Introduction

In spatial statistics and time series analysis, stationarity is a fundamental assumption. For Gaussian processes, there is a famous and extremely useful result:

Strong stationarity (strict sense) is equivalent to weak stationarity (second-order / wide sense).

This document provides a concise mathematical proof of this equivalence.

Definitions

Let \(\{Y(s) : s \in D\}\) be a stochastic process indexed by spatial locations \(s\) in a domain \(D \subseteq \mathbb{R}^d\).

(Note: The proof is identical if we use time index \(t \in \mathbb{R}\) instead of space; we use \(Y(s)\) here to align with common geostatistics notation.)

1. Strong Stationarity (Strict Sense)

The process is strictly stationary if for any finite set of locations \(s_1, \dots, s_n \in D\) and any spatial shift vector \(h\) such that \(s_i + h \in D\), the joint distribution is shift-invariant:

\[ (Y(s_1 + h), \dots, Y(s_n + h)) \stackrel{d}{=} (Y(s_1), \dots, Y(s_n)) \]

2. Weak Stationarity (Second-Order / Wide Sense)

The process is weakly stationary if:

  1. The mean is constant: \(\mathbb{E}[Y(s)] = \mu\) for all \(s\).
  2. The covariance depends only on the spatial lag (difference) between two points: \[ \text{Cov}(Y(s_i), Y(s_j)) = C(s_i - s_j) \]
  3. The second moments are finite: \(\mathbb{E}[Y(s)^2] < \infty\).

The Proof

Part 1: Strong Stationarity \(\Rightarrow\) Weak Stationarity

This implication holds for ANY process (no Gaussian assumption is needed).

If all finite-dimensional distributions are shift-invariant, then:

  • The marginal distribution of \(Y(s)\) is identical for all \(s\). Therefore, \(\mathbb{E}[Y(s)] = \mu\) (constant) and the variance is constant (and finite if it exists).
  • The joint distribution of \((Y(s_i + h), Y(s_j + h))\) is identical to that of \((Y(s_i), Y(s_j))\). Consequently, the covariance between them must be the same: \[ \text{Cov}(Y(s_i + h), Y(s_j + h)) = \text{Cov}(Y(s_i), Y(s_j)) \] This implies the covariance depends only on the lag \(s_i - s_j\).

Thus, strong stationarity implies weak stationarity (provided second moments exist).

Part 2: Weak Stationarity \(\Rightarrow\) Strong Stationarity

This implication REQUIRES the Gaussian assumption.

Let \(\{Y(s)\}\) be a Gaussian process that is weakly stationary. By definition of a Gaussian process, all finite-dimensional distributions are multivariate normal.

Take any finite set of locations \(s_1, \dots, s_n\) and any shift vector \(h\). Consider the two random vectors:

\[ \mathbf{Y} = (Y(s_1), \dots, Y(s_n))^T \] \[ \mathbf{Y}_h = (Y(s_1 + h), \dots, Y(s_n + h))^T \]

Since the process is Gaussian, both \(\mathbf{Y}\) and \(\mathbf{Y}_h\) are jointly multivariate normal. A multivariate normal distribution is completely and uniquely determined by its first two moments (mean vector and covariance matrix). Therefore, to show \(\mathbf{Y}\) and \(\mathbf{Y}_h\) have the same distribution, we just need to show they have identical means and covariances.

Step 1: Compare the Mean Vectors

Weak stationarity gives \(\mathbb{E}[Y(s)] = \mu\) for all \(s\). Hence:

\[ \mathbb{E}[\mathbf{Y}] = \begin{pmatrix} \mu \\ \mu \\ \vdots \\ \mu \end{pmatrix}, \quad \mathbb{E}[\mathbf{Y}_h] = \begin{pmatrix} \mu \\ \mu \\ \vdots \\ \mu \end{pmatrix} \]

They are identical.

Step 2: Compare the Covariance Matrices

The \((i,j)\)-th entry of the covariance matrix of \(\mathbf{Y}\) is:

\[ \text{Cov}(Y(s_i), Y(s_j)) = C(s_i - s_j) \]

The \((i,j)\)-th entry of the covariance matrix of \(\mathbf{Y}_h\) is:

\[ \text{Cov}(Y(s_i + h), Y(s_j + h)) \]

By the weak stationarity assumption, this covariance depends only on the lag between the two points:

\[ \text{Cov}(Y(s_i + h), Y(s_j + h)) = C((s_i + h) - (s_j + h)) \]

Simplify the algebra inside the parentheses:

\[ (s_i + h) - (s_j + h) = s_i + h - s_j - h = s_i - s_j \]

Therefore:

\[ \text{Cov}(Y(s_i + h), Y(s_j + h)) = C(s_i - s_j) \]

This is exactly the same as the covariance for \(\mathbf{Y}\). Since this holds for all \(i, j\), the two covariance matrices are identical.

Step 3: Conclude Equal Distributions

Because \(\mathbf{Y}\) and \(\mathbf{Y}_h\) are multivariate normal with the same mean vector and the same covariance matrix, they must have the identical joint distribution.

Since we chose an arbitrary finite set of locations and an arbitrary shift \(h\), this holds for all finite-dimensional distributions. Hence, the process is strongly stationary.

Conclusion

We have proven both directions:

  1. Strong \(\Rightarrow\) Weak (holds for all processes).
  2. Weak \(\Rightarrow\) Strong (holds specifically for Gaussian processes).

Therefore, for Gaussian processes:

\[ \boxed{\text{Strong Stationarity} \iff \text{Weak Stationarity}} \]

This is a profoundly useful result because weak stationarity is much easier to check (we only need to estimate the mean and the covariance function), yet for Gaussian data, it guarantees the full distributional invariance required by strong stationarity.