Classes of compactly supported covariance functions for multivariate random fields

Abstract

The paper combines simple general methodologies to obtain new classes of matrix-valued covariance functions that have two important properties: (i) the domains of the compact support of the several components of the matrix-valued functions can vary between components; and (ii) the overall differentiability at the origin can also vary. These models exploit a class of functions called here the Wendland–Gneiting class; their use is illustrated via both a simulation study and an application to a North American bivariate dataset of precipitation and temperature. Because for this dataset, as for others, the empirical covariances exhibit a hole effect, the turning bands operator is extended to matrix-valued covariance functions so as to obtain matrix-valued covariance models with negative covariances.

Appendices

Appendix 1: Concerning the Montée Operator

We collect some facts used in the proof of the main result of the paper.

It is convenient to write $I f(t) :=\int _t^\infty u f(u)\,{\mathrm{d}}u$ for an unnormalized operator for the same functions $f$, not least because the Wendland function $\psi _{\nu ,k}$ is defined by $I^k\psi _{\nu ,0}$ and the Wendland–Gneiting function by its normalized version ${\widetilde{\psi }}_{\nu ,k}(t) = I^k \psi _{\nu ,0}(t)/I^k \psi _{\nu ,0}(0)$ (and this quantity equals ${\widetilde{I}}^k\psi _\nu (t)\,)$. Now

$$\begin{aligned} \big (I^2 f\big )(t)&= \int _t^1 v\,{\mathrm{d}}v\int _v^1 u\,f(u)\,{\mathrm{d}}u \,=\, \int _t^1 u\,f(u)\,{\mathrm{d}}u\int _t^u v\,{\mathrm{d}}v \\&= \int _t^1 u\,f(u)\,{\mathrm{d}}u\int _{t^2}^{u^2} {\textstyle \frac{1}{2}}{\mathrm{d}}w \, =\, {\textstyle \frac{1}{2}}\int _t^1 u(u^2-t^2)\,f(u)\,{\mathrm{d}}u, \end{aligned}$$

and routine induction shows that for any function $f:{{\mathrm{I}}\!{\mathrm{R}}_+} \, \mapsto \, {\mathrm{I}}\!{\mathrm{R}}$ with support bounded by $[0,1]$ and for which the integrals concerned are well defined,

$$\begin{aligned} I^k f (t) = \int _t^1 u(u^2-t^2)^{k-1}\,f(u)\,{\mathrm{d}}u\big /2^{k-1}\Gamma (k), \quad k=1,2,\ldots ; \end{aligned}$$

for $k=0$ define $I^0$ as the identity operator, so that $I^0f(t)$ is identically equal to $f(t)$ for all $t \ge 0$. From this it follows that for $k\ge 1$ the Wendland–Gneiting function $\widetilde{I}^k \psi _{\nu }={\widetilde{\psi }}_{\nu ,k}$ is given by

$$\begin{aligned} {\widetilde{\psi }}_{\nu ,k}(t) = \frac{\big (I^k(1-\cdot )_+^\nu \big ) (t)}{\big (I^k(1-\cdot )_+^\nu \big ) (0)} = \int _t^1 \frac{u(u^2-t^2)^{k-1}\,(1-u)_+^{\nu }}{B(2k,\nu +1)} \,{\mathrm{d}}u \qquad (0\le t\le 1). \end{aligned}$$

(18)

Analogous arguments in Zastavnyi and Trigub (2002) give the equivalent expression valid for $\nu >0$,

$$\begin{aligned} {\widetilde{\psi }}_{\nu ,k}(t) = \int _t^1 (u^2-t^2)^k (1-u)^{\nu -1}\,{\mathrm{d}}u \big /B(2k+1,\nu ); \end{aligned}$$

(19)

integration by parts in (18) yields (19).

