Covariance And Stationarity Of The Ornstein-Uhlenbeck Process

From Mean Reversion to Covariance

The Ornstein–Uhlenbeck process is mean-reverting, but mean reversion alone does not make its distribution stationary. The initial value matters: a process started from a fixed point generally has time-dependent variance, while a suitable random initial value makes its mean and covariance invariant under time shifts. We derive the covariance in both cases and identify precisely when the process is stationary.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Fix $\kappa>0$, $\theta\in\mathbb{R}$, and $\sigma\in\mathbb{R}$. Consider the SDE $dX_t=\kappa(\theta-X_t)\,dt+\sigma\,dB_t$, where $\kappa(\theta-X_t)$ is the drift and $\sigma$ is the diffusion coefficient. Assume $X_0$ is $\mathcal{F}_0$-measurable, independent of the Brownian motion, and has finite second moment. Independence ensures that the initial value and future Brownian noise contribute no cross-covariance.

For every finite time horizon $T$, the deterministic integrand $u\mapsto e^{-\kappa(t-u)}\mathbf{1}_{[0,t]}(u)$ is progressively measurable and belongs to $\mathcal{L}^2$, since its squared integral is finite. Thus the Ito integral below is defined. The integrating-factor solution, established in the earlier material on linear SDEs, is

X_t=\theta+e^{-\kappa t}(X_0-\theta)+\sigma\int_0^t e^{-\kappa(t-u)}\,dB_u.

Write $m_0=\mathbb{E}[X_0]$, $v_0=\operatorname{Var}(X_0)$, and $q=\sigma^2/(2\kappa)$. The Ito integral has mean zero and, by the Ito isometry, variance $\int_0^t e^{-2\kappa(t-u)}\,du=(1-e^{-2\kappa t})/(2\kappa)$. Consequently

\mathbb{E}[X_t]=\theta+(m_0-\theta)e^{-\kappa t}, \qquad \operatorname{Var}(X_t)=v_0e^{-2\kappa t}+q(1-e^{-2\kappa t}).

The mean approaches $\theta$, and the variance approaches $q$. The variance need not equal $q$ at finite times: it does so for all $t$ precisely when $v_0=q$. To describe dependence across time, define $\operatorname{Cov}(X_s,X_t)=\mathbb{E}[(X_s-\mathbb{E}X_s)(X_t-\mathbb{E}X_t)]$, for square-integrable variables.

The Covariance Function

Theorem: Under the assumptions above, for $0\le s\le t$, $$ \operatorname{Cov}(X_s,X_t) =e^{-\kappa(t-s)}\left(v_0e^{-2\kappa s}+q(1-e^{-2\kappa s})\right). $$ Equivalently, the covariance is $e^{-\kappa(t-s)}\operatorname{Var}(X_s)$. For $t<s$, interchange $s$ and $t$.

Proof. Center the solution at its mean. For $s\le t$, its centered values are $X_s-\mathbb{E}X_s=e^{-\kappa s}(X_0-m_0)+\sigma\int_0^s e^{-\kappa(s-u)}\,dB_u$ and $X_t-\mathbb{E}X_t=e^{-\kappa t}(X_0-m_0)+\sigma\int_0^t e^{-\kappa(t-u)}\,dB_u$. The covariance of the initial-value terms is $v_0e^{-\kappa(s+t)}$. The initial value is independent of the Brownian motion, so the cross terms between initial-value terms and stochastic integrals have expectation zero. For deterministic square-integrable integrands, polarization of the Ito isometry gives the covariance of the two integrals as the integral of the product of their integrands. The two integrands overlap on $[0,s]$, so this contribution is $\sigma^2\int_0^s e^{-\kappa(s-u)}e^{-\kappa(t-u)}\,du$. Taking out $e^{-\kappa(t-s)}$ leaves $\sigma^2 e^{-\kappa(t-s)}\int_0^s e^{-2\kappa(s-u)}\,du =q e^{-\kappa(t-s)}(1-e^{-2\kappa s})$. Adding the initial contribution, and using $e^{-\kappa(s+t)}=e^{-\kappa(t-s)}e^{-2\kappa s}$, proves the formula. The covariance is symmetric in its two arguments by its definition, which gives the $t<s$ case. $\square$

This expression separates the two sources of fluctuation. The initial uncertainty contributes $v_0e^{-\kappa(s+t)}$, which decays as both times increase. The noise accumulated up to the earlier time contributes $q e^{-\kappa(t-s)}(1-e^{-2\kappa s})$. Unless the initial variance is already $q$, the covariance depends on the absolute time $s$, not just the time gap $t-s$.

