show/hide this revision's text 3 Edited for clarity

Beskos and Roberts' whole approach relies on being able to transform the SDE to one with unit diffusion coefficient. If you can't do that, then the bridged process law isn't equivalent to the law of a Brownian bridge. This means the Radon-Nikodym derivative isn't bounded and so you can't do rejection sampling.

In the

Beskos et al. submitted a discussion section of thispaper , on this topic to the journal of the royal statistical society (available here). Dan Crisan suggests that a similar approach might work if one were to use other tractable bridges - say, Bessel bridges. In that case, it looks like you don't have to transform your SDE to one with unit diffusion coefficient. However, the authors show that this approach is equivalent to one using Brownian bridges.

show/hide this revision's text 2 added 1 characters in body

Besos

Beskos and Roberts' whole approach relies on being able to transform the SDE to one with unit diffusion coefficient. If you can't do that, then the bridged process law isn't equivalent to the law of a Brownian bridge. This means the Radon-Nikodym derivative isn't bounded and so you can't do rejection sampling.

In the discussion section of this paper, Dan Crisan suggests that a similar approach might work if one were to use other tractable bridges - say, Bessel bridges. In that case, it looks like you don't have to transform your SDE to one with unit diffusion coefficient. However, the authors show that this approach is equivalent to one using Brownian bridges.

show/hide this revision's text 1

Besos and Roberts' whole approach relies on being able to transform the SDE to one with unit diffusion coefficient. If you can't do that, then the bridged process law isn't equivalent to the law of a Brownian bridge. This means the Radon-Nikodym derivative isn't bounded and so you can't do rejection sampling.

In the discussion section of this paper, Dan Crisan suggests that a similar approach might work if one were to use other tractable bridges - say, Bessel bridges. In that case, it looks like you don't have to transform your SDE to one with unit diffusion coefficient. However, the authors show that this approach is equivalent to one using Brownian bridges.