New decouplers of fractal dimension and Hurst effects

Abstract

A new parametric class of covariance functions of wide-sense homogeneous and isotropic random fields on Euclidean spaces is presented. While the new class possesses uncoupled fractal and Hurst effects, it is shown to encompass the Generalized Cauchy and Dagum models. On the basis of the new functions, product-based covariance models are developed, along with parametric conditions guaranteeing a decoupling of fractal and Hurst effects.

Appendix: Proofs

Some background is necessary for a self-contained exposition. A function $f:[0,\infty ) \rightarrow {\mathbb {R}}$ is called a Bernstein function [21] if its first derivative is completely monotonic.

Proof of Proposition 3.1:

According to the Schoenberg theorem [18], a candidate mapping $\varphi : [0,\infty ) \rightarrow {\mathbb {R}}$ belongs to the class $\varPhi _{\infty }$ if and only if the mapping $0 \le x \mapsto \varphi (x)$ is completely monotonic. Hence, our assertion is proved by a direct inspection that allows to check for complete monotonicity. To provide a proof based on direct construction, we define

$$\begin{aligned} \varphi _{\alpha }(x):= \frac{(1+x)^{\alpha -1}}{(1+x)^{\alpha }-x^{\alpha }}, \qquad x \ge 0, \end{aligned}$$

and

$$\begin{aligned} \theta _{\beta }(x):= \frac{1}{(1+x)^{\beta }}, \qquad x \ge 0. \end{aligned}$$

Direct inspection shows that the function $\psi _{\alpha ,\beta }(\cdot )$ can be expressed as follows:

$$\begin{aligned} \psi _{\alpha ,\beta }(x)&= (1-(1+x^{-1})^{-\alpha })^{\beta }\\&= \left( \frac{(1+x)^{\alpha }-x^{\alpha }}{(1+x)^{\alpha -1}}\right) ^\beta \frac{1}{(1+x)^\beta }\\&= \varphi _{\alpha }^{-\beta }(x)\theta _{\beta }(x), \qquad x \ge 0. \end{aligned}$$

We now invoke the fact that completely monotone functions are a convex cone that is closed under pointwise convergence [21]. Hence, the product of two completely monotone functions is still completely monotonic. We now observe that the function $\theta _{\beta }(\cdot )$ is completely monotonic for $\beta \ge 0$. Hence, we need prove that $\varphi _{\alpha }^{-\beta }(\cdot )$ is completely monotonic, implying that $\psi _{\alpha ,\beta }(x)$ is completely monotonic as well.

For a completely monotonic function, g and a Bernstein function, f, the composition $g\circ f$ is a completely monotonic function [21]. Direct inspection shows that the function $g(z):=z^{-\beta }$ is completely monotonic for $\beta \ge 0$. We now show that $\varphi _{\alpha }(\cdot )$ is a Bernstein function, and this fact will complete the proof.

Consider the following function:

$$\begin{aligned} \rho _{\alpha } (x) = \frac{x^{\alpha }-x(1+x)^{\alpha -1}}{(1+x)^\alpha -x^\alpha }, \qquad x \ge 0. \end{aligned}$$

For $0<\alpha <1$, $\rho _{\alpha }(x)$ is a complete Bernstein function ([22], p. 220). Now, consider the following expression:

$$\begin{aligned} 1+\rho _{\alpha }(x)&= \frac{x^{\alpha }-x(1+x)^{\alpha -1}}{(1+x)^\alpha -x^\alpha } + 1\\&= \frac{(1+x)^{\alpha -1}}{(1+x)^{\alpha }-x^{\alpha }}\\&= \varphi _{\alpha } (x), \qquad x \ge 0. \end{aligned}$$

Here, $\varphi _{\alpha } (x) = 1+\rho _{\alpha }(x) > 0$ is a Bernstein function for $0<\alpha <1$ as $\rho _{\alpha }(x)$ is a Bernstein function for $0<\alpha <1$. It can be easily verified that $\psi _{1,\beta }(x)=\theta _{\beta }(x)$ and therefore $\psi _{\alpha ,\beta }$ is completely monotonic for the case $\alpha =1$ as well. $\square$