Worked Example: Covariance from a Fixed Initial Value

Take $\kappa=2$, $\theta=3$, $\sigma=2$, and $X_0=1$. Compute the mean at time $t$, the variance at time $1$, and $\operatorname{Cov}(X_1,X_2)$.

Here $m_0=1$, $v_0=0$, and $q=\sigma^2/(2\kappa)=4/4=1$. Substitution in the mean formula gives $\mathbb{E}[X_t]=3+(1-3)e^{-2t}=3-2e^{-2t}$. The variance at $1$ is $0\cdot e^{-4}+1(1-e^{-4})=1-e^{-4}$. Since $2-1=1$, the covariance formula gives $\operatorname{Cov}(X_1,X_2)=e^{-2(2-1)}(1-e^{-4})=e^{-2}(1-e^{-4})$.

Check: The transition from time $1$ to time $2$ retains the earlier centered value with factor $e^{-2}$; multiplying the variance $1-e^{-4}$ at time $1$ by this factor reproduces the covariance. The variance at time $0$ is $0$, whereas at time $1$ it is $1-e^{-4}>0$, so this process is not stationary.

The Stationary Gaussian Distribution

Theorem: If $X_0$ is independent of $B$ and has distribution $N(\theta,q)$, then the OU process is strictly stationary. Here $N(\theta,0)$ means the point mass at $\theta$, covering $\sigma=0$. Conversely, any initial distribution independent of $B$ that is invariant under the OU transitions must be $N(\theta,q)$. Thus this is the unique invariant distribution.

Proof. First suppose $X_0\sim N(\theta,q)$, independent of $B$. Each finite vector of process values is jointly Gaussian: the solution expresses its values as linear combinations of the Gaussian variable $X_0$ and Brownian integrals with deterministic integrands. Its mean at every time is $\theta$. Since $v_0=q$, the covariance formula reduces, for all $s,t\ge0$, to $\operatorname{Cov}(X_s,X_t)=q e^{-\kappa|t-s|}$. Now shift any finite list of times $t_1,\ldots,t_n$ by the same $h\ge0$. The shifted vector and the original vector have the same mean vector, with every entry $\theta$, and the same covariance matrix, since $|(t_i+h)-(t_j+h)|=|t_i-t_j|$. Gaussian vectors are determined by their mean vector and covariance matrix, including when the covariance matrix is degenerate. Hence the two vectors have the same distribution. This is strict stationarity.

For the converse, let $\nu$ be an invariant initial distribution and let $\phi(u)=\mathbb{E}[e^{iu(X_0-\theta)}]$ be the characteristic function of its centered version. Invariance means that evolving for any $t>0$ leaves this distribution unchanged. The solution and independence of $X_0$ and $B$ therefore imply $\phi(u)=\phi(e^{-\kappa t}u)\exp[-q(1-e^{-2\kappa t})u^2/2]$. As $t\to\infty$, continuity of a characteristic function at zero gives $\phi(e^{-\kappa t}u)\to1$, while $1-e^{-2\kappa t}\to1$. Thus $\phi(u)=\exp(-qu^2/2)$ for every $u$, the characteristic function of $N(0,q)$. The centered initial law is therefore $N(0,q)$, so the original law is $N(\theta,q)$. This proves uniqueness, also when $q=0$. $\square$

The Gaussian qualification is essential. If $X_0$ is Gaussian and independent of the Brownian motion, the process is Gaussian; if $X_0$ is not Gaussian, the process need not be Gaussian, since its time-zero value is itself one of its process values. The covariance formula remains valid for any independent initial value with finite second moment, whether or not that value is Gaussian.

