Parameter Estimation in Rough Bessel Model

Abstract

In this paper, we construct consistent statistical estimators of the Hurst index, volatility coefficient, and drift parameter for Bessel processes driven by fractional Brownian motion with $H<1/2$. As an auxiliary result, we also prove the continuity of the fractional Bessel process. The results are illustrated with simulations.

1. Introduction

The Bessel process, defined as the square root $X=\sqrt{Z}$ of the solution to the stochastic differential equation (SDE)

$$\begin{array}{c}\hfill dZ\left(t\right)=kdt+2\sqrt{Z\left(t\right)}dW\left(t\right),Z\left(0\right)=x_0^2>0,k>0,\end{array}$$

(1)

is a well-known probabilistic model used in a wide range of fields including physics (e.g., in non-Abelian gauge field theories or for modeling Fermi acceleration and non-colliding particle systems, as can be seen in [1,2,3,4]) and finance (for modeling interest rates and stochastic volatility, as can be seen in [5,6,7,8]). If $k\in \mathbb{N}$, X can be interpreted as the Euclidean norm

$$X\left(t\right)=\sqrt{B_1^2\left(t\right)+\cdots +B_k^2\left(t\right)}$$

(2)

of a k-dimensional Brownian motion $(B_1,...,B_k)$ that is connected with W in (1) via the relation

$$W\left(t\right)=\sum _{i=1}^k{\int }_0^t\frac{B_i\left(s\right)}{X\left(s\right)}d{B}_i\left(s\right),$$

and therefore the parameter k is often referred to as the dimension of the Bessel process X. Moreover, as it is shown in Section 3 of [9] (also see [10]), for all real $k>1$, the Bessel process X is the unique non-negative strong solution to the SDE

$$dX\left(t\right)=\frac{k-1}{X\left(t\right)}dt+dW\left(t\right),X\left(0\right)=x_0>0.$$

(3)

For more details on the various properties of the Bessel processes, we refer the reader to Chapter XI of [11], the book [12] which considers general equations of the type (3) or [13,14,15] which deal with the case $0<k<1$.

However, in many applications, standard Brownian motion may not adequately capture the desired level of complexity observed in real-life phenomena. For example, a number of empirical studies [16,17,18,19,20] point out the presence of memory in financial markets. Other sources [21,22,23,24] indicate that models exhibiting very low Hölder regularity are better suited to reflect the behavior of market volatility. Given that Bessel-type processes are frequently employed in stochastic volatility modeling (see, e.g., Chapter 6 of [25]), there is a natural inclination to enhance them by incorporating the aforementioned memory or roughness. A common way to achieve such an effect is to replace (in some sense) the standard Brownian driver W with a fractional Brownian motion $B^H=\{B^H\left(t\right),t\ge 0\}$, i.e., a centered Gaussian process with the covariance function

$$\mathbb{E}\left[B^H\left(t\right)B^H\left(s\right)\right]=\frac{1}{2}\left(t^{2H}+s^{2H}-{|t-s|}^{2H}\right),s,t\ge 0.$$

For instance, Refs. [26,27,28] used the property (2) as the starting point of their modification and defined fractional Bessel process ${\rho }^H$ for $k\in \mathbb{N}$ as

$${\rho }^H\left(t\right)=\sqrt{{\left(B_1^H\left(t\right)\right)}^2+\cdots +{\left(B_k^H\left(t\right)\right)}^2}$$

with $B_1^H$,…, $B_k^H$ being k-independent fractional Brownian motions. In particular, they prove that ${\rho }^H$ admits a representation

$$d{\rho }^H\left(t\right)=H(k-1)\frac{t^{2H-1}}{{\rho }^H\left(t\right)}dt+\sum _{i=1}^k\frac{B_i^H\left(t\right)}{{\rho }^H\left(t\right)}d{B}_i^H\left(t\right),$$

(4)

where the integrals with respect to fractional Brownian motions are understood in the divergence sense.

Another possible “fractionalization” of the Bessel process may be achieved by replacing W with $B^H$ directly in (3). This approach was discussed in detail in the series of papers [29,30,31,32]: according to it, the fractional Bessel process $X^H=\{X^H\left(t\right),t\ge 0\}$ is defined as the a.s. point-wise limit

$$X^H\left(t\right):=\underset{\epsilon \downarrow 0}{lim}{X}_{\epsilon }^H\left(t\right)$$

(5)

of stochastic processes given by the SDE

$$\begin{array}{c}\hfill X_{\epsilon }^H\left(t\right)=x_0+{\int }_0^t\frac{a}{X_{\epsilon }^H\left(s\right){\mathbb{1}}_{\{X_{\epsilon }^H\left(s\right)>0\}}+\epsilon }ds+\sigma B^H\left(t\right).\end{array}$$

(6)

In the present paper, we consider the latter notion of the fractional Bessel process and assume that it is driven by a fractional Brownian motion with the Hurst index $H\in \left(0,\frac{1}{2}\right)$. Our goal is to perform the statistical estimation of three parameters: the Hurst index H mentioned above, the diffusion coefficient $\sigma >0$ regulating the magnitude of the driving noise as well as the drift parameter $a>0$ that serves as a direct counterpart of $k-1$ in (3).

