The Ornstein-Uhlenbeck Process from Its Transition Law

From Mean-Reverting Transitions to a Process

The Ornstein–Uhlenbeck process is a continuous-time model in which the conditional mean moves exponentially toward a fixed level, while each transition also adds Gaussian uncertainty. Its transition law is useful because it describes how the process evolves over any time interval, not just over an infinitesimal one. To specify the process completely, however, the transitions must be paired with a distribution for the initial value.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let the state space be \(S=\mathbb{R}\), with time \(t\geq 0\), and let \((\mathcal{F}_t)_{t\geq 0}\) be the natural filtration \(\mathcal{F}_t=\sigma(X_s:0\leq s\leq t)\). Fix a mean level \(m\in\mathbb{R}\), a mean-reversion rate \(\theta>0\), and a noise scale \(\sigma>0\). For a time increment \(h\geq 0\), define

Here \(N(r,w)\) denotes a normal distribution with mean \(r\) and variance \(w\). At \(h=0\), \(a_0=1\) and \(v_0=0\), so a zero-length transition leaves the state unchanged. For \(h>0\), the assumptions \(\theta>0\) and \(\sigma>0\) give \(0<a_h<1\) and \(v_h>0\). The conditional mean is a weighted average of the current state and the level \(m\); the conditional variance is the uncertainty added over the interval.

The transition law alone does not specify the distribution of \(X_0\). Different initial laws, used with the same parameters and transitions, give different process laws. Once \(\mu\) is chosen, the transitions determine the distributions at all finite collections of times.

Why the Transitions Are Consistent

Proof. Write \(a_u=e^{-\theta u}\), \(a_v=e^{-\theta v}\), and \(a_{u+v}=a_ua_v\). For independent standard normal variables \(Z_1,Z_2\), the two transitions can be represented as

Substituting the expression for \(Y\) gives \(W=m+a_ua_v(x-m)+a_v\sqrt{v_u}Z_1+\sqrt{v_v}Z_2\). Its mean is \(m+a_{u+v}(x-m)\). Since \(Z_1,Z_2\) are independent with variance \(1\), its variance is \(a_v^2v_u+v_v\). Using the definitions,

The composed transition therefore has the stated mean and variance for an interval of length \(u+v\). If either interval has length zero, its variance is zero and its multiplier is one, so the same calculation still applies. \(\square\)

This composition identity is the Chapman–Kolmogorov consistency condition for these transition kernels. It is what allows transitions over shorter intervals to agree with the transition over the combined interval. For strictly increasing times \(0\leq t_0<t_1<\cdots<t_r\), the finite-dimensional law is specified by

where \(\mu_{t_0}\) is the law at time \(t_0\), obtained from the initial law and the transition from time \(0\) to time \(t_0\), and \(K_h(x,\cdot)=N(m+a_h(x-m),v_h)\). When \(t_0=0\), \(\mu_{t_0}=\mu\). For positive gaps, the transition density at \(y\) from \(x\) is

Each such density integrates to one because it is a normal density with positive variance. Marginalizing an intermediate observation composes its two neighboring kernels; the proposition shows that the resulting kernel is exactly the one for the combined interval. Thus the finite-dimensional laws are consistent when times are omitted or repeated. The Kolmogorov extension theorem is the standard result that turns a consistent family of finite-dimensional distributions into a process; its proof is beyond this tutorial. This construction specifies finite-dimensional laws and does not, by itself, prove a path-continuity claim.

Gaussian Finite-Dimensional Distributions

Proof. For successive times \(t_0<t_1<\cdots<t_r\), set \(h_\ell=t_\ell-t_{\ell-1}\). Using independent standard normal variables \(Z_1,\ldots,Z_r\), the transitions have the recursive representation

At each step, \(X_{t_\ell}\) is an affine function of \(X_0,Z_1,\ldots,Z_\ell\). Therefore the whole vector is an affine transformation of the jointly Gaussian vector \((X_0,Z_1,\ldots,Z_r)\), and hence is jointly Gaussian. If \(X_0\) is not Gaussian, this argument does not apply; a mixture of Gaussian conditional laws need not be Gaussian. \(\square\)

The proposition describes how to obtain the full joint law without repeatedly deriving a new multivariate density from scratch: start with \(X_0\), then apply the transition recursion with independent normal innovations. Equivalently, the product of the initial law and successive transition densities specifies the joint law. These descriptions make explicit that the choice of initial distribution is part of the model.

What the Transition Law Does—and Does Not—Say

For a fixed initial value \(X_0=x\), the transition law determines every later finite-dimensional distribution through the product of kernels. For a random initial value, its law must also be supplied. Moreover, Gaussian conditional transitions do not mean that every possible initialization gives a Gaussian process. For instance, if \(X_0\) is non-Gaussian, then the one-time distribution at a later time is obtained by mixing the transition normals over the distribution of \(X_0\); such a mixture need not be normal.

A useful special case is the initial law \(N(m,\sigma^2/(2\theta))\). If \(X_s\) has this law, the transition over an interval \(h\) has mean \(m\), and its marginal variance is

Thus this initial Gaussian law is preserved by each transition. This verifies a one-time marginal property; the full finite-dimensional laws still come from the initial law together with the transition kernels.

Common mistake. A frequent error is to use \(\frac{\sigma^2}{2\theta}(1-a_h)\) for the transition variance instead of \(\frac{\sigma^2}{2\theta}(1-a_h^2)\). With \(\theta=1\), \(\sigma=2\), and \(h=\log 2\), the incorrect expression gives \(2(1-\tfrac12)=1\). The correct expression uses \(a_h^2=(1/2)^2=1/4\), giving \(2(1-\tfrac14)=\tfrac32\). The square is essential: it is also what makes the variance contributions compose correctly across successive intervals.