Worked Example: Comparing Covariances After a Time Shift

Take $\kappa=1$, $\theta=4$, $\sigma=\sqrt{2}$, and $X_0\sim N(4,1)$ independently of $B$. Compare the means, variances, and covariances at times $(1,3)$ and $(6,8)$.

Here $q=(\sqrt{2})^2/(2\cdot1)=2/2=1$, matching $v_0=1$. The mean formula gives $\mathbb{E}[X_t]=4+(4-4)e^{-t}=4$. The variance formula gives $1\cdot e^{-2t}+1(1-e^{-2t})=1$ at every time. Since each pair of times is separated by $2$, their covariances are $1\cdot e^{-1(2)}=e^{-2}$: specifically, $\operatorname{Cov}(X_1,X_3)=e^{-2}$ and $\operatorname{Cov}(X_6,X_8)=e^{-2}$.

Check: The two pairs have the same mean vector $(4,4)$, the same variances $1,1$, and the same covariance $e^{-2}$. More generally, the theorem shows that every finite vector retains its joint distribution under any common nonnegative time shift, not just these two pairs.

Worked Example: Gaussian but Not Stationary

Let $\kappa=1$, $\theta=0$, $\sigma=2$, and $X_0\sim N(0,1)$ independently of $B$. Compare $\operatorname{Cov}(X_0,X_1)$ with $\operatorname{Cov}(X_{\ln 2},X_{\ln 2+1})$.

Here $q=2^2/(2\cdot1)=2$, while $v_0=1$. The mean is zero for all $t$. At time $0$, the variance is $1$, so $\operatorname{Cov}(X_0,X_1)=e^{-1}\cdot1=e^{-1}$. At time $s=\ln 2$, $e^{-2s}=e^{-2\ln2}=1/4$, and the variance is $1\cdot(1/4)+2(1-1/4)=1/4+2(3/4)=1/4+3/2=7/4$. The covariance one time unit later is therefore $\operatorname{Cov}(X_{\ln2},X_{\ln2+1})=e^{-1}(7/4)=\tfrac74e^{-1}$.

Check: The two covariances differ because $7/4\ne1$; the process is not stationary, despite having constant mean and Gaussian finite-dimensional distributions. Its initial variance $1$ is not the invariant variance $q=2$. The displayed calculations also confirm that the covariance at a fixed time gap depends on the earlier time when the initial variance is not $q$.

Common Pitfalls and Interpretation

A tempting but incorrect calculation is to use $\operatorname{Cov}(X_s,X_t)=q e^{-\kappa(t-s)}$ for every initial condition. For the fixed-start example above, that would give $\operatorname{Cov}(X_1,X_2)=1\cdot e^{-2}=e^{-2}$. The correct value is $e^{-2}(1-e^{-4})$, because the variance at the earlier time is $1-e^{-4}$, not $q=1$. The shorter expression applies when the process has its stationary variance $q$, not merely because its drift is mean-reverting.

Likewise, a constant mean does not imply stationarity: the third example has zero mean at every time but a time-dependent variance and covariance. For a Gaussian process, equality of means and covariances under shifts does establish equality of all finite-dimensional distributions; for a non-Gaussian process, mean and covariance information alone does not generally establish strict stationarity.

The stationary covariance $q e^{-\kappa|t-s|}$ quantifies persistence: it equals $q$ when $s=t$ and decreases exponentially as the time separation grows. The rate $\kappa$ controls the decay, while $\sigma^2$ sets the scale $q=\sigma^2/(2\kappa)$. These formulas are useful for understanding the process’s dependence structure without confusing its long-run distribution with its transient behavior from an arbitrary start.

Check Your Understanding

Use the covariance formula and the stationarity result to answer the following questions.

For a deterministic initial value, what are $m_0$ and $v_0$, and how does the variance depend on time?
With $\kappa=2$ and $\sigma=2$, calculate the invariant variance $q$.
If $v_0=q$, simplify the covariance for $s\le t$. Which time difference does it depend on?
Can a Gaussian OU process have a constant mean and still fail to be stationary? Explain using its variance.
Why does a non-Gaussian initial value prevent the process from being a Gaussian process, even if the stochastic noise is Brownian?

Covariance And Stationarity Of The Ornstein-Uhlenbeck Process

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