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Construction of maximum likelihood estimator in the mixed fractional–fractional Brownian motion model with double long-range dependence
Volume 2, Issue 2 (2015), pp. 147–164
Yuliya Mishura   Ivan Voronov  

Authors

 
Placeholder
https://doi.org/10.15559/15-VMSTA28
Pub. online: 20 July 2015      Type: Research Article      Open accessOpen Access

Received
17 June 2015
Revised
7 July 2015
Accepted
7 July 2015
Published
20 July 2015

Abstract

We construct an estimator of the unknown drift parameter $\theta \in \mathbb{R}$ in the linear model
\[X_{t}=\theta t+\sigma _{1}{B}^{H_{1}}(t)+\sigma _{2}{B}^{H_{2}}(t),\hspace{0.2778em}t\in [0,T],\]
where ${B}^{H_{1}}$ and ${B}^{H_{2}}$ are two independent fractional Brownian motions with Hurst indices $H_{1}$ and $H_{2}$ satisfying the condition $\frac{1}{2}\le H_{1}<H_{2}<1$. Actually, we reduce the problem to the solution of the integral Fredholm equation of the 2nd kind with a specific weakly singular kernel depending on two power exponents. It is proved that the kernel can be presented as the product of a bounded continuous multiplier and weak singular one, and this representation allows us to prove the compactness of the corresponding integral operator. This, in turn, allows us to establish an existence–uniqueness result for the sequence of the equations on the increasing intervals, to construct accordingly a sequence of statistical estimators, and to establish asymptotic consistency.

1 Introduction

Consider the continuous-time linear model
(1)
\[X(t)=\theta t+\sigma _{1}{B}^{H_{1}}(t)+\sigma _{2}{B}^{H_{2}}(t),\hspace{0.2778em}t\in [0,T],\]
where ${B}^{H_{1}}$ and ${B}^{H_{2}}$ are two independent fractional Brownian motions with different Hurst indices $H_{1}$ and $H_{2}$ defined on some stochastic basis $(\varOmega ,\mathfrak{F},(\mathfrak{F})_{t},t\ge 0,\mathsf{P})$. We assume that the filtration is generated by these processes and completed by $\mathsf{P}$-negligible sets of $\mathfrak{F}_{0}$.
Recall that the fractional Brownian motion (fBm) ${B_{t}^{H}}$, $t\ge 0$, with Hurst index $H\in (0,1)$ is a centered Gaussian process with the covariance function
\[\mathsf{E}\left[{B}^{H}(t){B}^{H}(s)\right]=\frac{1}{2}\big({t}^{2H}+{s}^{2H}-|t-s{|}^{2H}\big).\]
From now on we suppose that the Hurst indices in (1) satisfy the inequality
\[\frac{1}{2}\le H_{1}<H_{2}<1,\]
and we consider the continuous modifications of both processes, which exist due to the Kolmogorov theorem. Assuming that the Hurst indices $H_{1}$, $H_{2}$ and parameters $\sigma _{1}\ge 0$, $\sigma _{2}\ge 0$ are known, we aim to estimate the unknown drift parameter θ by the continuous observations of the trajectories of X. Due to the long-range dependence property of fBm with $H>1/2$, we call our model the model with double long-range dependence.
In the case where $H_{1}=\frac{1}{2}$, the problem of drift parameter estimation in the model (1) was solved in [3], and in the case where $\frac{1}{2}<H_{1}<H_{2}<1$ and $H_{2}-H_{1}>1/4$, the estimator was constructed in [6]. The goal of the present paper is to generalize the results from [6] to arbitrary $\frac{1}{2}\le H_{1}<H_{2}<1$. The problem, more technical than principal, is that in the case where $H_{2}-H_{1}>1/4$ and $H_{1}>1/2$, the construction of the estimator is reduced to the question if the solution of the Fredholm integral equation of the 2nd kind with weakly singular kernel from $L_{2}[0,T]$ exists and is unique, but for $H_{2}-H_{1}\le 1/4$, the kernel does not belong to $L_{2}[0,T]$. Moreover, in this case, we can say that in the literature it is impossible to pick up for this kernel any suitable standard techniques for working with weak singular kernels, and it does not belong to any standard class of weak singular kernels. The matter lies in the fact that the kernel contains two power indices, $H_{1}$ and $H_{2}$, and they create more complex singularity than it usually happens. So, it is necessary to make many additional efforts in order to prove the compactness of the corresponding integral operator. Immediately after establishing the compactness of the corresponding integral operator, the problem of statistical estimation follows the same steps as in the paper [6], and we briefly present these steps for completeness.
The paper is organized as follows. In Section 2, we describe the model and explain how to reduce the solution of the estimation problem to the existence–uniqueness problem for the integral Fredholm equation of the 2nd kind with some nonstandard weakly singular kernel. In Section 3, we solve the existence–uniqueness problem. Section 4 is devoted to the basic properties of estimator, that is, we establish its form, consistency, and asymptotic normality. Section A contains the properties of hypergeometric function used in the proof of the existence–uniqueness result for the main Fredhom integral equation.