Appendix 2: Proofs of the technical Lemmas

1.1 2.1: An elementary inequality

The proofs of the lemmas in Section 3 invoke simple inequalities between functions $g: {\mathrm{I}}\!{\mathrm{R}}_+ \, \mapsto \, [0,1]$ (or a sub-interval of [0,1]) of the form

$$\begin{aligned} g(t) := g(t; \alpha , \gamma ,b,c) = c t^\alpha \psi _{\gamma } \Big (\frac{t}{b}\Big ), \end{aligned}$$

where $\alpha >0$, $\gamma \ge 0$, $b>0$ and $c>0$. Define the function $g_1 := g(t; \alpha _1, \gamma _1, b_1, c_1)$ similarly.

Proposition B

(a) Given any functions $g$ as at (4) and $g_1$ as described, if the inequality

$$\begin{aligned} g(t)\ge g_1(t)\qquad (\hbox {all }\,t\in {{\mathrm{I}}\!{\mathrm{R}}_+}) \quad \quad \quad {(*)} \end{aligned}$$

holds then necessarily $\alpha _1\ge \alpha$ and $b_1\le b$.

(b) For $(*)$ to hold it is necessary and sufficient that exactly one of the sets of conditions (A)–(F) below should hold; in these conditions, $\Delta _\alpha =\alpha _1-\alpha ,$ $\Delta _\gamma = \gamma _1-\gamma , \,$ $q_1(t) = \gamma b_1 - \gamma _1 b -\Delta _\alpha (b+b_1) + \Delta _\gamma t$ and $q_2(t) = \Delta _\alpha (bb_1+t^2) + t\,q_1(t)\!:$

• (A) $\alpha _1=\alpha$ and $b_1 = b$ and $c_1\le c$ and $\gamma _1 \ge \gamma;$

• (B) $\alpha _1=\alpha$ and $b_1 < b$ and $c_1\le c$ and $\gamma _1 b\ge \gamma b_1;$

• (C) $\alpha _1=\alpha$ and $b_1 < b$ and $c_1\le c$ and $\gamma _1 b < \gamma b_1$ and $c \ge c_1 \Big (\frac{\gamma _1}{b_1}\Big )^{\gamma _1} \Big (\frac{\gamma }{b}\Big )^{-\gamma } \Big (\frac{-\Delta _\gamma }{b-b_1}\Big )^{-\Delta _\gamma };$

• (D) $\alpha _1>\alpha$ and $b_1 = b$ and $\gamma _1=\gamma$ and $c\ge c_1 b_1^{\Delta _\alpha };$

• (E) $\alpha _1>\alpha$ and $b_1 = b$ and $\gamma _1>\gamma$ and $c\ge c_1b_1^{\Delta _\alpha } (\Delta _\gamma /\Delta _\alpha )^{\Delta _\gamma } \big / (1+\Delta _\gamma /\Delta _\alpha )^{\Delta _\alpha +\Delta _\gamma };$

• (F) $\alpha _1>\alpha$ and $b_1 < b$ and $k(t_0)\le 1,$ where $k(\cdot )$ is given by (20) and $t_0$ is the zero in $(0,b_1)$ of $q_2(t)=0.$

Proof

The function $g$ has support $\{t:g(t)>0\} = (0,b)$. so that when $(*)$ holds we must have $(0,b)\supseteq (0,b_1)$, and then $(*)$ is equivalent by virtue of continuity of $g$ and $g_1$ and the positivity of $g$ on $(0,b_1)$ to

$$\begin{aligned} k(t) := \frac{g_1(t)}{g(t)}\,=\, \frac{c_1 t^{\Delta _\alpha } (1 - t/b_1)^{\gamma _1}}{c(1-t/b)^\gamma } \le 1 \quad (0\,<\,t\,<\,b_1). \end{aligned}$$

(20)

Here $k$ inherits continuity from $g$ and $g_1$ on its range of interest. If $\alpha _1<\alpha$, then $k(t)>1$ for sufficiently small positive $t$, so for $(*)$, necessarily $\alpha _1\ge \alpha$. Part (a) is established.

Part (b) lists sets of conditions which, consistent with part (a), are exhaustive with respect to the possible parameter sets $\{\alpha _1\ge \alpha \}$ and $\{b_1\le b\}$. The list then stipulates sufficient conditions for $(c_1,\gamma _1)$, explicitly when either $\alpha _1=\alpha$ or $b_1=b$ (or both), and implicitly when neither of these holds. One of the conditions necessarily holds because, subject to the constraints on $(\alpha _1, b_1)$, the argument used to explore the possibilities in the parameter space for $(c_1,\gamma _1)$ is logically exhaustive.

