Kriging

Let $X=(X(s){)}_{s\in T}$ be a real random field in the sense of Random fields, stationarity, covariance, and variograms, and suppose the finite vector of field values considered below is square-integrable. Choose $n\ge 1$ observation sites $s_1,\dots ,s_n$ and a target site $s_0$. Write $Z=(X(s_1),\dots ,X(s_n){)}^{\mathrm{\top }}$, $Y=X(s_0)$, $\mathrm{\Sigma }=\mathrm{Cov}(Z,Z)$, $c_0=\mathrm{Cov}(Z,Y)$, and $v_0=\mathrm{Var}(Y)$. A positive-definite matrix $\mathrm{\Sigma }$ means $u^{\mathrm{\top }}\mathrm{\Sigma }u>0$ for every nonzero $u\in {\mathbb{R}}^n$.

If the mean vector $m=\mathbb{E}Z$ and target mean $m_0=\mathbb{E}Y$ are known, a simple-kriging linear predictor with weight vector $\lambda \in {\mathbb{R}}^n$ is ${\widehat{Y}}_{\mathrm{S}\mathrm{K}}(\lambda )=m_0+{\lambda }^{\mathrm{\top }}(Z-m)\mathrm{.}$ Its mean-square prediction error is $\mathbb{E}[(Y-{\widehat{Y}}_{\mathrm{S}\mathrm{K}}(\lambda ){)}^2]$.

For an ordinary-kriging model, the mean is an unknown common constant $\theta$: $\mathbb{E}Z=\theta {\mathbf{1}}_n$ and $\mathbb{E}Y=\theta$. The predictor is ${\widehat{Y}}_{\mathrm{O}\mathrm{K}}(\lambda )={\lambda }^{\mathrm{\top }}Z$. The admissible ordinary weights satisfy ${\mathbf{1}}_n^{\mathrm{\top }}\lambda =1$.

For a universal-kriging model, fix a trend map $p:T\to {\mathbb{R}}^q$, where $q\ge 1$, and an unrestricted coefficient vector $\beta \in {\mathbb{R}}^q$. Define the finite-site design matrix $P\in {\mathbb{R}}^{n\times q}$ by row $i$ equal to $p(s_i{)}^{\mathrm{\top }}$, and put $p_0=p(s_0)$. The model means are $\mathbb{E}Z=P\beta$ and $\mathbb{E}Y=p_0^{\mathrm{\top }}\beta$. The predictor is ${\widehat{Y}}_{\mathrm{U}\mathrm{K}}(\lambda )={\lambda }^{\mathrm{\top }}Z$; its unbiasedness for every $\beta \in {\mathbb{R}}^q$ is expressed by $P^{\mathrm{\top }}\lambda =p_0$. Ordinary kriging is the special case $q=1$, $P={\mathbf{1}}_n$, and $p_0=1$.

The covariance-based optimality statements on this page use the actual finite covariance matrix and vector. In particular, the inverse formula for simple kriging is not asserted when $\mathrm{\Sigma }$ is singular.

The covariance formulations above require finite second moments. A variogram formulation has a separate hypothesis: for an order-two intrinsically stationary field indexed by all of ${\mathbb{R}}^d$, its variogram $\gamma$ is conditionally negative definite over real coefficients, meaning ${\sum }_{i,j}{a}_i{a}_j\gamma (s_i-s_j)\le 0$ whenever ${\sum }_i{a}_i=0$ (Order-two intrinsic variograms are conditionally negative definite). This non-strict inequality does not by itself assert invertibility of a variogram augmented system; any such system needs its own full-rank and invertibility conditions. The theorem on this page proves the covariance systems.

The optimality claims are oracle claims for fixed mean and covariance quantities. Replacing those quantities by estimates defines a different procedure; no optimality or coverage guarantee for that plug-in procedure is asserted here.

Use the notation of Simple, ordinary, and universal kriging. For simple kriging, if $\mathrm{\Sigma }$ is positive definite, the unique mean-square minimizing weight vector is ${\lambda }_{\mathrm{S}\mathrm{K}}={\mathrm{\Sigma }}^{-1}{c}_0$, and its mean-square prediction error is $v_0-c_0^{\mathrm{\top }}{\mathrm{\Sigma }}^{-1}{c}_0\mathrm{.}$

For an unknown trend with design matrix $P$ and target vector $p_0$, ${\lambda }^{\mathrm{\top }}Z$ is unbiased for every $\beta \in {\mathbb{R}}^q$ if and only if $P^{\mathrm{\top }}\lambda =p_0$. Suppose $\mathrm{\Sigma }$ is positive definite and $P$ has full column rank (that is, $Py=0$ implies $y=0$ for $y\in {\mathbb{R}}^q$). Then the unique unbiased mean-square minimizing weights $\lambda \in {\mathbb{R}}^n$ and Lagrange multiplier $\nu \in {\mathbb{R}}^q$ solve $\left(\begin{array}{cc}\mathrm{\Sigma }& P\\ P^{\mathrm{\top }}& 0\end{array}\right)\left(\begin{array}{c}\lambda \\ \nu \end{array}\right)=\left(\begin{array}{c}{c}_0\\ p_0\end{array}\right)\mathrm{.}$ The augmented matrix is invertible under these hypotheses. The resulting mean-square prediction error is $v_0-{\lambda }^{\mathrm{\top }}{c}_0-{\nu }^{\mathrm{\top }}{p}_0\mathrm{.}$ For ordinary kriging, take $P={\mathbf{1}}_n$ and $p_0=1$, giving $\left(\begin{array}{cc}\mathrm{\Sigma }& {\mathbf{1}}_n\\ {\mathbf{1}}_n^{\mathrm{\top }}& 0\end{array}\right)\left(\begin{array}{c}\lambda \\ \nu \end{array}\right)=\left(\begin{array}{c}{c}_0\\ 1\end{array}\right)\mathrm{.}$

Assume the Axiom of Choice. Let $X$ be a Gaussian real field, meaning every finite coordinate vector has a possibly singular multivariate-normal law. For the target $Y=X(s_0)$ and observations $Z$ of Simple, ordinary, and universal kriging, suppose $\mathrm{\Sigma }$ is positive definite. Then the simple-kriging predictor $m_0+({\mathrm{\Sigma }}^{-1}{c}_0{)}^{\mathrm{\top }}(Z-m)$ equals $\mathbb{E}[Y\mid \sigma (Z)]$ almost surely, where $\sigma (Z)$ is the sigma-algebra generated by the observation vector.