2 Preliminaries. How to reduce the original problem to the integral equation

Since we suppose that the Hurst parameters $H_{1},H_{2}$ and scale parameters $\sigma _{1},\sigma _{2}$ are known, for technical simplicity, we consider the case where $\sigma _{1}=\sigma _{2}=1$ and, as it was mentioned before, $\frac{1}{2}\le H_{1}<H_{2}<1$. If we wish to include the unknown parameter θ into the fractional Brownian motion with the smallest Hurst parameter in order to apply Girsanov’s theorem for construction of the estimator, we consider a couple of processes $\{{\widetilde{B}}^{H_{1}}(t),{B}^{H_{2}}(t),\hspace{0.2778em}t\ge 0\}$, $i=1,2$, defined on the space $(\varOmega ,\mathfrak{F},(\mathfrak{F})_{t})$ and let $\mathsf{P}_{\theta }$ be a probability measure under which ${\widetilde{B}}^{H_{1}}$ and ${B}^{H_{2}}$ are independent, ${B}^{H_{2}}$ is a fractional Brownian motion with Hurst parameter $H_{2}$, and ${\widetilde{B}}^{H_{1}}$ is a fractional Brownian motion with Hurst parameter $H_{1}$ and drift θ, that is,
\[{\widetilde{B}}^{H_{1}}(t)=\theta t+{B}^{H_{1}}(t).\]
The probability measure $\mathsf{P}_{0}$ corresponds to the case $\theta =0$. Our main problem is the construction of maximum likelihood estimator for $\theta \in \mathbb{R}$ by the observations of the process $Z(t)=\theta t+{B}^{H_{1}}(t)+{B}^{H_{2}}(t)={\widetilde{B}}^{H_{1}}(t)+{B}^{H_{2}}(t),\hspace{0.2778em}t\in [0,T]$. As in [6], we apply to Z the linear transformation in order to reduce the construction to the sum with one term being the Wiener process. So, we take the kernel $l_{H}(t,s)={(t-s)}^{1/2-H}{s}^{1/2-H}$ and construct the integral
(2)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle Y(t)=& \displaystyle {\int _{0}^{t}}l_{H_{1}}(t,s)dZ(s)=\theta \mathrm{B}\bigg(\frac{3}{2}-H_{1},\frac{3}{2}-H_{1}\bigg){t}^{2-2H_{1}}+{M}^{H_{1}}(t)\\{} & \displaystyle +{\int _{0}^{t}}l_{H_{1}}(t,s)d{B}^{H_{2}}(s),\end{array}\]
where $\mathrm{B}(\alpha ,\beta )={\int _{0}^{1}}{x}^{\alpha -1}{(1-x)}^{\beta -1}dx$ is the beta function, and ${M}^{H_{1}}$ is a Gaussian martingale (Molchan martingale), admitting the representations
\[{M}^{H}(t)={\int _{0}^{t}}l_{H}(t,s)d{B}^{H}(s)=\gamma _{H}{\int _{0}^{t}}{s}^{1/2-H}dW(s)\]
with $\gamma _{H}={(2H(\frac{3}{2}-H)\varGamma {(3/2-H)}^{3}\varGamma (H+\frac{1}{2})\varGamma {(3-2H)}^{-1})}^{\frac{1}{2}}$ and a Wiener process W. According to [6], the linear transformation (2) is well defined, and the processes Z and Y are observed simultaneously. This means that we can reduce the original problem to the equivalent problem of the construction of maximum likelihood estimator of $\theta \in \mathbb{R}$ basing on the linear transformation Y. For simplicity, denote $\mathcal{B}_{H_{1}}:=\mathrm{B}(\frac{3}{2}-H_{1},\frac{3}{2}-H_{1})$. Now the main problem can be formulated as follows. Let $\frac{1}{2}\le H_{1}<H_{2}<1$,
\[\Bigg\{\widetilde{X}_{1}(t)={\widetilde{M}}^{H_{1}}(t),X_{2}(t):={\int _{0}^{t}}l_{H_{1}}(t,s)d{B}^{H_{2}}(s),\hspace{0.2778em}t\ge 0\Bigg\},\]
$i=1,2$, be a couple of processes defined on the space $(\varOmega ,\mathfrak{F})$, and $\mathsf{P}_{\theta }$ be a probability measure under which $\widetilde{X}_{1}$ and $X_{2}$ are independent, ${B}^{H_{2}}$ is a fractional Brownian motion with Hurst parameter $H_{2}$, and $\widetilde{X}_{1}$ is a martingale with square characteristics $\langle \widetilde{X}_{1}\rangle (t)=\frac{{\gamma _{H_{1}}^{2}}}{2-2H_{1}}{t}^{2-2H_{1}}$ and drift $\theta \mathcal{B}_{H_{1}}{t}^{2-2H_{1}}$, that is,
\[\widetilde{X}_{1}(t)={\widetilde{M}}^{H_{1}}(t)=\theta \mathcal{B}_{H_{1}}{t}^{2-2H_{1}}+{M}^{H_{1}}(t).\]
Also, denote $X_{1}(t)={M}^{H_{1}}(t)$. Our main problem is the construction of maximum likelihood estimator for $\theta \in \mathbb{R}$ by the observations of the process
\[Y(t)=\theta \mathcal{B}_{H_{1}}{t}^{2-2H_{1}}+X_{1}(t)+X_{2}(t)=\widetilde{X}_{1}(t)+X_{2}(t).\]
Note that, under the measure $\mathsf{P}_{\theta }$, the process
\[\widetilde{W}(t):=W(t)+\frac{\theta (2-2H_{1})\mathcal{B}_{H_{1}}}{\gamma _{H_{1}}(\frac{3}{2}-H_{1})}{t}^{\frac{3}{2}-H_{1}}\]
is a Wiener process with drift. Denote $\delta _{H_{1}}=\frac{(2-2H_{1})\mathcal{B}_{H_{1}}}{\gamma _{H_{1}}}$.
By Girsanov’s theorem and independence of $X_{1}$ and $X_{2}$,
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \frac{dP_{\theta }}{dP_{0}}& \displaystyle =\exp \Bigg\{\theta \delta _{H_{1}}{\int _{0}^{T}}{s}^{\frac{1}{2}-H_{1}}d\widetilde{W}(s)-\frac{{\theta }^{2}{\delta _{H_{1}}^{2}}}{4(1-H_{1})}{T}^{2-2H_{1}}\Bigg\}\\{} & \displaystyle =\exp \bigg\{\theta \delta _{H_{1}}\widetilde{X}_{1}(T)-\frac{{\theta }^{2}{\delta _{H_{1}}^{2}}}{4(1-H_{1})}{T}^{2-2H_{1}}\bigg\}.\end{array}\]
As it was mentioned in [3], the derivative of such a form is not the likelihood ratio for the problem at hand because it is not measurable with respect to the observed σ-algebra
\[{\mathfrak{F}_{T}^{Y}}:=\sigma \big\{Y(t),t\in [0,T]\big\}={\mathfrak{F}_{T}^{X}}:=\sigma \big\{X(t),t\in [0,T]\big\},\]
where $X(t)=X_{1}(t)+X_{2}(t)$.
We shall proceed as in [3]. Let $\mu _{\theta }$ be the probability measure induced by Y on the space of continuous functions with the supremum topology under probability $\mathsf{P}_{\theta }$. Then for any measurable set A, $\mu _{\theta }(A)=\int _{A}\varPhi (x)\mu _{0}(dx)$, where $\varPhi (x)$ is a measurable functional such that $\varPhi (X)=E_{0}(\frac{dP_{\theta }}{dP_{0}}|{\mathfrak{F}_{T}^{X}})$. This means that $\mu _{\theta }\ll \mu _{0}$ for any $\theta \in \mathbb{R}$. Taking into account that $\widetilde{X}_{1}=X_{1}$ under $\mathsf{P}_{0}$ and the fact that the vector process $(X_{1},X)$ is Gaussian, we get that the corresponding likelihood function is given by
\[\begin{array}{c}\displaystyle L_{T}(X,\theta )=\mathsf{E}_{0}\bigg(\frac{d\mathsf{P}_{\theta }}{d\mathsf{P}_{0}}|{\mathfrak{F}_{T}^{X}}\bigg)=\mathsf{E}_{0}\bigg(\exp \bigg\{\theta \delta _{H_{1}}X_{1}(T)-\frac{{\theta }^{2}{\delta _{H_{1}}^{2}}}{4(1-H_{1})}{T}^{2-2H_{1}}\bigg\}|{\mathfrak{F}_{T}^{X}}\bigg)\\{} \displaystyle =\exp \bigg\{\theta \delta _{H_{1}}\mathsf{E}_{0}\big(X_{1}(T)|{\mathfrak{F}_{T}^{X}}\big)+\frac{{\theta }^{2}{\delta _{H_{1}}^{2}}}{2}\bigg(V(T)-\frac{{T}^{2-2H_{1}}}{2-2H_{1}}\bigg)\bigg\},\end{array}\]