When $\alpha _1=\alpha$, (20) gives $k(0+)=c_1/c >0$ and for $(*)$ to hold we must have $c_1\le c$. Also from (20), $k(b_1-) = 0$ for $b_1<b$ and also for $b_1=b$ and $\gamma _1>\gamma$; for $b_1=b$ and $\gamma _1=\gamma$ we have $k(t)=c_1/c$ identically. If $b_1=b$ and $\gamma _1 < \gamma$ then $k(t)>1$ for $1-t/b_1$ sufficiently small and positive so when $(*)$ holds, $\alpha _1=\alpha$ and $b_1=b$ implies both $c_1\le c$ and $\gamma _1\ge \gamma$. Conditions (A) are established.

In the rest of the proof we use the derivative $k'(\cdot )$; observe that $k(t)$ is differentiable with $k'$ continuous and finite on $(0,b_1)$, and given by

$$\begin{aligned} k'(t)&= \bigg (\frac{c_1 t^{\Delta _\alpha -1} (1 - t/b_1)^{\gamma _1-1}}{c (1-t/b)^{\gamma +1}} \bigg ) \bigg [\Delta _\alpha \Big (1-\frac{t}{b}\Big )\Big (1-\frac{t}{b_1}\Big ) - \frac{\gamma _1 t}{b_1}\Big (1-\frac{t}{b}\Big ) + \frac{\gamma t}{b}\Big (1-\frac{t}{b_1}\Big )\bigg ] \\&= \Big (\cdots \Big )q_2(t)\big /b b_1 \end{aligned}$$

(21)

where $q_2$ is as stated in the proposition.

This quadratic term $q_2$ in (21) determines the sign and zeroes of $k'(t)$ for $0<t<b_1$. When $\Delta _\alpha =0$, $q_2(t) = t\,q_1(t)$ and the factor $t$ eliminates $t^{\Delta _\alpha }$ from $\big (\cdots \big )$, leaving the linear factor $q_1(t)$ to determine the sign and any zero of $k'$ on $(0,b_1)$.

Because $k(0+)>0=k(b_1-)$, if $k'(0+) \le 0$ then $(*)$ holds; since $k'(0+)=(c_1/c)(\gamma b_1-\gamma _1 b)$, Conditions (B) are established. If on the other hand $k'(0+)>0$, hence $\gamma > \gamma _1 b/b_1 \ge \gamma _1$, linearity of $q_1(t)$ shows that $k'(t)$ can have at most one zero on $(0,b_1)$, at $t_0$ satisfying $q_1(t_0)=0$, hence $t_0 = (\gamma b_1 - \gamma _1 b)/(\gamma - \gamma _1)$; straightforward algebra checks that such $t_0<b_1$ and, with more effort, that $k(t_0)\le 1$ is equivalent to the rest of Conditions (C).

It remains to consider cases where $\alpha _1>\alpha$. Since then $k(0+)=0$, necessarily $k'(0+)>0$ for $(*)$ to hold. Consider first $b_1=b$ so that $\gamma _1>\gamma$ (else $k(t)>1$ for $1-t/b_1$ sufficiently small positive). Either $\gamma _1=\gamma$ and then from (2), $k(b_1-) = \sup _{0<t\le b_1} k(t) = (c_1/c)b_1^{\Delta _\alpha }$ which establishes Conditions D; or $k(b_1-)=k(0+)=0$ and we require $k(t_0)\le 1$ where $t=t_0 = \arg \sup _{0<t<b_1} k(t)$, so $\Delta _\alpha (1-t_0/b_1) = \Delta _\gamma t_0/b_1$. Substitution in $k(t_0)\le 1$ yields the rest of Conditions (E).

