Fractional Brownian Motion with Variable Hurst Parameter: Definition and Properties

Abstract

A class of Gaussian processes generalizing the usual fractional Brownian motion for Hurst indices in (1/2,1) and multifractal Brownian motion introduced in Ralchenko and Shevchenko (2010 Theory Probab Math Stat 80:119–130) and Boufoussi et al. (Bernoulli 16(4):1294–1311, 2010) is presented. Any measurable function assuming values in this interval can now be chosen as a variable Hurst parameter. These processes allow for modeling of phenomena where the regularity properties can change with time either continuously or through jumps, such as in the volatility of a stock or in Internet traffic. Some properties of the sample paths of the new process class, including different types of continuity and long-range dependence, are discussed. It is found that the regularity properties of the Hurst function chosen directly correspond to the regularity properties of the sample paths of the processes. The long-range dependence property of fractional Brownian motion is preserved in the larger process class. As an application, Fokker–Planck-type equations for a time-changed fractional Brownian motion with variable Hurst parameter are found.

Appendices

Appendix 1

Lemma 6.1

Let $\alpha <2, \gamma >1/2,\delta >1/2$ and $I=\int _0^{a}u^{1-\alpha }(b-u)^{\gamma -\frac{3}{2}}(a-u)^{\delta -\frac{3}{2}}\,\mathrm{d}u$, then

• (i)
  $$\begin{aligned} I=(b-a)^{\gamma +\delta -2}\int \limits _{\frac{b}{a}}^{\infty }(x-1)^{\alpha -\delta -\gamma } x^{\gamma -\frac{3}{2}}(ax-b)^{1-\alpha }\,\mathrm{d}x. \end{aligned}$$

• (ii) If additionally $\alpha =\gamma +\delta$, then

  $$\begin{aligned} I\!=\!(b\!-\!a)^{\alpha -2}\int \limits _{\frac{b}{a}}^{\infty }x^{\gamma -\frac{3}{2}} (ax\!-\!b)^{1-\alpha }\,\mathrm{d}x\!=\!(b\!-\!a)^{\alpha -2}b^{\frac{1}{2}-\delta } a^{\frac{1}{2}-\gamma }\beta (\delta \!-\!1/2,2\!-\!\alpha ). \end{aligned}$$

• (iii) If $\alpha ,\gamma \in (-1,1)$ arbitrary, $0<a<b$ and $J=\int _a^by^{-\alpha }\int _0^az^{\alpha }(y-z)^{-\gamma }\,\mathrm{d}z\,\mathrm{d}y,$ then

  $$\begin{aligned} J&= \frac{a^{2-\gamma }}{2-\gamma }\int \limits _{\frac{a}{b}}^1v^{\alpha +\gamma -2} (1-v)^{-\gamma }\,\mathrm{d}v\\&\quad +\frac{b^{2-\gamma }}{2-\gamma }\int \limits _0^{\frac{a}{b}}v^{\alpha } (1-v)^{-\gamma }\,\mathrm{d}v-\frac{a^{2-\gamma }}{2-\gamma }\beta (\alpha +1,1-\gamma ). \end{aligned}$$

Proof

Part (i) is obtained via substituting $x=\frac{u-b}{u-a}$. Part (ii) follows by substituting $y=\frac{b}{ax}.$ For (iii) $z=yv$ is substituted and it follows that

$$\begin{aligned} J&= \int \limits _a^by^{1-\gamma }\int \limits _0^{\frac{a}{y}}v^{\alpha }(1-v)^{-\gamma }\,\mathrm{d}v\,\mathrm{d}y\\&= \int \limits _{\frac{a}{b}}^1v^{\alpha }(1-v)^{-\gamma }\int \limits _a^{\frac{a}{v}}y^{1-\gamma }\,\mathrm{d}y\, \mathrm{d}v+\int \limits _0^{\frac{a}{b}}v^{\alpha }(1-v)^{-\gamma }\int \limits _a^{b}y^{1-\gamma }\,\mathrm{d}y\,\mathrm{d}v\\&= \frac{a^{2-\gamma }}{2-\gamma }\int \limits _{\frac{a}{b}}^1v^{\alpha +\gamma -2}(1-v)^{-\gamma }\,\mathrm{d}v+\frac{b^{2-\gamma }}{2-\gamma } \int \limits _0^{\frac{a}{b}}v^{\alpha }(1-v)^{-\gamma }\,\mathrm{d}v\\&\quad -\frac{a^{2-\gamma }}{2-\gamma }\beta (\alpha +1,1-\gamma ). \end{aligned}$$

$\square$

Appendix 2

Formulas (i)–(iv) below can be found in The Handbook of mathematical functions, p. 559 by Abramowitz and Stegun [1]. Formula (v) is from the Integrals and Series handbook by Prudnikov at. al. [28].

