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• Dear ai-geostats A question about simulation I understand in the simulation methods SGS that each node the data a kriging is performed than an value is drawn
Message 1 of 1 , Jun 9, 2003
Dear ai-geostats

I understand in the simulation methods SGS that each node the data a kriging
is performed than an value is drawn from a normal distribution with the
local mean (from data and previously simulated nodes captured in a search
neigbourhood) and variance defined by the kriging variance.

My understanding of kriging variance is that this variance is really an
index of data configuration independent of the data values. This leads to
my question.

How a set of multiple simulations capture information about the variability
of conditioning data values when the kriging variance is only a data
location index. Specifically, it is possible to have the same conditioning
data configuration but the variability of the values attached to the data
can be quite different. How is this recognised in the simulation process.

Thanks

Joe

Hi Joe,

After the backtransform, the distribution of simulated values
at each node is not Gaussian anymore and its variance can be
used as a local index of uncertainty, which accounts for both
the range of surrounding values and their closeness in
terms of data configuration.

Regards,

Pierre Goovaerts

<><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><>

Dr. Pierre Goovaerts
President of PGeostat, LLC
Chief Scientist with Biomedware Inc.
710 Ridgemont Lane
Ann Arbor, Michigan, 48103-1535, U.S.A.

E-mail: goovaert@...
Phone: (734) 668-9900
Fax: (734) 668-7788
http://alumni.engin.umich.edu/~goovaert/

<><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><>

A couple of additional observations about SGS. This algorithm is based on
properties of the multivariate normal distribution. Let Z0, Z1,....., Zn be
jointly distributed with all having finite variances then the best estimator
of Z0 given Z1,...., Zn is the conditional expectation of Z0 given Z1,...,
Zn. "Best" in this case means unbiased and with minimal estimation variance.
IF in addition the joint distribution is multivariate normal then the
conditional expectation has a particularly simple form, i.e. it is the same
as the simple kriging estimator. MOREOVER the conditional distribution of Z0
given the Z1,..., Zn is normal AND the mean of this distribution is the
conditional mean and its variance is the estimation variance. All of the
above is true without any reference to spatial problems (except in drawing
the analogy to simple kriging).

One critical practical problem then that arises is how does one know that
the assumption of multivariate normality is satisfied? In practice one does
not, it is simply an assumption. Note that doing a statistical test on the
data does not answer the question, the data is not the right kind to test
for multivariate distributional properties.

In the SGS algorithm we sort of turn things around. The simple kriging
equations are derived without any distributional assumptions and we know how
to compute the kriging variance and the kriged estimate without the
distributional assumptions. IF the multivariate normality assumption were
true then the simple kriging value would be the conditional mean, i.e., the
mean of the conditional distribution and the kriging variance would be the
variance of the conditional distribution. As you have noted the kriging
variance is computed without using the data values and is completely
determined by (1) the covariance model, (2) the pattern of the data
locations. Under the multivariate normality assumption however this is the
right quantity. There is no way to get away from that important assumption.
A histogram and other descriptive statistics are useful in deciding whether
that assumption is "reasonable" but you can't absolutely answer the
question.

Donald E. Myers
http://www.u.arizona.edu/~donaldm

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