Thanks for your reply,

my comments;

> Digby

+ + ++ +

>

> I don't understand some of your comments:

>

> 1. Stationarity is not a property of data, it is a property of the

> underlying random function (no matter which form of stationarity you are

> considering).

>

+ + ++ + + + +

+ + + + +

+ + +

Michelle David describes this data as having local drift but overall

stationarity, so the mean changes with location.

>

My apologies I mistyped the email, I intended to say estimation

>

> 2. The actual kriging error is unknown so how can you compare the

> kriging error with the kriging variance?

>

variance vs. sample variance.

>

Is it possible you could have some type of drift as above, yet have a

> 3. In the absence of spatial correlation, i.e., pure nugget effect

> variogram, the kriged value at each location will be the sample mean.

>

pure nugget effect variogram?

> 4. In what sense might the sample mean be "better" than the kriged

I was meaning to say that if you have a drift in the mean is it possible

> estimate? If one uses the sample mean for all estimates then you will

> not have an exact interpolator hence at the data locations the sample

> mean could not be "better" than the kriged estimate. Except at data

> locations you will not know the true value and hence you can't compare

> which is "better", the sample mean or the kriged estimate.

>

you may have a kriging variance equal to the sample variance,

but a large difference between the local mean and the sample mean

(I should think about this), so your local kriged estimate would be far

more accurate than the sample mean. Hence you should not adopt your

sample mean if you have drift.

I was really mixing terms of what Pierre described as trend which may be

appropriate for universal kriging, but I meant trends on a small scale

which I described as drift, where such methods may be too complex,

or inconvenient to apply.

(I should have read the book, Michelle does incidentally recommend

universal kriging for the above sketch example, but assummed that in

cases software would not be available to carry out universal kriging

the drift may be ignored).

>

Thankyou all for your patience,

> If one argues that the sample mean is better than kriged estimates then

> you are really arguing that there is no spatial correlation (and hence

> that your variogram estimation and modeling step was not adequate).

>

> The sample mean is in fact a special case of the kriging estimator,

> i.e., where all the weights are the same. The kriging equatiions are

> derived by minimizing the estimation variance, hence an equal weight

> estimator has to be one of the possibilities. Of course this is all

> predicated on several assumptions that are not really testable

>

> a. The underlying random function satisfies an appropriate form of

> stationarity

> b. The variogram has been adequately estimated and modeled (the kriging

> variance does not capture any uncertainty related to inadequate

> estimation and modeling of the variogram)

> c. You are restricted to linear estimators (the conditional expectation

> would be the optimal estimator in general, it coincides with the simple

> kriging estimator in the case of multivariate normality)

>

> Ultimately however the question of whether using the sample mean (at all

> locations) or using kriging (assuming a non-pure nugget variogram) is

> "better" may depend on what you going to use the results for, that is

> not exactly a statistical question.

>

>

> Donald E. Myers

> http://www.u.arizona.edu/~donaldm

>

Best Regards Digby Millikan

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