Now let $\alpha _1>\alpha$ and $b_1< b$. Since $k(0+)=0= k(b_1-)$, $q_2(0+)>0$ and $q_2(b_1) = -\gamma b_1(b+b_1) < 0$ there is one zero $t_0$ as stated in Conditions (F).

From our preliminary remarks to part (b) of the proof the Proposition is now established. $\square$

1.2 Appendix 2.2: Proof of Lemma 1

Proof

For any given $t$, ${\mathbf{G}}(t)$ is positive definite if and only if for all real $z$, $g_{11}(t) + 2zg_{12}(t)+z^2g_{22}(t) \ge 0$. This inequality holds if and only if the discriminant of this quadratic form is not positive, i.e. $[2g_{12}(t)]^2 \le 4g_{11}(t)g_{22}(t)$, equivalently, much as in the proof of Proposition B in the Appendix,

$$\begin{aligned} k(t) := \frac{c_{12}^2 t^{2\alpha _{12}} (1-t/{b_{12}})_+^{2\gamma _{12}}}{c_{11}c_{22} t^{\alpha _{11}+\alpha _{22}} (1 - t/b_{11})_+^{\gamma _{11}} (1-t/b_{22})_+^{\gamma _{22}}}\le 1 \qquad (0 < t<b_{12}). \end{aligned}$$

(22)

Here, the support of the numerator cannot lie outside the support of the denominator; hence, Condition (1). The terms $t^\alpha$ cancel by assumption on the $\alpha _{ij}$s, and then $1\ge k(0+) = c_{12}^2/c_{11}c_{22}$, implying Condition (2).

For Condition (3), observe that $k$ is continuous and has a continuous derivative $k'$ on $(0,b_{12})$ where the sign and any zeroes of $k'(\cdot )$ are in fact determined by the quadratic expression $q(t) := -2\gamma _{12}(b_{11}-t)(b_{22}-t) + \gamma _{11}(b_{12}-t)(b_{22}-t) + \gamma _{22}(b_{11}-t)(b_{12}-t)$.

Since $k(b_{12}^{}-)=0$, (22) holds when $k'(0+)\le 0$, and $k'(0+)$ has the same sign as $q(0+)/(b_{11}b_{12}b_{22}) = Q$ as in the statement of the lemma. Otherwise, i.e. $k'(0+)>0$, the continuity of $k'$ and the fact that $k(b_{12}^{}-)=0$ imply the existence of a root $t_0$ in $(0,b_{12})$, and (22) holds when $k(t_0)\le 1$; the quadratic nature of $q$ and the properties of $k$ as stated ensure the uniqueness of such a root $t_0$. $\square$

Remark 1

(Case $m=2$ but with different $\alpha _{ij}$) For ${\mathbf{G}}(t)$ to be positive definite the inequality at (22) must still hold, so as in the proof of the necessary conditions in Proposition B we must have $2\alpha _{12} \ge \alpha _{11}+\alpha _{22}$. Suppose that strict inequality holds; then we are in the realm of Conditions (D)–(F) of Proposition B. Analogues of Conditions (D) and (E) are simpler to find, albeit detailed; we consider only an analogue of Condition (F).

Since $k(0+) = 0 = k(b_{ij}^{}-)$ and $k(t) > 0$ for $0\,<\,t\,<\,b_{ij}$ where the derivative $k'$ is continuous, differentiation shows that $k'$ has the same sign and zeroes on $(0,b_{12})$ as

$$\begin{aligned} q(t) = \frac{2\alpha _{12}-\alpha _{11}-\alpha _{22}}{t} - \frac{2\gamma _{12}}{b_{12} - t} + \frac{\gamma _{11}}{b_{11}-t} + \frac{\gamma _{22}}{b_{22}-t}\,. \end{aligned}$$

The right-hand side has a unique zero in $(0,b_{12})$, at $t_0$ say. Then ${\mathbf{G}}(t)$ is positive definite for all $t\in (0,b_{12})$ under the inequality condition on the $\alpha _{ij}$s provided $k(t_0) \le 1$.

