Random fields and variograms

Put $W:={\sum }_{i=1}^n{a}_{i}X(s_i)$. The pointwise inequality $W^2\le n{\sum }_{i=1}^n{a}_i^{2}X(s_i{)}^2$ shows that $W\in L^2$, since each of the finitely many terms on the right is integrable by the given square-integrability. Thus $\mathrm{Var}(W)$ is defined by [F2].

Covariance bilinearity in [F3] gives $\mathrm{Var}(W)=\mathrm{Cov}(W,W)={\sum }_{i=1}^n{\sum }_{j=1}^n{a}_i{a}_j\mathrm{Cov}(X(s_i),X(s_j))={\sum }_{i=1}^n{\sum }_{j=1}^n{a}_i{a}_{j}K(s_i,s_j)\mathrm{.}$ The variance is nonnegative because its defining integrand is a square. The same covariance symmetry in [F3] gives $K(s,t)=K(t,s)$.

For each nonempty finite $F\subseteq T$, use AC to choose an enumeration $e_F:\{1,\dots ,n_F\}\to F$. Put $t_i=e_F(i)$, $m_F=(m(t_i){)}_{i=1}^{n_F}$, and $K_F=[K(t_i,t_j){]}_{i,j=1}^{n_F}$. Let $r_F:{\mathbb{R}}^{n_F}\to {\mathbb{R}}^F$ relabel coordinates by $r_F(x)(e_F(i))=x_i$. The matrix $K_F$ is symmetric positive semidefinite by the hypothesis on $K$, so [F1] supplies $N_{n_F}(m_F,K_F)$; define the probability measure on the actual coordinate space by ${\mu }_F:=(r_F{)}_{\mathrm{\#}}{N}_{n_F}(m_F,K_F)$. If $F=\mathrm{\varnothing }$, use the unique probability law on the singleton empty product. Changing the enumeration permutes $m_F$ and both indices of $K_F$ by the same permutation. The characteristic-function formula in [F2] shows that the corresponding permutation of ${\mathbb{R}}^{n_F}$ carries one Gaussian law to the other; after relabelling, the resulting measure on ${\mathbb{R}}^F$ is unchanged. Thus ${\mu }_F$ is independent of the chosen enumeration.

The intervals $(p,q)$ with rational endpoints form a countable family by the countability of $\mathbb{Q}$ and [F4]. They form a base: if $U$ is open and $x\in U$, choose $r>0$ with $(x-r,x+r)\subseteq U$; rational density supplies $p,q\in \mathbb{Q}$ with $x-r<p<x<q<x+r$, so $x\in (p,q)\subseteq U$. Thus $\mathbb{R}$ is second-countable by [F4]. If $x\mathrm{\ne }y$, the open balls about $x$ and $y$ of radius $\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}3$ are disjoint: a point in both would give $\mathrm{\mid }x-y\mathrm{\mid }<2\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}3$ by the triangle inequality. Hence the metric topology is Hausdorff. It is locally compact since each $x$ has the compact interval $[x-1,x+1]$ as a neighborhood by [F4]. Therefore [F3] makes $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ standard Borel.

Let $F\subseteq H$ be finite and nonempty, and use the enumerations fixed in step 1.1. Write ${\pi }_{H,F}:{\mathbb{R}}^H\to {\mathbb{R}}^F$ for restriction. For $u\in {\mathbb{R}}^F$, define $\tilde{u}\in {\mathbb{R}}^{n_H}$ by ${\tilde{u}}_j=u(e_H(j))$ when $e_H(j)\in F$, and ${\tilde{u}}_j=0$ otherwise. Under the coordinate relabelling $r_H$, the characteristic function of $({\pi }_{H,F}{)}_{\mathrm{\#}}{\mu }_H$ at $u$ is the characteristic function of $N_{n_H}(m_H,K_H)$ at $\tilde{u}$. By [F2] this is $\exp \text{\hspace{0.17em}⁣}(i{\sum }_{t\in F}u(t)m(t)-\frac{1}{2}{\sum }_{s,t\in F}u(s)u(t)K(s,t))\mathrm{.}$ This is also the characteristic function of ${\mu }_F=(r_F{)}_{\mathrm{\#}}{N}_{n_F}(m_F,K_F)$ at $u$, by [F2]. Identifying ${\mathbb{R}}^F$ with ${\mathbb{R}}^{n_F}$ through $r_F$, uniqueness of characteristic functions in [F2] gives $({\pi }_{H,F}{)}_{\mathrm{\#}}{\mu }_H={\mu }_F$. Empty sets give the unique singleton marginal. Hence the family is consistent in the sense of [F5].

The consistent family from step 2.1 now satisfies the standard-Borel hypotheses of [F7]. Kolmogorov extension gives a unique probability measure $\mu$ on ${\mathcal{C}}_T$ with the specified finite-coordinate marginals.

