Exact sampling of an Ornstein-Uhlenbeck process

Ornstein-Uhlenbeck process

{c}
{title2=$X_t\mid X_0=x_0\sim\mathcal N(m(t),v(t))$}

For a fixed initial value $X_0=x_0$, the <Ornstein-Uhlenbeck process> has the <normal distribution>
$$
X_t\sim\mathcal N\!\left(\mu+(x_0-\mu)e^{-\theta t},\ \frac{\sigma^2}{2\theta}(1-e^{-2\theta t})\right),\qquad t>0.
$$
The two parameters of $\mathcal N(m,v)$ are the <expected value> and <variance>, not the <standard deviation>. At $t=0$ the <variance> is zero and the value is fixed. More generally, its <transition kernel> over any interval of length $h>0$ has <conditional expected value> $\mu+(x-\mu)e^{-\theta h}$ and conditional <variance> $\sigma^2(1-e^{-2\theta h})/(2\theta)$ when the current value is $x$. The coefficients depend on elapsed time; the initial <probability distribution> determines whether the resulting <stochastic process> is stationary.<Applied Stochastic Differential Equations by Simo Särkkä and Arno Solin>[Applied Stochastic Differential Equations, Example 6.2, printed p. 81, equations (6.27)–(6.30); centered model with diffusion constant q]{ref}

In the stipulated example $\theta=1$, $\mu=2$, $\sigma=\sqrt2$, $x_0=6$, the <expected value> at time two is $2+4e^{-2}\approx2.5413$ and the <variance> is $1-e^{-4}\approx0.9817$. The initial distance from two has fallen by a factor $e^{-2}$, but the <standard deviation> has grown to about $0.9908$. Approaching the reference level in <expected value> is compatible with growing uncertainty from a perfectly known initial value.

\Image[/ornstein-uhlenbeck-transition-moments.png]
{title=Ornstein-Uhlenbeck transition densities, mean relaxation and variance growth from a fixed initial value}

The solid <probability density functions> use times $0.25$, $1$ and $2$; the dashed curve is the limiting <normal distribution> with <expected value> two and <variance> one. The curves are exact model predictions. There is no ordinary density for the fixed value at time zero.

A short derivation uses the <integrating factor> $e^{\theta t}$. The <Itô product rule> gives $d[e^{\theta t}(X_t-\mu)]=\sigma e^{\theta t}\,dW_t$, hence
$$
X_t=\mu+(x_0-\mu)e^{-\theta t}+\sigma\int_0^t e^{-\theta(t-u)}\,dW_u.
$$
The <stochastic integral> has zero <expected value>. The <Itô isometry> gives its <variance> as $\sigma^2\int_0^t e^{-2\theta(t-u)}du$. Its deterministic integrand makes it a <Gaussian random variable>, yielding the displayed <normal distribution>. Shifting the integration interval gives the same <transition kernel> from any later time.


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