• In order to estimate H and $\sigma$, we use the standard technique based on quadratic variations of fractional Brownian motion. The main challenge arises in the limit $L^H\left(t\right):={lim}_{\epsilon \downarrow 0}{\int }_0^t\frac{a}{X_{\epsilon }^H\left(s\right){\mathbb{1}}_{\{X_{\epsilon }^H\left(s\right)>0\}}+\epsilon }ds$: when $H<\frac{1}{2}$, Corollary 3.3 of [32] only guarantees its continuity in t almost everywhere with respect to the Lebesgue measure. It is not enough for the analysis of quadratic variations of $L^H$, so we prove that $L^H$ is continuous at every $t\ge 0$ and utilize this fact to obtain consistent estimators of H and $\sigma$.
• Our estimator of a is, in turn, based on the unconventional technique tailored for our specific model. The problem here is that it is currently not known whether the fractional Bessel process exhibits any ergodic properties typically utilized for drift parameter estimation. In our analysis, we exploit the explicit dynamics of $X^H$ instead to compare the behavior of $X^H\left(T\right)$ and ${\int }_0^T\frac{1}{X^H\left(t\right)}dt$ when $T\to \infty$.

This paper is structured as follows. In Section 2, we provide the definition of the fractional Bessel process and prove the continuity of the latter. Section 3 is devoted to the estimation of H and $\sigma$. In Section 4, we present our estimator of the drift and prove its consistency. Section 5 contains simulations. In Appendix A, we prove a technical result related to the finiteness of the limit (5).

2. Rough Bessel Processes

Let $B^H=\{B^H\left(t\right),t\ge 0\}$ be a fractional Brownian motion with Hurst index $H\in \left(0,\frac{1}{2}\right)$.

Remark 1. 

Using the Kolmogorov–Chentsov theorem, it is possible to prove that $B^H$ has a modification with paths that are locally Hölder continuous up to the order H, i.e., for any $T>0$ and $\lambda \in (0,H)$, there exists a positive random variable $\Lambda ={\Lambda }_{T,\lambda }$ such that

$$|B^H\left(t_1\right)-B^H\left(t_2\right)|\le \Lambda |t_1-t_2{|}^{\lambda },t_1,t_2\in [0,T].$$

(7)

Moreover, by [33], the random variable Λ can be chosen in such a way that for any $p>0$

$$\mathbb{E}\left[{\Lambda }^p\right]<\infty .$$

In what follows, we always consider this modification of $B^H$.

For any $\epsilon >0$, $x_0>0$, $a>0$, $b\ge 0$, and $\sigma >0$, consider a random process $X_{\epsilon }^H=\{X_{\epsilon }^H\left(t\right),t\ge 0\}$ given by a stochastic differential Equation (6). Note that the SDE (6) has Lipschitz continuous drift and additive noise and hence, by the standard Picard iteration argument applied pathwise, (6) has a unique solution for each $\epsilon >0$. Moreover, by Lemma 2.1 of [32] (see also Lemma 1.6 in [29]), for any ${\epsilon }_1<{\epsilon }_2$,

$$X_{{\epsilon }_1}^H\left(t\right)\ge X_{{\epsilon }_2}^H\left(t\right),t\in [0,T],\omega \in \Omega ,$$

(8)

and hence, for any $t\ge 0$, one can define the limit

$$X^H\left(t\right):=\underset{\epsilon \downarrow 0}{lim}{X}_{\epsilon }^H\left(t\right).$$

(9)

Definition 1. 

The process $X^H$ defined by (9) will be called a fractional or rough Bessel process.

Remark 2. 

This construction of rough Bessel processes was introduced and extensively studied in [32]. In particular, it was shown that

(1) 

The limit (9) is finite, i.e., with probability 1,

$$X^H\left(t\right)<\infty$$

(10)

for all $t\ge 0$;

(2) 

With probability 1, $X^H\left(t\right)\ge 0$ for all $t\ge 0$ and, moreover, $X^H\left(t\right)>0$ for almost all $t\ge 0$.

It should be noted that one of the possible cases is missing in Step 2 of the proof of Theorem 2.1 in [32] concerning (10). For the reader’s convenience, we provide the completed version of it in Appendix A.

