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A point about "linear" krigingIt has been claimed that ordinary kriging with a linear semivariogram model produces the same results as a linear interpolator. Despite the similar names, the techniques produce different results. This page provides an example to show why. It concerns kriging with three neighbors using a linear semivariogram. Suppose we have point data supported at corners of a triangle. After a preliminary affine transformation in the plane and an independent affine transformation of the data (to recenter and scale them) we may suppose that the triangle is equilateral with vertices at (sqrt(3), 0), (0,1), and (0, -1) and that the data values are z, 1, and 0, respectively. These transformations, because they are linear, will not alter the linearity or non-linearity of the contours presented below. They can change the anisotropy of the semivariogram, but it will still be linear in every direction. The triangle's vertices are shown by the yellow dots at the bottom left, top left, and middle right parts of the figure below. To make the computations simple, we will suppose that the semivariogram is isotropic after all these preliminary transformations. Since multiplying the semivariogram by a constant does not change interpolated values, we may also suppose that the semivariogram is given by the formula gamma(h) = |h| for all lags h. Because the semivariogram is isotropic, the ordinary kriging estimator for the value at any point x in the plane depends on x only through its distances a, b, and c to the three vertices, respectively. A calculation based on the kriging equations for this semivariogram shows that the contour ("isopleth") at any level r is given by the locus of points x such that
In this equation, r and z are treated as constants, because r is the desired contour level and z is determined by the data. The left hand side is a linear combination of distances from x to three fixed points. It generalizes the more familiar definitions of an ellipse
and an hyperbola
(actually, that's just one limb of an hyperbola). In the case of kriging with three neighbors, the contours are all infinite in extent, because we cannot simultaneously have all three coefficients 1-2z, z-2, and 1+z greater than or equal to zero. The figure shows some contours for z = 4. None are linear, although sections of them within the interior of the triangle do appear to be approximately linear over short distances. They do not even interpolate linearly between the data supports (that is, along the edges of the triangle).
These curves are defined by fairly complicated rational functions. But, for special values of z--the ones for which the data are in arithmetic progression--one of the terms in the equation drops out. For example, for z = 2 the equation becomes
which, for r <> 1, describes one limb of an hyperbola with foci at the right and bottom vertices. The r = 1 contour is the perpendicular bisector of the points with data values z = 0 and z = 2. The r = 2 contour is a ray which extends the side of the triangle going from the point z = 0 to the point z = 2; the r = 0 contour is a similar ray in the other direction. Furthermore, since the sides of the triangle have length 2, c - a must lie between -2 and 2. This implies that r (the contour level) must lie between 0 and 2, proving that in this case the kriged estimates cannot be more extreme than the data. (This is not true in general for all semivariograms.) We can view the contours in the general case as being slight perturbations of hyperbolic shapes whose foci are determined by the locations where z takes on its extreme values. You can see this happening in the figure, where the contours close to the z=4 and z=0 vertices look hyperbolic and the contours near the perpendicular bisector of the z=4 and z=0 vertices are close to linear. It is evident from the form of equation [1] that the kriged surface is continuous everywhere, but it is not necessarily differentiable everywhere: in the example with z=2, the contours for r=0 and r=2 cover the points where the surface does not have a unique slope. Similar calculations show that linear interpolation is accomplished by using a parabolic semivariogram; that is, one of the form gamma(h) = <h,h> = |h|^2 for small values of |h|. --William Huber, 7 January 1999 |
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