Proof of Positive definiteness of $\psi _{\alpha ,\beta }$ in ${\mathbb {R}}^3$ The sufficient conditions for the positive definiteness can be obtained by applying Pólya theorem [20] for radial functions and the corresponding extension to ${\mathbb {R}}^n$ as given in [23]. Define the following function:

$$\begin{aligned} \eta _1(x) = \left( -\frac{d}{du}\right) ^k \psi (u^{1/2})\vert _{u=x^2}, \end{aligned}$$

for $x>0$ and k a nonnegative integer. From Gneiting’s theorem, if there exists $m \ge \frac{1}{2}$ such that

$$\begin{aligned} \eta _2(x) = \left( -\frac{d}{dx}\right) ^{k+l-1} [-\eta _1^{\prime }(x^m)] \end{aligned}$$

is convex for $x>0$ and l a nonnegative integer, then $\psi$ is positive definite on ${\mathbb {R}}^n$. The above-mentioned arguments are taken from [24]. By setting $k=0$, $l=1$, and $m=1$, if we can show that

$$\begin{aligned} -\eta ^{\prime }_1(x) = -\psi _{\alpha ,\beta }^{\prime }(x) = \alpha \beta (1-(1+x^{-1})^{-\alpha })^{\beta -1}(1+x^{-1})^{-\alpha -1}x^{-2}=: f_{\alpha ,\beta }(x) \end{aligned}$$

(A.1)

is convex, then (3.1) is valid on ${\mathbb {R}}^3$. To show this, we take the second derivative of Eq. (A.1), which has the following expression:

$$\begin{aligned} f_{\alpha ,\beta }^{\prime \prime }(x) =f_{\alpha ,\beta }(x)(1+x^{-1})^{\alpha }\frac{P(x,\alpha ,\beta )+(1+x^{-1})^{\alpha } Q(x,\alpha ,\beta )+(1+x^{-1})^{-\alpha }R(x,\alpha ,\beta )}{x^2((1+x^{-1})^\alpha - 1)^2(x + 1)^2} \end{aligned}$$

Here,

$$\begin{aligned} P(x,\alpha ,\beta )&:= \alpha ^2 + 3\alpha - 4 + 3\alpha \beta (1-\alpha ) + (6\alpha (1+\beta )-12)x - 12x^2\\ Q(x,\alpha ,\beta )&:= \alpha ^2 - 3\alpha + 2 + (1-\alpha )6r + 6r^2\\ R(x,\alpha ,\beta )&:= \alpha ^2\beta ^2 - 3\alpha \beta + 2 + (1-\alpha \beta )6x + 6x^2 \end{aligned}$$

$f_{\alpha ,\beta }^{\prime \prime }(x)$ has to be positive for $f_{\alpha ,\beta }(x)$ to be convex. It can be easily shown that $f_{\alpha ,\beta }(x)$ is positive when $\alpha >0$ and $\beta >0$. For $f_{\alpha ,\beta }^{\prime \prime }(x)$ to be positive, the expression,

$$\begin{aligned} S(x,\alpha ,\beta ):=P(x,\alpha ,\beta )+(1+x^{-1})^{\alpha }Q(x,\alpha ,\beta )+(1+x^{-1})^{-\alpha }R(x,\alpha ,\beta ) \end{aligned}$$

has to be positive when $\alpha >0$ and $\beta >0$.

The expression $Q(x,\alpha ,\beta )$ is quadratic in x with a positive coefficient of $x^2$. $Q(x,\alpha ,\beta )$ is always positive if its discriminant is negative.

$$\begin{aligned} \Delta _Q&= (6(1-\alpha ))^2-4(6)(\alpha ^2 - 3\alpha + 2)\\&= 12(\alpha ^2-1)\\&<0 \quad \text {for } \alpha <1 \end{aligned}$$

Similarly, the expression $R(x,\alpha ,\beta )$ is quadratic in x with a positive coefficient of $x^2$. $R(x,\alpha ,\beta )$ is always positive if its discriminant is negative.

$$\begin{aligned} \Delta _R&= (6(1-\alpha \beta ))^2-4(6)(\alpha ^2\beta ^2 - 3\alpha \beta + 2)\\&= 12(\alpha ^2\beta ^2-1)\\&<0 \quad \text {for } \alpha \beta <1 \end{aligned}$$

The expression $P(x,\alpha ,\beta )$ as well is quadratic in x, however, with a negative coefficient of $x^2$. So, $P(x,\alpha ,\beta )$ can either be completely negative or mostly negative.