Since the limit (9) is well defined and finite, the limit

$$\begin{array}{cc}\hfill \underset{\epsilon \downarrow 0}{lim}{\int }_0^t\frac{a}{X_{\epsilon }^H\left(s\right){\mathbb{1}}_{\{X_{\epsilon }^H\left(s\right)>0\}}+\epsilon }ds& =\underset{\epsilon \downarrow 0}{lim}\left(X_{\epsilon }^H\left(t\right)-x_0-\sigma B^H\left(t\right)\right)\hfill \\ \hfill & =X^H\left(t\right)-x_0-\sigma B^H\left(t\right)\hfill \end{array}$$

(11)

also exists and is finite. In what follows, we will use the notation

$$L^H\left(t\right):=\underset{\epsilon \downarrow 0}{lim}{\int }_0^t\frac{1}{X_{\epsilon }^H\left(s\right){\mathbb{1}}_{\{X_{\epsilon }^H\left(s\right)>0\}}+\epsilon }ds,$$

(12)

i.e., $X^H$ satisfies the equation

$$X^H\left(t\right)=x_0+a{L}^H\left(t\right)+\sigma B^H\left(t\right),t\ge 0.$$

(13)

By Fatou’s lemma, for any $t\ge 0$,

$${\int }_0^t\frac{1}{X^H\left(s\right)}ds\le \underset{\epsilon \downarrow 0}{lim\; inf}{\int }_0^t\frac{1}{X_{\epsilon }^H\left(s\right){\mathbb{1}}_{\{X_{\epsilon }^H\left(s\right)>0\}}+\epsilon }ds=L^H\left(t\right)<\infty .$$

Denote

$$R^H\left(t\right):=L^H\left(t\right)-{\int }_0^t\frac{1}{X^H\left(s\right)}ds$$

and re-write (13) as

$$X^H\left(t\right)=x_0+a{\int }_0^t\frac{1}{X^H\left(s\right)}ds+\sigma B^H\left(t\right)+a{R}^H\left(t\right).$$

(14)

At the moment, it is not clear whether $R^H\left(t\right)=0$ for all $t\ge 0$. However, we have the following result.

Proposition 1. (1) 

With a probability of 1, there exists $\tau >0$ such that for all $t\ge \tau$

$$X^H\left(t\right)>0.$$

(2) 

Let $\tau >0$ be such that $X^H\left(\tau \right)>0$. Then, there exists a neighborhood $({\tau }_1,{\tau }_2)\ni \tau$ such that, for all $t_1,t_2\in ({\tau }_1,{\tau }_2)$, $R^H\left(t_1\right)-R^H\left(t_2\right)=0$, i.e.,

$$L^H\left(t_2\right)-L^H\left(t_1\right)={\int }_{t_1}^{t_2}\frac{1}{X^H\left(s\right)}ds.$$

(15)

Proof. 

Item (1) directly follows from Theorem 3.3 in [30] so let us prove the claim (2). Let $\tau >0$ be such that $X^H\left(\tau \right)>0$. Since $X_{\epsilon }^H\left(\tau \right)\uparrow X^H\left(\tau \right)$ as $\epsilon \downarrow 0$, there exists ${\epsilon }_0>0$ such that for all $\epsilon \le {\epsilon }_0$, $X_{\epsilon }^H\left(\tau \right)>0$. Moreover, the process $X_{{\epsilon }_0}^H$ is continuous with respect to t; hence, there exists a neighborhood $({\tau }_1,{\tau }_2)\ni \tau$ and some positive value $x>0$ such that for all $t\in ({\tau }_1,{\tau }_2)$

$$X_{{\epsilon }_0}^H\left(t\right)>x.$$

In particular, (8) implies that, for all $\epsilon \le {\epsilon }_0$

$$X_{\epsilon }^H\left(t\right)>x,t\in ({\tau }_1,{\tau }_2).$$

Therefore, for any $\epsilon \le {\epsilon }_0$ and $t\in ({\tau }_1,{\tau }_2)$,

$$\begin{array}{cc}\hfill \frac{1}{X_{\epsilon }^H\left(t\right){\mathbb{1}}_{\{X_{\epsilon }^H\left(t\right)>0\}}+\epsilon }& <\frac{1}{x}\hfill \end{array}$$

and hence, by the dominated convergence theorem, for any $t_1,t_2\in ({\tau }_1,{\tau }_2)$, $t_1<t_2$,

$$\begin{array}{cc}\hfill L^H\left(t_2\right)-L^H\left(t_1\right)& =\underset{\epsilon \downarrow 0}{lim}{\int }_{t_1}^{t_2}\frac{1}{X_{\epsilon }^H\left(s\right){\mathbb{1}}_{\{X_{\epsilon }^H\left(s\right)>0\}}+\epsilon }ds\hfill \\ \hfill & ={\int }_{t_1}^{t_2}\frac{1}{X^H\left(s\right)}ds.\hfill \end{array}$$

□

Corollary 1. 

With a probability of 1, there exists $\tau >0$ such that, for all $t_2>t_1\ge \tau$

$$R^H\left(t_1\right)=R^H\left(t_2\right)$$

and

$$L^H\left(t_2\right)-L^H\left(t_1\right)={\int }_{t_1}^{t_2}\frac{1}{X^H\left(s\right)}ds.$$

Note that Corollary 3.3 from [32] only establishes the continuity of $L^H$almost everywhere on ${\mathbb{R}}_{+}$ with respect to the Lebesgue measure. It is not enough for our purposes: in order to estimate H and $\sigma$, we want to utilize the behavior of quadratic variations of $X^H$ and any possible discontinuities of $L^H$ would create substantial obstacles for our analysis. It turns out, however, that $L^H$ (and hence $X^H$) is continuous at all points $t\ge 0$. The corresponding theorem is provided below.