where $V(t)=\mathsf{E}_{0}{(X_{1}(t)-\mathsf{E}_{0}(X_{1}(t)|{\mathfrak{F}_{t}^{X}}))}^{2},\hspace{0.2778em}t\in [0,T]$.
The next reasonings repeat the corresponding part of [6]. We have to solve the following problem: to find the projection $P_{X}X_{1}(T)$ of $X_{1}(T)$ onto
\[\left\{X(t)=X_{1}(t)+X_{2}(t),\hspace{2.5pt}t\in [0,T]\right\}.\]
According to [4], the transformation formula for converting fBm into a Wiener process is of the form
\[W_{i}(t)={\int _{0}^{t}}\big({\big({K_{H_{i}}^{\ast }}\big)}^{-1}\mathbf{1}_{[0,t]}\big)(s)d{B}^{H_{i}}(s),\hspace{2.5pt}i=1,2,\]
where
\[\big({K_{H}^{\ast }}f\big)(s)={\int _{s}^{T}}f(t)\partial _{t}K_{H}(t,s)dt=\beta _{H}{s}^{1/2-H}{\int _{s}^{T}}f(t){t}^{H-1/2}{(t-s)}^{H-3/2}dt,\]
$\beta _{H}={(\frac{H(2H-1)}{\mathrm{B}(H-1/2,2-2H)})}^{\frac{1}{2}}$, and the square-integrable kernel $K_{H}(t,s)$ is of the form
\[K_{H}(t,s)=\beta _{H}{s}^{1/2-H}{\int _{s}^{t}}{(u-s)}^{H-3/2}{u}^{H-1/2}du.\]
We have that $W_{i},\hspace{2.5pt}i=1,2$, are standard Wiener processes, which are obviously independent. Also, we have
(3)
\[X_{1}(t)=\gamma _{H_{1}}{\int _{0}^{t}}{s}^{1/2-H_{1}}dW_{1}(s),\hspace{0.2778em}{B}^{H_{2}}(t)={\int _{0}^{t}}K_{H_{2}}(t,s)dW_{2}(s).\]
Then
\[X_{2}(t)={\int _{0}^{t}}K_{H_{1},H_{2}}(t,s)dW_{2}(s),\]
where
(4)
\[K_{H_{1},H_{2}}(t,s)=\beta _{H_{2}}{s}^{1/2-H_{2}}{\int _{s}^{t}}{(t-u)}^{1/2-H_{1}}{u}^{H_{2}-H_{1}}{(u-s)}^{H_{2}-3/2}du.\]
For an interval $[0,T]$, denote by ${L_{H}^{2}}[0,T]$ the completion of the space of simple functions $f:[0,T]\to \mathbb{R}$ with respect to the scalar product
\[{\left\langle f,g\right\rangle _{H}^{2}}:=\alpha _{H}{\int _{0}^{T}}{\int _{0}^{T}}f(t)g(s)|t-s{|}^{2H-2}dsdt,\]
where $\alpha _{H}=H(2H-1)$. Note that this space contains both functions and distributions. For functions from ${L_{H_{2}}^{2}}[0,T]$, we have that
\[{\int _{0}^{T}}f(s)dX_{2}(s)={\int _{0}^{T}}\big({K_{H_{1},H_{2}}^{\ast }}f\big)(s)dW_{2}(s),\]
where
\[\big({K_{H_{1},H_{2}}^{\ast }}f\big)(s)={\int _{s}^{T}}f(t)\partial _{t}K_{H_{1},H_{2}}(t,s)dt.\]
The projection of $X_{1}(T)$ onto $\{X(t),t\in [0,T]\}$ is a centered X-measurable Gaussian random variable and, therefore, is of the form
\[P_{X}X_{1}(T)={\int _{0}^{T}}h_{T}(t)dX(t)\]
with $h_{T}\in {L_{H_{1}}^{2}}[0,T]$. Note that $h_{T}$ still may be a distribution. However, as we will further see, it is a continuous function. The projection for all $u\in [0,T]$ must satisfy
(5)
\[\mathsf{E}\left[X(u)P_{X}X_{1}(T)\right]=\mathsf{E}\left[X(u)X_{1}(T)\right].\]
Using (5) together with independency of $X_{1}$ and $X_{2}$, we arrive at the equation
(6)
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle \mathsf{E}\left[X_{1}(u){\int _{0}^{T}}h_{T}(t)dX_{1}(t)+X_{2}(u){\int _{0}^{T}}h_{T}(t)dX_{2}(t)\right]\\{} & \displaystyle \hspace{1em}=\mathsf{E}\left[X_{1}(u)X_{1}(T)\right]=\varepsilon _{H_{1}}{u}^{2-2H_{1}},\end{array}\]
where $\varepsilon _{H}={\gamma _{H}^{2}}/(2-2H)$. Finally, from (3)–(6) we get the prototype of a Fredholm integral equation
(7)
\[\displaystyle \varepsilon _{H_{1}}{u}^{2-2H_{1}}={\gamma _{H_{1}}^{2}}{\int _{0}^{u}}h_{T}(s){s}^{1-2H_{1}}ds+{\int _{0}^{T}}h_{T}(s)r_{H_{1},H_{2}}(s,u)ds,\hspace{0.2778em}u\in [0,T],\]
where
\[r_{H_{1},H_{2}}(s,u)={\int _{0}^{s\wedge u}}\partial _{s}K_{H_{1},H_{2}}(s,v)K_{H_{1},H_{2}}(u,v)dv.\]
Differentiating (7), we get the Fredholm integral equation of the 2nd kind,
(8)
\[{\gamma _{H_{1}}^{2}}h_{T}(u){u}^{1-2H_{1}}+{\int _{0}^{T}}h_{T}(s)k(s,u)ds={\gamma _{H_{1}}^{2}}{u}^{1-2H_{1}},\hspace{1em}u\in (0,T],\]
where
(9)
\[k(s,u)={\int _{0}^{s\wedge u}}\partial _{s}K_{H_{1},H_{2}}(s,v)\partial _{u}K_{H_{1},H_{2}}(u,v)dv\]
with the function $K_{H_{1},H_{2}}$ defined by (4).
We will establish in Remark 2 that for the case $H_{1}=\frac{1}{2}$, Eq. (8) can be reduced to the corresponding equation from [3]:
(10)
\[h_{T}(u)+H_{2}(2H_{2}-1){\int _{0}^{T}}h_{T}(s)|s-u{|}^{2H_{2}-2}ds=1,\hspace{1em}u\in [0,T],\]
but the difference between (10) and (8) lies in the fact that (10) can be characterized as the equation with standard kernel, whereas (8) with two different power exponents is more or less nonstandard, and, therefore, it requires an unconventional approach. On the one hand, it is known from the paper [6] that if the conditions $H_{2}-H_{1}>\frac{1}{4}$ and $H_{1}>1/2$ are satisfied, then Eq. (8) has a unique solution $h_{T_{n}}$ with $h_{T_{n}}(t){t}^{\frac{1}{2}-H_{1}}\in L_{2}[0,T_{n}]$ on any sequence of intervals $[0,T_{n}]$ except, possibly, a countable number of $T_{n}$ connected to eigenvalues of the corresponding integral operator (the meaning of this sentence will be specified later because, finally, we will get a similar result but in more general situation). On the other hand, the existence–uniqueness result for Eq. (10) in [3] is proved without any restriction on Hurst index $H_{2}$ while $H_{1}=\frac{1}{2}$. The difference between these results can be explained so that in [3] the authors state the existence and uniqueness of the continuous solution, whereas in [6] the solution is established in the framework of $L_{2}$-theory.
In this paper, we propose to consider Eq. (8) in the space $C[0,T]$ again. This means that we consider the corresponding integral operator as an operator from $C[0,T]$ into $C[0,T]$ and establish an existence–uniqueness result in $C[0,T]$. This approach has the advantage that we do not need anymore the assumption $H_{2}-H_{1}>\frac{1}{4}$ and can include the case $H_{1}=1/2$ again into the consideration.
We say that two integral equations are equivalent if they have the same continuous solutions. In this sense, Eqs. (7) and (8) are equivalent, and both are equivalent to the equation
(11)
\[h_{T}(u)+\frac{1}{{\gamma _{H_{1}}^{2}}}{\int _{0}^{T}}h_{T}(s)\kappa (s,u)ds=1,\hspace{1em}u\in [0,T],\]
with continuous right-hand side, where
(12)
\[\kappa (s,u)={u}^{2H_{1}-1}k(s,u),\hspace{1em}s,u\in [0,T].\]
We get that the main problem (i.e., the MLE construction for the drift parameter) is reduced to the existence–uniqueness result for the integral equation (7).

