Dear Gali Sirkis,

If you know the variogram 2gamma of the values without error and the variance

of the observation errors sigma^2_x, this can be used fairly easy by setting

2gamma_new(x-y) = 2gamma(x-y)+sigma^2_x+sigma^2_y

if x and y are to observation locations and

2gamma_new(x-y) = 2gamma(x-y)+sigma^2_x

if y is the prediction location (since you want to predict the correct value

and not the one with error)

and then build the kriging system with the gamma_new instead of gamma. However

that only modifies the kriging error and not the weights.

This theory is more or less discribed in the books e.g. Cressie.

For your practical applications three problems:

--> How to estimate the 2gamma without error?

You could in principle use the same formula to correct the observed squared

increments by substracting the sgima^2_x.

--> You don't really have error variances.

You more or less have error bounds. You could consider Bayesian Maximum

Entropy here, but that would increase complexity of your work dramatically.

--> How to do the computation?

Thats a problem if you favourit software doesn't do it.

However If you would assume that sigma^2_x is a constant and accept the

kriging error to be for the observation and not for the true mean value, you

could also compute your variogram based on the observations with measurement

error directly estimating 2gamma_new and do you kriging in the usual way.

So in conclusion: If your situation is nice there is nothing to do. If your

situation is not such nice there is plenty of work.

Best regards,

Gerald v.d. Boogaart

Am Sonntag, 18. September 2005 02:58 schrieb Gali:

> Dear list members,

>

> Question: At each observation point we do know an

> error of measurement. How can we incorporate this

> knowledge when krigging sparce observation values?

>

> The specific example: You create depth markers by

> puting mark over the zone of transition between

> different lithologies on the lithology log. You can be

> wrong as much as the thickness of transition zone, so

> you can calculate your observation error quite

> precisely. As you can understand, those errors are not

> spatially correlated, neither they are correlated with

> marker value.

>

> Many thanks in advance,

>

> Gali Sirkis.

>

>

>

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