In spatial statistics one often hears the statements like the following:

unaccounted for spatial autocorrelation may lead to spurious significance / understimated uncertainty / too narrow confidence intervals

and so on. The general idea being that ignoring spatial autocorrelation that is unexplained by your covariates is a big no-no. I'm talking here about models in which there are parameters of fixed effects (such as known spatial covariates) on which we seek to make inference.

I am looking for a mathematical description of this principle

I have heard this principle explained by analogy to pseudo-replicates. By ignoring pseudo-replicates we artificially increase our sample size and so our uncertainty is underestimated. Spatial autocorrelation is like the 'continuous' version of this principle.

Obviously I'm unsatisfied with leaving it at this analogy but haven't been able to find a mathematical description of this issue anywhere.


Okay, let's zoom back from spatial, and think about time-series (1D spatial).

Suppose you are looking at the sum of a statistic over a region:

$s = \sum_{i=1}^N x_i$


$\text{var}_{true}(s) = \sum_{i=1}^N \text{var}(x_i) + 2\sum_{i<j}\text{cov}(x_i,x_j)$

If you pretend that the $x_i$ are independent, i.e. ignore the spatial dependency, then you'll mistakenly think that:

$\text{var}_{iid}(s) = \sum_{i=1}^N \text{var}(x_i)$

Which, if the $\{x_i\}$ are postively correlated (not guaranteed, but this the most common scenario in reality) we will have:

$\text{var}_{iid}(s) \le \text{var}_{true}(s)$

Then, because you believe the variance of your statistic to be lower than it should be, you find it 'easier' to achieve significance. But, of course, such results are not accurate.

  • $\begingroup$ Thanks, I understand this point. I'm probably being stupid here but struggling to relate it to understanding $\text{var}(\beta_i)$ where $\beta_i$ is a fixed effect parameter. Say I model a gaussian response $y(t) = \beta_0 + \beta_1 x(t)$ but the 'true' model is $y(t) = \beta_0 + \beta_1 x(t) + f(t)$ where f is some smooth thing (e.g. b-spline GAM). How does missing out $f$ affect my estimated variance for $\beta_1$ ? $\endgroup$ – ASeaton Dec 13 '18 at 14:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.