Choosing a spatial weight matrix

Video lesson · 01:37 Declare who counts as whose neighbour

Vera compares contiguity, distance-band, and k-nearest-neighbour constructions, showing that each W encodes a different spatial mechanism. The lesson establishes the choice logic before the note covers standardisation, asymmetry, and robustness checks.

The lesson compares common definitions of neighbourhood; theoretical justification and sensitivity checks remain in the full note.

Every spatial statistic — Moran’s I, a spatial lag, a spatial error term — rests on one prior decision: who is whose neighbour? The spatial weights matrix $W$ is where you answer it. It is an $n \times n$ matrix whose entry $w_{ij}$ encodes how connected unit $i$ is to unit $j$, with $w_{ii}=0$ by convention. Once fixed, the spatial lag $Wz$ is just the (weighted) average of each unit’s neighbours — the object every spatial method actually operates on.

The data do not hand you W. You choose its structure:

• Contiguity. $w_{ij}=1$ if areas $i$ and $j$ share a border. Rook contiguity requires a shared edge; Queen also counts a shared corner (so more neighbours). Natural for areal units like districts or census tracts.
• Distance band. $w_{ij}=1$ if the distance $d_{ij}$ is within a cutoff $\delta$, else 0. Simple, but the cutoff is delicate: too small and some units become islands with no neighbours; too large and dense areas acquire enormous neighbour sets.
• k-nearest-neighbours. Each unit links to its k closest others, guaranteeing exactly k neighbours everywhere. Robust when density is uneven (a crowded centre, a sparse edge), at the cost of being asymmetric by construction — i can be a neighbour of j without the reverse.

Then row-standardise: divide each row by its sum so $\sum_j w_{ij}=1$. Now $Wz$ is a true average — comparable across units with different neighbour counts — and the spatial parameters ($\rho$, $\lambda$) land in an interpretable, bounded range. This is standard for both Moran’s I and spatial regression.

The point to internalise: W is a modelling assumption, not a measurement. It is exogenous and pre-specified — you do not estimate it from the outcome y. And it has teeth: a denser W generally inflates measured autocorrelation, and a badly chosen W can flip the diagnostic tests that decide which spatial model you fit. So the discipline is:

1. Pick a structure with a substantive rationale — why these neighbours?
2. Fix it before looking at results, so you can’t tune W toward a wanted answer.
3. Re-run the analysis over alternative W (different k, rook vs queen, other cutoffs) and report whether the conclusions survive.

If the story only holds for one hand-picked weighting, it is a story about your W, not about space.