Some Variogram FAQ's

Sometimes my sample variograms are messy and appear to have no sill. I'm told that this is due to the presence of a trend in my data. Is this true?

Sample variograms calculated in the direction of a trend may be significantly impacted by the presence of a trend. For example, if the trend can be recognized on a contour map it is likely strong enough to significantly influence the sample variogram calculations. To understand how a trend may impact sample variogram calculations, let's consider a linear trend in the direction of the x coordinate:

baxxZ +=)( ……….(1)

Using the traditional variogram estimator, the estimated sample variogram of the trend is given by:

22

22

2

2

][2

][21

)]()([21)(

x

hii

hii

hii

ha

xxN

a

baxbaxN

xZxZN

h

=

−=

−−+=

−=

∑

∑

∑

+

+

+γ

………..(2)

Thus, we can see from Equation 2 that the sample variogram of a linear trend appears as a parabolic curve in the form 2xy = that has no sill, e.g.:

Can I model the sample variogram of a trend using a model such as 22)( hah =γ and then proceed with other geostatistical procedures such

as kriging or conditional simulation etc.?

No. Geostatistics is based on the Random Function model with some rather restrictive limitations. First, the expected value of the random function must be stationary. In other words, a trend in the mean is not allowed. Second, the variance of the increment between Z(x) and Z(x+h) must also be stationary. Thus, the variogram model for a non stationary random function is not defined and any attempt to apply such a model to kriging or conditional simulation etc. will provide undefined or meaningless results.

So what should I do when there are significant trends in my sample data?

One very effective way to deal with a trends is to filter the trend and calculate the sample variogram on the residuals. There are several ways to do this, so lets look at one of the more efficient ways for removing trends. Consider the following sample variogram pairs for the separation vector h.

In the figure above, TZ are the “tail” members and HZ are the “head” members of N pairs separated by the vector h. Now let’s standardize the sample values at the tail of the separation vector as follows:

T

TTT

mZRσ−

= ………….(3)

where Tm is the mean or average of TZ , and Tσ is the standard deviation of

TZ . This is an old trick that has been around for some time. By applying Equation (3) to TZ , we have created a new variable TR that has a mean of 0.0 and a standard deviation of 1.0. Note that the distribution of TR is not necessarily Normal unless the distribution of TZ is Normal.

Next, we standardize the sample values at the head of the separation vector using the same trick:

H

HHH

mZRσ−

=………(4)

Note that if there is a trend in Z in the direction of h, Tm will be either larger or smaller than Hm . If there is no trend in Z, then Tm should be more or less equal to Hm . More importantly the mean of TR is equal to the mean of HR since both are equal to 0.0 whether or not there is a trend in Z.

Similarly, the variance of TR is equal to the variance of HR since both variances are equal to 1.0 whether or not a trend exists in the variance of Z. Thus, the transformations given by Equations 3 and 4 not only remove the trends in the mean, they also remove any trend that may be present in the variance of Z.

Next, we proceed to calculate the variogram of the residual, R as follows:

∑

∑

+−=

−=

]2[21

][21)(

22

2

HHTT

HT

RRRRN

RRN

hγ…………(5)

Next, we want to express Equation (5) in terms of Z. To do this we will expand Equation (5) term by term. For example, the first term of Equation (5) can be expanded as follows:

21

/21

/][21

/]2[21

21

21

22

222

222

22

=

=

−=

+−=

−=

∑

∑

∑ ∑

TT

TTT

TTTTT

T

TTT

mZN

mmZZN

mZN

RN

σσ

σ

σ

σ

……(6)

Similarly, the last term of Equation (5) is equivalent to:

21

21 2 =∑ HRN

………..(7)

The second term of Equation (5) can be expanded as

)(

/][1

/][1

1

221

HT

HTHTHT

HTHTTHHTHT

H

HH

T

TT

HT

ZZ

mmZZN

mmZmZmZZN

mZmZN

RRN

ρ

σσ

σσ

σσ

−=

−−=

+−−−=

−

−−=

−

∑

∑

∑

∑

….…(8)

which is nothing less than the correlation coefficient between TZ and HZ . Finally, we collect the terms obtained in Equations 6, 7, and 8 to obtain:

HT

HT

ZZ

ZZ

HT RRN

h

ρ

ρ

γ

−=

+−=

−= ∑

0.121

21

][21)( 2

…..(9)

Thus, the transformations used to detrend the mean and the variance of Z in Equations 3 and 4 are equivalent to calculating 1.0 - the correlogram directly.

To summarize, the correlogram can be used to estimate the continuity of a spatially distributed attribute whether or not trends are present. If a trend in the mean or variance is present, the correlogram will filter the trends and estimate the spatial continuity of the residuals. When no trends are present, the correlogram estimates are similar to the standardized traditional variogram estimator. Experience shows that this estimator provides excellent results whether or not a trend is present in the original spatially distributed attribute. The correlogram is one of several sample variogram estimators provided by SAGE2001.