3 Compactness of integral operator. Existence–uniqueness result for the Fredholm integral equation

Consider the integral operator K generated by the kernel K bearing in mind that the notations of the kernel and of the corresponding operator will always coincide:
\[(Kx)(u)={\int _{0}^{T}}K(s,u)x(s)ds,\hspace{1em}x\in C[0,T].\]
Now we are in position to establish the properties of the kernel $\kappa (s,u)$ defined by (12) and (9). Introduce the notation ${[0,T]_{0}^{2}}={[0,T]}^{2}\setminus \{(0,0)\}$.
Lemma 1.
Up to a set of Lebesgue measure zero, the kernel $\kappa (s,u)$, $s,u\in [0,T]$, admits the following representation on $[0,T]$:
(13)
\[\kappa (s,u)=\left\{\begin{array}{l@{\hskip10.0pt}l}\kappa _{0}(s,u)\varphi (s,u),\hspace{1em}& s\ne u,\\{} 0,\hspace{1em}& s=u,\end{array}\right.\]
where $\varphi (s,u)={(s\wedge u)}^{1-2H_{1}}{u}^{2H_{1}-1}|s-u{|}^{2H_{2}-2H_{1}-1}$, and the function $\kappa _{0}$ is bounded and belongs to $C({[0,T]_{0}^{2}})$.
Proof.
We take (9) and first present the derivative of $K_{H_{1},H_{2}}(t,s)$, defined by (4), in an appropriate form. To start, put $u=s+(t-s)z$. This allows us to rewrite $K_{H_{1},H_{2}}(t,s)$ as
(14)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle K_{H_{1},H_{2}}(t,s)& \displaystyle =\beta _{H_{2}}{s}^{\frac{1}{2}-H_{2}}{(t-s)}^{H_{2}-H_{1}}\\{} & \displaystyle \hspace{1em}\times {\int _{0}^{1}}{(1-z)}^{\frac{1}{2}-H_{1}}{\big(s+(t-s)z\big)}^{H_{2}-H_{1}}{z}^{H_{2}-\frac{3}{2}}dz.\end{array}\]
Differentiating (14) w.r.t. t for $0<s<t\le T$, we get
(15)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \partial _{t}K_{H_{1},H_{2}}(t,s)& \displaystyle =(H_{2}-H_{1})\beta _{H_{2}}{s}^{\frac{1}{2}-H_{2}}{(t-s)}^{H_{2}-H_{1}-1}\\{} & \displaystyle \hspace{1em}\times {\int _{0}^{1}}{(1-z)}^{\frac{1}{2}-H_{1}}{\big(s+(t-s)z\big)}^{H_{2}-H_{1}}{z}^{H_{2}-\frac{3}{2}}dz\\{} & \displaystyle \hspace{1em}+(H_{2}-H_{1})\beta _{H_{2}}{s}^{\frac{1}{2}-H_{2}}\\{} & \displaystyle \hspace{1em}\times {(t-s)}^{H_{2}-H_{1}}{\int _{0}^{1}}{(1-z)}^{\frac{1}{2}-H_{1}}{\big(s+(t-s)z\big)}^{H_{2}-H_{1}-1}{z}^{H_{2}-\frac{1}{2}}dz\\{} & \displaystyle =(H_{2}-H_{1})\beta _{H_{2}}{s}^{\frac{1}{2}-H_{2}}{(t-s)}^{H_{2}-H_{1}-1}\\{} & \displaystyle \hspace{1em}\times \Bigg({\int _{0}^{1}}{(1-z)}^{\frac{1}{2}-H_{1}}{\big(s+(t-s)z\big)}^{H_{2}-H_{1}}{z}^{H_{2}-\frac{3}{2}}dz\\{} & \displaystyle \hspace{1em}+(t-s){\int _{0}^{1}}{(1-z)}^{\frac{1}{2}-H_{1}}{\big(s+(t-s)z\big)}^{H_{2}-H_{1}-1}{z}^{H_{2}-\frac{1}{2}}dz\Bigg)\\{} & \displaystyle =(H_{2}-H_{1})\beta _{H_{2}}{s}^{\frac{1}{2}-H_{2}}{(t-s)}^{H_{2}-H_{1}-1}\\{} & \displaystyle \hspace{1em}\times \Bigg({s}^{H_{2}-H_{1}}{\int _{0}^{1}}{z}^{H_{2}-\frac{3}{2}}{(1-z)}^{\frac{1}{2}-H_{1}}{\bigg(1-\frac{s-t}{s}z\bigg)}^{H_{2}-H_{1}}dz\\{} & \displaystyle \hspace{1em}\hspace{0.1667em}+\hspace{0.1667em}(t\hspace{0.1667em}-\hspace{0.1667em}s){s}^{H_{2}-H_{1}-1}\hspace{-0.1667em}\hspace{-0.1667em}{\int _{0}^{1}}\hspace{-0.1667em}{(1\hspace{0.1667em}-\hspace{0.1667em}z)}^{\frac{1}{2}-H_{1}}{\bigg(1\hspace{0.1667em}-\hspace{0.1667em}\frac{s-t}{s}z\bigg)}^{H_{2}-H_{1}-1}\hspace{-0.1667em}{z}^{H_{2}-\frac{1}{2}}dz\Bigg)\hspace{-0.1667em}.\end{array}\]
Denote for technical simplicity $\alpha _{i}=H_{i}-\frac{1}{2},\hspace{2.5pt}i=1,2$. Then, according to the definition and properties of the Gauss hypergeometric function (see Eqs. (31) and (32)), the terms in the right-hand side of (15) can be rewritten as follows. For the first term, thats is, for
(16)
\[I_{1}(t,s):={s}^{H_{2}-H_{1}}{\int _{0}^{1}}{z}^{\alpha _{2}-1}{(1-z)}^{-\alpha _{1}}{\bigg(1-\frac{s-t}{s}z\bigg)}^{H_{2}-H_{1}}dz,\]
the values of parameters for the underlying integral equal $a=H_{1}-H_{2}$, $b=\alpha _{2}$, $c=H_{2}-H_{1}+1$, and $x=\frac{s-t}{s}<1$, respectively; therefore, $\frac{x}{x-1}=\frac{t-s}{t}$, $c-b=1-\alpha _{1}$, and
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I_{1}(t,s)& \displaystyle =\mathrm{B}(1-\alpha _{1},\alpha _{2}){s}^{H_{2}-H_{1}}F\bigg(H_{1}-H_{2},\alpha _{2},1-H_{1}+H_{2};\frac{s-t}{s}\bigg)\\{} & \displaystyle =\mathrm{B}(1\hspace{0.1667em}-\hspace{0.1667em}\alpha _{1},\alpha _{2}){s}^{H_{2}-H_{1}}{\bigg(\frac{t}{s}\bigg)}^{H_{2}-H_{1}}\hspace{-0.1667em}F\bigg(\hspace{-0.1667em}H_{1}\hspace{0.1667em}-\hspace{0.1667em}H_{2},1\hspace{0.1667em}-\hspace{0.1667em}\alpha _{1},1\hspace{0.1667em}-\hspace{0.1667em}H_{1}\hspace{0.1667em}+\hspace{0.1667em}H_{2};\frac{t\hspace{0.1667em}-\hspace{0.1667em}s}{t}\hspace{-0.1667em}\bigg)\\{} & \displaystyle =\mathrm{B}(1-\alpha _{1},\alpha _{2}){t}^{H_{2}-H_{1}}F\bigg(H_{1}-H_{2},1-\alpha _{1},1-H_{1}+H_{2};\frac{t-s}{t}\bigg).\end{array}\]
Similarly, for the second term, that is, for
(17)
\[I_{2}(t,s):=(t-s){s}^{H_{2}-H_{1}-1}{\int _{0}^{1}}{z}^{\alpha _{2}}{(1-z)}^{-\alpha _{1}}{\bigg(1-\frac{s-t}{s}z\bigg)}^{H_{2}-H_{1}-1}dz,\]
the values of parameters for the underlying integral equal $a=H_{1}-H_{2}+1$, $b=\alpha _{2}+1$, $c=H_{2}-H_{1}+2$, and $x=\frac{s-t}{s}$, respectively; therefore, $\frac{x}{x-1}=\frac{t-s}{t}$, $c-b=1-\alpha _{1}$, and
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I_{2}(t,s)& \displaystyle =(t-s){s}^{H_{2}-H_{1}-1}\mathrm{B}(1-\alpha _{1},\alpha _{2}+1)\\{} & \displaystyle \hspace{1em}\times F\bigg(H_{1}-H_{2}+1,\alpha _{2}+1,H_{2}-H_{1}+2;\frac{s-t}{s}\bigg)\\{} & \displaystyle =(t-s){s}^{H_{2}-H_{1}-1}{\bigg(\frac{t}{s}\bigg)}^{H_{2}-H_{1}-1}\\{} & \displaystyle \hspace{1em}\times B(1-\alpha _{1},\alpha _{2}+1)F\bigg(H_{1}-H_{2}+1,1-\alpha _{1},2-H_{1}+H_{2};\frac{t-s}{t}\bigg)\\{} & \displaystyle =(t-s){t}^{H_{2}-H_{1}-1}\mathrm{B}(1-\alpha _{1},\alpha _{2}+1)\\{} & \displaystyle \hspace{1em}\times F\bigg(H_{1}-H_{2}+1,1-\alpha _{1},2-H_{1}+H_{2};\frac{t-s}{t}\bigg).\end{array}\]
It is easy to see from the initial representations (16) and (17) that $I_{1}(t,s)$ and $I_{2}(t,s)$ are continuous on the set $0<s\le t\le T$.
Now, introduce the notations
\[\varPsi _{1}(t,s)=\mathrm{B}(1-\alpha _{1},\alpha _{2})F\bigg(H_{1}-H_{2},1-\alpha _{1},1-H_{1}+H_{2};\frac{t-s}{t}\bigg)\]
and
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \varPsi _{2}(t,s)& \displaystyle ={\bigg(\frac{t-s}{t}\bigg)}^{1-H_{2}+H_{1}}\mathrm{B}(1-\alpha _{1},\alpha _{2}+1)\\{} & \displaystyle \hspace{1em}\times F\bigg(H_{1}-H_{2}+1,1-\alpha _{1},2-H_{1}+H_{2};\frac{t-s}{t}\bigg),\end{array}\]
so that $I_{1}(t,s)={t}^{H_{2}-H_{1}}\varPsi _{1}(t,s)$ and $I_{2}(t,s)={(t-s)}^{H_{2}-H_{1}}\varPsi _{2}(t,s)$. Note that $\frac{t-s}{t}\in [0,1)$; therefore,