Theorem 1. 

1. 

The process $L^H$ defined by (12) is non-decreasing.

2. 

The processes $X^H=\{X^H\left(t\right),t\ge 0\}$ and $L^H=\{L^H\left(t\right),t\ge 0\}$ have continuous paths.

Proof. 

First of all, observe that for any $0\le t_1<t_2\le T$

$$\begin{array}{cc}\hfill L^H\left(t_1\right)& =\underset{\epsilon \downarrow 0}{lim}{\int }_0^{t_1}\frac{1}{X_{\epsilon }^H\left(s\right){\mathbb{1}}_{\{X_{\epsilon }^H\left(s\right)>0\}}+\epsilon }ds\hfill \\ \hfill & \le \underset{\epsilon \downarrow 0}{lim}{\int }_0^{t_2}\frac{1}{X_{\epsilon }^H\left(s\right){\mathbb{1}}_{\{X_{\epsilon }^H\left(s\right)>0\}}+\epsilon }ds\hfill \\ \hfill & =L^H\left(t_2\right),\hfill \end{array}$$

i.e., the process $L^H$ is indeed non-decreasing. Next, fix an arbitrary deterministic $T>0$, choose an arbitrary $\lambda \in (0,H)$ and let $\Lambda ={\Lambda }_{T,\lambda }$ be such that, for all $s,t\in [0,T]$,

$$|B^H\left(t\right)-B^H{\left(s\right)|\le \Lambda |t-s|}^{\lambda }.$$

(16)

Note that the monotonicity of $L^H$ implies that its discontinuities can only take the form of positive jumps and, moreover, left and right limits

$$L^H(t-):=\underset{\delta \downarrow 0}{lim}{L}^H(t-\delta ),L^H(t+):=\underset{\delta \downarrow 0}{lim}{L}^H(t+\delta )$$

are well defined at any point $t\in (0,T)$. Next, observe that the limit process $X^H$ defined by (9) satisfies the equation

$$X^H\left(t\right)=x_0+a{L}^H\left(t\right)+\sigma B^H\left(t\right),t\in [0,T],$$

and hence points of discontinuity of $X^H$ coincide with the ones of $L^H$, and can only occur in the form of positive jumps of the same size as the corresponding jumps of $L^H$ and, finally, one can define the limits

$$X^H(t-):=\underset{\delta \downarrow 0}{lim}{X}^H(t-\delta ),X^H(t+):=\underset{\delta \downarrow 0}{lim}{X}^H(t+\delta ).$$

Assume that $\tau \in [0,T]$ is a point of discontinuity of $L^H$ (and hence $X^H$) for some $\omega \in \Omega$ and observe that $X^H\left(\tau \right)=0$ since, otherwise, $\tau$ cannot be a point of discontinuity of $X^H$ and $L^H$ by (15). Let $\alpha >0$ be such that

$$X^H(\tau +)=L^H(\tau +)=\alpha$$

and $\delta >0$ be such that for all $t\in (\tau ,\tau +\delta )$

$$X^H\left(t\right)>\frac{9\alpha }{10}.$$

Next, choose ${\delta }_0<\delta$ such that

$$\frac{4a}{\alpha }{\delta }_0+\sigma \Lambda {\delta }_0^{\lambda }\le \frac{\alpha }{4},$$

where $\lambda \in (0,H)$ and $\Lambda >0$ are from (16). Since $X_{\epsilon }^H(\tau +{\delta }_0)\uparrow X^H(\tau +{\delta }_0)$ as $\epsilon \downarrow 0$, there exists ${\epsilon }_0>0$ such that for all $\epsilon \in (0,{\epsilon }_0]$

$$X_{\epsilon }^H(\tau +{\delta }_0)>\frac{4\alpha }{5},X_{\epsilon }^H\left(\tau \right)\le 0$$

Furthermore, $X_{\epsilon }^H\left(\tau \right)\uparrow X^H\left(\tau \right)=0$ as $\epsilon \downarrow 0$ hence $X_{\epsilon }^H\left(\tau \right)\le 0$ for all $\epsilon >0$, and therefore, for any $\epsilon \in (0,{\epsilon }_0]$ one can define

$${\tau }_{\epsilon }^{-}:=sup\left\{s\in (\tau ,\tau +{\delta }_0):X_{\epsilon }^H\left(s\right)=\frac{\alpha }{4}\right\}$$

and

$${\tau }_{\epsilon }^{+}:=sup\left\{s\in ({\tau }_{\epsilon }^{-},\tau +{\delta }_0):X_{\epsilon }^H\left(s\right)=\frac{3\alpha }{4}\right\}.$$

