Anisotropy of Hölder Gaussian random fields: characterization, estimation, and application to image textures

Abstract

The characterization and estimation of the Hölder regularity of random fields has long been an important topic of Probability theory and Statistics. This notion of regularity has also been widely used in image analysis to measure the roughness of textures. However, such a measure is rarely sufficient to characterize textures as it does not account for their directional properties (e.g., isotropy and anisotropy). In this paper, we present an approach to further characterize directional properties associated with the Hölder regularity of random fields. Using the spectral density, we define a notion of asymptotic topothesy which quantifies directional contributions of field high-frequencies to the Hölder regularity. This notion is related to the topothesy function of the so-called anisotropic fractional Brownian fields, but is defined in a more generic framework of intrinsic random fields. We then propose a method based on multi-oriented quadratic variations to estimate this asymptotic topothesy. Eventually, we evaluate this method on synthetic data and apply it for the characterization of historical photographic papers.

Appendix: Proofs

Proof

(Proposition 2) For some positive constant c, we have

$$\begin{aligned} \Delta _A(x)= & {} \vert K_{A,f}(x) - K_{A,f^*}(x) \vert \le c\\&\int _{\vert w \vert \ge A} \vert f(w) - f^*(w) \vert \mathrm{d}w. \end{aligned}$$

Since the field density satisfies Condition (6), we further obtain

$$\begin{aligned} \Delta _A(x)\le & {} c \int _{\vert w \vert \ge A} \vert f(w) - g_{\tau ,\eta }(w) \vert \mathrm{d}w \nonumber \\&\quad + \int _{\vert w \vert \ge A} \vert f^*(w) - g_{\tau ,\eta }(w) \vert \mathrm{d}w, \nonumber \\\le & {} \tilde{c} \int _{\vert w \vert \ge A} \vert w \vert ^{-2H-d-\gamma } \mathrm{d}w\nonumber \\&\quad + \int _{\vert w \vert \ge A} \vert f^*(w) - g_{\tau ,\eta }(w) \vert \mathrm{d}w, \nonumber \\\le & {} o(A^{-2H}) + \int _{\vert w \vert \ge A} \vert f^*(w) - g_{\tau ,\eta }(w) \vert \mathrm{d}w, \end{aligned}$$

as A tends to $+\infty$.

Now, in directions s of the set $E_0=\{s,\tau ^*(s)>0\}$, we notice that $\eta (s)=H$ and $\tau ^*(s)=\tau (s)$. Hence, $f^*(w)=g_{\tau ,\eta }(w)$ whenever $\arg (w)$ is in $E_0$, $f^*(w)=g_{\tau ,\eta }(w)$, and 0 otherwise. Consequently, in polar coordinate, we have

$$\begin{aligned} \int _{\vert w \vert \ge A} \vert f^*(w) - g_{\tau ,\eta }(w) \vert \mathrm{d}w= & {} \int _{E_0^c} \int _A^{+\infty } g_{\tau ,\eta }(\rho s) \rho ^{d-1} \mathrm{d}\rho \, \mathrm{d}s, \nonumber \\\le & {} c \int _{E_0^c} \int _A^{+\infty } \rho ^{-2\eta (\theta )-1} \mathrm{d}\rho \, \mathrm{d}s. \end{aligned}$$

Then, let us decompose the integral over $E_0^c$ into the sum of two integrals, one over a set $F_\delta =\{s, \eta (s)-H>\delta /2 \}$ defined for $\delta >0$, and the other over a set $E_\delta =\{s, 0<\eta (s)-H<\delta /2 \}$. It follows that

$$\begin{aligned} \int _{\vert w \vert \ge A} \vert f^*(w) - g_{\tau ,\eta }(w) \vert \mathrm{d}w = O(A^{-2H}) (A^{-\delta } + \mu (F_\delta )). \end{aligned}$$

where $\mu (F_\delta )$ is the measure of $F_\delta$ on the unit sphere of $\mathbb {R}^d$. But, as shown by Richard (2016b), $\lim _{\delta \rightarrow 0} \mu (F_\delta ) =0$. Hence, letting $\delta =\log (A)^{1-\alpha }$ for some $0<\alpha <1$, we obtain

$$\begin{aligned} \int _{\vert w \vert \ge A} \vert f^*(w) - g_{\tau ,\eta }(w) \vert \mathrm{d}w = o(A^{-2H}). \end{aligned}$$

Consequently, $\Delta _A(x)= o(A^{-2H})$ as A tends to $+\infty$. $\square$

Proof

(Proposition 3) When the increment field is mono-directional, the expression of $\varGamma _{H,v}$ of Eq. (16) reduces to

$$\begin{aligned} \varGamma _{H,v}(\theta ) = \int _{\mathbb {R}^+} \vert \hat{v}_1(\rho \cos (\theta )) \vert ^2 \rho ^{-2H-1} \mathrm{d}\rho . \end{aligned}$$

The expression of $\beta _{H,\tau ^*}$ in Eq. (28) follows from the simple coordinate change $u=\rho \cos (\theta )$. $\square$

Proof

(Theorem 2) We aim at estimating the solution $\tau ^*$ of a linear system $L_H \, \, \tau = \beta$ with a random vector

$$\begin{aligned} \tilde{\tau }_\lambda ^*= (B+\lambda R)^{-1} L^T_H\Sigma ^{-1}\tilde{\beta } \end{aligned}$$

where $B=L^T_H \Sigma ^{-1} L_H$, R is a diagonal matrix, $\lambda >0$, and $\tilde{\beta }$ is an unbiased estimate of $\beta$ of variance $V(\beta )$.

