Stationarity and Isotropy in Spatial Covariance

Why Spatial Covariance Assumptions Matter

In a spatial GLMM, a latent effect can represent dependence among observations taken at different locations. A covariance model describes how that dependence changes across pairs of sites. Two useful simplifying assumptions are stationarity, which removes absolute location from the covariance, and isotropy, which removes direction as well. They reduce a covariance that could vary separately for every pair of sites to one described by a displacement vector, or even just a distance.

These assumptions are substantive, not automatic consequences of using a spatial random effect. A region may have a strong trend, different variability in different subregions, or dependence that is stronger along one direction than another. The assumptions should therefore be stated and assessed in the setting being modeled.

Definitions in Terms of a Spatial Latent Effect

Let $Z(\mathbf{s})$ be a second-order spatial random field at location $\mathbf{s}\in D\subseteq\mathbb{R}^d$. Second-order means that the field has finite means and variances. Write its covariance at two locations as $C(\mathbf{s},\mathbf{t})=\operatorname{Cov}\{Z(\mathbf{s}),Z(\mathbf{t})\}$. The definitions below concern this latent field; they do not require the observed GLMM responses themselves to have constant means or variances.

Definition: The field is second-order stationary if its mean is constant, $E\{Z(\mathbf{s})\}=m$, and there is a function $C$ such that $$ \operatorname{Cov}\{Z(\mathbf{s}),Z(\mathbf{t})\}=C(\mathbf{t}-\mathbf{s}) $$ for all locations in its domain. Thus covariance depends on the displacement $\mathbf{h}=\mathbf{t}-\mathbf{s}$, not on the absolute locations. In particular, the variance is constant and equals $C(\mathbf{0})$.

Covariance symmetry requires $C(\mathbf{h})=C(-\mathbf{h})$: exchanging the two locations reverses the displacement but does not change their covariance. A stationary covariance is not an arbitrary function of displacement. For every finite selection of sites, the resulting covariance matrix must be positive semidefinite; this is the validity requirement for a covariance structure.

Definition: A stationary field is isotropic in covariance if its covariance is unchanged by every rotation or reflection of the displacement vector. Equivalently, there is a function $c$ such that $$ C(\mathbf{h})=c(\|\mathbf{h}\|). $$ Here $\|\mathbf{h}\|$ is Euclidean distance. Isotropy is therefore a special case of stationarity: pairs at the same distance have the same covariance, regardless of direction.

Covariance isotropy is a second-order property; by itself, it does not assert that all joint distributions are unchanged by rotations. Nor does observing similar sample covariances in several directions prove isotropy. With data from a finite region, there are only a limited number of pairs at each displacement or distance, and empirical estimates are noisy.

From Displacement Invariance to Distance Invariance

The distinction can be expressed using a translation and an orthogonal transformation. Translating both sites by the same vector preserves their displacement. Isotropy adds invariance when the displacement itself is rotated or reflected. The short argument below explains why this gives a distance-only covariance.

Use stationarity.
For locations $\mathbf{s}$ and $\mathbf{t}$, write $\mathbf{h}=\mathbf{t}-\mathbf{s}$. Their covariance is $C(\mathbf{h})$.

Use isotropy.
For any orthogonal matrix $Q$, isotropy gives $C(Q\mathbf{h})=C(\mathbf{h})$. Orthogonal transformations preserve length, so $\|Q\mathbf{h}\|=\|\mathbf{h}\|$.

Compare equal-length displacements.
Any two vectors with the same Euclidean length can be related by an orthogonal transformation. Their covariance is therefore equal, so covariance can be written as a function $c$ of length alone.

Conversely, if $C(\mathbf{h})=c(\|\mathbf{h}\|)$, then translating both locations leaves $\mathbf{h}$ unchanged, and rotating or reflecting it leaves its norm unchanged. The covariance is consequently stationary and isotropic, provided the function defines valid covariance matrices. This algebraic implication does not establish that a proposed distance function is valid; positive semidefiniteness remains necessary.

