Testing Spatial Autocorrelation in Weighted Networks: The Modes Permutation Test

Abstract

Permutation tests of spatial autocorrelation are justified under exchangeability, that is the premise that the observed scores follow a permutation-invariant joint distribution. Yet, in the frequently encountered case of geographical data collected on regions differing in importance, the variance of a regional score is expected to decrease with the size of the region, in the same way that the variance of an average is inversely proportional to the size of the sample in elementary statistics: heteroscedasticity holds in effect, already under spatial independence, thus weakening the rationale of the celebrated spatial autocorrelation permutation test (e.g. Cliff and Ord 1973; Besag and Diggle 1977) in the case of a weighted network.

Appendix

Proof of (4.7) U being orthogonal, $\sum _{i}f_{i}c_{i\alpha }c_{i\beta } =\sum _{i}u_{i\alpha }u_{i\beta } =\delta _{\alpha \beta }$ and $\sum _{i}f_{i}c_{i\alpha } =\sum _{i}\sqrt{f_{i}}u_{i\alpha } =\sum _{i}u_{i0}u_{i\alpha } =\delta _{\alpha 0}$.

Proof of (4.9) independence implies the functional form σ _ij  = δ _ij  g( f _i ) where g(f) expresses a possible size dependence. Consider the aggregation of regions j into super-region J, with aggregated field $X_{J} =\sum _{j\in J}f_{j}X_{j}/f_{J}$, where f _J : = ∑ _j ∈ J f _j . By construction,

$$\displaystyle{g(\,f_{J}) = \mbox{ Var}(X_{J}) = \frac{1} {f_{J}^{2}}\sum _{i,j\in J}f_{i}f_{j}\sigma _{ij} = \frac{1} {f_{J}^{2}}\sum _{j\in J}f_{j}^{2}g(\,f_{ j})}$$

that is f _J ² g( f _J ) = ∑ _j ∈ J f _j ² g( f _j ), with unique solution $g(\,f_{j}) =\sigma ^{2}/f_{j}$ (and $g(\,f_{J}) =\sigma ^{2}/f_{J}$), where $\sigma ^{2} = \mbox{ Var}(\bar{X})$.

Proof of (4.10) $\hat{\sigma }_{\alpha \beta }:= \mbox{ Cov}(\hat{X}_{\alpha },\hat{X}_{\beta }) =\sum _{ij}f_{i}f_{j}c_{i\alpha }c_{j\beta }\mbox{ Cov}(X_{i},X_{j}) =\sigma ^{2}\sum _{i}f_{i}c_{i\alpha }c_{i\beta } =\sigma ^{2}\sum _{i}u_{i\alpha }u_{i\beta } =\sigma ^{2}\delta _{\alpha \beta }$.

Proof of (4.11) $\sum _{\alpha \geq 1}\hat{x}_{\alpha }^{2} =\sum _{ij}\sqrt{f_{i } f_{j}}x_{i}x_{j}\sum _{\alpha \geq 0}u_{i\alpha }u_{j\alpha } -\hat{ x}_{0}^{2} =\sum _{i}f_{i}x_{i}^{2} -\bar{ x}^{2} = \mbox{ var}(x)$. Also, $\mbox{ var}_{\mbox{ loc}}(x) = \frac{1} {2}\sum _{ij}e_{ij}(x_{i}-x_{j})^{2} =\sum _{ i}f_{i}x_{i}^{2}-\sum _{ ij}e_{ij}x_{i}x_{j} =\sum _{i}f_{i}x_{i}^{2}-\bar{x}^{2}-\sum _{\alpha \geq 1}\lambda _{\alpha }\sum _{i}c_{i\alpha }x_{i}\sum _{j}c_{j\alpha }x_{j} = \mbox{ var}(x)-\sum _{\alpha \geq 1}\lambda _{\alpha }\hat{x}_{\alpha }^{2}.$

Proof of (4.12) and (4.13) define

$$\displaystyle{a_{\alpha }:= \frac{\hat{x}_{\alpha }^{2}} {\sum _{\beta \geq 1}\hat{x}_{\beta }^{2}}\qquad \mbox{ with}\qquad \sum _{\alpha \geq 1}a_{\alpha } = 1\qquad \mbox{ and}\qquad I(\hat{x}) =\sum _{\alpha \geq 1}\lambda _{\alpha }a_{\alpha }.}$$

Under H ₀, the distribution of the non-trivial modes is exchangeable, i.e. f(a) = f(π(a)). By symmetry, $E_{\pi }(a_{\alpha }) = 1/(n - 1)$, $E_{\pi }(a_{\alpha }^{2}) = s(x)/(n - 1)^{2}$ where $s(x) =\sum _{\beta \geq 1}a_{\beta }^{2}/(n - 1)$ and $E_{\pi }(a_{\alpha }a_{\beta }) = (1 - s(x)/(n - 1))/[(n - 1)(n - 2)]$ for α ≠ β. Further substitution proves the result.

Proof of the Semi-Negative Definiteness of Q in (4.20) for any vector h,

$$\displaystyle{0 \leq \frac{1} {2}\sum _{ij}\varepsilon _{ij}(h_{i} - h_{j})^{2} =\sum _{ i}\sigma _{i}h_{i}^{2} -\sum _{ ij}\epsilon _{ij}h_{i}h_{j} = -\sum _{ij}(\epsilon _{ij} -\delta _{ij}\sigma _{j})h_{i}h_{j}.}$$

Relation Between the Eigen-Decompositions of E ^s (t) and Q in (4.20) in matrix notation, $Q = \Pi ^{\frac{1} {2} }R\Pi ^{-\frac{1} {2} }$, and hence $Q\sqrt{f} = 0$ by (4.19), showing $u_{0} = \sqrt{f}$ with μ ₀ = 0. Consider another, non-trivial eigenvector u _α of Q, with eigenvalue μ _α , orthogonal to $\sqrt{ f}$ by construction. Identity $E(t) = \Pi \exp (tR)$ together with (4.5) yield

$$\displaystyle{E^{s}(t) =\sum _{ k\geq 0}\frac{t^{k}} {k!}Q^{k} -\sqrt{f}\sqrt{f}'\qquad \qquad E^{s}(t)u_{\alpha } =\sum _{ k\geq 0}\frac{t^{k}\mu _{\alpha }^{k}} {k!} u_{\alpha } =\exp (\mu _{\alpha }t)u_{\alpha }.}$$