Since $\tilde{\beta }$ is unbiased, the expectation of $\tilde{\tau }_\lambda ^*$ satisfies

$$\begin{aligned} (B+\lambda R) \mathbb {E}(\tilde{\tau }_\lambda ^*) = L^T_H \Sigma ^{-1} \beta . \end{aligned}$$

Moreover, $B \tau = L^T_H \Sigma ^{-1} \beta$. Hence,

$$\begin{aligned} (B+\lambda R) ( \mathbb {E}(\tilde{\tau }_\lambda ^*)-\tau ^*) = - \lambda R \tau ^*. \end{aligned}$$

Thus,

$$\begin{aligned} \vert \mathbb {E}(\tilde{\tau }_\lambda ^*)-\tau ^*\vert _2 =\vert (B+\lambda R)^{-1} \lambda R \tau ^*\vert _2, \end{aligned}$$

where $\vert \cdot \vert _2$ denotes the 2-norm. Hence, using norm properties, we get

$$\begin{aligned} \vert \mathbb {E}(\tilde{\tau }_\lambda ^*)-\tau ^*\vert _2 \le \lambda \vert (B+\lambda R)^{-1} \vert _2 \, \vert R \vert _2 \, \vert \tau ^*\vert _2. \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{\vert \mathbb {E}(\tilde{\tau }_\lambda ^*)-\tau ^*\vert _2}{ \vert \tau ^*\vert _2}\le & {} \lambda \vert B^{-1} \vert _2 \, \vert (I+\lambda B^{-1} R)^{-1} \vert _2 \, \vert R \vert _2 \nonumber \\\le & {} \frac{\lambda \vert R \vert _2 }{\nu _-} \, \vert (I+\lambda B^{-1} R)^{-1} \vert _2, \end{aligned}$$

(30)

where $\nu _-$ is the lowest eigenvalue of B.

Next, we establish a bound for $\vert (I+\lambda B^{-1} R)^{-1} \vert _2$. For any vector u, we have

$$\begin{aligned} \vert B^{-1} R u \vert ^2 \ge \frac{1}{\nu _+^2} \vert R u \vert ^2 \ge \left( \frac{1}{\nu _+}\right) ^2 \vert u \vert ^2. \end{aligned}$$

Therefore, the lowest eigenvalue of $B^{-1} R$ is above $\nu _+^{-1}$. Thus, the one of $I+\lambda B^{-1} R$ is above $1+\lambda \nu _+^{-1}$. Consequently,

$$\begin{aligned} \vert (I+\lambda A^{-1} R)^{-1} \vert _2 \le \frac{\nu _+}{(\nu _++\lambda )}. \end{aligned}$$

(31)

Using Eq. (30), we eventually obtain the inequality (24).

Now, we turn to the variance of the estimator. We have

$$\begin{aligned} V(\tilde{\tau }^*_\lambda )= & {} \mathbb {E}\left( (\tilde{\tau }^*_\lambda -\mathbb {E}(\tilde{\tau }^*_\lambda ))(\tilde{\tau }^*_\lambda -\mathbb {E}(\tilde{\tau }^*_\lambda ))^T\right) ) \\= & {} (B+\lambda R)^{-1} B (B+\lambda R)^{-1}. \end{aligned}$$

Hence,

$$\begin{aligned} \mathrm {trace}(V(\tilde{\tau }^*_\lambda )) = \mathrm {trace}\left( (I+ \lambda B^{-1} R)^{-2} B^{-1}\right) . \end{aligned}$$

Moreover, since $B^{-1}$ is a covariance matrix, any term $B_{ij}^{-1}$ of $B^{-1}$ is bounded by $\sqrt{B_{ii}^{-1}B_{jj}^{-1}}$. Hence, it follows that

$$\begin{aligned} \mathrm {trace}(V(\tilde{\tau }^*_\lambda ))\le & {} \vert (I+ \lambda B^{-1} R)^{-1} \Delta \vert _2^2 \\\le & {} \vert (I+ \lambda B^{-1} R)^{-1} L_H^T \vert _2^2 \, \vert \Delta \vert _2^2, \end{aligned}$$

where $\Delta$ is a vector formed by terms $\sqrt{(B)_{ii}^{-1}}$. Noticing that $\vert \Delta \vert _2^2=\mathrm {trace}(B^{-1})$ and using Eq. (31), we get

$$\begin{aligned} \sqrt{\mathrm {trace}(V(\tilde{\tau }^*_\lambda ))} = \frac{\nu _+}{\nu _++ \lambda } \sqrt{\mathrm {trace}(B^{-1})}. \end{aligned}$$

(32)

Besides, since $B \tau ^*= L^T_H \Sigma ^{-1} \beta$, we have

$$\begin{aligned} \vert B \vert _2 \, \vert \tau ^*\vert _2 \ge \vert B^{-1/2} \Sigma ^{-1/2} \beta \vert _2 \ge \sqrt{\nu _-} \sqrt{\langle \Sigma ^{-1}\beta ,\beta \rangle _2}. \end{aligned}$$

Hence,

$$\begin{aligned} \frac{1}{\vert \tau ^*\vert _2} \le \frac{\nu _+}{\sqrt{\nu _-} \sqrt{\langle \Sigma ^{-1}\beta ,\beta \rangle _2}}. \end{aligned}$$

Combined with Eqs. (32), (25) follows. $\square$

Proof

(Corollary 1) Using expressions of bias and variance in Theorem 2, we clearly see that the function g bounds the relative mean square error. Then, a simple variation analysis of this function suffices to show that it reaches a global minimum at $\lambda ^*$. $\square$