By continuity, it is evident that $X_{\epsilon }^H\left({\tau }_{\epsilon }^{-}\right)=\frac{\alpha }{4}$, $X_{\epsilon }^H\left({\tau }_{\epsilon }^{+}\right)=\frac{3\alpha }{4}$ and $X_{\epsilon }^H\left(t\right)\ge \frac{\alpha }{4}$ for any $t\in [{\tau }_{\epsilon }^{-},{\tau }_{\epsilon }^{+}]$. Therefore for any $\epsilon \in (0,{\epsilon }_0]$

$$\begin{array}{cc}\hfill \frac{\alpha }{2}& =X_{\epsilon }^H\left({\tau }_{\epsilon }^{+}\right)-X_{\epsilon }^H\left({\tau }_{\epsilon }^{-}\right)\hfill \\ \hfill & ={\int }_{{\tau }_{\epsilon }^{-}}^{{\tau }_{\epsilon }^{+}}\frac{a}{X_{\epsilon }^H\left(s\right){\mathbb{1}}_{\{X_{\epsilon }^H\left(s\right)>0\}}+\epsilon }ds+\sigma \left(B^H\left({\tau }_{\epsilon }^{+}\right)-B^H\left({\tau }_{\epsilon }^{-}\right)\right)\hfill \\ \hfill & \le \frac{4a}{\alpha }({\tau }_{\epsilon }^{+}-{\tau }_{\epsilon }^{-})+\sigma \Lambda {({\tau }_{\epsilon }^{+}-{\tau }_{\epsilon }^{-})}^{\lambda }\hfill \\ \hfill & \le \frac{4a}{\alpha }{\delta }_0+\sigma \Lambda {\delta }_0^{\lambda }\hfill \\ \hfill & <\frac{\alpha }{4},\hfill \end{array}$$

which gives a contradiction. Therefore, $\alpha =0$ and $X^H\left(\tau \right)=X^H(\tau +)$, i.e., $X^H$ (and hence $L^H$) is right-continuous.

It remains to notice that $X^H(t-)=X^H\left(t\right)$ for any $t\in (0,T]$. Indeed, as mentioned above, $X^H$ is continuous at t if $X^H\left(t\right)>0$. If $X^H\left(t\right)=0$, since $X^H\left(t\right)\ge 0$ can only have positive jumps,

$$X^H\left(t\right)-X^H(t-)=-X^H(t-)$$

and, since $X^H$ can potentially have only positive jumps, $X^H(t-)=0$. □

Note that the pre-limit processes $X_{\epsilon }^H$ given by (6) are monotonically non-decreasing with respect to $\epsilon$. Moreover, with probability 1, their point-wise limit is a continuous function by Theorem 1. Therefore, by Dini’s theorem, we immediately obtain uniform convergence which is summarized in the following corollary.

Corollary 2. 

For any $T>0$,

$$\underset{t\in [0,T]}{sup}|X^H\left(t\right)-X_{\epsilon }^H\left(t\right)|\to 0a.s.,\epsilon \downarrow 0.$$

3. Estimation of Hurst Index and Volatility Coefficient

Let us now move on to the parameter estimation for the rough Bessel processes. Our first goal is to estimate the Hurst index $H\in \left(0,\frac{1}{2}\right)$ and the volatility parameter $\sigma >0$. Note that, by Theorem 1, the process $L^H$ is continuous and non-decreasing and hence has zero quadratic variation. Therefore, we can utilize the standard estimation technique based on power variations. For more details on this method, we refer the reader to [34].

Throughout this section, we assume that the rough Bessel process $X^H=\{X^H\left(t\right),t\ge 0\}$ satisfies Equation (14) with (unknown) parameters $a,\sigma >0$, $H<1/2$ and is observed on a discrete uniform partition $0=t_0<t_1<...<t_n=T$ of a fixed compact $[0,T]$, $t_k:=\frac{kT}{n}$.

3.1. Estimation of H

Let us start with some useful notation for quadratic variations.

Notation 1. 

For a stochastic process $\xi =\left\{\xi \right(t),t\in [0,T\left]\right\}$ observed on a discrete uniform partition $0=t_0<t_1<...<t_n=T$, $t_k:=\frac{kT}{n}$, denote

$$V_{1,2}^n\left(\xi \right):=\sum _{k=0}^{n-1}{(\xi \left(t_{k+1}\right)-\xi \left(t_k\right))}^2$$

(17)

and

$$V_{2,2}^n\left(\xi \right):=\sum _{k=0}^{n-2}{(\xi \left(t_{k+2}\right)-2\xi \left(t_{k+1}\right)+\xi \left(t_k\right))}^2.$$

(18)

In order to estimate the unknown Hurst parameter, we utilize the following result from Proposition 4.2 in [35] and Lemma 2.10 in [34].

Theorem 2. 

Let $H\in (0,1)$, $\sigma >0$ and $V_{1,2}^n\left(\sigma B^H\right)$, $V_{2,2}^n\left(\sigma B^H\right)$ be as defined in Notation 1. Then, with the probability of 1,

$${\left(\frac{n}{T}\right)}^{-1+2H}{V}_{1,2}^n\left(\sigma B^H\right)\to {\sigma }^{2}T$$

and

$${\left(\frac{n}{T}\right)}^{-1+2H}{V}_{2,2}^n\left(\sigma B^H\right)\to (4-2^{2H}){\sigma }^{2}T$$

as $n\to \infty$.