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle F\bigg(H_{1}-H_{2},1-\alpha _{1},1-H_{1}+H_{2};\frac{t-s}{t}\bigg)\\{} & \displaystyle \hspace{1em}=\frac{1}{\mathrm{B}(1-\alpha _{1},\alpha _{2})}\times {\int _{0}^{1}}{z}^{-\alpha _{1}}{(1-z)}^{\alpha _{2}-1}{\bigg(1-\frac{t-s}{t}z\bigg)}^{H_{2}-H_{1}}dz\\{} & \displaystyle \hspace{2em}\le \frac{1}{\mathrm{B}(1-\alpha _{1},\alpha _{2})}{\int _{0}^{1}}{z}^{-\alpha _{1}}{(1-z)}^{\alpha _{2}-1}dz=1,\end{array}\]
whence the function $\varPsi _{1}(t,s)$ is bounded by $\mathrm{B}(1-\alpha _{1},\alpha _{2})$. In order to establish that $\varPsi _{2}(t,s)$ is bounded, we use Proposition 1. Its conditions are satisfied: $a=H_{1}-H_{2}+1\in (0,1)$, $b=1-\alpha _{1}>0$, $c-b=\alpha _{2}+1>1$, and $x=\frac{t-s}{t}\in [0,1)$. Therefore,
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle {x}^{1-H_{2}+H_{1}}F(H_{1}-H_{2}+1,1-\alpha _{1},2-H_{1}+H_{2};x)\le {x}^{1-H_{2}+H_{1}}\\{} & \displaystyle \hspace{1em}\times {\bigg(1-\frac{1-\alpha _{1}}{1-H_{1}+H_{2}}x\bigg)}^{-1-H_{1}+H_{2}}={\bigg(\frac{1}{x}-\frac{1-\alpha _{1}}{1-H_{1}+H_{2}}\bigg)}^{-1-H_{1}+H_{2}}\\{} & \displaystyle \hspace{1em}\le {\bigg(1-\frac{1-\alpha _{1}}{1-H_{1}+H_{2}}\bigg)}^{-1-H_{1}+H_{2}}={\bigg(\frac{1-H_{1}+H_{2}}{\alpha _{2}}\bigg)}^{H_{1}-H_{2}+1},\end{array}\]
whence $\varPsi _{2}(t,s)\le \mathrm{B}(1-\alpha _{1},\alpha _{2}+1){(\frac{1-H_{1}+H_{2}}{\alpha _{2}})}^{H_{1}-H_{2}+1}$. Additionally, both functions are homogeneous:
\[\varPsi _{i}(at,as)=\varPsi _{i}(t,s)\hspace{0.2778em}\text{for}\hspace{0.2778em}a>0,\hspace{2.5pt}i=1,2.\]
Introduce the notation
(18)
\[\varPhi (t,s)=I_{1}(t,s)+I_{2}(t,s)={t}^{H_{2}-H_{1}}\varPsi _{1}(t,s)+{(t-s)}^{H_{2}-H_{1}}\varPsi _{2}(t,s)\]
and note that $\varPhi \in C({[0,T]_{0}^{2}})$ is bounded and homogeneous:
(19)
\[\varPhi (at,as)={a}^{H_{2}-H_{1}}\varPhi (t,s),a>0.\]
In terms of notation (18), the representation (15) for $\partial _{t}K_{H_{1},H_{2}}(t,s)$ can be rewritten as
(20)
\[\partial _{t}K_{H_{1},H_{2}}(t,s)=\beta _{H_{2}}(H_{2}-H_{1}){s}^{\frac{1}{2}-H_{2}}{(t-s)}^{H_{2}-H_{1}-1}\varPhi (t,s).\]
In turn, the kernel $k(s,u)$ from (9) can be rewritten as
(21)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle k(s,u)& \displaystyle ={\big(\beta _{H_{2}}(H_{2}-H_{1})\big)}^{2}\\{} & \displaystyle \hspace{1em}\times {\int _{0}^{s\wedge u}}{v}^{1-2H_{2}}{(s-v)}^{H_{2}-H_{1}-1}{(u-v)}^{H_{2}-H_{1}-1}\varPhi (s,v)\varPhi (u,v)dv.\end{array}\]
Consider the kernel $k(s,u)$ for $s>u$. Then it evidently equals
\[\begin{array}{r@{\hskip0pt}l}\displaystyle k(s,u)& \displaystyle ={\big(\beta _{H_{2}}(H_{2}-H_{1})\big)}^{2}\\{} & \displaystyle \hspace{1em}\times {\int _{0}^{u}}{v}^{1-2H_{2}}{(s-v)}^{H_{2}-H_{1}-1}{(u-v)}^{H_{2}-H_{1}-1}\varPhi (s,v)\varPhi (u,v)dv.\end{array}\]
Put $z=\frac{u-v}{s-u}$ and transform $k(s,u)$ to
\[\begin{array}{r@{\hskip0pt}l}\displaystyle k(s,u)& \displaystyle ={\big(\beta _{H_{2}}(H_{2}-H_{1})\big)}^{2}{(s-u)}^{2H_{2}-2H_{1}-1}{\int _{0}^{\frac{u}{s-u}}}{z}^{H_{2}-H_{1}-1}{(1+z)}^{H_{2}-H_{1}-1}\\{} & \displaystyle \hspace{1em}\times {\big(u-z(s-u)\big)}^{1-2H_{2}}\varPhi \big(s,u-z(s-u)\big)\varPhi \big(u,u-z(s-u)\big)dz\\{} & \displaystyle =:\frac{k_{0}(s,u)}{{(s-u)}^{1-2H_{2}+2H_{1}}},\end{array}\]
where
\[\begin{array}{r@{\hskip0pt}l}\displaystyle k_{0}(s,u)& \displaystyle ={\big(\beta _{H_{2}}(H_{2}-H_{1})\big)}^{2}{\int _{0}^{\frac{u}{s-u}}}{z}^{H_{2}-H_{1}-1}{(1+z)}^{H_{2}-H_{1}-1}\\{} & \displaystyle \hspace{1em}\times {\big(u-z(s-u)\big)}^{1-2H_{2}}\varPhi \big(s,u-z(s-u)\big)\varPhi \big(u,u-z(s-u)\big)dz.\end{array}\]
In turn, transform $k_{0}(s,u)$ with the change of variables $tu=z$ and apply (19):
(22)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle k_{0}(s,u)& \displaystyle ={\big(\beta _{H_{2}}(H_{2}-H_{1})\big)}^{2}{\int _{0}^{\frac{1}{s-u}}}{(tu)}^{H_{2}-H_{1}-1}{(1+tu)}^{H_{2}-H_{1}-1}\\{} & \displaystyle \hspace{1em}\times {\big(u-tu(s-u)\big)}^{1-2H_{2}}\varPhi \big(s,u-tu(s-u)\big)\varPhi \big(u,u-tu(s-u)\big)udt\\{} & \displaystyle ={\big(\beta _{H_{2}}(H_{2}-H_{1})\big)}^{2}{u}^{1-2H_{1}}{\int _{0}^{\frac{1}{s-u}}}{\big(1-t(s-u)\big)}^{1-2H_{2}}{(1+tu)}^{H_{2}-H_{1}-1}\\{} & \displaystyle \hspace{1em}\times {t}^{H_{2}-H_{1}-1}\varPhi \big(s,u-tu(s-u)\big)\varPhi \big(1,1-t(s-u)\big)dt.\end{array}\]
Introducing the kernel $\kappa _{0}(s,u)=k_{0}(s,u){u}^{2H_{1}-1}$, we can present $k(s,u)$ as
(23)
\[k(s,u)=\frac{\kappa _{0}(s,u)}{{(s-u)}^{1-2H_{2}+2H_{1}}{u}^{2H_{1}-1}},\]
where, for $s>u>0$,
(24)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \kappa _{0}(s,u)& \displaystyle ={\big(\beta _{H_{2}}(H_{2}-H_{1})\big)}^{2}{\int _{0}^{\frac{1}{s-u}}}{\big(1-(s-u)t\big)}^{1-2H_{2}}{(1+ut)}^{H_{2}-H_{1}-1}\\{} & \displaystyle \hspace{1em}\times {t}^{H_{2}-H_{1}-1}\varPhi \big(s,u-tu(s-u)\big)\varPhi \big(1,1-t(s-u)\big)dt\\{} & \displaystyle ={\big(\beta _{H_{2}}(H_{2}-H_{1})\big)}^{2}{\int _{0}^{\infty }}1_{t\le \frac{1}{s-u}}{\big(1-(s-u)t\big)}^{1-2H_{2}}{(1+ut)}^{H_{2}-H_{1}-1}\\{} & \displaystyle \hspace{1em}\times {t}^{H_{2}-H_{1}-1}\varPhi \big(s,u-tu(s-u)\big)\varPhi \big(1,1-t(s-u)\big)dt.\end{array}\]
For the case $u>s>0$, we can replace s and u in formulas (23) and (24). Substituting formally $u=s$ into (24), for $s>0$, we get
(25)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \kappa _{0}(s,s)& \displaystyle ={\big(\beta _{H_{2}}(H_{2}-H_{1})\big)}^{2}\varPhi (s,s)\varPhi (1,1){\int _{0}^{\infty }}{(1+st)}^{H_{2}-H_{1}-1}{t}^{H_{2}-H_{1}-1}dt\\{} & \displaystyle ={\big(\beta _{H_{2}}(H_{2}-H_{1})\big)}^{2}{s}^{H_{2}-H_{1}}\varPhi {(1,1)}^{2}{\int _{0}^{\infty }}{(1+st)}^{H_{2}-H_{1}-1}{t}^{H_{2}-H_{1}-1}dt.\end{array}\]
Note that $\varPhi (1,1)=\mathrm{B}(1-\alpha _{1},\alpha _{2})$ and ${\int _{0}^{\infty }}{(1+st)}^{H_{2}-H_{1}-1}{t}^{H_{2}-H_{1}-1}dt={s}^{H_{1}-H_{2}}\mathrm{B}(H_{2}-H_{1},1-2H_{2}+H_{1})$. The former equation holds due to (34). We get that $\kappa _{0}(s,s)$ does not depend on s and equals some constant $C_{H}:={(\beta _{H_{2}}(H_{2}-H_{1})\mathrm{B}(1-\alpha _{1},\alpha _{2}))}^{2}\mathrm{B}(H_{2}-H_{1},1-2H_{2}+H_{1})$. Therefore, we define $\kappa _{0}(s,s)=C_{H}$, $s>0$.
Now the continuity of $\kappa _{0}$ on ${(0,T]}^{2}$ follows from the Lebesgue dominated convergence theorem supplied by representation (24), Eq. (25), and its consequence $\kappa _{0}(s,s)=C_{H}$, $s>0$, together with the facts that $\varPhi \in C({[0,T]_{0}^{2}})$ and is bounded. Consider $\kappa _{0}(s,u)$ for $u\downarrow 0$ and let $s>0$ be fixed. Then
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \underset{u\downarrow 0}{\lim }\kappa _{0}(s,u)& \displaystyle ={C_{H}^{1}}:={\big(\beta _{H_{2}}(H_{2}-H_{1})\big)}^{2}\varPhi (1,0)\\{} & \displaystyle \hspace{1em}\times {\int _{0}^{1}}{(1-y)}^{1-2H_{2}}{y}^{H_{2}-H_{1}-1}\varPhi (1,1-y)dy<\infty ,\end{array}\]