Consider $S(x,\alpha ,\beta )$ when $0<\alpha <1$ and $0<\alpha \beta <1$ as $x\rightarrow 0$ and as $x\rightarrow \infty$

$$\begin{aligned} \lim _{x\rightarrow 0}S(x,\alpha ,\beta )&= \infty >0 \end{aligned}$$

$$\begin{aligned} \lim _{x\rightarrow \infty }S(x,\alpha ,\beta )&= \alpha ^2(\beta -1)(\beta -2)\\&>0 \quad \text {for }\beta <1 \end{aligned}$$

Therefore, consider $S(x,\alpha ,\beta )$ when $0<\alpha <1$ and $0<\beta <1$:

$$\begin{aligned} S(x,\alpha ,\beta )&=P(x,\alpha ,\beta )+(1+x^{-1})^{\alpha }Q(x,\alpha ,\beta )+(1+x^{-1})^{-\alpha }R(x,\alpha ,\beta )\\&\ge P(x,\alpha ,\beta )+2\sqrt{Q(x,\alpha ,\beta )R(x,\alpha ,\beta )}=:T(x,\alpha ,\beta ) \end{aligned}$$

The expression $T(x,\alpha ,\beta )$ is much simpler compared to $S(x,\alpha ,\beta )$. It is minimized using Wolfram Mathematica with the constrains $\{0<\alpha<1; 0<\beta <1\}$.

$$\begin{aligned} \text {min } T(x,\alpha ,\beta )&= \alpha ^2-1+\sqrt{(\alpha ^2-1)(\alpha ^2\beta ^2-1)}\\&> 0 \end{aligned}$$

$$\begin{aligned} \implies S(x,\alpha ,\beta )\ge T(x,\alpha ,\beta ) >0 \end{aligned}$$

$\square$

Proof of Proposition 3.2:

Consider the following volume integral for the case $n=3$ over $V\in {\mathbb {R}}^3$ to check $L^1-$ integrability,

$$\begin{aligned} \iiint _V \psi _{\alpha ,\beta }({\Vert {\textbf{x}}\Vert }) \,d{\textbf{x}}&= \int \limits _{0}^{2\pi }\int \limits _{0}^{\pi }\int \limits _{0}^{R}\vert (1-(1+x^{-1})^{-\alpha })^{\beta }\vert x^2\sin {(\theta )} \,dx\,d\theta \,d\varPhi \\&= \int \limits _{0}^{2\pi }\int \limits _{0}^{\pi }\int \limits _{0}^{R}\frac{((1+x^{-1})^{\alpha }-1)^{\beta }}{(1+x^{-1})^{\alpha \beta }} x^2\sin {(\theta )} \,dx\,d\theta \,d\varPhi \\&= 4\pi \int \limits _{0}^{R}\frac{((1+x^{-1})^{\alpha }-1)^{\beta }}{(1+x^{-1})^{\alpha \beta }} x^2 \,dx. \end{aligned}$$

For $R\le 1$

$$\begin{aligned} \iiint _V \psi _{\alpha ,\beta }(\Vert {\textbf{x}}\Vert ) \,d{\textbf{x}}&\le 4\pi \int \limits _{0}^{R} x^2 \,dx\\&=4\pi \frac{R^3}{3}\\&\le 4\pi /3\\&< \infty . \end{aligned}$$

For $R> 1$,

$$\begin{aligned} \iiint _V \psi _{\alpha ,\beta }({\Vert {\textbf{x}}\Vert }) \,d{\textbf{x}}&\le \frac{4\pi }{3}+4\pi \int \limits _{1}^{R}((1+x^{-1})^{\alpha }-1)^{\beta } x^2 \,dx\\&\le \frac{4\pi }{3}+4\pi \int \limits _{1}^{R}((1+\alpha x^{-1})-1)^{\beta } x^2 \,dx\\&\le \frac{4\pi }{3}+4\pi \int \limits _{1}^{R}\alpha ^{\beta } x^{-\beta +2} \,dx\\&=\frac{4\pi }{3} + 4\pi \frac{\alpha ^{\beta }}{-\beta +3}(R^{-\beta +3}-1)\\&< \infty \quad \text {for } \beta >3. \end{aligned}$$

