A failed stationary covariance and a valid quadratic variogram

If $C$ were a stationary covariance, its matrix at the two sites $0$ and $1$ would be $M=\left(\begin{array}{cc}C(0)& C(-1)\\ C(1)& C(0)\end{array}\right)=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)\mathrm{.}$ The covariance-kernel lemma [F2] would make this matrix positive semidefinite, but for $v=(1,-1{)}^T$, $v^{T}Mv=-2<0$. Hence $C(h)=1+h^2$ is not a stationary covariance.

For any real $a_1,\dots ,a_n$ with ${\sum }_i{a}_i=0$ and sites $s_1,\dots ,s_n$, ${\sum }_{i,j}{a}_i{a}_j(s_i-s_j{)}^2=-2{\left({\sum }_i{a}_i{s}_i\right)}^2\le 0.$ This directly verifies the zero-sum CND inequality for $\gamma (h)=h^2$ and is consistent with [F3].

Define $Z(\omega ):=\sqrt{2}\omega$ on $\mathrm{\Omega }=\{-1,+1\}$. By the finite space and expectation definitions [F4]–[F6], $\mathbb{E}[Z]=\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}=0,\mathbb{E}[Z^2]=\frac{1}{2}(2)+\frac{1}{2}(2)=2.$

Define $X(t):=tZ$, so every coordinate has finite range and is square-integrable. For any admissible $t,h$, $X(t+h)-X(t)=hZ$, and $\frac{1}{2}\mathbb{E}[(X(t+h)-X(t){)}^2]=\frac{1}{2}{h}^2\mathbb{E}[Z^2]=h^2,$ independent of $t$. Its increment mean is also zero, so the stated intrinsic stationarity and variogram claim hold by [F1].

The field mean is zero and $\mathrm{Var}(X(t))=2t^2$ by [F6]–[F7], so its variance is not constant in $t$. Thus this intrinsically stationary field is not wide stationary, as claimed.