1.3 Appendix 2.3: Proof of Lemma 2

Proof

For (8) to hold, the support set $(0,\max _{j\ne i} b_{ij})$ of the right-hand side must be contained by the support set $(0,b_{ii})$ of the left-hand side, i.e. $b_{ii} \ge \max _{j\ne i} b_{ij}$. Further, the inequality (8) is certainly satisfied when

$$\begin{aligned} c_{ii} \ge \sum _{j\ne i} c_{ij} \sup _{0<t<b_{ij}} \frac{t^{\alpha _{ij}-\alpha _{ii}} (1-t/b_{ij})_+^{\gamma _{ij}}}{(1 - t/b_{ii})_+^{\gamma _{ii}}} \,=:\, \sum _{j\ne i} c_{ij} \sup _{0\,<\,t\,<\,b_{ij}} h_{ij}(t) \qquad \hbox {say}. \end{aligned}$$

(23)

when $\alpha _{ij}=\alpha _{ii}$ for all $j$, $h_{ij}(\cdot )$ simplifies and the supremum of $h_{ij}$ on $(0,b_{ij})$ is found much as in the argument establishing Conditions (B) and (C) of Proposition B: $\sup _{0<t<b_{ij}} h_{ij}(t) = h_{ij}(0+) = 1$ when $h'_{ij}(0+)\le 0$, i.e. when either $\gamma _{ii}=0$ or $\gamma _{ij}/\gamma _{ii} \ge b_{ij}/b_{ii}$. Otherwise we find $\sup _{t\in (0,b_{ij})} h_{ij}(t) = h_{ij}(t_{ij})$ via the unique root $t_{ij}$ in $(0,b_{ij})$ of $h'_{ij}(t_{ij})=0$. Combine these facts to obtain the last term in the inequality at Condition (2) of Lemma 2 where we have summarized this discussion. $\square$

Appendix 3: Multivariate Wendland functions: proof of Theorem 1

Proof

Inspection of Lemma 2 shows that when ${\mathbf{G}}(t)$ as defined satisfies Conditions (1) and (2) there, so too does ${\mathbf{G}}_k(t)$, which is thus diagonally dominant and hence positive definite. Then Theorem A shows that ${\varvec{\Psi}}$ as defined is a covariance matrix. It remains to establish (11).

The right-hand side of (11) in the case $k=0$, with $0<\xi =\Vert {\varvec{x}}\Vert <b_{ij}$, equals

$$ \int _{{\mathrm{I}}\!{\mathrm{R}}_+}\psi _\nu \Big (\frac{\xi }{t}\Big )\, g_{ij}(t; \nu ,\gamma _{ij},b_{ij},c_{ij})\,{\mathrm{d}}t= c_{ij}\int _{{\mathrm{I}}\!{\mathrm{R}}_+}\Big (1 - \frac{\xi }{t}\Big )_+^\nu \, t^\nu \Big (1 - \frac{t}{b_{ij}}\Big )_+^{\gamma _{ij}}\,{\mathrm{d}}t = \frac{c_{ij}}{b_{ij}^{\gamma _{ij}}} \int _\xi ^{b_{ij}} (t - \xi )^\nu (b_{ij}-t)^{\gamma _{ij}}\,{\mathrm{d}}t = \frac{c_{ij}}{b_{ij}^{\gamma _{ij}}} \int _0^{b_{ij}-\xi } u^\nu (b_{ij}-\xi -u)^{\gamma _{ij}}\,{\mathrm{d}}u = \frac{c_{ij}}{b_{ij}^{\gamma _{ij}}} \int _0^1 (b_{ij} - \xi )^{\nu +\gamma _{ij}+1} v^\nu (1-v)^{\gamma _{ij}}\,{\mathrm{d}}v = c_{ij}\,b_{ij}^{\nu +1}\Big (1 - \frac{\xi }{b_{ij}}\Big )^{\nu +\gamma _{ij}+1} \,B(\nu +1,\gamma _{ij}+1) = c_{ij}\,b_{ij}^{\nu +1} \,B(\nu +1,\gamma _{ij}+1) \,\psi _{\nu +\gamma _{ij}+1}\Big (\frac{\xi }{b_{ij}}\Big ) , $$

so (10) and (11) agree when $k=0$.

