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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.

The universal kriging model has a linear drift in $y$ and a zero-mean Gaussian residual field:The Gau range parameter in this software is $a=1/\varphi$, since its semivariogram uses $1-e^{-(h/a{)}^2}$. The output gives ${\hat{\tau }}^2=0$, ${\hat{\sigma }}^2=0.4726257$, and $\hat{a}=1.112235$, hence $\hat{\varphi }\simeq 0.899091$. The exact Gaussian range is infinite; the 95-percent practical range is approximately $1.92508$ in the coordinate distance units.

The BLUE=TRUE option asks for the best linear unbiased estimator of the drift, rather than the full field kriging prediction. Thus the two printed fitted trends give ${\hat{\beta }}_0=20.13793$ at $y=0$ and ${\hat{\beta }}_0+{\hat{\beta }}_1=22.08399$ at $y=1$. ThereforeThere is no $x$ coefficient in the drift. The reported variances $0.09456496$ and $0.06097459$ are the variances of the corresponding estimated trend values, not full-field prediction variances. They do not determine the slope variance without its covariance with the intercept. The interpolated map can still vary with $x$ through the correlated residual prediction, even though the fitted deterministic drift depends only on $y$.