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Conditioning of a Simulation

Consider the regionalized variable $z(/boma{x})$ at the locations $/boma{x}_{i}, /; i /; = /; 1,/ldots,n,$ to be known. We interpret it as a realization of the random function $Z(/boma{x})$ defined with $E(Z(/boma{x})) = m$ and semi-variogram $/gamma(/boma{h}).$

Suppose $y(/boma{x})$ denotes a realization of a random function $Y(/boma{x})$ which has the same spatial law as $Z$. The values of $y(/boma{x}_{i})$, of course, will in general be different from those of $z(/boma{x}_{i})$. They also might be interpreted as simulated values, however, they are not conditioned.

Using the Krige estimator and the values of $z(/boma{x}_{i})$ we can estimate values $z^{/ast}(/boma{x})$ at any location $/boma{x}$. We know that $z^{/ast}(/boma{x}_{i}) = z(/boma{x}_{i}), /; i=1,/ldots,n$. The same is true for kriged values $y^{/ast}(/boma{x})$ on the basis of $y(/boma{x}_{i})$, namely, $y^{/ast}(/boma{x}_{i}) = y(/boma{x}_{i}), /;
i=1,/ldots,n$.

As a condition for simulated values we would like to have them the same values $z(/boma{x}_{i})$ at the locations $/boma{x}_{i}$, $i=1,/ldots,n$. This may be achieved with the setting.

/begin{displaymath}y_{s}(/boma{x})=z^{/ast}(/boma{x}) + [y(/boma{x}) - y^{/ast}(/boma{x})]/
/ / ,/end{displaymath}

because

/begin{displaymath}y_{s}(/boma{x}_{i})=z^{/ast}(/boma{x}_{i}) + [y(/boma{x}_{i}) -
y^{/ast}(/boma{x}_{i})]= z^{/ast}(/boma{x}_{i})/ / / ./end{displaymath}

It is obvious that the expectation $EY_{s}$ of the values of $y_{s}$ is also $EZ = m$. It is also true that the variogram $/gamma_{Y_{s}}(/boma{h})$ von $Y_{s}$ is equal to the variogram $/gamma_{Z}(/boma{h})$ of $Z$ (for a proof see e.g. David, 1977[5]).


next up previous contents
Next: Simulation of an Unconditional Up: Conditioning Previous: Conditioning   Contents
Rudolf Dutter 2003-03-13