To show that (10) and (11) agree for positive integers $k$, we exploit the exposition in Sect. 2.1 from which we know (see (19)) that for $0<w\le 1$,

$$\begin{aligned} {\widetilde{\psi }}_{\nu ,k}(w) = {\widetilde{I}}^k \psi _\nu (w) = I^k\psi _\nu (w)\big /I^k\psi _\nu (0) \,= \int _w^1 \frac{(1-u)^{\nu -1} (u^2-w^2)^k\,{\mathrm{d}}u}{B(\nu ,2k+1)}\,. \end{aligned}$$

In particular, for $0<y<t\le 1$,

$$\begin{aligned} {\widetilde{I}}^k \psi _\nu \Big (\frac{y}{t}\Big )&= \frac{1}{B(\nu ,2k+1)} \int _{y/t}^1 (1 - u)^{\nu -1} \Big (u^2 - \frac{y^2}{t^2}\Big )^k\,{\mathrm{d}}u \\&= \frac{1}{B(\nu ,2k+1)} \int _y^t \Big (1 - \frac{v}{t}\Big )^{\nu -1} (v^2 - y^2)^k\,\frac{{\mathrm{d}}v}{t^{2k+1}}\,. \\ \end{aligned}$$

It follows from (10) that, for $0 < \xi =\Vert {\varvec{x}}\Vert < b_{ij}$ as for $k=0$,

$$\Psi _{ij}({\varvec{x}})= c_{ij}\int _\xi ^{b_{ij}} {\widetilde{I}}^k\psi _\nu \left(\frac{\xi }{t}\right) \,t^{\nu +2k}\,\left(1 - \frac{t}{b_{ij}}\right)^{\gamma _{ij}}\,{\mathrm{d}}t = \frac{c_{ij}}{B(\nu ,2k+1)} \int _\xi ^{b_{ij}} \,t^{\nu +2k}\,\left(1 - \frac{t}{b_{ij}}\right)^{\gamma _{ij}}\,{\mathrm{d}}t \int _\xi ^t \left(1 - \frac{v}{t}\right)^{\nu -1} (v^2 - \xi ^2)^k\,\frac{{\mathrm{d}}v}{t^{2k+1}}\,=\,\frac{c_{ij}\,b_{ij}^{-\gamma _{ij}}}{B(\nu ,2k+1)} \int _\xi ^{b_{ij}} (v^2-\xi ^2)^k\,{\mathrm{d}}v \int _v^{b_{ij}} (b_{ij}-t)^{\gamma _{ij}} (t-v)^{\nu -1}\,{\mathrm{d}}t = \frac{c_{ij}\,b_{ij}^{-\gamma _{ij}}}{B(\nu ,2k+1)} B(\nu ,\gamma _{ij}+1) \int _\xi ^{b_{ij}} (b_{ij}-v)^{\nu +\gamma _{ij}} (v^2-\xi ^2)^k\,{\mathrm{d}}v = c_{ij}\,b_{ij}^{\nu +2k+1} \frac{B(\nu ,\gamma _{ij}+1)}{B(\nu ,2k+1)} \int _{\xi /b_{ij}}^1 (1-u)^{\nu +\gamma _{ij}} \left(u^2-\frac{\xi ^2}{b_{ij}^2}\right)^k\,{\mathrm{d}}u = c_{ij}\,b_{ij}^{\nu +2k+1} \frac{B(\nu ,\gamma _{ij}+1)}{B(\nu ,2k+1)}\, B(\nu +\gamma _{ij}+1,2k+1)\, {\widetilde{I}}^k \psi _{\nu +\gamma _{ij}+1}\left(\frac{\Vert {\varvec{x}}\Vert }{b_{ij}}\right) = c_{ij}\,b_{ij}^{\nu +2k+1}\, B(\nu +2k+1,\gamma _{ij}+1)\, {\widetilde{\psi }}_{\nu +\gamma _{ij}+1, k}\left(\frac{\Vert {\varvec{x}}\Vert }{b_{ij}}\right),$$

so (10) and (11) agree as asserted. $\square$