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If $W(s)$ is stationary, then $Z(s)=W(s)-W(0)$ has exactly the same increments and semivariogram, but $\mathrm{Var}Z(s)=2{\gamma }_W(s)$ is generally nonconstant. Thus existence of a stationary covariance model for a semivariogram does not imply stationarity of every process with that semivariogram. Adding an independent random constant also leaves the semivariogram unchanged while changing the covariogram.

The covariogram is the displacement-dependent covariance $C(h)=\mathrm{Cov}[Z(s+h),Z(s)]$ of a second-order stationary field. It is an even positive-definite kernel. A semivariogram determines differences $C(0)-C(h)$, but not an arbitrary constant covariance component.

Geostatistics models spatial dependence for estimation and prediction. A mean or drift describes systematic variation, while a covariogram or semivariogram describes residual dependence. Kriging combines the two for spatial prediction.

Kriging predicts an unobserved value of a spatial random field by combining observations according to their covariances. Nearest-neighbor interpolation uses the closest sample; kriging also asks how strongly each sample is related to the target and how much information the samples share with one another. Two nearby, highly correlated samples need not provide twice the information of one. The different forms of kriging depend on whether the mean is known, an unknown constant, or an unknown regression drift.

Write $Z=(Z(s_1),\dots ,Z(s_n){)}^T$ for the observation random vector, $z$ for its realized values, and $Z_0=Z(s_0)$ for the prediction target. Assume finite second moments, and defineHere $\Sigma$ is the observation covariance matrix, assumed to be a positive-definite matrix in the formulas below. A covariance function supplies its entries and the target covariance vector $c$. For a second-order stationary random field, the covariogram $C(h)$ depends on displacement, and the semivariogram satisfies $\gamma (h)=C(0)-C(h)$. The intrinsically stationary random field formulation permits kriging using a semivariogram even when a stationary covariogram is unavailable; the finite covariance formulation below is a convenient important case.

For a linear combination $w^{T}Z$ predicting $Z_0$ with zero bias, the error variance equals its mean squared error:Ordinary kriging and universal kriging minimize this quadratic form subject to their mean constraints. Their optimality is among the specified linear predictors with zero bias under the chosen statistical model; covariance information alone does not establish optimality among arbitrary nonlinear predictors. The Gaussian random field interpretation supplies additional conclusions in simple kriging.

The reported error variance assumes a chosen covariance function, its statistical parameters, observation locations, and a mean structure. It omits uncertainty in fitted covariance parameters and does not validate the trend. A repeated-sampling mean squared error need not equal the error variance conditional on each realized data vector.^[ref]

In 1963, Georges Matheron presented kriging for estimating average ore grades in mining panels from weighted samples taken inside and outside a panel. This setting in geostatistics explains why prediction targets can be averages over regions as well as values at points.^[ref]

Ordinary kriging assumes $\mathbb{E}[Z(s)]=\mu$ for an unknown constant $\mu$. A predictor $w^{T}Z$ has zero bias for every $\mu$ precisely when $1^{T}w=1$, where $1$ 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 isThe sign of the last term follows from the displayed convention $\Sigma w+\eta 1=c$. To derive the system, differentiate $Q(w)+2\eta (1^{T}w-1)$. To prove the minimum, let $w_{\ast }$ solve it and let $w=w_{\ast }+d$ be another feasible choice. Then $1^{T}d=0$, soA 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 $X=1$ 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 functionThe 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 areSolving the displayed system givesThese values can be checked directly: the weights sum to one and $\Sigma w=(41/116,35/58,35/58{)}^T=c+(3/29)1$. 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.

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 $C(h)=2^{-h^2}$. ThenThe 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.

There is a literal domain error in the printed formula. A semivariogram indexed by a signed displacement must satisfy $\gamma (h)=\gamma (-h)$, since reversing the two observations leaves the squared increment unchanged. For positive ${\sigma }^2$ or ${\tau }^2$, the displayed positive-lag value is nonzero whereas the displayed negative-lag value is zero. Thus the claim on all of $\mathbb{R}$ is false as written.

Interpret the argument as a nonnegative distance, or use the even extension of the Gaussian semivariogram. For ${\sigma }^2,{\tau }^2\ge 0$ and $\varphi >0$, the intended model isIt admits the covariogramThe Gaussian kernel is a positive-definite kernel, as can be seen from the Fourier transform of a Gaussian: in one dimensionand the nonnegative spectral density makes every finite covariance quadratic form nonnegative. In higher dimensions use the product Gaussian density. Adding independent Gaussian white noise of variance ${\tau }^2$ adds the diagonal nugget covariance. Thus a zero-mean stationary Gaussian random field exists with this covariance, and $C(0)-C(h)$ gives the corrected semivariogram.

For nonzero structured variance, the nugget effect is the right limit ${\tau }^2$ at zero, the sill of a semivariogram is ${\tau }^2+{\sigma }^2$, and the range of a semivariogram is infinite if defined as the distance at which the sill is exactly reached. A 95-percent practical range solves $1-e^{-{\varphi }^2{h}^2}=0.95$:The scale $1/\varphi$ is sometimes called the range parameter; it is neither the exact range nor the 95-percent practical range. If ${\sigma }^2=0$, there is no nontrivial structured range.

The fact that a semivariogram does not determine stationarity is a further distinction: a semivariogram specifies increment variation, not a unique process or absolute covariance. Adding an independent random constant adds a constant to $C$ without changing $\gamma$. More strongly, subtracting $W(0)$ from a stationary field $W(s)$ preserves its semivariogram but generally makes its variance depend on $s$. The corrected statement is that this semivariogram admits a second-order stationary model, with the displayed covariance using the usual convention $C(h)\to 0$ at large distance; it does not force every compatible process to be stationary.