As established in Theorem 1, the process $a{L}^H$ in the right-hand side of (14) is a continuous process of bounded variation and hence, with a probability of 1,

$$V_{1,2}^n\left(a{L}^H\right)\to 0,V_{2,2}^n\left(a{L}^H\right)\to 0,n\to \infty .$$

Therefore, we immediately obtain the following corollary.

Corollary 3. 

Let $X^H$ be a rough Bessel process given by (13) observed on a discrete uniform partition $0=t_0<t_1<...<t_n=T$, $t_k=\frac{kT}{n}$. Then, with probability 1,

$${\left(\frac{n}{T}\right)}^{-1+2H}{V}_{1,2}^n\left(X^H\right)\to {\sigma }^{2}T$$

and

$${\left(\frac{n}{T}\right)}^{-1+2H}{V}_{2,2}^n\left(X^H\right)\to (4-2^{2H}){\sigma }^{2}T$$

as $n\to \infty$.

Next, define

$$\widehat{H}:=\frac{log\left(4-\frac{V_{2,2}^n\left(X^H\right)}{V_{1,2}^n\left(X^H\right)}\right)}{2log2}.$$

(19)

Theorem 3. 

The estimator $\widehat{H}$ given by (19) is a (strongly) consistent estimator of the Hurst index H, i.e., with a probability of 1,

$$\widehat{H}\to H,n\to \infty .$$

Proof. 

The statement immediately follows from Corollary 3. □

Remark 3. 

By Corollary 3, with probability 1,

$$4-\frac{V_{2,2}^n\left(X^H\right)}{V_{1,2}^n\left(X^H\right)}\to 2^{2H}>1,n\to \infty ,$$

and hence the logarithm in (19) is well defined for large enough values of n. Moreover,

$$\begin{array}{cc}\hfill V_{2,2}^n\left(X^H\right)& =\sum _{k=0}^{n-2}{(X^H\left(t_{k+2}\right)-2X^H\left(t_{k+1}\right)+X^H\left(t_k\right))}^2\hfill \\ \hfill & =\sum _{k=0}^{n-2}{\left((X^H\left(t_{k+2}\right)-X^H\left(t_{k+1}\right))-(X^H\left(t_{k+1}\right)-X^H\left(t_k\right))\right)}^2\hfill \\ \hfill & \le 2\sum _{k=0}^{n-2}{(X^H\left(t_{k+2}\right)-X^H\left(t_{k+1}\right))}^2+2\sum _{k=0}^{n-2}{(X^H\left(t_{k+1}\right)-X^H\left(t_k\right))}^2\hfill \\ \hfill & \le 4V_{1,2}^n\left(X^H\right),\hfill \end{array}$$

(20)

i.e.,

$$4-\frac{V_{2,2}^n\left(X^H\right)}{V_{1,2}^n\left(X^H\right)}\ge 0,$$

and the equality in (20) only occurs if $x_0=X^H\left(t_0\right)=X^H\left(t_1\right)=...=X^H\left(t_n\right)$. It remains an open question whether such an event can occur with positive probability; however, in practice, it is never the case.

3.2. Estimation of $\sigma$

Next, assume that the Hurst index H is known and the goal is to estimate the volatility coefficient $\sigma$. Using Corollary 3, it can be performed in a straightforward manner. Namely, we have the following result.

Theorem 4. 

Let the Hurst index H be known. Then, the estimator

$$\widehat{\sigma }=\frac{1}{T^H}\sqrt{n^{-1+2H}{V}_{1,2}^n\left(X^H\right)}$$

(21)

is a (strongly) consistent estimator of the volatility coefficient σ, i.e., with a probability of 1,

$$\widehat{\sigma }\to \sigma ,n\to \infty .$$

Proof. 

The statement immediately follows from Corollary 3. □

4. Drift Parameter Estimation

Let us move to the estimation of the drift parameter a in (14). In contrast to the high-frequency setting considered in Section 3, we will now assume that we observe a continuous path of $X^H$ on $[0,T]$ with $T\to \infty$. We study the following estimator of the parameter a:

$$\widehat{a}\left(T\right):=\frac{X^H\left(T\right)}{{\int }_0^T\frac{1}{X^H\left(t\right)}dt}.$$

(22)

Remark 4. 

Note that we do not assume any prior knowledge of the parameters H or σ.

Let us start by presenting a well-known result related to the growth of fractional Brownian motion (for more details, see [36], Section B 3.5 of [34] or Corollary 2.2 in [30]).

Theorem 5. 

For any $\delta >0$, there exist random time ${\tau }_{\delta }$ and positive random variable ξ such that, with a probability of 1,

$$\underset{t\in [0,T]}{max}\left|B^H\left(t\right)\right|\le \xi T^{H+\delta }$$

(23)

for all $T>{\tau }_{\delta }$.

Lemma 1. 

