Is there an intuitive interpretation of a negative variogram “nugget” value?

A variogram plots the variance of the difference between sample pairs on a field (any dimensionality) against spatial separation (the "lag") of those samples.

The extrapolation from observed small-lag variances to a zero-lag variogram value is generally termed the "nugget" value.

Variograms I've come across before (geostatistics) always had positive "nuggets", and these values were generally interpreted as arising from (and being some sort of guide to the magnitude of) a combination of measurement error and sub-observation-scale variation in the observed field.

However, I'm just now looking at variograms of some imaging data and any reasonable extrapolation of the low-lag variogram clearly results in a negative "nugget" value. Of course it's always possible to "force" a non-negative nugget by, say, fitting an exponential model... but this doesn't look very convincing when plotted.

Is there any simple interpretation of negative nugget values (or a simple explanation of how they arise) as there is for positive ones ?

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The nugget actually is not the extrapolation to zero: it is the (necessarily nonnegative) jump at zero. When extrapolation of a fitted ("model") variogram based on the data is negative, this suggests that the spatial process is smooth, which for certain kinds of images might be no surprise, and you should be using a smooth model. This behavior also arises from data with finite (rather than point) support, which is usually the case with digital images, whose values are pixel averages. The true underlying variance is "smeared" out, frequently resulting in what you describe. – whuber♦ Sep 6 '12 at 17:11

1 Answer

The empirical variogram plots squared values, so I don't see how they can be negative. In the simple (Geostatistical) model:

$Y_i = S(x_i) + Z_i,$

where $Z_i \sim N(0,\tau^2)$ are assumed to be mutually independent, often assumed to represent measurement error. Here $\tau^2$ is the nugget variance, which is what your estimate from an empirical variogram gives, since for two measurements at the same location:

$Y_{i1} - Y_{i2} \sim N(0,2\tau^2)$

The variogram gives $\frac{1}{2} \text{Var}\{Y_{i1} - Y_{i2} \}$, which is $\tau^2$ (the nugget).

So the nugget is a variance, which means it can't be negative. The empirical variogram plots $\frac{1}{2}(Y_{i1} - Y_{i2})^2$, which is an estimate for the variance (since this is a zero-mean process), so an estimate for the nugget $\tau^2$, and again can't be negative.

So for your problem, if the variogram is showing negative values then it's probably something else masquerading as a variogram. And if it just looks like it might dip below zero, but doesn't actually, then it's a trick of the eye I think. Hope that helps.

(Some of this answer was taken from Section 2.5 of the book Model-based Geostatistics, by Diggle and Ribeiro (2000). Any mistakes though are mine not theirs!)

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