingale.We will show that unless Y lives on one or two points, there can exist no stopping time τ′ with {B(tΛτ′); t≥0} uniformly integrable and B(τ′) distributed as Y, such that whenever (X,Y) is a martingale pair there exist τ with τ≤τ′ a.s. and B(τ) distributed as X.