Worked Examples

Worked Example: Checking Isotropy on a Four-Site Square

Consider hypothetical sites at $(0,0)$, $(1,0)$, $(1,1)$, and $(0,1)$, with coordinates measured in kilometres. Suppose a proposed covariance assignment gives variance $4$ at each site, covariance $1.5$ for each side of length $1$, and covariance $1$ for each diagonal of length $\sqrt{2}$. The covariance is in squared units of the latent effect.

In the stated site order, the matrix is $$ \begin{pmatrix} 4&1.5&1&1.5\\ 1.5&4&1.5&1\\ 1&1.5&4&1.5\\ 1.5&1&1.5&4 \end{pmatrix}. $$ Each pair at the same distance has the same assigned covariance: every side has covariance $1.5$, and both diagonals have covariance $1$. The assignment is consistent with isotropy on these four sites. It is also consistent with stationarity on this set: repeated displacement vectors have the same covariance, and reversing a displacement preserves it.

To check validity on this finite set, the matrix is circulant, with first row $(4,1.5,1,1.5)$. Its eigenvalues are $$ 4+1.5+1+1.5=8,\qquad 4-1=3,\qquad 4-1.5+1-1.5=2,\qquad 4-1=3. $$ Thus the eigenvalues are $8,3,3,2$, all positive. The matrix is positive definite, so it is a valid covariance matrix for these four sites. This finite check does not prove that a covariance rule extending the assignment to every location in the plane is valid.

Worked Example: Comparing Directional Pair Covariances

Suppose hypothetical residual products from pairs one kilometre apart are used to estimate covariance in two directions. For five east–west pairs, let the products, in squared outcome units, be $3,2,1,2,$ and $2.5$. Their average is $$ \widehat C_{\mathrm{east-west}}(1)=\frac{3+2+1+2+2.5}{5}=\frac{10.5}{5}=2.1. $$ For five north–south pairs, let the products be $0.5,1,0.8,1.2,$ and $0.5$. Then $$ \widehat C_{\mathrm{north-south}}(1)=\frac{0.5+1+0.8+1.2+0.5}{5}=\frac{4}{5}=0.8. $$

Both sets of pairs have distance $1$ kilometre, but their estimated covariances differ by $2.1-0.8=1.3$ squared outcome units. If the population covariance were isotropic, those two directional covariances would be equal. The discrepancy is a reason to investigate isotropy, not conclusive evidence against it: with only five pairs in each direction, sampling variability may be substantial, and pairs may also share sites. A useful assessment compares directional estimates across multiple distance bands and considers their uncertainty.

Worked Example: A Valid Covariance That Is Not Stationary

Let locations be $\mathbf{s}_i=(i,0)$ for $i=0,1,2$, with distances measured in kilometres. Define a hypothetical latent field by $$ Y(\mathbf{s}_i)=a_i U+\epsilon_i,\qquad a_i=1+i, $$ where $E(U)=E(\epsilon_i)=0$, $\operatorname{Var}(U)=1$, $\operatorname{Var}(\epsilon_i)=1$, and the $\epsilon_i$ are mutually independent and independent of $U$. This construction gives a valid covariance because it is the covariance of explicitly defined random variables.

The variance at site $i$ is $a_i^2+1$, giving $2,5,$ and $10$ at sites $0,1,$ and $2$. For distinct sites $i,j$, independence of the errors gives $\operatorname{Cov}\{Y(\mathbf{s}_i),Y(\mathbf{s}_j)\}=a_i a_j$. The two pairs separated by one kilometre therefore have covariances $$ \operatorname{Cov}\{Y(\mathbf{s}_0),Y(\mathbf{s}_1)\}=1\cdot2=2,\qquad \operatorname{Cov}\{Y(\mathbf{s}_1),Y(\mathbf{s}_2)\}=2\cdot3=6. $$

The same displacement has different covariance, and the variances also differ by location. Both facts violate second-order stationarity. The example shows why validity and stationarity are separate: a covariance can be valid without depending only on displacement.

