Exact sampling of an Ornstein-Uhlenbeck process ($X_{t+\Delta }=\mu +a(X_t-\mu )+\sqrt{v_{\infty }(1-a^2)}Z$)

For any step $\Delta >0$, set $a=e^{-\theta \Delta }$. The Ornstein-Uhlenbeck transition distribution gives the exact updateusing a fresh independent standard normal random variable for each interval. The update reproduces the joint probability distribution on the selected time grid. It can start from a fixed value or use Ornstein-Uhlenbeck stationary initialization. On an equally spaced grid it is an autoregressive process of order one after subtracting $\mu$. Its stationary variance follows directly from $v=a^{2}v+v_{\infty }(1-a^2)$.

For the stipulated example with $\Delta =0.1$, $a=e^{-0.1}\approx 0.90484$ and the innovation variance is $1-e^{-0.2}\approx 0.18127$. Given $X_t=6$, the next conditional expected value is about $5.61935$ and the conditional standard deviation is about $0.42576$.

The Euler-Maruyama method instead approximates the same stochastic differential equation byFor $0<\theta \Delta <2$ its stationary variance solves $v=(1-\theta \Delta {)}^{2}v+{\sigma }^2\Delta$, givingAt $\Delta =0.1$ in the example, this is $1/0.95\approx 1.05263$, about $5.26\%$ above the continuous model's stationary variance. If $\theta \Delta \ge 2$, this Euler recurrence has no finite stationary variance despite the stable continuous model. Exact transitions avoid this discretization error at sampled times; neither update alone determines the unresolved motion between samples.