Statistika Spasial, Teknik Geodesi, Itenas, Rian Nurohman 23-2014-126
geoestadisticaDescripción completa
kriging
kriging
Descripción: Evaluación de Yacimientos
Como obtener kriggin simple y ordinarioDescripción completa
Kriging Example The main result in kriging is concerned with estimation of the value
Z (r0) (also referred to as Z 0 ) based on the observed values T
{Z 1 , · · · , Z N N } =
Z
We will consider linear estimators only N
ˆ0 = Z
wi Z i = wT Z
i=1
We demand that the estimator is unbiased and and get T
w 1 =
wi = 1
The kriging estimate is obtained by choosing w so that the estimation variance 2 ˆ0 ]2} σE = E {[Z 0 − Z
is minimised. Let
D{Z } =
= = C {Z , Z 0 (r)} =
=
V {Z 1 }
C {Z 1 , Z 0(r)}
C 10 10
C {Z N N , Z 0
C N N 0
.. .
.. .
C {Z N · ·· N , Z 1 } C 11 11 .. .
···
C 1N .. .
· · · C N N C N N 1
C
D
}
· · · C {Z 1 , Z N N }
.. .
V {Z N N
= ( )} r
Using this, the mean squared error expression can be written 2 ˆ0 (r))2 } σE = E {(Z 0 (r) − Z ˆ0 (r)} {Z 0 (r) − Z = V { T T = C 00 00 + w Cw − 2w D
.. .
The problem is now to minimise
C 00 + wT Cw − 2wT D with the constraint that T
w 1 =
1
We introduce a Lagrange multiplier −2λ and minimise
F = C 00 + wT Cw − 2w T D + 2λ(wT 1 − 1) without constraints. By differentiation we get
∂F = 2Cw − 2D + 2λ1 = ∂ w ∂F = 2w T 1 − 2 = 0 ∂λ
0
or
C
1
T
0
1
w
λ
=
D
1
,
which is the so-called ordinary kriging system. By direct solution we get Cw
+ λ1 =
D
i.e. i.e.
+ λC −1 1 = C −1 D −1 −1 T T T 1 w + λ1 C 1 = 1 C D w
Since 1 T w = 1 we have
λ =
T
−1
1 C D−1 −1 T 1 C 1
=
T
−1
D C 1−1 −1 T 1 C 1
If we insert this above we get 2
w
= C −1 [D − λ1]
The weights can all be determined from the spatial lay-out and the semivariogram. The matrix inversion includes the observations we want to krige from only and not the point we want to krige to. This point is included via there covariances D linearly in λ and in w. The minimal variance is the so-called ordinary kriging variance, which is
V {Z 0 − wT Z } = C 00 + wT C w − 2wT D
= C 00 + wT [D − λ1] − 2wT D = 2 = C 00 − wT D − λ σOK
Some results on kriging weights We consider the following situation Z
Z
1
-2
2
-1
Z
Z
0
0
1
2
3
3
ˆ0 = wT Z = w 1Z 1 + w2 Z 2 + w3 Z 3 Z We want to krige a value for Z 0 based on the observations { Z 1 , Z 2 , Z 3 }. We assume that we have the semivariogram
γˆ (h) =
0 + + C 0 C 0
C 1 C 1
3 h 2 R
1
−2
( ) h
R
3
h = 0 0 < h < R R ≤ h.
i.e. a spherical model with nugget effect. We assume that R = 6, and for C 0 = 0 and C 1 = 1 we get 3
h 0 1 2 3 4 5 6 γ (h) 0 .2477 .4815 .6875 .8519 .9606 1.0000 This gives the correlation function
h 0 1 2 3 4 5 6
C (h) C 0 + C 1 C 1 0.7523 C 1 0.5185 C 1 0.3125 C 1 0.1481 C 1 0.0394 0
λ = [DT C −1 1 − 1]/[1T C −1 1] −1 w = C [D − λ1] 2 σOK = C 0 + C 1 − wT D − λ 4
1
These values have been determined for different values of C 0 and C 1 and are shown in the table below. It is seen that we for a pure nugget effect, i.e. C 1 = 0 have the weights 13 , 13 , 13 , i.e. just a smoothing. The kriging variance V {Z 0 − wT Z } = σ 2 + σ 2 /3 = 43 σ 2 is also seen. For neglectable nugget effect we notice that the weights converge towards -.04
.80
-2
-1
Z
.25
0
0
1
2
3
If we have equal amounts of nugget effect og “dependent variance” we get the weights .27
.43
-2
-1
Z
.30
0
0
1
2
3
We also notice that the weights do not depend upon the scaling of the variances. The dependence is only on the ratio between the variances C 0 and C 1 .