By similar computations, it can be shown that $\psi _{\alpha ,\beta }(x)$ is $L^1-$ and $L^2-$ integrable, respectively, under the conditions $\beta > n$ and $\beta >n/2$ for $n=1,2,3$. $\square$

Proof of Proposition 3.3:

To estimate the fractal dimension for $\psi _{\alpha ,\beta }$, consider the following:

$$\begin{aligned} \lim _{x\rightarrow 0}\frac{1-\psi _{\alpha ,\beta }(x)}{x^{\delta }}&= \lim _{x\rightarrow 0}\frac{1-(1-(1+x^{-1})^{-\alpha })^{\beta }}{x^{\delta }}\\&= \lim _{x\rightarrow 0}\frac{1-(1-x^{\alpha }(1+x)^{-\alpha })^{\beta }}{x^{\delta }}\\&= \lim _{x\rightarrow 0}\frac{1-(1-\beta x^{\alpha }(1+x)^{-\alpha })}{x^{\delta }}\\&= \lim _{x\rightarrow 0}\frac{\beta x^{\alpha }}{x^{\delta }}\lim _{x\rightarrow 0}(1+x)^{-\alpha }\\&= \lim _{x\rightarrow 0}\frac{\beta x^{\alpha }}{x^{\delta }}. \end{aligned}$$

Setting $\delta = \alpha$ implies

$$\begin{aligned} \lim _{x\rightarrow 0}\frac{1-\psi _{\alpha ,\beta }(x)}{x^{\delta }}&= \beta > 0. \end{aligned}$$

Therefore, the fractal dimension is identically equal to $D = n+1-\alpha /2$. To estimate the Hurst parameter for the above covariance function, consider the following:

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{\psi _{\alpha ,\beta }(x)}{x^{-\omega }}&= \lim _{x\rightarrow \infty }\frac{((1+x^{-1})^{\alpha }-1)^{\beta }}{(1+x^{-1})^{\alpha \beta }x^{-\omega }}\\&= \lim _{x\rightarrow \infty }\frac{((1+\alpha x^{-1})-1)^{\beta }}{x^{-\omega }}\\&= \lim _{x\rightarrow \infty }\frac{\alpha ^{\beta } x^{-\beta }}{x^{-\omega }}. \end{aligned}$$

Consider $\omega = \beta$, this implies that

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{\psi _{\alpha ,\beta }(x)}{x^{-\omega }}&= \alpha ^{\beta } > 0 \end{aligned}$$

Therefore, the Hurst parameter is $H = 1-\beta /2$. $\square$

Proof of Proposition 3.6:

To obtain the fractal dimension for the MOSP model corresponding to $\psi _{\alpha ,\beta ,\gamma }$ as in Eq. (3.2), consider the following

$$\begin{aligned} \lim _{x\rightarrow 0}\frac{1-{\tilde{\psi }}_{\alpha _1,\beta _1,\gamma _1,\alpha _2,\beta _2,\gamma _2}(x)}{x^{\delta }}&= \lim _{x\rightarrow 0}\frac{1-(1-(1+x^{-\gamma _1})^{-\alpha _1})^{\beta _1}(1-(1+x^{-\gamma _2})^{-\alpha _2})^{\beta _2}}{x^{\delta }} \\&= \lim _{x\rightarrow 0}\frac{1-(1-{\beta _1}(1+x^{-\gamma _1})^{-\alpha _1})(1-{\beta _2}(1+x^{-\gamma _2})^{-\alpha _2})}{x^{\delta }}\\&= \lim _{x\rightarrow 0}\frac{1-(1-{\beta _1}(1+x^{-\gamma _1})^{-\alpha _1}-{\beta _2}(1+x^{-\gamma _2})^{-\alpha _2})}{x^{\delta }}\\&-\frac{{\beta _1}{\beta _2}(1+x^{-\gamma _1})^{-\alpha _1}(1+x^{-\gamma _2})^{-\alpha _2}}{x^{\delta }}\\&= \lim _{x\rightarrow 0}\frac{{\beta _1}x^{\gamma _1\alpha _1}(1+x^{\gamma _1})^{-\alpha _1} +{\beta _2}x^{\gamma _2\alpha _2}(1+x^{\gamma _2})^{-\alpha _2}}{x^{\delta }}\\&-\frac{{\beta _1}{\beta _2}x^{\gamma _1\alpha _1+\gamma _2\alpha _2} (1+x^{\gamma _1})^{-\alpha _1}(1+x^{\gamma _2})^{-\alpha _2}}{x^{\delta }}\\&= \lim _{x\rightarrow 0}\frac{{\beta _1}x^{\gamma _1\alpha _1}+{\beta _2}x^{\gamma _2\alpha _2}+ {\beta _1}{\beta _2}x^{\gamma _1\alpha _1+\gamma _2\alpha _2}}{x^{\delta }}. \end{aligned}$$