Let $X^H$ be a rough Bessel process satisfying (14). Then, with a probability of 1,

$$\underset{T\to \infty }{lim\; inf}\frac{1}{\sqrt{T}}{\int }_0^T\frac{1}{X^H\left(t\right)}dt>0.$$

(24)

Proof. 

Take an arbitrary $\delta \in \left(0,\frac{1}{2}-H\right)$ and fix an $\omega \in \Omega$ such that the corresponding path of $B^H$ is continuous, (23) holds and $R^H$ becomes constant after some random time point $\tau$ as described in Corollary 1.

Assume that (24) does not hold, i.e., there exists an increasing sequence $\{T_n,n\ge 1\}$ such that $T_n\to \infty$ as $n\to \infty$, $T_1$ exceeds ${\tau }_{\delta }$ from Theorem 5, and

$$\frac{1}{\sqrt{T_n}}{\int }_0^{T_n}\frac{1}{X^H\left(t\right)}dt\to 0,n\to \infty .$$

Integrating both sides of (14) from 0 to $T_n$, we obtain that, for all $n\ge 1$

$$\begin{array}{cc}\hfill {\int }_0^{T_n}{X}^H\left(t\right)dt& =x_0{T}_n+a{\int }_0^{T_n}{\int }_0^t\frac{1}{X^H\left(s\right)}dsdt\hfill \\ \hfill & +\sigma {\int }_0^{T_n}{B}^H\left(t\right)dt+a{\int }_0^{T_n}{R}^H\left(t\right)dt.\hfill \end{array}$$

(25)

Let us prove that the equality (25) cannot hold by comparing the asymptotics of its left- and right-hand sides.

On the one hand, using the Cauchy–Schwartz inequality, it is easy to see that, for all $T>0$,

$${\int }_0^T{X}^H\left(t\right)dt{\int }_0^T\frac{1}{X^H\left(t\right)}dt\ge T^2,$$

so

$$\frac{1}{T_n^{3/2}}{\int }_0^{T_n}{X}^H\left(s\right)ds\to \infty ,n\to \infty .$$

On the other hand,

$$\frac{x_0{T}_n}{T_n^{3/2}}\to 0,n\to \infty ,$$

$$\begin{array}{cc}\hfill \frac{1}{T_n^{3/2}}{\int }_0^{T_n}{\int }_0^t\frac{1}{X^H\left(s\right)}dsdt& =\frac{1}{T_n^{3/2}}{\int }_0^{T_n}\frac{1}{X^H\left(t\right)}(T_n-t)dt\hfill \\ \hfill & \le \frac{T_n}{T_n^{3/2}}{\int }_0^{T_n}\frac{1}{X^H\left(t\right)}dt\hfill \\ \hfill & \to 0,n\to \infty ,\hfill \end{array}$$

and, by (23),

$$\begin{array}{cc}\hfill \frac{1}{T_n^{3/2}}\left|{\int }_0^{T_n}{B}^H\left(t\right)dt\right|& \le \frac{T^{1+H+\delta }}{T_n^{3/2}}\frac{\xi }{1+H+\delta }\to 0,n\to \infty .\hfill \end{array}$$

Finally, since $R^H$ is constant after some time point $\tau$,

$$\frac{1}{T_n^{3/2}}{\int }_0^{T_n}{R}^H\left(t\right)dt\to 0,n\to \infty .$$

In other words, if such a sequence $\{T_n,n\ge 1\}$ exists, the left-hand side of (25) divided by $T_n^{3/2}$ converges to ∞ whereas the right-hand side divided by $T_n^{3/2}$ converges to zero. We obtain a contradiction that proves (24). □

We are now ready to move to the main result of this section.

Theorem 6. 

With a probability of 1,

$$\widehat{a}\left(T\right)\to a,T\to \infty .$$

Proof. 

First of all, note that, by Lemma 1,

$$\frac{x_0}{{\int }_0^T\frac{1}{X^H\left(s\right)}ds}\to 0\mathrm{a}.\mathrm{s}.$$

when $T\to \infty$. Next, since there exists $\tau >0$ such that $R^H\left(t\right)=0$ for all $t\ge \tau$, Lemma 1 also implies that

$$\frac{a{R}^H\left(t\right)}{{\int }_0^T\frac{1}{X^H\left(s\right)}ds}\to 0\mathrm{a}.\mathrm{s}.$$

when $T\to \infty$. Finally, fixing $\delta \in \left(0,\frac{1}{2}-H\right)$ and using (23), we can deduce that

$$\begin{array}{cc}\hfill \frac{|B^H\left(T\right)|}{{\int }_0^T\frac{1}{X^H\left(s\right)}ds}& \le \frac{1}{T^{\frac{1}{2}-H-\delta }}\frac{\xi }{\frac{1}{\sqrt{T}}{\int }_0^T\frac{1}{X^H\left(s\right)}ds}\to 0\mathrm{a}.\mathrm{s}.\hfill \end{array}$$

as $T\to \infty$. Therefore, with probability 1,

$$\begin{array}{cc}\hfill {\widehat{a}}_1\left(T\right)& =\frac{x_0+a{\int }_0^t\frac{1}{X^H\left(s\right)}ds+\sigma B^H\left(t\right)+a{R}^H\left(t\right)}{{\int }_0^T\frac{1}{X^H\left(s\right)}ds}\hfill \\ \hfill & =a+\frac{x_0}{{\int }_0^T\frac{1}{X^H\left(s\right)}ds}+\frac{\sigma B^H\left(t\right)}{{\int }_0^T\frac{1}{X^H\left(s\right)}ds}+\frac{a{R}^H\left(t\right)}{{\int }_0^T\frac{1}{X^H\left(s\right)}ds}\hfill \\ \hfill & \to a,T\to \infty .\hfill \end{array}$$

