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

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Did you mean $dZ_t = \hat{\mu}(Z_t)dt + \hat{\sigma} dW_t$, where $\hat{\sigma}$ is constant? – Simon Lyons Feb 1 '11 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. – Alekk Feb 1 '11 at 16:09