Setting $\delta = \gamma _1\alpha _1 \le \gamma _2\alpha _2$ implies

$$\begin{aligned} \lim _{x\rightarrow 0}\frac{1-{\tilde{\psi }}_{\alpha _1,\beta _1,\gamma _1,\alpha _2,\beta _2,\gamma _2}(x )}{x^{\delta }}&= {\left\{ \begin{array}{ll} \beta _1, \qquad \gamma _1\alpha _1<\gamma _2\alpha _2\\ \beta _1+\beta _2, \qquad \gamma _1\alpha _1=\gamma _2\alpha _2. \end{array}\right. } \end{aligned}$$

Therefore, the fractal dimension is identically equal to $D=n+1-\text {min}(\gamma _1\alpha _1,\gamma _2\alpha _2)/2$. To obtain the Hurst parameter, consider the following:

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{{\tilde{\psi }}_{\alpha _1,\beta _1,\gamma _1,\alpha _2,\beta _2,\gamma _2}(x)}{x^{-\omega }}&= \lim _{x\rightarrow \infty } \frac{(1-(1+x^{-\gamma _1})^{-\alpha _1})^{\beta _1} (1-(1+x^{-\gamma _2})^{-\alpha _2})^{\beta _2}}{x^{-\omega }} \\&= \lim _{x\rightarrow \infty } \frac{(1-(1+x^{-\gamma _1})^{-\alpha _1})^{\beta _1}}{x^{-\omega _1}} \lim _{x\rightarrow \infty } \frac{(1-(1+x^{-\gamma _2})^{-\alpha _2})^{\beta _2}}{x^{-\omega _2}}\\&= \lim _{x\rightarrow \infty } \frac{((1+x^{-\gamma _1})^{\alpha _1}-1)^{\beta _1}}{(1+x^{-\gamma _1})^{\alpha _1}x^{-\omega _1}} \lim _{x\rightarrow \infty } \frac{((1+x^{-\gamma _2})^{\alpha _2}-1)^{\beta _2}}{(1+x^{-\gamma _2})^{\alpha _2}x^{-\omega _2}}\\&= \lim _{x\rightarrow \infty } \frac{((1+\alpha _1x^{-\gamma _1})-1)^{\beta _1}}{x^{-\omega _1}} \lim _{x\rightarrow \infty } \frac{((1+\alpha _2x^{-\gamma _2})-1)^{\beta _2}}{x^{-\omega _2}}\\&= \lim _{x\rightarrow \infty } \frac{\alpha _1^{\beta _1}x^{-\gamma _1\beta _1}}{x^{-\omega _1}} \lim _{x\rightarrow \infty } \frac{\alpha _2^{\beta _2}x^{-\gamma _2\beta _2}}{x^{-\omega _2}} \end{aligned}$$

Considering $\omega _1 = \gamma _1\beta _1$ and $\omega _2 = \gamma _2\beta _2$, we get

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{{\tilde{\psi }}_{\alpha _1,\beta _1,\gamma _1,\alpha _2,\beta _2,\gamma _2}(x)}{x^{-\omega }}&= \alpha _1^{\beta _1} \alpha _2^{\beta _2}>0 \end{aligned}$$

Therefore, the Hurst parameter is $H = 1 - (\gamma _1\beta _1 + \gamma _2\beta _2)/2$. $\square$