CHAPTER 4: Estimation
CALCULATION
OF GAMMA-BAR TERMS
Having produced a formula for the extension variance, it only remains to
explain how to calculate such terms as 
(S,A) in practice. For the sake of our
(too) simplistic approach, we will consider for the moment only simple
idealistic cases, and these only in one or two dimensions. Generalisation will
be discussed later.
Fig.4.3. Example of using a peripheral point to
estimate the average value of the line segment.
Consider,
as an example, the setup in Fig. 4.3. There is a length of, say, drive, l m long, whose grade is unknown.
We have
at our disposal a single sample, perhaps at a development heading, whose value
is known. In our previous notation T is the average grade over l, T* is the grade at the sample
position, A is the length and S is the single sample point. The reliability of this
estimator is given by:
(S,S) is the semi-variogram between the sample
point and itself, which is zero because the sample is a ‘point’. (A,A) is none other than the F(l) function encountered in Chapter 3. Our problem
arises with (S,A) which has been defined as the average
semi-variogram between the sample point and every point in the line. That is,
we take M as a fixed point (the sample) and M’ can be anywhere on the line. We
take all such pairs that are possible, calculate the value of the
semi-variogram for each pair, sum these (using an integration), and average
this sum. Because the ‘sum’ is being performed over a continuous length, we
cannot divide it by the ‘number of points’ in the sum. Instead we divide by the
length of the line itself, l. This produces another auxiliary function
which is called c(l) and deals with the specific case
of points on the end of lines. Thus our extension variance becomes:
It remains only to determine the function c(l) for the particular model in use and the
standard error is immediately available. The one-dimensional auxiliary
functions are given below for the three ‘common’ models. Semi-variograms
comprising more than one component model are easily handled. The auxiliary
function for each component is evaluated and then the component auxiliary
functions added together.
Auxiliary functions
Linear
model for the semi-variogram;
Exponential model for the semi-variogram;
Spherical model for the semi-variogram:
Thus in our example above, if we have a linear
semi-variogram, the extension variance for the setup in Fig. 4.3 becomes:
For any specific problem, we need specify only the length
of the line l and the slope of the semi-variogram, p.
Let us now consider a slightly more interesting example,
such as that shown in Fig. 4.4.
Fig 4.4. Example of using a central point to estimate the average value of the line segment.
Here the
point sample is in the middle of the line, but otherwise the situation
remains the same. In:
only the first term (S,A) has changed. Rather than invent a new
auxiliary function, or have to do the integration all over again, we can use
the existing c(l) function to produce the required
term.
The term
we require is as follows:
|
|
= the average semi-variogram value between the sample point and every point along the line = (sum of all semi-variogram values between the sample point and every
point along the line)/l = (sum of all the semi-variogram values between the sample point and
every point in the left hand half of the line + sum of all the values between
the sample and the right hand half of the line)/l |
Fig 4.5. Simplifying the central point problem
to allow the use of auxiliary functions.
Figure 4.5 illustrates the ‘splitting’ of
the line so as to put the sample point at the end of two shorter lines. Now, c(l/2) would give us the average of all the semi-variogram
values between M (the sample point) and the M’ on the left hand half of the
line. Returning to the definition of the c function, it can easily be seen
that the sum of all the semi-variogram values between M and M’
will be the average multiplied by the length of line under
consideration.
Thus:
so that
In a particular case the user may substitute his own model
for the semi-variogram, and hence the appropriate auxiliary functions. Before
moving on let us compare this result with the previous situation, where the
sample lay at the end of the
line. In the former case the extension variance was:
By definition c(l) must be greater than (or at least equal
to) c(l/2). The conclusion? If you can only
take one sample, it is better to take it in the middle of what you are trying
to estimate. It is reassuring to find that so-called common sense has a sound
mathematical background.
Fig. 4.6 Generalisation of the ‘central’ point problem.
Using the
same sort of logic on Fig. 4.6, you should be able to deduce that:
so that
Fig. 4.7. Extrapolation of the peripheral point
problem.
Figure 4.7 at first sight seems to be a different kettle of
fish. However, let us follow the same procedure and see where it leads.
The point lies on the end of a ‘line’ of length l+b. The expression (l+b) c(l+b) would give us the sum of all the
semi-variogram values between the sample and the length l+b. However we do not require the points
corresponding to M’ within the length b, so we may subtract those in the
form c(b). That is:
so that
For the linear model, for example, this would be:
This is obviously larger than the expression when the point
was on the end of the line, as would be expected.
One last example before we abandon one-dimensional
examples: Fig. 4.8 shows the ‘same’ line, which now contains three samples.
Fig. 4.8. More complex problem when three samples are available to estimate the line segment.
We shall
use the arithmetic mean of the three grades to estimate the length, i.e. T*=(g1+g2+g3)÷3. Then our extension variance is
where S is now a set of three points. (A,A) remains unchanged, equal to F(l) since we have not changed the length to
be estimated at all. However, (S,A) is now the average semi-variogram value
between each of the three points and the line, so that
where S1 represents sample 1 and so on. Now
(S1,A) is simply c(l), as is (S3,A). The term (S2,A) is the same situation as that in Fig.
4.4, so this equals c(l/2). Thus,
The middle term of the variance (S,S) requires us to take each point in the
sample set with each point in the sample set. Since there are three points in
the set, there are nine such pairs of points:
Each of the individual terms is simply the semi-variogram
between a pair of points. Three of the terms, g(S1,S1), g(S2,S2)
and g(S3,S3)
are automatically zero since the samples are points. The terms g(S1,S2), g( S2,S1), g(S2,S3)
and g(S3,S2)
are all equal to g(l/2), whilst g(S1,S3)
and g( S3,S1)
are equal to g(l). Thus:
so that
For our linear example, this reduces to:
This is considerably less than the other evaluations, as seems
only sensible with three times as many samples.