Exact simulation of SDE - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:43:59Z http://mathoverflow.net/feeds/question/53975 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53975/exact-simulation-of-sde Exact simulation of SDE Alekk 2011-02-01T11:23:18Z 2011-02-01T16:32:25Z <p>Consider a one dimensional SDE of the form $dX_t = \mu(X_t) dt + \sigma dW_t$, where $\sigma>0$ is a <strong>constant</strong>. Under mild regularity assumptions on $\mu(\cdot)$, one can exactly simulate trajectories of this SDE: because $\sigma$ is constant, one can first exactly simulate a (scaled) Brownian motion $dY_t = \sigma dW_t$ and use the fact that (Girsanov) $\text{Law}(x)$ and $\text{Law}(Y)$ are equivalent to do some kind of <a href="http://en.wikipedia.org/wiki/Rejection_sampling" rel="nofollow">rejection sampling</a> on the Wiener space. See <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.aoap/1133965767" rel="nofollow">here</a> for more details.</p> <p>If $\sigma(\cdot)$ is not constant, in the one dimensional case, one can always find a function $\Psi$ such that $Z_t = \Psi(W_t)$ is of the form $dZ_t = \hat{\mu}(Z_t) dt + \sigma(Z_t) dW_t$: this follows from the fact that any $1$-dimensional continuous function is a derivative. This shows that a large class of $1$-dimensional SDE can be exactly simulated.</p> <p><strong>Question</strong>: the situation is quite different in $\mathbb{R}^d$ for $d \geq 2$: is there any diffusion $dX_t = \mu(X_t)dt + \sigma(X_t) \cdot dW_t$ that can be exactly simulated and that cannot be obtained through rejection sampling based on the process $Z_t = \Psi(W_t)$ for a well chosen function $\Psi:\mathbb{R}^n \to \mathbb{R}^d$.</p> http://mathoverflow.net/questions/53975/exact-simulation-of-sde/53992#53992 Answer by Simon Lyons for Exact simulation of SDE Simon Lyons 2011-02-01T15:23:42Z 2011-02-01T16:32:25Z <p>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.</p> <p>Beskos et al. submitted a discussion paper on this topic to the journal of the royal statistical society <a href="http://onlinelibrary.wiley.com/doi/10.1111/j.1467-9868.2006.00552.x/abstract" rel="nofollow">(available here)</a>. 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.</p>