= Ordinary kriging
<Ordinary kriging> assumes $\mathbb E[Z(s)]=\mu$ for an unknown constant $\mu$. A predictor $w^TZ$ has zero <bias> for every $\mu$ precisely when $\mathbf1^Tw=1$, where $\mathbf1$ is the all-ones vector. <Ordinary kriging> minimizes the error <variance> $Q(w)$ subject to this constraint, giving the <best linear unbiased predictor>. Its explicit system is
$$
\begin{pmatrix}
\Sigma&\mathbf1\\
\mathbf1^T&0
\end{pmatrix}
\begin{pmatrix}w\\\eta\end{pmatrix}
=\begin{pmatrix}c\\1\end{pmatrix},
\qquad
\widehat Z_0=w^Tz,
\qquad
s_{\mathrm{OK}}^2=v_0-w^Tc-\eta.
$$
The sign of the last term follows from the displayed convention $\Sigma w+\eta\mathbf1=c$. To derive the system, differentiate $Q(w)+2\eta(\mathbf1^Tw-1)$. To prove the minimum, let $w_*$ solve it and let $w=w_*+d$ be another feasible choice. Then $\mathbf1^Td=0$, so
$$
Q(w)-Q(w_*)=d^T\Sigma d\geq0.
$$
A <positive-definite matrix> $\Sigma$ makes equality possible only for $d=0$. This argument uses <second moments>, not a <normal distribution> assumption. <Ordinary kriging> is the special case of <universal kriging> with $X=\mathbf1$ and $x_0=1$.
For an original one-dimensional teaching example, observe $z=(1,3,2)^T$ at $s=(0,1,3)$, and predict at $s_0=2$. Choose unit marginal <variance>, exact observations without measurement noise, an unknown constant <mean>, and the <exponential covariance function>
$$
C(h)=2^{-|h|}=e^{-(\log2)|h|}.
$$
The <exponential covariance function> corresponds to a stationary <Ornstein-Uhlenbeck process>, so it gives a valid <covariance> model. Its <semivariogram> is $1-2^{-|h|}$. The resulting <matrix> and vector are
$$
\Sigma=\begin{pmatrix}1&1/2&1/8\\1/2&1&1/4\\1/8&1/4&1\end{pmatrix},
\qquad c=\begin{pmatrix}1/4\\1/2\\1/2\end{pmatrix},\qquad v_0=1.
$$
Solving the displayed system gives
$$
w=\frac1{29}\begin{pmatrix}2\\13\\14\end{pmatrix},\quad
\eta=-\frac3{29},\quad
\widehat Z_0=\frac{69}{29}\approx2.3793,\quad
s_{\mathrm{OK}}^2=\frac{18}{29}\approx0.6207.
$$
These values can be checked directly: the weights sum to one and $\Sigma w=(41/116,35/58,35/58)^T=c+(3/29)\mathbf1$. Substitution into $Q(w)$ gives $18/29$. <Nearest-neighbor interpolation> must choose between the two equally close samples at one and three, returning either three or two; their error <variances> under this model are both $1+1-2(1/2)=1$. Thus the benefit is quantified under the chosen model, without knowing the unobserved target value.
\Image[/kriging-spatial-prediction-uncertainty.png]
{title=Ordinary kriging combines three samples and reports uncertainty under a fixed exponential covariance model}
The upper panel applies the same <ordinary kriging> calculation at each target location. The lower panel shows the square root of its error <variance>, in the units of the observations. It is zero at the exact observation locations and rises between and beyond them. No <confidence interval> is inferred from a <variance> alone; a <probability distribution> assumption is needed to attach a coverage level.
The weights are not required to be positive. Keep the same observations and target, but replace the <covariogram> by the smoother <Gaussian kernel> $C(h)=2^{-h^2}$. Then
$$
w=\frac1{4199}\begin{pmatrix}-736\\2663\\2272\end{pmatrix},\qquad
\widehat Z_0=\frac{11797}{4199}\approx2.8095,\qquad
s_{\mathrm{OK}}^2=\frac{2115}{4199}\approx0.5037.
$$
The negative coefficient compensates for dependence between samples; the constraint eliminating <bias> still holds. <Kriging> is not generally a <convex combination>, and its prediction need not stay between the observed values. A smaller model-based error <variance> does not establish that one <covariance function> is better supported by these three observations.
\Image[/kriging-covariance-model-weights.png]
{title=Changing the covariance model changes the ordinary kriging weights and can make a weight negative}
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