Diagnosing the Assumptions and Choosing Alternatives

A practical assessment starts with the latent spatial structure after accounting for the modeled mean and other covariates. A trend left in the mean can create apparent spatial dependence in residuals; it should not automatically be interpreted as evidence about the covariance. The model’s constant-mean assumption concerns the latent field $Z(\mathbf{s})$, not necessarily the GLMM response, whose mean may vary with covariates and the link function.

Check for location-dependent variation. Compare residual or latent-effect variability across subregions. Persistent changes in spread are inconsistent with a constant-variance stationary field.
Check repeated displacements. Compare pairs with similar displacement vectors but different absolute locations. Large systematic differences suggest nonstationarity.
Check directions at a fixed distance. Compare pair summaries in different directions. Persistent directional differences challenge isotropy, though finite-sample noise must be considered.
Check the domain. A boundary, barrier, or disconnected region can make Euclidean closeness a poor description of spatial connection. Pair comparisons may be sparse or incomparable across parts of the domain.

Stationarity and isotropy are alternatives to a fully location-specific covariance, not to all spatial modeling. If direction matters but translation invariance remains plausible, a stationary but anisotropic covariance allows covariance to depend on the direction as well as the displacement length. If variability or dependence changes across the region, a nonstationary covariance is more appropriate. Spatial models based on neighbourhood structure provide another modeling framework when the data are organized by areas and adjacency rather than by continuous locations; their assumptions and interpretation differ from a distance-based covariance.

Each alternative costs flexibility or requires additional structure. A more flexible covariance can be harder to estimate, especially with few spatial locations or limited pairs at particular distances. Conversely, imposing isotropy can hide meaningful directional dependence, while imposing stationarity can obscure spatially varying uncertainty. A diagnostic plot or comparison is evidence to guide model choice, not a guarantee that a chosen covariance is correct. Any proposed structure must still yield valid covariance matrices for the locations being modeled.

Common Pitfall

Wrong reasoning: “Pairs at the same distance have similar estimated covariances, so the process is stationary and isotropic.” Similar estimates do not establish either assumption. Isotropy concerns population covariance across directions; stationarity also requires that covariance not change with absolute location and that the latent mean and variance be constant. A finite set of noisy pair estimates cannot verify those conditions everywhere.

Correction: Treat equal-distance comparisons as a diagnostic for isotropy, and separately examine repeated displacements across locations and the latent mean and variance for stationarity. Also distinguish those assumptions from positive definiteness: matching distance patterns does not by itself make a proposed covariance valid.

Key takeaway: Stationarity makes covariance a function of displacement; isotropy further makes it a function of distance alone. Both are modeling assumptions about the latent field, and neither replaces the requirement that the covariance be valid.

Check Your Understanding

Answer each question, then compare with the solutions below.

For a stationary field, what feature of two locations determines their covariance? What additional restriction does isotropy impose?
Two pairs are each $3$ kilometres apart, but one pair is east–west and the other north–south. What equality does isotropy require of their population covariances?
A latent field has constant mean but its variance increases from west to east. Can it be second-order stationary? Explain.
A proposed covariance matrix for observed sites is positive definite. Does this prove that an assumed distance-based covariance rule is valid for all locations? Explain.

Solutions. (1) Under stationarity, covariance depends on the displacement vector from one site to the other, not the absolute locations; isotropy further requires equal covariance for all displacements of equal length. (2) The population covariances must be equal because the distances match. (3) No. Second-order stationarity requires constant variance as well as constant mean and displacement-based covariance. (4) No. Positive definiteness verifies the matrix for those sites only; validity of a rule over all locations requires positive semidefinite covariance matrices for every finite set of locations.