Lemma 6.2

Let $a,b,c$ be real numbers and $z\in \mathbb C$.

• (i) For $|\mathrm{arg}(z)|, |\mathrm{arg}(1-z)|<\pi$ and when all terms are defined,

  $$\begin{aligned} \,_2F_1(a,b;c;z)&= \frac{\varGamma (c)\varGamma (c\!-\!a\!-\!b)}{\varGamma (c\!-\!a) \varGamma (c\!-\!b)}z^{-a}\,_2F_1\left( a,a\!-\!c\!+\!1;a\!+\!b\!-\!c\!+\!1;1\!-\!\frac{1}{z}\right) \\&\quad +\frac{\varGamma (c)\varGamma (a\!+\!b\!-\!c)}{\varGamma (a)\varGamma (b)} (1\!-\!z)^{c-a-b}z^{a-c}\,_2F_1\left( c\!-\!a,1\!-\!a;c\!-\!a\!-\!b\!+\!1;1\!-\!\frac{1}{z}\right) . \end{aligned}$$

• (ii) If $(1-z)^{-a}$ is defined,

  $$\begin{aligned} _2F_1(a,b;c;z)=(1-z)^{-a}\,_2F_1\left( a,c-b;c;\frac{z}{z-1}\right) . \end{aligned}$$

• (iii) For $|\mathrm{arg}(1-z)|<\pi$ and when all terms are defined,

  $$\begin{aligned} \,_2F_1(a,b;c;z)&= \frac{\varGamma (c)\varGamma (c-a-b)}{\varGamma (c-a)\varGamma (c-b)}\,_2F_1(a,b;a+b-c+1;1-z)\\&\quad +\frac{\varGamma (c)\varGamma (a\!+\!b\!-\!c)}{\varGamma (a)\varGamma (b)}(1\!-\!z)^{c-a-b}\,_2F_1(c\!-\!a,c\!-\!b;c\!-\!a\!-\!b\!+\!1;1\!-\!z). \end{aligned}$$

• (iv) If $(1-z)^{c-a-b}$ is defined,

  $$\begin{aligned} _2F_1(a,b;c;z)=(1-z)^{c-a-b}\,_2F_1(c-a,c-b;c;z). \end{aligned}$$
• (v)
  $$\begin{aligned}&\int \limits _a^b(x-a)^{\alpha -1}(b-x)^{\delta -1}(cx+d)^{\gamma }\,\mathrm{d}x\\&\quad =\beta (\alpha ,\delta )(b-a)^{\alpha +\delta -1}(ac+d)^{\gamma } \,_2F_1\left( \alpha ,-\gamma ;\alpha +\delta ;\frac{c(a-b)}{ac+d}\right) \end{aligned}$$

  if $\mathrm{Re}(\alpha )>0,\ \mathrm{Re}(\delta )>0$ and $|\mathrm{arg}((d+cb)/(d+ca))|<\pi .$

• (vi) Under the assumptions of (ii), (iii), and (iv)

  $$\begin{aligned}&\,_2F_1(a,b;c;z)={1\!\!1}_{\{a=0\vee b=0\}}\\&\!\!\!\!\!\!\!\!\quad +{1\!\!1}_{\{a\ne 0\wedge b\ne 0\}}(1\!-\!z)^{-a}\Bigg [\frac{\varGamma (c)\varGamma (b\!-\!a)}{\varGamma (c\!-\!a)\varGamma (b)} \,_2F_1\left( a,c\!-\!b;a\!-\!b\!+\!1;\frac{1}{1\!-\!z}\right) \\&\!\!\!\!\!\!\!\!\quad +\frac{\varGamma (c)\varGamma (a\!-\!b)}{\varGamma (a)\varGamma (c\!-\!b)} \left( \frac{z}{z\!-\!1}\right) ^{1-c}\left( \frac{1}{1\!-\!z}\right) ^{b-a}\,_2 F_1\left( b\!-\!c\!+\!1,1\!-\!a;b\!-\!a\!+\!1;\frac{1}{1\!-\!z}\right) \Bigg ]. \end{aligned}$$

Proof

Part (vi): The equality is obtained by consecutively applying parts (ii) and (iii) of the Lemma to $\,_2F_1(a,b;c;z)$ and then applying part (iv) to the second term of what was obtained in the first two steps. The indicator functions make up for the case that $a=0$ or $b=0$, i.e., when (iii) cannot be applied. $\square$