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?

  • $\begingroup$ 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. $\endgroup$
    – whuber
    Oct 15, 2012 at 15:33
  • $\begingroup$ More complicated than I would have liked but thems the breaks. $\endgroup$
    – jay
    Oct 20, 2012 at 0:07
  • $\begingroup$ thanks for taking the time to answer though $\endgroup$
    – jay
    Oct 20, 2012 at 0:07
  • $\begingroup$ 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. $\endgroup$
    – whuber
    Aug 16, 2020 at 15:04


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