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I have some spatially autocorrelated vegetation data, and would like to know the how well tree size measured at one location can predict tree size in plots 100m away.

I've made a semivariogram of the data, but am wondering if I can calculate an R squared value from this? From my (basic) understanding, I'm thinking the semivariance at the sill gives the total variance in the data, whereas the semivariance at 100m measures the variance at that distance; so $\small 1-\frac{\text{semivariance at 100m}}{\text{semivariance at sill}}$ might be equivalent to an $R^2$ value for the relationship between plots separated by 100m?

For example, if I have a semivariance of 10 at the sill and 2 at 100m, would this suggest that 1-2/10 = 80% of the variation in tree size across all plots can be explained by tree size in plots 100m away?

kjetil b halvorsen
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jay
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  • Semivariograms do not predict. They are used to model spatial correlation within a "random field" or spatial stochastic model, which in turn is applied to the prediction problem via *kriging.* You could--at least in principle, by severely limiting your options--krige the data using neighborhoods of a single tree and cross-validate that kriging (with a standard leave-one-out jackknife procedure) to obtain an *estimation variance.* – whuber Oct 15 '12 at 15:33
  • More complicated than I would have liked but thems the breaks. – jay Oct 20 '12 at 00:07
  • thanks for taking the time to answer though – jay Oct 20 '12 at 00:07
  • The sill (or, more generally, the sill plus the nugget plus the measurement variance) does not give the total semivariance in the data: it tends to overestimate it. It gives the *fitted* value of the *limiting* semivariance at large distances. – whuber Aug 16 '20 at 15:04

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