Parabolic Hausdorff dimension

In fractal geometry, the parabolic Hausdorff dimension is a restricted version of the genuine Hausdorff dimension.Template:Sfn Only parabolic cylinders, i. e. rectangles with a distinct non-linear scaling between time and space are permitted as covering sets. It is useful to determine the Hausdorff dimension of self-similar stochastic processes, such as the geometric Brownian motionTemplate:Sfn or stable Lévy processesTemplate:Sfn plus Borel measurable drift function $f$.

Definitions

We define the $\alpha$-parabolic $\beta$-Hausdorff outer measure for any set $A\subseteq {ℝ}^{d+1}$ as

${𝒫}^{\alpha }-{\mathscr{H}}^{\beta }(A):=\underset{\delta \downarrow 0}{lim}inf\{\sum _{k=1}^{\mathrm{\infty }}{\left|P_k\right|}^{\beta }:A\subseteq \bigcup _{k=1}^{\mathrm{\infty }}{P}_k,P_k\in {𝒫}^{\alpha },\left|P_k\right|\le \delta \}.$

where the $\alpha$-parabolic cylinders ${\left(P_k\right)}_{k\in \mathbb{N}}$ are contained in

${𝒫}^{\alpha }:=\{[t,t+c]\times \prod _{i=1}^d[x_i,x_i+c^{1/\alpha }];t,x_i\in ℝ,c\in (0,1]\}.$

We define the $\alpha$-parabolic Hausdorff dimension of $A$ as

${𝒫}^{\alpha }-\dim A:=inf\{\beta \ge 0:{𝒫}^{\alpha }-{\mathscr{H}}^{\beta }(A)=0\}.$

The case $\alpha =1$ equals the genuine Hausdorff dimension $\dim$.

Application

Let ${\phi }_{\alpha }:={𝒫}^{\alpha }-\dim {𝒢}_T(f)$. We can calculate the Hausdorff dimension of the fractional Brownian motion $B^H$ of Hurst index $1/\alpha =H\in (0,1]$ plus some measurable drift function $f$. We get

$\dim {𝒢}_T(B^H+f)={\phi }_{\alpha }\wedge \frac{1}{\alpha }\cdot {\phi }_{\alpha }+(1-\frac{1}{\alpha })\cdot d$

and

$\dim {ℛ}_T(B^H+f)={\phi }_{\alpha }\wedge d.$

For an isotropic $\alpha$-stable Lévy process $X$ for $\alpha \in (0,2]$ plus some measurable drift function $f$ we get

$\dim {𝒢}_T(X+f)=\{\begin{array}{cc}{\phi }_1,& \alpha \in (0,1],\\ {\phi }_{\alpha }\wedge \frac{1}{\alpha }\cdot {\phi }_{\alpha }+(1-\frac{1}{\alpha })\cdot d,& \alpha \in [1,2]\end{array}$

and

$\dim {ℛ}_T(X+f)=\{\begin{array}{cc}\alpha \cdot {\phi }_{\alpha }\wedge d,& \alpha \in (0,1],\\ {\phi }_{\alpha }\wedge d,& \alpha \in [1,2].\end{array}$

Inequalities and identities

For ${\varphi }_{\alpha }:={𝒫}^{\alpha }-\dim A$ one has

$\dim A\le \{\begin{array}{cc}{\varphi }_{\alpha }\wedge \alpha \cdot {\varphi }_{\alpha }+1-\alpha ,& \alpha \in (0,1],\\ {\varphi }_{\alpha }\wedge \frac{1}{\alpha }\cdot \alpha +(1-\frac{1}{\alpha })\cdot d,& \alpha \in [1,\mathrm{\infty })\end{array}$

and

$\dim A\ge \{\begin{array}{cc}\alpha \cdot {\varphi }_{\alpha }\vee {\varphi }_{\alpha }+(1-\frac{1}{\alpha })\cdot d,& \alpha \in (0,1],\\ {\varphi }_{\alpha }+1-\alpha ,& \alpha \in [1,\mathrm{\infty }).\end{array}$

Further, for the fractional Brownian motion $B^H$ of Hurst index $1/\alpha =H\in (0,1]$ one has

${𝒫}^{\alpha }-\dim {𝒢}_T\left(B^H\right)=\alpha \cdot \dim T$

and for an isotropic $\alpha$-stable Lévy process $X$ for $\alpha \in (0,2]$ one has

${𝒫}^{\alpha }-\dim {𝒢}_T\left(X\right)=(\alpha \vee 1)\cdot \dim T$

and

$\dim {ℛ}_T(X)=\alpha \cdot \dim T\wedge d.$

For constant functions $f_C$ we get

${𝒫}^{\alpha }-\dim {𝒢}_T\left(f_C\right)=(\alpha \vee 1)\cdot \dim T.$

If $f\in C^{\beta }(T,{ℝ}^d)$, i. e. $f$ is $\beta$-Hölder continuous, for ${\phi }_{\alpha }={𝒫}^{\alpha }-\dim {𝒢}_T(f)$ the estimates

${\phi }_{\alpha }\le \{\begin{array}{cc}\dim T+(\frac{1}{\alpha }-\beta )\cdot d\wedge \frac{\dim T}{\alpha \cdot \beta }\wedge d+1,& \alpha \in (0,1],\\ \alpha \cdot \dim T+(1-\alpha \cdot \beta )\cdot d\wedge \frac{\dim T}{\beta }\wedge d+1,& \alpha \in [1,\frac{1}{\beta }],\\ \alpha \cdot \dim T+\frac{1}{\beta }(\dim T-1)+\alpha \wedge d+1,& \alpha \in [\frac{1}{\beta },\mathrm{\infty })]\end{array}$

hold.

Finally, for the Brownian motion $B$ and $f\in C^{\beta }(T,{ℝ}^d)$ we get

$\dim {𝒢}_T(B+f)\le \{\begin{array}{cc}d+\frac{1}{2},& \beta \le \frac{\dim T}{d}-\frac{1}{2d},\\ \dim T+(1-\beta )\cdot d,& \frac{\dim T}{d}-\frac{1}{2d}\le \beta \le \frac{\dim T}{d}\wedge \frac{1}{2},\\ \frac{\dim T}{\beta },& \frac{\dim T}{d}\le \beta \le \frac{1}{2},\\ 2\cdot \dim T\wedge \dim T+\frac{d}{2},& \text{ else}\end{array}$

and

$\dim {ℛ}_T(B+f)\le \{\begin{array}{cc}\frac{\dim T}{\beta },& \frac{\dim T}{d}\le \beta \le \frac{1}{2},\\ 2\cdot \dim T\wedge d,& \frac{\dim T}{d}\le \frac{1}{2}\le \beta ,\\ d,& \text{ else}.\end{array}$

Sources

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