# Exact simulation of SDE

Consider a one dimensional SDE of the form $$dX_t = \mu(X_t) dt + \sigma dW_t$$, where $$\sigma>0$$ is a constant. 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 rejection sampling on the Wiener space. See here for more details.

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.

Question: 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 $$\Psi:\mathbb{R}^n \to \mathbb{R}^d$$.

• Did you mean $dZ_t = \hat{\mu}(Z_t)dt + \hat{\sigma} dW_t$, where $\hat{\sigma}$ is constant? Feb 1, 2011 at 15:10
• there are two approaches: or $X$ has a non constant volatility function $\sigma(\cdot)$, and one can find a good function $\Psi$ such that $\Psi(X_t)$ has a constant volatility function (also known as Lamperti transform). Or one can take a Brownian motion $W$ and find a good function $\Psi$ such that $\Psi(W_t)$ has $\sigma(\cdot)$ as volatility function. These are indeed essentially the same thing. Feb 1, 2011 at 16:09