□

5. Simulations

In this section, we illustrate our results with simulations. In all cases, we use the standard Euler scheme to simulate the pre-limit processes $X_{\epsilon }^H$ given by (6) with $\epsilon =0.0001$. Each estimator was tested on 1000 samples. All simulations were performed using R programming language; in order to simulate the trajectories of fractional Brownian motion, the package somebm was used.

5.1. Simulation of $\widehat{H}$ and $\widehat{\sigma }$

We start by testing the estimators $\widehat{H}$ and $\widehat{\sigma }$ given by (19) and (21), respectively. The real values of the parameters are chosen to be $H=0.3$ and $\sigma =1$ whereas the time horizon is $T=1$.

At first, we analyze $\widehat{H}$ for different sizes n of the partition. For each n, 1000 trajectories of $X^H$ were generated and the value of $\widehat{H}$ was computed for each of the generated paths.

Table 1. Performance of the estimator $\widehat{H}$ given by (19) for different sizes of the partition. The real value is $H=0.3$.

	$\mathit{n}=100$	$\mathit{n}=1000$	n = 10,000	n = 100,000
Mean	0.3074612	0.3018724	0.299684	0.2999879
Variance	0.007608759	0.0006979747	0.00007438351	0.000007596678
CV	0.2837047	0.08751781	0.02877894	0.009187727

contains the mean, variance, and coefficient of variation (CV, i.e., the ratio of the standard deviation to the mean) for $\widehat{H}$ and Figure 1 depicts the corresponding box-and-whisker plots.

Next, we perform the same procedure for the estimator $\widehat{\sigma }$ given by (21) under the assumption that the real value of H is known. The results are given in

Table 2. Performance of the estimator $\widehat{\sigma }$ given by (21) for different sizes of the partition under assumption that H is known. The real value is $\sigma =1$.

	$\mathit{n}=100$	$\mathit{n}=1000$	n = 10,000	n = 100,000
Mean	0.9964649	0.9999765	0.9996666	0.9999794
Variance	0.005895953	0.0006022594	0.00005529735	0.000005777412
CV	0.07705752	0.02454155	0.007438699	0.002403674

and Figure 2.

Finally, we consider a more realistic situation when neither H nor $\sigma$ are known. In this case, we first estimate H using the estimator $\widehat{H}$ given by (19) and then plug in the result in the estimator $\widehat{\sigma }$ defined by (21). The results of this estimation of $\sigma$ are given in

Table 3. Performance of the estimator $\widehat{\sigma }$ given by (21) for different sizes of the partition; H is unknown and estimated using (19). The real value is $\sigma =1$.

	$\mathit{n}=100$	$\mathit{n}=1000$	n = 10,000	n = 100,000
Mean	1.098972	1.021549	0.9991251	1.001703
Variance	0.1783677	0.03501052	0.006181774	0.000860674
CV	0.3843009	0.183164	0.0786931	0.02928738

and Figure 3.

Overall, in all cases, the simulations show the convergence of the estimators to the true values. This also applies to the least favorable situation of estimation of $\sigma$ when H is unknown.

5.2. Simulation of $\widehat{a}\left(T\right)$

Next, we test the performance of the estimator $\widehat{a}\left(T\right)$ for different time horizons T using the same numerical techniques as in Section 5.1. The real values of the parameters are once again chosen to be $H=0.3$, $\sigma =1$ and the integral ${\int }_0^T\frac{1}{X^H\left(t\right)}dt$ is approximated by the integral sum

$$\frac{1}{n}\sum _{i=0}^{\left[Tn\right]-1}\frac{1}{max\left\{X^H\left(\frac{i}{n}\right),0.001\right\}}$$

with $n=10,000$. For each T, 1000 simulations of $\widehat{a}\left(T\right)$ were performed. The results are given in

Table 4. Performance of the estimator $\widehat{a}\left(T\right)$ for different time horizons T. The real value is $a=2$.

	$\mathit{T}=1$	$\mathit{T}=10$	$\mathit{T}=100$	$\mathit{T}=1000$
Mean	3.872642	2.409692	2.109389	2.029289
Variance	4.09064	0.5914294	0.1749705	0.06310528
CV	0.5222618	0.3191463	0.1983014	0.1237909

and Figure 4.

Simulations confirm the convergence of $\widehat{a}\left(T\right)$ to a. However, they also indicate that relatively high values of T are required to guarantee the reasonable variance of the estimator.