Hi Digby
Are we back on this one again?
The variogram is the variance of increments Z(x)-Z(x+h), yes. The
semi-variogram is one-half of this. We assume no trend or drift in values, so
that the (population) mean increment is zero. Assuming the mean increment is
known means that we do not lose any degrees of freedom and can divide the
variance by N rather than N-1. The variogram then becomes:
2 * gamma(h) = average (Z(x) - Z(x+h))^2
If you expand this, it becomes:
2 * gamma(h) = average[Z(x)^2] + average[Z(x+h)^2] - 2 average [Z(x)Z(x+h)]
If there is no trend, then:
(a) average[Z(x)^2] - mu^2 = average[Z(x+h)^2] - mu^2 = population variance
sigma^2
(b) average [Z(x)Z(x+h)] - mu^2 = covariance[Z(x),Z(x+h)]
so that
2 * gamma(h) = 2 * population variance sigma^2 - 2 * covariance[Z(x),Z(x+h)]
and the semi-variogram is one-half of this:
gamma(h) = population variance sigma^2 - covariance[Z(x),Z(x+h)]
Note: now you know why I insist on the proper name semi-variogram!
In effect, in the absence of trend and with stationarity if the increments,
the semi-variogram is simply the population sample variance less the covariance
at that lag. The kriging can then be expressed in either semi-variogram or
covariance function and lead to identical answers.
When the semi-variogram reaches its sill, we assume that the covariance (or
relationship) between Z(x) and Z(x+h) has become zero. So, the ultimate sill
should be:
gamma(h) = population variance sigma^2 - 0
The semi-variogram can go above this value if the covariance goes negative,
as in the hole effect model which cycles around the sigma^2 level.
It should be borne in mind that this is a theoretical relationship. The sill
on the calculated semi-variogram is a 'better' estimate of sigma^2 than the
ordinary statistical s^2, since the latter assumes random and independent
sampling. In practice, the sill and s^2 should be roughly the same, but not
identical.
Hope this helps
Isobel
http://www.kriging.com/books
