CHAPTER 5: Kriging
KRIGING EXAMPLES
Earlier
in this chapter we took the setup in Fig.5.1, a semi-variogram of the form g(h)=ph, and allocated a set of weights to each
of the three samples. We showed that in that case the estimation variance was
0.0729pl. Now let us see if we can find the ‘best’
set of weights using the kriging system. Since we have three weights, there are
four equations which in the general form are:
and the kriging variance would be:
The right hand side of the equations are 
(S1,A) , (S2,A)
and (S3,A). (S1,A) is the average semi-variogram value
between a line of length l and a point on the end of it. This
is the definition of c(l). So is (S3,A). (S2,A) is the average semi-variogram between the
line and a central point. This example was tackled in Chapter 4 and found to be
c(l/2). Thus we have the right hand side
of the equations and most of the variance. The left hand side of the equations
is made up of the individual sample-with-sample terms. Since all of the samples
in this case are supposed to be points, all of these ‘left hand side’
relationships are given by the semi-variogram model. It remains only to
calculate the distances between the pairs and use the model to produce the
terms. The diagonal terms, (S1,S1), (S2,S2) and (S3,S3) are all zero, since g(0) =0 by definition, (S1,S2) is equal to (S2,S1), and is g(h) ( l/2) . So is (S2,S3) and (S3,S2). (S1,S3) is g (l) as is (S3,S1). Finally, (A,A) is F(l) by definition. Putting these
together gives:
|
|
w2 γ(l/2) |
+ w2 γ(l) |
+ l |
= |
c(l) |
|
w1
γ(l/2) |
|
+ w2 γ(l/2) |
+ l |
= |
c(l/2) |
|
w1
γ(l) |
+ w2 γ(l/2) |
|
+ l |
= |
c(l) |
|
w1
|
+ w2 |
+ w3 |
|
= |
1 |
and the
kriging variance is
Up until this point the system does not depend on the
actual model for the semi-variogram. Our model was g(h) =ph and for this example we will take p=4. For the linear semi-variogram, c(l) =pl/2, so in this case c(l) =2l. Similarly, F(l) =4l/3. Substituting in the above system:
|
|
2lw2 |
+ 4lw2 |
+ l |
= |
2l |
(1) |
|
2lw1 |
|
+ 2lw2 |
+ l |
= |
l |
(2) |
|
4lw1 |
+ 2lw2 |
|
+ l |
= |
2l |
(3) |
|
w1
|
+ w2 |
+ w3 |
|
= |
1 |
(4) |
and
Adding equations (1)
and (3) gives
whilst equation (4)
gives
These two together show that in this case l=0. Thus we can eliminate l from the first three equations,
and divide all of them through by l. This suggests that the results
--- the final values of the weights --- do not rely on the length being
estimated. We have then:
|
|
2w2 |
+ 4w2 |
= |
2 |
(5) |
|
2w1 |
|
+ 2w2 |
= |
1 |
(6) |
|
4w1 |
+ 2w2 |
|
= |
2 |
(7) |
Subtracting
equation (5) from equation (7) gives:
so that equation (6) gives:
Therefore, w3 = 1/4 and w2 = 1/2. The ‘optimal’ set of weights for the
problem in Fig. 5.1 is therefore (1/4, 1/2, 1/4) and this estimator gives a kriging variance
of:
The final result is therefore that:
the BLUE has weights of (1/4, 1/2, 1/4) and a kriging
variance of l/6.
In our previous studies of this particular setup, we found
that the arithmetic mean gave an extension variance of pl/18 and that the set of weights (1/8, 3/4,
1/8) gave 7pl/96. To match the example above we should
set p=4, so that the variances become 2l/9 and 7l/24 respectively. The kriging
procedure has improved over the arithmetic mean by about 25%, this being the
difference in magnitude between the extension variance and the kriging
variance. The spurious set of weights which we tried earlier in this chapter
give a variance almost twice that of the optimal estimator. Note that in this
case, since we have used a linear
semi-variogram model, the set of weights is independent of the length being
estimated, but the variance is directly proportional to it. For an exercise,
see if you can show that the set of weights is also independent of the slope of the semi-variogram, p. Show also that the kriging variance is directly
proportional to p, which seems only sensible.
TWO-DIMENSIONAL EXAMPLE
Now let us
return to the familiar uranium example shown in Fig. 5.2. The data --- position
co-ordinates and grade values --- were given in Table 4.1. We have five
samples, so there will be six equations in the kriging system. The (Si,Sj) terms on the left hand side of the
equations are all ‘point-with-point’ relationships, and can be evaluated
directly from the semi-variogram model. The model was spherical, with a range
of influence of 100ft, a sill of 700 (p.p.m.)² and a nugget effect of 100
(p.p.m.)². The (Si,A) terms are all ‘point-with-panel’
relationships and hence are simple combinations of the H(l,b) auxiliary function. The (A,A) term is, of course, F(l,b).
The kriging system turns out to be:
and the kriging variance would be:
Solving this system of equations (by computer) gives:
This kriging standard deviation compares favourably with
the standard error of the ‘inverse distance’ weights. The main difference in
the weighting is perhaps a surprising one at first sight. Sample 2 is allocated
an almost zero weight. In fact, it receives only 20% of the weight on sample 5,
which is considerably farther from the block centre. This is because the
kriging system automatically takes account of the relationship amongst the
samples. Sample 2 provides little extra information about the block in the
presence of sample 1, and to a lesser extent samples 3 and 4. To illustrate
this point further, let us consider the situation in Fig. 5.3, where sample 3
was moved to the south of the block. The kriging system now becomes:
Note that the only difference between this
and the previous system is in row 3 and column 3 on the left hand side. The
right hand side is unchanged since we have not changed the relationship of each
individual sample to the panel. The new weights are:
The weight on the third sample is now less
than that of sample 2. The main movement in weight has been towards the
‘northern’ samples, 1, 2 and 5, although the largest increase is, naturally, in
sample 1. The really striking change is in the value of the estimator which has dropped by 17 p.p.m.
As a
third example let us consider the same sampling situation as before, but now
with the block rotated through 90 degrees, as in Fig. 5.4.
Fig. 5.4. Block to be estimated is
rotated through 90º.
The
kriging system for this setup has the same left hand side as the first
situation in Fig. 5.2. However, all of the terms on the right hand side have
changed, to:
The weights allocated to the samples change drastically:
The
estimator T* takes the value 371.9 p.p.m., and has a
standard error of 10.7 p.p.m. Sample 2 has been completely screened out by
sample 1, and the influence of sample 5 has also declined somewhat. The biggest
surprise, perhaps, is that sample 1 no longer has anything like the importance
it had in the previous two examples. There is no method other than kriging in
which this shift in ‘information value’ can be quantified and utilised.