On $({\mathbb{R}}^T,{\mathcal{C}}_T,\mu )$ define $X(t)(\omega ):=\omega (t)$. Each coordinate is measurable by [F6], and for every finite set $F$ its joint law is ${\mu }_F$ by [F7]. The means and covariances are therefore $m(t)$ and $K(s,t)$ by [F1]. For any finite list $t_1,\dots ,t_n$, possibly with repeated sites, and coefficients $a_1,\dots ,a_n$, collect the distinct sites in $F=\{t_1,\dots ,t_n\}$ and put $b_t:={\sum }_{j:t_j=t}{a}_j$. Then ${\sum }_j{a}_{j}X(t_j)={\sum }_{t\in F}{b}_{t}X(t)$ is a linear projection of the $F$-coordinate Gaussian vector, so it has a normal law by [F1]. This is the Gaussian-process condition [F8], and the coordinates form a real field by [F9]. The extension supplies only the cylinder sigma-algebra; no continuity or joint-measurability conclusion was used.

Suppose first that $X$ is wide stationary. For every $s,h$, its increments have mean zero because the field mean is constant. Wide stationarity gives $C(h)=\mathrm{Cov}(X(s),X(s+h))$ and $C(-h)=\mathrm{Cov}(X(s+h),X(s))$; covariance symmetry [F3] therefore gives $C(-h)=C(h)$. Both level variances equal $C(0)$, and the cross-covariance in the increment variance is $C(-h)=C(h)$. Thus $\mathbb{E}[(X(s+h)-X(s){)}^2]=\mathrm{Var}(X(s+h)-X(s))=2C(0)-2C(h)\mathrm{.}$ Dividing by two gives $\gamma (h)=C(0)-C(h)$.

For arbitrary sites $s_1,\dots ,s_n$, set $Y_i:=X(s_i)-X(0)$. Each $Y_i$ is square-integrable by order-two intrinsic stationarity, since it is an increment from $0$ to $s_i$. Moreover, $Y_i-Y_j=X(s_i)-X(s_j)$, so [F1] gives $2\gamma (s_i-s_j)=\mathbb{E}[(Y_i-Y_j{)}^2]\mathrm{.}$ All cross products are integrable because $2\mathrm{\mid }{Y}_i{Y}_j\mathrm{\mid }\le Y_i^2+Y_j^2$.

Expanding the finite sum from step 1.2 and using ${\sum }_i{a}_i=0$ yields $\begin{array}{rl}{\displaystyle \sum _{i,j}{a}_i{a}_j\gamma (s_i-s_j)}& {\displaystyle =\frac{1}{2}\sum _{i,j}{a}_i{a}_j\mathbb{E}[Y_i^2+Y_j^2-2Y_i{Y}_j]}\\ {\displaystyle }& {\displaystyle =-\mathbb{E}\text{\hspace{0.17em}⁣}\left[{\left(\sum _i{a}_i{Y}_i\right)}^2\right]\le 0.}\end{array}$ The two sums containing $Y_i^2$ or $Y_j^2$ vanish because each contains the factor ${\sum }_i{a}_i$; the remaining term is the negative expectation of a square. No choice principle is used.

By wide stationarity [F1], $C(h)=\mathrm{Cov}(X(0),X(h))$ and $C(-h)=\mathrm{Cov}(X(h),X(0))$. Covariance symmetry in [F2] gives $C(-h)=C(h)$. For any finite $h_1,\dots ,h_n\in {\mathbb{R}}^d$, the matrix $M_{ij}:=C(h_i-h_j)$ is the covariance matrix of $(X(h_1),\dots ,X(h_n))$: indeed, $\mathrm{Cov}(X(h_i),X(h_j))=C(h_j-h_i)=C(h_i-h_j)$ by evenness. Thus $M$ is real symmetric positive semidefinite by [F2].

Write any complex coefficients as $z_i=u_i+i{v}_i$ with $u_i,v_i\in \mathbb{R}$. Since $M$ is real symmetric, ${\sum }_{i,j}{z}_i\overline{z_j}{M}_{ij}={\sum }_{i,j}{u}_i{u}_j{M}_{ij}+{\sum }_{i,j}{v}_i{v}_j{M}_{ij}\ge 0.$ The imaginary cross terms cancel after interchanging $i$ and $j$, and each real quadratic form is nonnegative by step 1.1. Therefore $C:G\to \mathbb{C}$ is positive definite in the complex-coefficient sense of [F3].

The given covariance is continuous, so it satisfies the continuity hypothesis of [F5]. The given usual topology on $G$ is locally compact, Hausdorff, and abelian, so [F4] identifies the dual and its evaluation pairing required there. Applying [F5] with the hypotheses [A1] and [A2] gives a unique finite positive Radon measure $\mu$ on $\widehat{G}$ with the displayed representation and $\mu (\widehat{G})=C(0)$. The zero-covariance case is included: then the mass is zero and uniqueness gives the zero measure.