Is it true that we're just using a different set of weight values than the 1/distance in IDW?
Yes, both IDW interpolation and Ordinary Kriging (OK) will calculate weights based on distance, but with different criteria. In both methods, weights do not depend on sample values. The answer from Dahn Jahn in Ordinary kriging example step by step? is very clarifying about this.
The main difference is that in Ordinary Kriging distances are used to measure the correlation structure among sample points (i.e. how similar they are based on lags of distances) and this correlation structure is captured in the form of the variogram model. The variogram model is used to calculate covariances (or semi variances): i) among sample points and; ii) between sample points and prediction points. These covariances are then, used to calculate weights (see how below).
How does Ordinary Kriging form weights?
The main processing steps in Ordinary Kriging are:
- Calculate experimental variogram.
- Fit theoretical variogram model.
- Calculate weights (using the Lagrange multiplier method).
- Predict (and estimate kriging variance).
For a detailed mathematical explanation and theoretical insights about Ordinary Kriging refer to materials linked in 'References'.
Assume from bullet number 2 our variogram model is of type exponential:
C(h) = a + (σ² - a), if |h| = 0
C(h) = (σ² - a)exp(-3|h|/r), if |h| > 0
C(h) = covariance;
a = nugget effect;
σ² = sill (
σ² - a = partial sill);
r = range.
We want to predict a value so that
Ŷ0 = wiyi where
Ŷ0 is the predicted value at location
wi is the weight from sample
yi is the observed value from sample
The idea of kriging is that its estimator
Ŷ0 is unbiased (i) and the estimation variance is minimized (ii):
E(Ŷ0) = E(Y0). For this to happen,
∑wi = 1 and the mean is stationary (i.e.
E(Yi) = u, given any
σ²_ε = E[Y0 - Ŷ0]² = Var(Y0 - Ŷ0) is minimized.
In order to minimize
σ²_ε, the Lagrange multiplier method is used:
L = Var(Y0 - Ŷ0) + 2λ(∑wi - 1)
= σ² + ∑∑wiwjCij - 2∑wiCi0 + 2λ(∑wi - 1)
L is the Lagrange function,
λ is the Lagrange multiplier and
2λ(∑wi - 1) is the part that guarantees
∑wi = 1.
Cij is the covariance among sample points
Ci0 is the covariance between sample point
i and prediction point.
Now, taking partial derivatives from
L with respect to
λ, setting the equations to zero, and solving them, will result in (skipped all calculations through the last step):
[w1] [C11 ... C1n 1]-1 [C10]
|. | |. . .| |. |
|. | = |. . .| |. |
|. | |. . .| |. |
|wn| |Cn1 ... Cnn 1| |Cn0|
[λ ] [1 ... 1 0] [1 ]
Assume the following sample points (1, 2 and 3) and the prediction point (0), their respective values and distances:
point | x_coordinate y_coordinate | value | dist_to_1 dist_to_2 dist_to_3 dist_to_0
1 | 0 0 | 10 | 0 2 3.16 1.41
2 | 0 2 | 7 | 2 0 3.16 1.41
3 | 3 1 | 3 | 3.16 3.16 0 2
0 | 1 1 | ??? | 1.41 1.41 2 0
Assume the covariance function with nugget
a = 1, sill
σ² = 4 and range
r = 6:
C(h) = σ² = 4, if |h| = 0
C(h) = (σ² - a)exp(-3|h|/r) = (4 - 1)exp(-3|h|/6), if |h| > 0
Calculate covariances among sample points:
[C11 C12 C13] [4 1.1 0.62]
|C21 C22 C23| = |1.1 4 0.62|
[C31 C32 C33] [0.62 0.62 4 ]
Calculate covariances between sample points and prediction point:
|C20| = |1.48|
[C30] [1.1 ]
[w1] [4 1.1 0.62 1]-1 [1.48] [ 0.353]
|w2| = |1.1 4 0.62 1| |1.48| = | 0.353|
|w3| |0.62 0.62 4 1| |1.1 | | 0.293|
[λ ] [1 1 1 0] [1 ] [-0.505]
w1 = 0.353,
w2 = 0.353,
w3 = 0.293 and the Lagrange multiplier =
-0.505. See that
∑wi = 0.353 + 0.353 + 0.293 = 1.
Predict value in point 0 (
Ŷ0 = 0.353*10 + 0.353*7 + 0.293*3 = 6.88
Calculate kriging variance (aka minimized estimation variance):
σ²_ε = sill - [w1 w2 w3 λ] [C10]
σ²_ε = 3.14
Same example, in R:
p = SpatialPoints(cbind(c(0,0,3),c(0,2,1)))
p$z = c(10,7,3)
krige(z~1, p, SpatialPoints(cbind(1,1)), vgm(psill=3, nugget=1, range=2, "Exp"))
# psill = partial sill.
# the vgm function takes the 'range' argument equal to
#C(h) = (σ² - a)*exp(-|h|/r) (or S(h) = a + (σ² - a)*(1-exp(-[h]/r)
#that is why range = 2 (and not equal 6 as in the previous example)
# which means approximate 1/3 of the distance where the curve flattens.
[using ordinary kriging]
coordinates var1.pred var1.var
1 (1, 1) 6.887426 3.136307
How can Ordinary Kriging solve the well-know summit problem for IDW? (whereby IDW cannot predict a higher value outside the range of the control points).
Besides Kazuhito's answer, this is also addressed in Edzer Pebesma's answer to Kriging values greater than the most extreme value of the input layer?.