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I have a large gridded dataset for the globe (i.e a spherical, wraparound surface) that I'm applying spatial regression to (using a CAR model). I've been using the default autocorrelation function, however as my data is global (point 0,0 lies next to max(x),0 and 0,max(y)), I wondered if a spherical autocorrelation function would work better.

Though I've alot of references on its use, I've been unable to find a simple answer or diagram as to what a spherical autocorrelation function actually is! Is it as I presumed, an autocorrelation where data wraps around a sphere, or a specific mathematical function of the variogram shape, or something else entirely?

alopex
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    Do you have a reference? Please note that the so-called "spherical" variogram model has *nothing* to do with spatial correlation on the sphere! Also, where you write "spatial autocorrelation function," did you really mean "spherical autocorrelation function"? The former is a very general question whereas the latter is more consistent with the rest of your question. – whuber Nov 28 '11 at 18:22
  • Indeed I did. Thanks for pointing that out, and apologies for any confusion in terminology. – alopex Nov 29 '11 at 15:04
  • For a discussion of autocorrelation on the sphere, see the technical report at http://www.stat.indiana.edu/files/TR/TR-09-01.pdf. – whuber Nov 29 '11 at 15:48

1 Answers1

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I'll make a leap of faith, and assume that you are referring to a spherical spatial correlation structure.

A spherical spatial correlation structure has two parameters: $n$, the "nugget" effect, which acts to reduce all the correlations between two observations more than 0 distance apart, and $d$, the range (distance) over which the correlations will be nonzero. Slightly rephrasing the documentation from R's nlme package:

The correlation between two observations a distance $r < d$ apart is, if the nugget effect is zero, $1 - 1.5(r/d) + 0.5(r/d)^3$. If $r\ge d$ the correlation is zero. If there is a nugget effect $n$, the correlation is just $(1-n)(1 - 1.5(r/d) + 0.5(r/d)^3)$ for all observations for which $r > 0$.

The "spherical" refers to its symmetry with respect to direction, like a sphere with respect to the origin, rather than to the shape of the surface from which the data was collected, although, confusingly, it is also the name of the structure. However, it does seem to me that, if there is correlation between your observations that is related to distance between them regardless of direction, a spherical spatial correlation structure would be a reasonable first try. There are other spatial correlation structures, though, e.g., Gaussian or exponential (which also are symmetric with respect to direction.)

A reference is "Statistics for Spatial Data", by N. A. C. Cressie, 1993.

jbowman
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    Some points: (1) the nugget is not part of the definition of the spherical variogram. (2) "Spherical" does not refer to the isotropy, which is a separate concept and characteristic of spatial processes. (3) "Spherical" refers to a particular model that yields this variogram. (4) One does not "try" various variogram structures; there are principled procedures to *identify* them and estimate their parameters; this is generally called "variography." – whuber Nov 29 '11 at 15:48