Common mistake 1 with Spatial autocorrelation

Reading clustering as causation. Neighbouring places are similar for three quite different reasons - a spillover between them, a shared unobserved cause, or units drawn at the wrong scale - and this test cannot tell them apart. Spatial regression starts on that question; it does not finish it.

Common mistake 2 with Spatial autocorrelation

Choosing the weights matrix until the answer looks good. Report the scheme, and say what happens under a different one.

Common mistake 3 with Spatial autocorrelation

Using the analytical p-value for LISA. Local statistics are strongly non-normal at small neighbour counts, which is why the permutation is the default here.

Common mistake 4 with Spatial autocorrelation

Forgetting the multiple-comparison problem in LISA. Testing 200 places at 5% gives you about ten false hot spots for free.

EconStatsCraft › Spatial autocorrelation

Spatial autocorrelation

Is this variable clustered in space, or scattered at random over the map? Moran's I answers it globally, LISA answers it place by place and names the hot spots, the cold spots and the outliers. Run it before any regression on geographic data: if the answer is yes, ordinary standard errors are too small.

What it needs

Variable

The intuition

Tobler's first law: everything is related to everything else, but near things more than distant things. Before fitting any spatial model, the question is whether that is true of your data - whether districts with high values sit next to other districts with high values more than chance would produce. Moran's I answers it with a single number that behaves like a correlation coefficient for space.

When to use it

Before any regression on geographic units - districts, states, countries, firms with addresses. If the dependent variable or the residuals are spatially clustered, ordinary standard errors are too small, because two neighbouring observations are not two independent pieces of information.

Where the estimator comes from

Given a spatial weights matrix W with w_{ij} describing how connected i and j are, and z_i=x_i-\bar{x}, I=\frac{n}{\sum_i\sum_j w_{ij}}\cdot\frac{\sum_i\sum_j w_{ij}z_iz_j}{\sum_i z_i^2}. The second factor is a ratio of a weighted cross-product to a variance, which is why I reads like a correlation. Under the null of no spatial association E[I]=-1/(n-1) - NOT zero, which matters in small samples - and a variance is available analytically or by permutation. Geary\'s C uses squared differences instead of cross-products and is more sensitive to local variation; the two usually agree in direction and are inversely related.

The model

I=\frac{n}{S_0}\cdot\frac{\sum_i\sum_j w_{ij}(y_i-\bar y)(y_j-\bar y)}{\sum_i (y_i-\bar y)^2}\qquad E[I]=\frac{-1}{n-1}

What it assumes

You have defined what "neighbour" means. This is not a detail: every number here is a statement about the weights matrix as much as about the data, and changing k or the distance band can change the conclusion. Choose the scheme for a reason and report it.
Every unit has a value. A missing one is not just a lost row - its neighbours' statistics depend on it.
For Getis-Ord G, the variable has a meaningful zero and cannot be negative. It is a ratio of cross-products, so a variable that goes negative makes it meaningless rather than merely odd.
The weights matrix is a modelling choice made before looking at the result. Nearest neighbours, contiguity and distance bands give different statistics on the same data, and choosing among them after seeing which rejects is not a test.

Statistical properties

The expected value under the null is -1/(n-1), not 0. With 30 units that is -0.034, and treating a small positive I as evidence without comparing against the correct null is a common error.
Everything depends on W, which is a modelling choice rather than data. Contiguity, k-nearest-neighbours and distance bands give different answers, and the choice should be justified and its sensitivity checked.
Row-standardising W makes I comparable across units with different numbers of neighbours, and is the usual convention.
Permutation inference - shuffling values across locations and recomputing I - makes no distributional assumption and is preferable to the analytic normal approximation for irregular geographies.
Global Moran's I averages over the whole map and can be near zero when strong clustering of opposite signs cancels out. Local indicators (LISA) decompose it and reveal exactly that.
Spatial autocorrelation in a REGRESSION residual is the thing that matters for inference. Testing the raw variable tells you geography exists; testing residuals tells you the model has missed it.

How to read the output

Compare I with E[I] = −1/(n−1), not with zero. On a small sample that expectation is visibly negative, and a small positive I can be nothing at all.
The Moran scatterplot IS the statistic: I is the slope of the fitted line. The four quadrants are the four LISA categories, which is what makes the local map readable.
A high-low or low-high unit is a spatial outlier - a rich place surrounded by poor ones. Those are often the interesting cases, and they are invisible in the global statistic.
Geary's C weights local differences rather than covariances, so it disagrees with Moran when the pattern is driven by a few extreme pairs. Read them together.
LISA needs a reasonable n to say anything. With a few dozen units and few neighbours each, a random rearrangement often reproduces the pattern, so nothing will be significant - that is honest, not a failure.

A worked reading

District GDP per capita over 641 Indian districts, with five nearest neighbours and row-standardised weights. Moran's I = 0.7337 against an expected value under the null of -0.0016, giving z = 30.97: emphatically clustered. Note the reference point is -0.0016 and not zero - the expectation of Moran's I under no spatial pattern is -1/(n-1), so with 641 units it is very close to zero but with 16 it would be -0.067, and on a small map that difference is the whole test. Geary's C = 0.2618 against its own null of 1 says the same thing in the other direction. The local decomposition is where the interpretation lives: 89 districts are high-value surrounded by high, 92 are low surrounded by low, exactly two are outliers against their neighbourhoods, and 458 - more than two thirds - are not distinguishable from chance. A map of that statistic shows clusters in a fifth of the country and nothing in the rest, which is a more honest picture than the single global number.

Symptom → diagnosis → what to do

If you see this	It means	Do this
Moran I near zero	No global clustering, or opposite-sign clusters cancelling	Check local indicators before concluding
I negative and significant	A chequerboard: neighbours differ systematically	Real, and much rarer than positive clustering
Results changing with the weights matrix	Expected. W is an assumption	Report several and say which was chosen and why
Some units having no neighbours	Islands under a contiguity rule	They contribute nothing. Use k-nearest instead
I positive on the raw variable only	Geography, not model misspecification	Test the residuals
Very small n	The -1/(n-1) null matters	Use permutation inference

Common mistakes

Reading clustering as causation. Neighbouring places are similar for three quite different reasons - a spillover between them, a shared unobserved cause, or units drawn at the wrong scale - and this test cannot tell them apart. Spatial regression starts on that question; it does not finish it.
Choosing the weights matrix until the answer looks good. Report the scheme, and say what happens under a different one.
Using the analytical p-value for LISA. Local statistics are strongly non-normal at small neighbour counts, which is why the permutation is the default here.
Forgetting the multiple-comparison problem in LISA. Testing 200 places at 5% gives you about ten false hot spots for free.

Where to read this properly

Moran (1950), "Notes on Continuous Stochastic Phenomena", Biometrika 37(1/2).
Anselin (1995), "Local Indicators of Spatial Association", Geographical Analysis 27(2).
Cliff and Ord, Spatial Processes (1981).
LeSage and Pace, Introduction to Spatial Econometrics (2009), ch. 1 and 4.

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Spatial autocorrelation

What it needs

The intuition

When to use it

Where the estimator comes from

The model

What it assumes

Statistical properties

How to read the output

A worked reading

Symptom → diagnosis → what to do

Common mistakes

Where to read this properly

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