and we can put $\kappa _{0}(s,0)=\kappa _{0}(0,u)={C_{H}^{1}}$, $s>0$, $u>0$, thus extending the continuity of $\kappa _{0}$ to ${[0,T]_{0}^{2}}$.
It is easy to see that the values $\kappa _{0}(s,s)$ and $\kappa _{0}(s,0)$ do not depend on $s>0$ and do not coincide: $C_{H}\ne {C_{H}^{1}}$. Consequently, the limit
\[\underset{(s,u)\to (0,0)}{\lim }\kappa _{0}(s,u)\]
does not exist and depends on the way the variables s and u tend to zero. We can equate $\kappa _{0}(0,0)$ to any constant; for example, let $\kappa _{0}(0,0)=0$.
In order to prove that $\kappa _{0}$ is bounded, we consider the case $s>u$ (the opposite case is treated similarly) and put $z=(s-u)t$. Then
(26)
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle {\int _{0}^{\frac{1}{s-u}}}{\big(1-(s-u)t\big)}^{1-2H_{2}}{(1+ut)}^{H_{2}-H_{1}-1}{t}^{H_{2}-H_{1}-1}\varPhi \big(s,u-tu(s-u)\big)\\{} & \displaystyle \hspace{1em}\times \varPhi \big(1,1-t(s-u)\big)dt=\frac{1}{{(s-u)}^{H_{2}-H_{1}}}{\int _{0}^{1}}{(1-z)}^{1-2H_{2}}\\{} & \displaystyle \hspace{1em}\times {\bigg(1+\frac{u}{s-u}z\bigg)}^{H_{2}-H_{1}-1}{z}^{H_{2}-H_{1}-1}\varPhi \big(s,u(1-z)\big))\varPhi (1,1-z)dz=:I_{3}(s,u).\end{array}\]
It follows from (19) that, for $s\ne 0$,
\[\varPhi \big(s,u(1-z)\big)={s}^{H_{2}-H_{1}}\varPhi \bigg(1,\frac{u}{s}(1-z)\bigg).\]
Denote $r=\frac{s}{s-u}$ and put $t=\frac{1-z}{1-(1-r)z}$. Then
\[\frac{u}{s-u}=r-1,\hspace{0.2778em}t<1,\hspace{0.2778em}z=\frac{1-t}{1-t(1-r)}\in (0,1),\]
and the right-hand side of (26) can be rewritten as
(27)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I_{3}(s,u)& \displaystyle ={r}^{H_{2}-H_{1}}{\int _{0}^{1}}{(1-z)}^{1-2H_{2}}{\big(1-(1-r)z\big)}^{H_{2}-H_{1}-1}{z}^{H_{2}-H_{1}-1}\\{} & \displaystyle \hspace{1em}\times \varPhi \bigg(1,\frac{u}{s}(1-z)\bigg)\varPhi (1,1-z)dz={r}^{1-2H_{1}}{\int _{0}^{1}}{t}^{1-2H_{2}}{(1-t)}^{H_{2}-H_{1}-1}\\{} & \displaystyle \hspace{1em}\times {\big(1-(1-r)t\big)}^{2H_{1}-1}\varPhi \bigg(1,\frac{u}{s}\frac{rt}{1-(1-r)t}\bigg)\varPhi \bigg(1,\frac{rt}{1-(1-r)t}\bigg)dt.\end{array}\]
Finally, put $y=1-t$. Then the right-hand side of (27) is transformed to
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I_{3}(s,u)& \displaystyle ={r}^{1-2H_{1}}{r}^{2H_{1}-1}{\int _{0}^{1}}{(1-y)}^{1-2H_{2}}{y}^{H_{2}-H_{1}-1}{\bigg(1-y\frac{r-1}{r}\bigg)}^{2H_{1}-1}\\{} & \displaystyle \hspace{1em}\times \varPhi \bigg(1,\frac{u}{s}\frac{r(1-y)}{r-y(r-1)}\bigg)\varPhi \bigg(1,\frac{r(1-y)}{r-y(r-1)}\bigg)dy.\end{array}\]
Recall that $r=\frac{s}{s-u}$. Then it follows from the boundedness of Φ that there exists a constant ${C_{H}^{1}}$ such that, for $s>u$,
(28)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \kappa _{0}(s,u)& \displaystyle ={\big(\beta _{H_{2}}(H_{2}-H_{1})\big)}^{2}{\int _{0}^{1}}{(1-y)}^{1-2H_{2}}{\bigg(1-\frac{u}{s}y\bigg)}^{2H_{1}-1}{y}^{H_{2}-H_{1}-1}\\{} & \displaystyle \hspace{1em}\times \varPhi \bigg(1,\frac{u(1-y)}{s-uy}\bigg)\varPhi \bigg(1,\frac{s(1-y)}{s-uy}\bigg)dy\\{} & \displaystyle \hspace{1em}\le {C_{H}^{1}}{\int _{0}^{1}}{(1-y)}^{1-2H_{2}}{y}^{H_{2}-H_{1}-1}dy,\end{array}\]
so $\kappa _{0}$ is bounded, and the lemma is proved.  □
Remark 1.
Figure 1 demonstrates the graph of $\kappa _{0}(s,u)$ for $H_{1}=0.7$ and $H_{2}=0.9$.
vmsta-2-2-vmsta28-g001.jpg
Fig. 1.
Function $\kappa _{0}(s,u)$
Now, consider the properties of the function
\[\varphi (s,u)={(s\wedge u)}^{1-2H_{1}}{u}^{2H_{1}-1}|s-u{|}^{2H_{2}-2H_{1}-1}\]
participating in the kernel representation (13).
Lemma 2.
The function φ has the following properties:
  • (i) for any $u\in [0,T]$, $\varphi (\cdot ,u)\in L_{1}[0,T]$ and $\underset{u\in [0,T]}{\sup }\| \varphi (\cdot ,u)\| _{L_{1}}<\infty $
  • (ii) for any $u_{1}\in [0,T]$, ${\int _{0}^{T}}|\varphi (s,u)-\varphi (s,u_{1})|ds\to 0$ as $u\to u_{1}$.
Proof.
(i) It follows from the evident calculations that
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle {\int _{0}^{T}}|\varphi (s,u)|ds={\int _{0}^{T}}\varphi (s,u)ds={\int _{0}^{u}}\frac{{u}^{2H_{1}-1}ds}{{s}^{2H_{1}-1}{(u-s)}^{1+2H_{1}-2H_{2}}}\\{} & \displaystyle \hspace{1em}+{\int _{u}^{T}}\frac{ds}{{(s-u)}^{1+2H_{1}-2H_{2}}}={u}^{2H_{2}-2H_{1}}\mathrm{B}(2-2H_{1},2H_{2}-2H_{1})\\{} & \displaystyle \hspace{1em}+\frac{{(T-u)}^{2H_{2}-2H_{1}}}{2H_{2}-2H_{1}}\le C_{H_{1},H_{2}}{T}^{2H_{2}-2H_{1}}<\infty \hspace{2.5pt}\text{for all}\hspace{2.5pt}u\in [0,T].\end{array}\]
(ii) First, let $u_{1}=0$ and $u\downarrow 0$. Note that $\varphi (s,0)={s}^{2H_{2}-2H_{1}-1}$. Therefore,
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle {\int _{0}^{T}}|\varphi (s,u)-\frac{1}{{s}^{1+2H_{1}-2H_{2}}}|ds={\int _{0}^{u}}|\frac{{u}^{2H_{1}-1}}{{s}^{2H_{1}-1}{(u-s)}^{1+2H_{1}-2H_{2}}}-\frac{1}{{s}^{1+2H_{1}-2H_{2}}}|ds\\{} & \displaystyle \hspace{2em}\hspace{0.1667em}+{\int _{u}^{T}}\hspace{-0.1667em}\frac{ds}{{(s-u)}^{1+2H_{1}-2H_{2}}}\hspace{0.1667em}-{\int _{u}^{T}}\hspace{-0.1667em}\frac{ds}{{s}^{1+2H_{1}-2H_{2}}}ds\hspace{0.1667em}\le {\int _{0}^{u}}\hspace{-0.1667em}\frac{{u}^{2H_{1}-1}ds}{{s}^{2H_{1}-1}{(u-s)}^{1+2H_{1}-2H_{2}}}\\{} & \displaystyle \hspace{2em}+{\int _{0}^{u}}\frac{ds}{{s}^{1+2H_{1}-2H_{2}}}+\frac{1}{2H_{2}-2H_{1}}\big({(s-u)}^{2H_{2}-2H_{1}}-{s}^{2H_{2}-2H_{1}}\big){|_{s=u}^{s=T}}\\{} & \displaystyle \hspace{1em}=\mathrm{B}(2-2H_{1},2H_{2}-2H_{1}){u}^{2H_{2}-2H_{1}}\\{} & \displaystyle \hspace{2em}+\frac{1}{2H_{2}-2H_{1}}\big(2{u}^{2H_{2}-2H_{1}}+{(T-u)}^{2H_{2}-2H_{1}}-{T}^{2H_{2}-2H_{1}}\big)\to 0,\hspace{2.5pt}\text{as}\hspace{2.5pt}u\to 0.\end{array}\]
From now on suppose that $u_{1}>0$ is fixed. Without loss of generality, suppose that $u\uparrow u_{1}$. Then
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle \underset{0}{\overset{T}{\int }}|\varphi (s,u)-\varphi (s,u_{1})|ds=\underset{0}{\overset{u}{\int }}|\varphi (s,u)-\varphi (s,u_{1})|ds+\underset{u}{\overset{u_{1}}{\int }}|\varphi (s,u)-\varphi (s,u_{1})|ds\\{} & \displaystyle \hspace{1em}+{\int _{u_{1}}^{T}}|\varphi (s,u)-\varphi (s,u_{1})|ds=:I_{1}(u,u_{1})+I_{2}(u,u_{1})+I_{3}(u,u_{1}).\end{array}\]
Consider the terms separately. First, we establish that $\varphi (s,\cdot )$ is decreasing in the second argument. Indeed, for $0<s<u<u_{1}$,
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \varphi (s,u_{1})& \displaystyle =\frac{{u_{1}^{2H_{1}-1}}}{{s}^{2H_{1}-1}{(u_{1}-s)}^{1+2H_{1}-2H_{2}}}=\frac{1}{{s}^{2H_{1}-1}{(1-\frac{s}{u_{1}})}^{1+2H_{1}-2H_{2}}{u_{1}^{2-2H_{2}}}}\\{} & \displaystyle \hspace{1em}\le \frac{1}{{s}^{2H_{1}-1}{(1-\frac{s}{u})}^{1+2H_{1}-2H_{2}}{u}^{2-2H_{2}}}=\varphi (s,u).\end{array}\]
