Regularisation and Deregularisation of Covariances and Variograms for Univariate Random Functions in One-dimensional Space

Abstract

The regularisation on a segment of a stationary or intrinsic variogram model is usually expressed as an integral. By introducing an auxiliary function, it is shown that this integral is a linear combination of elementary terms. This paper shows how these terms are arranged and provides a simple and safe procedure to calculate them. An important application of this procedure is the deregularisation problem, that is, the conversion of a regularised variogram to a point variogram. This ill-posed problem is tackled using an iterative optimisation algorithm designed to produce a family of point variograms that best fit the initial regularised variogram. Subsequently, each of these point variograms can be regularised on any segment, which results in a fully coherent family of simple and cross regularised variograms.

Appendices

Appendix 1: Proofs in the Stationary Case

Let us recall that C is a function of positive type. If it is continuous at the origin (no nugget effect), then a theorem by Bochner (1955) states that C is the Fourier transform of a positive measure dF (called the spectral measure of C)

$$\begin{aligned} C (t) = \int _{\mathbb {R}}e^{i u t} \, d F (u) . \end{aligned}$$

There exists a relation between the auxiliary function and the spectral measure. To obtain it, we start with definition (1) and replace C by its Bochner’s expression. After changing the order of integration, we obtain

$$\begin{aligned} \vartheta (s) = \int _{{\mathbb {R}}} \int _0^s e^{i u t } \, (s - t ) \, d t \, d F (u). \end{aligned}$$

The inner integral can be calculated as

$$\begin{aligned} \int _0^s e^{i u t } \, (s - t ) \, d t = - \frac{e^{i u s} - 1 + i u s}{u^2} , \end{aligned}$$

and since the spectral measure is even, the ius term cancels out, and we get

$$\begin{aligned} \vartheta (s) = \int _{{\mathbb {R}}} \frac{ 1 - e^{i u s}}{u^2} \, d F (u) , \end{aligned}$$

(16)

which shows that $\vartheta$ is a variogram with spectral measure F (Chilès and Delfiner 1999). More explicitly, $\vartheta$ is nothing but the variogram of the intrinsic process

$$\begin{aligned} T (x) = \int _0^x Z (y) \, d y \qquad x \in {\mathbb {R}}. \end{aligned}$$

We now turn to the calculation of $Cov \{ Z_\lambda (x) , Z_\mu (y) \}$. Using the spectral representation of C, we have

$$\begin{aligned} Cov \{ Z_\lambda (x) , Z_\mu (y) \}&= \frac{1}{\lambda \, \mu } \int _{{\check{x}}}^{{\hat{x}}} \int _{{\check{y}}}^{{\hat{y}}} C ( s - t ) \, d s \, d t \\&= \frac{1}{\lambda \, \mu } \int _{{\check{x}}}^{{\hat{x}}} \int _{{\check{y}}}^{{\hat{y}}} \int _{\mathbb {R}}e^{ i u ( s - t ) } d F (u) \, d s \, d t \\&= \frac{1}{\lambda \, \mu } \int _{\mathbb {R}}\int _{{\check{x}}}^{{\hat{x}}} e^{ i u s} \, d s \int _{{\check{y}}}^{{\hat{y}}} e^{ - i u t} \, d t \, d F (u) \\&= \frac{1}{\lambda \, \mu } \int _{{\mathbb {R}}} \, \frac{ e^{i u {\hat{x}}} - e^{i u {\check{x}}}}{i u} \, \frac{e^{- i u {\hat{y}}} - e^{i u {\check{y}}}}{-i u} \, d F (u) \\&= \frac{1}{\lambda \, \mu } \int _{{\mathbb {R}}} \, \frac{ e^{i u ({\hat{x}} - {\hat{y}})} - e^{i u ({\hat{x}} - {\check{y}})} - e^{i u ({\check{x}} - {\hat{y}})} + e^{i u ({\check{x}} - {\check{y}})}}{u^2} \, d F (u). \end{aligned}$$

Of course, there is no inconvenience in inserting one value $+1$ or $-1$ next to each exponential term. It then suffices to apply formula (16) to complete the proof of formula (2).

Appendix 2: Proofs in the Intrinsic Case

Exactly as in the stationary case, the auxiliary function

$$\begin{aligned} \vartheta ( s ) = \int _0^s \gamma ( t ) \, ( s - t ) \, d t \end{aligned}$$

is non-negative and even, increasing and convex.

In what follows, the variogram is supposed without a drift. Then it admits a spectral representation of the form

$$\begin{aligned} \gamma ( h ) = \int _{\mathbb {R}}\frac{ 1 - \cos (u h)}{u^2} \, d F (u), \end{aligned}$$

where F is a positive measure without an atom at the origin and satisfying

$$\begin{aligned} \int _{\mathbb {R}}\frac{ d F ( u )}{1 + u^2} < \infty \, \cdot \end{aligned}$$

Replacing $\gamma$ by its spectral representation, and permuting integrals, we obtain

$$\begin{aligned} \vartheta ( s ) = \int _{\mathbb {R}}\int _0^s \frac{ 1 - \cos (u t)}{u^2} \, ( s - t ) \, d t \, d F (u). \end{aligned}$$

Moreover, we have

$$\begin{aligned} \int _0^s \frac{ 1 - \cos (u t)}{u^2} \, ( s - t ) \, d t = \frac{ \cos (u s) - 1 + u^2 s^2 / 2}{u^4} , \end{aligned}$$

from which we derive

$$\begin{aligned} \vartheta ( s ) = \int _{\mathbb {R}}\frac{ \cos (u s) - 1 + u^2 s^2 / 2}{u^4} \, d F (u) , \end{aligned}$$

(17)

which is the spectral representation of a generalised covariance of order 1.

Regarding the proof of formula (11), we have

$$\begin{aligned} \int _0^\lambda \gamma ( t + h ) \, ( \lambda - t ) \, d t&= \int _h^{\lambda + h} \gamma ( t ) \, ( \lambda + h - t ) \, d t = \vartheta ( \lambda + h ) - \int _0^h \gamma ( t ) \, ( \lambda + h - t ) \, d t \\ \int _0^\lambda \gamma ( t - h ) \, ( \lambda - t ) \, d t&= \int _{-h}^{\lambda - h} \gamma ( t ) \, ( \lambda - h - t ) \, d t = \vartheta ( \lambda - h ) - \int _0^{-h} \gamma ( t ) \, ( \lambda - h - t ) \, d t. \end{aligned}$$

Moreover, we have

$$\begin{aligned} \int _0^h \gamma ( t ) \, ( \lambda + h - t ) \, d t&+ \int _0^{-h} \gamma ( t ) \, ( \lambda - h - t ) \, d t \\&= \int _0^h \gamma ( t ) \, ( \lambda + h - t ) \, d t - \int _0^h \gamma ( t ) \, ( \lambda - h + t ) \, d t \\&= 2 \vartheta ( h ), \end{aligned}$$

which completes the proof.