= Exact sampling of an Ornstein-Uhlenbeck process
{title2=$X_{t+\Delta}=\mu+a(X_t-\mu)+\sqrt{v_\infty(1-a^2)}Z$}
For any step $\Delta>0$, set $a=e^{-\theta\Delta}$. The <Ornstein-Uhlenbeck transition distribution> gives the exact update
$$
X_{t+\Delta}=\mu+a(X_t-\mu)+\sqrt{\frac{\sigma^2}{2\theta}(1-a^2)}\,Z,\qquad Z\sim\mathcal N(0,1),
$$
using 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 $v=a^2v+v_\infty(1-a^2)$.
For the stipulated example with $\Delta=0.1$, $a=e^{-0.1}\approx0.90484$ and the innovation <variance> is $1-e^{-0.2}\approx0.18127$. 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> by
$$
X_{t+\Delta}^{\mathrm E}=\mu+(1-\theta\Delta)(X_t^{\mathrm E}-\mu)+\sigma\sqrt\Delta\,Z.
$$
For $0<\theta\Delta<2$ its stationary <variance> solves $v=(1-\theta\Delta)^2v+\sigma^2\Delta$, giving
$$
v_{\mathrm E}=\frac{\sigma^2}{2\theta-\theta^2\Delta}=\frac{v_\infty}{1-\theta\Delta/2}.
$$
At $\Delta=0.1$ in the example, this is $1/0.95\approx1.05263$, about $5.26\%$ above the continuous model's stationary <variance>. If $\theta\Delta\ge2$, 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.
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