Therefore,
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I_{1}(u,u_{1})& \displaystyle ={\int _{0}^{u}}\big(\varphi (s,u)-\varphi (s,u_{1})\big)ds={\int _{0}^{u}}\varphi (s,u)ds-{\int _{0}^{u_{1}}}\varphi (s,u_{1})ds\\{} & \displaystyle \hspace{1em}+\underset{u}{\overset{u_{1}}{\int }}\varphi (s,u_{1})ds\le \mathrm{B}(2-2H_{1},2H_{2}-2H_{1})\big({u}^{2H_{2}-2H_{1}}-{u_{1}^{2H_{2}-2H_{1}}}\big)\\{} & \displaystyle \hspace{1em}+\frac{{u_{1}^{2H_{1}-1}}{(u_{1}-u)}^{2H_{2}-2H_{1}}}{2H_{2}-2H_{1}}\to 0,\hspace{2.5pt}\text{as}\hspace{2.5pt}u\uparrow u_{1}.\end{array}\]
The second integral vanishes as well:
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I_{2}(u,u_{1})& \displaystyle \le \underset{u}{\overset{u_{1}}{\int }}\varphi (s,u)ds+\underset{u}{\overset{u_{1}}{\int }}\varphi (s,u_{1})ds\\{} & \displaystyle \le \bigg(\frac{1}{2H_{2}-2H_{1}}+{\bigg(\frac{u_{1}}{u}\bigg)}^{2H_{1}-1}\bigg){(u_{1}-u)}^{2H_{2}-2H_{1}}\to 0\end{array}\]
as $u\uparrow u_{1}$. Finally,
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I_{3}(u,u_{1})& \displaystyle ={\int _{u_{1}}^{T}}\frac{ds}{{(s-u_{1})}^{1+2H_{1}-2H_{2}}}-{\int _{u_{1}}^{T}}\frac{ds}{{(s-u)}^{1+2H_{1}-2H_{2}}}=\frac{1}{2H_{2}-2H_{1}}\\{} & \displaystyle \hspace{1em}\times \big({(T-u_{1})}^{2H_{2}-2H_{1}}-{(T-u)}^{2H_{2}-2H_{1}}+{(u_{1}-u)}^{2H_{2}-2H_{1}}\big)\to 0\\{} & \displaystyle \hspace{1em}\hspace{2em}\hspace{2em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{2.5pt}\text{as}\hspace{2.5pt}u\uparrow u_{1}.\end{array}\]
The lemma is proved.  □
Lemma 3.
The kernel κ generates a compact integral operator $\kappa :C[0,T]\to C[0,T]$.
Proof.
According to [2], it suffices to prove that the kernel κ defined by (13) satisfies the following two conditions:
  • (iii) for any $u\in [0,T]$, $\kappa (\cdot ,u)\in L_{1}[0,T]$ and $\underset{u\in [0,T]}{\sup }\| \kappa (\cdot ,u)\| _{L_{1}}<\infty $;
  • (iv) For any $u_{1}\in [0,T]$, ${\int _{0}^{T}}|\kappa (s,u)-\kappa (s,u_{1})|ds\to 0$ as $u\to u_{1}$.
The first condition follows directly from fact that $\kappa _{0}(s,u)$ is bounded (see Lemma 1) and from Lemma 2 (i).
In order to check (iv), consider
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle {\int _{0}^{T}}|\kappa (s,u)-\kappa (s,u_{1})|ds={\int _{0}^{T}}|\kappa _{0}(s,u)\varphi (s,u)-\kappa _{0}(s,u_{1})\varphi (s,u_{1})|ds\\{} & \displaystyle \hspace{1em}\le {\int _{0}^{T}}\kappa _{0}(s,u)|\varphi (s,u)-\varphi (s,u_{1})|ds+{\int _{0}^{T}}\varphi (s,u_{1})|\kappa _{0}(s,u)-\kappa _{0}(s,u_{1})|ds.\end{array}\]
Again, Lemma 1 in the part that states that $\kappa _{0}(s,u)$ is bounded, together with Lemma 2 (ii), guarantees that the first term converges to zero as $u\to u_{1}$. Furthermore, Lemma 1 in the part that states that $\kappa _{0}\in C({[0,T]_{0}^{2}})$ guarantees that $\kappa _{0}(s,u)$ converges to $\kappa _{0}(s,u_{1})$ as $u\to u_{1}$ for a.e. $s\in [0,T]$. Since
\[\varphi (s,u_{1})|\kappa _{0}(s,u)-\kappa _{0}(s,u_{1})|\le C\varphi (s,u_{1})\in L_{1}[0,T],\]
the proof follows from the Lebesgue dominated convergence theorem.  □
Remark 2.
In the case where $H_{1}=\frac{1}{2}$, the kernel $\kappa (s,u)$ can be simplified to
\[\kappa (s,u)=H_{2}(2H_{2}-1)|s-u{|}^{2H_{2}-2},\]
and Eq. (8) coincides with (10). Indeed, let $H_{1}=\frac{1}{2}$. Then the function $\kappa _{0}(s,u)$ equals $H_{2}(2H_{2}-1)$. Consider the function $\varPhi (s,v)$ defined by (18):
(29)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \varPhi (t,s)& \displaystyle ={t}^{H_{2}-\frac{1}{2}}\Bigg({\int _{0}^{1}}{\bigg(1-\frac{t-s}{t}z\bigg)}^{H_{2}-\frac{1}{2}}{(1-z)}^{H_{2}-\frac{3}{2}}dz\\{} & \displaystyle \hspace{1em}+\frac{t-s}{t}{\int _{0}^{1}}{(1-z)}^{H_{2}-\frac{1}{2}}{\bigg(1-\frac{t-s}{t}z\bigg)}^{H_{2}-\frac{3}{2}}dz\Bigg)\\{} & \displaystyle =-\frac{{t}^{H_{2}-\frac{1}{2}}}{H_{2}-\frac{1}{2}}{\int _{0}^{1}}{\bigg({\bigg(1-\frac{t-s}{t}z\bigg)}^{H_{2}-\frac{1}{2}}{(1-z)}^{H_{2}-\frac{1}{2}}\bigg)_{z}^{{^{\prime }}}}dz=\frac{{t}^{H_{2}-\frac{1}{2}}}{H_{2}-\frac{1}{2}}.\end{array}\]
Combining (28) and (29), we get
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle \kappa _{0}(s,u)\\{} & \displaystyle \hspace{1em}={\big(\beta _{H_{2}}(H_{2}\hspace{0.1667em}-\hspace{0.1667em}H_{1})\big)}^{2}\hspace{-0.1667em}{\int _{0}^{1}}{(1-t)}^{1-2H_{2}}{t}^{H_{2}-\frac{3}{2}}\varPhi \bigg(1,\frac{u(1-t)}{s-ut}\bigg)\varPhi \bigg(1,\frac{s(1-t)}{s-ut}\bigg)dt\\{} & \displaystyle \hspace{1em}={\beta _{H_{2}}^{2}}\hspace{-0.1667em}{\int _{0}^{1}}\hspace{-0.1667em}{(1\hspace{0.1667em}-\hspace{0.1667em}t)}^{1-2H_{2}}{t}^{H_{2}-\frac{3}{2}}dt\hspace{0.1667em}=\hspace{0.1667em}{\beta _{H_{2}}^{2}}\mathrm{B}\bigg(H_{2}\hspace{0.1667em}-\hspace{0.1667em}\frac{1}{2},2-2H_{2}\bigg)=H_{2}(2H_{2}-1).\end{array}\]
Theorem 1.
There exists a sequence $T_{n}\to \infty $ such that the integral equation (11) has a unique solution $h_{T_{n}}(u)\in C[0,T_{n}]$.
Proof.
We work on the space $C([0,T])$. Recall that (11) is of the form
\[h_{T}(u)+\frac{1}{{\gamma _{H_{1}}^{2}}}{\int _{0}^{T}}h_{T}(s)\kappa (s,u)ds=1,\hspace{1em}u\in [0,T].\]
The corresponding homogeneous equation is of the form
(30)
\[{\int _{0}^{T}}h_{T}(s)\kappa (s,u)ds=-{\gamma _{H_{1}}^{2}}h_{T}(u),\hspace{1em}u\in [0,T].\]
Since the integral operator κ is compact, classical Fredholm theory states that Eq. (11) has a unique solution if and only if the corresponding homogeneous equation (30) has only the trivial solution. Now, it is easy to see that, for any $a>0$, the following equalities hold:
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \kappa _{0}(sa,ua)& \displaystyle =\kappa _{0}(s,u),\\{} \displaystyle \varphi (sa,ua)& \displaystyle ={a}^{2H_{2}-2H_{1}-1}\varphi (s,u).\end{array}\]
Consequently, $\kappa (sa,ua)={a}^{2H_{2}-2H_{1}-1}\kappa (s,u)$. We can change the variable of integration $s={s^{\prime }}T$ and put $u={u^{\prime }}T$ in (30). Therefore, the equation will be reduced to the equivalent form
\[{\int _{0}^{1}}h_{T}(Ts)\kappa (s,u)ds=-{\gamma _{H_{1}}^{2}}{T}^{2H_{1}-2H_{2}}h_{T}(Tu),\hspace{1em}u\in [0,1].\]
Denote $\lambda =-{\gamma _{H_{1}}^{2}}{T}^{2H_{1}-2H_{2}}$. Note that λ depends continuously on T. At the same time, the compact operator κ has no more than countably many eigenvalues. Therefore, we can take the sequence $T_{n}\to \infty $ in such a way that
\[\lambda _{n}=-{\gamma _{H_{1}}^{2}}{T_{n}^{2H_{1}-2H_{2}}}\]
will be not an eigenvalue. Consequently, the homogeneous equation has only the trivial solution, whence the proof follows.  □

