Gaussian Processes: Weak Stationarity Implies Strict Stationarity

Why Gaussianity Changes the Stationarity Question

Weak stationarity describes a process through its first two moments: its mean stays constant over time, and its covariance depends only on the separation between observations. Strict stationarity asks for more: the entire joint distribution of any finite set of observations must remain unchanged when all their time indices are shifted by the same amount. In general, knowing only means and covariances does not determine a distribution. For a Gaussian process, it does. That is the key reason the two notions coincide in this setting.

The implication is useful because many time-series models use Gaussian assumptions to make their joint distributions tractable. Once a Gaussian process has constant mean and lag-dependent covariance, one need not separately verify every feature of every finite-dimensional distribution. The proof applies the same mean-and-covariance characterization to each such distribution.

The Gaussian Stationarity Result

A multivariate normal distribution is completely determined by its mean vector and covariance matrix, including when that covariance matrix is singular. Therefore, to establish strict stationarity for a Gaussian process, it is enough to show that shifting a finite set of time indices leaves both quantities unchanged.

To see the result directly, choose any \(k\) time indices \(t_1,\ldots,t_k\), and write \(Y=(X_{t_1},\ldots,X_{t_k})^\mathsf{T}\). Weak stationarity gives \(E[Y]=(\mu,\ldots,\mu)^\mathsf{T}\). The covariance matrix of \(Y\) has entry \(\gamma(t_i-t_j)\) in row \(i\), column \(j\). After shifting every index by \(c\), the corresponding entry becomes \(\gamma((t_i+c)-(t_j+c))=\gamma(t_i-t_j)\). Thus the shifted vector has the same mean vector and covariance matrix as \(Y\). Both vectors are multivariate normal by Gaussianity, so they have the same distribution. The indices and shift were arbitrary, which is exactly strict stationarity.

This argument does not require observations at different times to be independent. Their dependence is represented by the off-diagonal entries of the covariance matrix, and those entries are preserved by a common shift. Nor does the proof require an invertible covariance matrix: equality of the mean and covariance still identifies a possibly degenerate multivariate normal distribution.

Verification for a Stationary Gaussian AR(1)

A stationary Gaussian AR(1) process can be written as \(X_t-\mu=\phi(X_{t-1}-\mu)+\varepsilon_t\), where \(|\phi|<1\) and the innovations \(\varepsilon_t\) are independent \(N(0,\sigma_\varepsilon^2)\) random variables. For a process defined at all integer times, take the stationary solution

The series converges in mean square because \(\sum_{j=0}^{\infty}\phi^{2j}<\infty\). Each finite vector of observations is a limit of linear combinations of Gaussian innovations and is jointly Gaussian. Its mean is \(\mu\). The variance and autocovariance, obtained by summing the variances and shared innovation terms, are

The mean does not depend on \(t\), and the covariance depends only on the lag \(h\). Thus the process is weakly stationary. Since it is Gaussian, the theorem implies that it is strictly stationary as well.

A Second Gaussian Construction

The result is not limited to autoregressive models. A simple construction also shows how a process can have dependence across time while still meeting the theorem’s conditions. Let \(Z\) and \(W\) be independent \(N(0,1)\) random variables, and define \(X_t=Z\cos(\omega t)+W\sin(\omega t)\), for integer \(t\) and a fixed real \(\omega\). Every finite vector of observations is a linear transformation of \((Z,W)^\mathsf{T}\), and hence is jointly Gaussian.

Why Gaussianity Matters

For non-Gaussian processes, equal means and covariances do not generally imply equal joint distributions. A common mistake is to reason that a process must be strictly stationary whenever its mean and autocovariance depend only on the lag. That conclusion omits the Gaussian assumption: outside the Gaussian setting, two distributions can agree in their first two moments but differ in shape or dependence structure.

One illustration uses the oscillating construction above, but replaces \(Z\) with a standard normal variable and \(W\) with an independent random variable \(Y\) that equals \(1\) or \(-1\) with equal probability. Both variables have mean zero and variance one. The process \(X_t=Z\cos(\omega t)+Y\sin(\omega t)\) still has mean zero and covariance \(\cos(\omega h)\), so it is weakly stationary. Yet, for \(\omega=\pi/2\), \(X_0=Z\) has a continuous normal distribution while \(X_1=Y\) takes only the values \(-1\) and \(1\). The one-time marginal distributions differ, so the process is not strictly stationary. This example pinpoints what the Gaussian assumption contributes: it turns mean-and-covariance equality into full distributional equality.