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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.

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.