4 Statistical results: The form of a maximum likelihood estimator, its consistency, and asymptotic normality

The following result establishes the way MLE for the drift parameter θ can be calculated. The proof of the theorem is the same as the proof of the corresponding statement from [6], so we omit it.
Theorem 2.
The likelihood function is of the form
\[L_{T_{n}}(X,\theta )=\exp \bigg\{\theta \delta _{H_{1}}N(T_{n})-\frac{1}{2}{\theta }^{2}{\delta _{H_{1}}^{2}}\langle N\rangle (T_{n})\bigg\},\]
and the maximum likelihood estimator is of the form
\[\widehat{\theta }(T_{n})=\frac{N(T_{n})}{\delta _{H_{1}}\langle N\rangle (T_{n})},\]
where $N(t)=E_{0}(X_{1}(t)|{\mathfrak{F}_{t}^{X}})$ is a square-integrable Gaussian ${\mathfrak{F}_{t}^{X}}$-martingale, $N(T_{n})={\int _{0}^{T_{n}}}h_{T_{n}}(t)dX(t)$ with $h_{T_{n}}(t){t}^{\frac{1}{2}-H_{1}}\in L_{2}[0,T_{n}]$, $h_{T_{n}}(t)$ is a unique solution to (11), and $\langle N\rangle (T_{n})={\gamma _{H_{1}}^{2}}{\int _{0}^{T_{n}}}h_{T_{n}}(t){t}^{1-2H_{1}}dt$.
The next two results establish basic properties of the estimator; their proofs repeat the proofs of the corresponding statements from [6] and [3].
Theorem 3.
The estimator $\widehat{\theta }_{T_{n}}$ is strongly consistent, and
\[\underset{T_{n}\to \infty }{\lim }{T_{n}}^{2-2H_{2}}E_{\theta }{(\widehat{\theta }_{T_{n}}-\theta )}^{2}=\frac{1}{{\textstyle\int _{0}^{1}}h_{0}(u){u}^{\frac{1}{2}-H_{1}}du},\]
where the function $h_{0}(u)$ is the solution of the integral equation
\[\kappa h(u)={\gamma _{H_{1}}^{2}}.\]
Theorem 4.
The estimator $\widehat{\theta }_{T_{n}}$ is unbiased, and the corresponding estimation error is normal
\[\widehat{\theta }_{T_{n}}-\theta \sim N\bigg(0,\frac{1}{{\textstyle\int _{0}^{T_{n}}}h_{T_{n}}(s){s}^{1-2H_{1}}ds}\bigg).\]

A Appendix. Some properties of the hypergeometric function

Recall the integral representation of the Gauss hypergeometric function and some of its properties.
For $c>b>0$ and $x<1$, the Gauss hypergeometric function is defined as the integral (see [1], formula 15.3.1)
(31)
\[\displaystyle F(a,b,c;x)=_{2}F_{1}(a,b,c;x)=\frac{1}{\mathrm{B}(b,c-b)}{\int _{0}^{1}}{t}^{b-1}{(1-t)}^{c-b-1}{(1-xt)}^{-a}dt.\]
For the same values of parameters, the following equality holds (see [1], 15.3.4):
(32)
\[F(a,b,c;x)={(1-x)}^{-a}F\bigg(a,c-b,c;\frac{x}{x-1}\bigg),\]
Evidently, $F(a,b,c;x)$ at $x=1$ is correctly defined for $c-a-b>1$ and in this case equals
(33)
\[F(a,b,c;1)=\frac{\varGamma (c)\varGamma (c-a-b)}{\varGamma (c-a)\varGamma (c-b)}.\]
Finally, it is easy to check with the help of (31) that
(34)
\[F(a,b,c;0)=F(0,b,c;x)=1.\]
The following result gives upper bounds for the hypergeometric function (see [5] Theorem 4 and 5, respectively).
Proposition 1.
(i) For $c>b>1$, $x>0$, and $0<a\le 1$, we have the inequality
\[F(a,b,c;-x)<\frac{1}{{(1+x(b-1)/(c-1))}^{a}}.\]
(ii) For $0<a\le 1$, $b>0$, $c-b>1$, and $x\in (0,1)$, we have the inequality
\[F(a,b,c;x)<\frac{1}{{(1-\frac{b}{c-1}x)}^{a}}.\]

References

[1] 
Abramowitz, M., Stegun, I.A.: Handbook of Mathematical Functions. Applied Mathematics Series 55, 62 (1966)
[2] 
Berezansky, Y.M., Sheftel, Z.G., Us, G.F.: Functional Analysis, vol. 1. Birkhäuser (2012)
[3] 
Cai, C., Chigansky, P., Kleptsyna, M.: Mixed fractional Brownian motion: The filtering perspective. To appear in Annals of Probability (2015)
[4] 
Jost, C.: Transformation formulas for fractional Brownian motion. Stochastic Processes and their Applications 116(10), 1341–1357 (2006) MR2260738. doi:10.1016/j.spa.2006.02.006
[5] 
Karp, D., Sitnik, S.: Two-sided inequalities for generalized hypergeometric function. Research report collection 10(2) (2007)
[6] 
Mishura, Y.: Maximum likelihood drift estimation for the mixing of two fractional Brownian motions. arXiv preprint arXiv:1506.04731v1 (2015)
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Table of contents
  • 1 Introduction
  • 2 Preliminaries. How to reduce the original problem to the integral equation
  • 3 Compactness of integral operator. Existence–uniqueness result for the Fredholm integral equation
  • 4 Statistical results: The form of a maximum likelihood estimator, its consistency, and asymptotic normality
  • A Appendix. Some properties of the hypergeometric function
  • References

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Keywords
Fractional Brownian motion maximum likelihood estimator integral equation with weakly singular kernel compact operator asymptotic consistency

MSC2010
60G22 62F10 62F12 62G12

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  • Figures
    1
  • Theorems
    4
vmsta-2-2-vmsta28-g001.jpg
Fig. 1.
Function $\kappa _{0}(s,u)$
Theorem 1.
Theorem 2.
Theorem 3.
Theorem 4.
vmsta-2-2-vmsta28-g001.jpg
Fig. 1.
Function $\kappa _{0}(s,u)$
Theorem 1.
There exists a sequence $T_{n}\to \infty $ such that the integral equation (11) has a unique solution $h_{T_{n}}(u)\in C[0,T_{n}]$.
Theorem 2.
The likelihood function is of the form
\[L_{T_{n}}(X,\theta )=\exp \bigg\{\theta \delta _{H_{1}}N(T_{n})-\frac{1}{2}{\theta }^{2}{\delta _{H_{1}}^{2}}\langle N\rangle (T_{n})\bigg\},\]
and the maximum likelihood estimator is of the form
\[\widehat{\theta }(T_{n})=\frac{N(T_{n})}{\delta _{H_{1}}\langle N\rangle (T_{n})},\]
where $N(t)=E_{0}(X_{1}(t)|{\mathfrak{F}_{t}^{X}})$ is a square-integrable Gaussian ${\mathfrak{F}_{t}^{X}}$-martingale, $N(T_{n})={\int _{0}^{T_{n}}}h_{T_{n}}(t)dX(t)$ with $h_{T_{n}}(t){t}^{\frac{1}{2}-H_{1}}\in L_{2}[0,T_{n}]$, $h_{T_{n}}(t)$ is a unique solution to (11), and $\langle N\rangle (T_{n})={\gamma _{H_{1}}^{2}}{\int _{0}^{T_{n}}}h_{T_{n}}(t){t}^{1-2H_{1}}dt$.
Theorem 3.
The estimator $\widehat{\theta }_{T_{n}}$ is strongly consistent, and
\[\underset{T_{n}\to \infty }{\lim }{T_{n}}^{2-2H_{2}}E_{\theta }{(\widehat{\theta }_{T_{n}}-\theta )}^{2}=\frac{1}{{\textstyle\int _{0}^{1}}h_{0}(u){u}^{\frac{1}{2}-H_{1}}du},\]
where the function $h_{0}(u)$ is the solution of the integral equation
\[\kappa h(u)={\gamma _{H_{1}}^{2}}.\]
Theorem 4.
The estimator $\widehat{\theta }_{T_{n}}$ is unbiased, and the corresponding estimation error is normal
\[\widehat{\theta }_{T_{n}}-\theta \sim N\bigg(0,\frac{1}{{\textstyle\int _{0}^{T_{n}}}h_{T_{n}}(s){s}^{1-2H_{1}}ds}\bigg).\]

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