1 Introduction
Consider the continuous-time linear model
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 $\mathfrak{F}_{0}$.
(1)
\[X(t)=\theta t+\sigma _{1}{B}^{H_{1}}(t)+\sigma _{2}{B}^{H_{2}}(t),\hspace{0.2778em}t\in [0,T],\]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
From now on we suppose that the Hurst indices in (1) satisfy the inequality
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,
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
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
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 $\frac{1}{2}\le H_{1}<H_{2}<1$,
(2)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle Y(t)=& \displaystyle {\int _{0}^{t}}l_{H_{1}}(t,s)dZ(s)=\theta \mathrm{B}\bigg(\frac{3}{2}-H_{1},\frac{3}{2}-H_{1}\bigg){t}^{2-2H_{1}}+{M}^{H_{1}}(t)\\{} & \displaystyle +{\int _{0}^{t}}l_{H_{1}}(t,s)d{B}^{H_{2}}(s),\end{array}\]
\[\Bigg\{\widetilde{X}_{1}(t)={\widetilde{M}}^{H_{1}}(t),X_{2}(t):={\int _{0}^{t}}l_{H_{1}}(t,s)d{B}^{H_{2}}(s),\hspace{0.2778em}t\ge 0\Bigg\},\]
$i=1,2$, 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,
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
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
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
We have that $W_{i},\hspace{2.5pt}i=1,2$, are standard Wiener processes, which are obviously independent. Also, we have
Then
where
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
(3)
\[X_{1}(t)=\gamma _{H_{1}}{\int _{0}^{t}}{s}^{1/2-H_{1}}dW_{1}(s),\hspace{0.2778em}{B}^{H_{2}}(t)={\int _{0}^{t}}K_{H_{2}}(t,s)dW_{2}(s).\](4)
\[K_{H_{1},H_{2}}(t,s)=\beta _{H_{2}}{s}^{1/2-H_{2}}{\int _{s}^{t}}{(t-u)}^{1/2-H_{1}}{u}^{H_{2}-H_{1}}{(u-s)}^{H_{2}-3/2}du.\]
\[{\left\langle f,g\right\rangle _{H}^{2}}:=\alpha _{H}{\int _{0}^{T}}{\int _{0}^{T}}f(t)g(s)|t-s{|}^{2H-2}dsdt,\]
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
where
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
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
Using (5) together with independency of $X_{1}$ and $X_{2}$, we arrive at the equation
where $\varepsilon _{H}={\gamma _{H}^{2}}/(2-2H)$. Finally, from (3)–(6) we get the prototype of a Fredholm integral equation
where
where
with the function $K_{H_{1},H_{2}}$ defined by (4).
(6)
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle \mathsf{E}\left[X_{1}(u){\int _{0}^{T}}h_{T}(t)dX_{1}(t)+X_{2}(u){\int _{0}^{T}}h_{T}(t)dX_{2}(t)\right]\\{} & \displaystyle \hspace{1em}=\mathsf{E}\left[X_{1}(u)X_{1}(T)\right]=\varepsilon _{H_{1}}{u}^{2-2H_{1}},\end{array}\](7)
\[\displaystyle \varepsilon _{H_{1}}{u}^{2-2H_{1}}={\gamma _{H_{1}}^{2}}{\int _{0}^{u}}h_{T}(s){s}^{1-2H_{1}}ds+{\int _{0}^{T}}h_{T}(s)r_{H_{1},H_{2}}(s,u)ds,\hspace{0.2778em}u\in [0,T],\]
\[r_{H_{1},H_{2}}(s,u)={\int _{0}^{s\wedge u}}\partial _{s}K_{H_{1},H_{2}}(s,v)K_{H_{1},H_{2}}(u,v)dv.\]
Differentiating (7), we get the Fredholm integral equation of the 2nd kind,
(8)
\[{\gamma _{H_{1}}^{2}}h_{T}(u){u}^{1-2H_{1}}+{\int _{0}^{T}}h_{T}(s)k(s,u)ds={\gamma _{H_{1}}^{2}}{u}^{1-2H_{1}},\hspace{1em}u\in (0,T],\](9)
\[k(s,u)={\int _{0}^{s\wedge u}}\partial _{s}K_{H_{1},H_{2}}(s,v)\partial _{u}K_{H_{1},H_{2}}(u,v)dv\]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]:
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.
(10)
\[h_{T}(u)+H_{2}(2H_{2}-1){\int _{0}^{T}}h_{T}(s)|s-u{|}^{2H_{2}-2}ds=1,\hspace{1em}u\in [0,T],\]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
with continuous right-hand side, where
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:
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 $[0,T]$:
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 $C({[0,T]_{0}^{2}})$.
(13)
\[\kappa (s,u)=\left\{\begin{array}{l@{\hskip10.0pt}l}\kappa _{0}(s,u)\varphi (s,u),\hspace{1em}& s\ne u,\\{} 0,\hspace{1em}& s=u,\end{array}\right.\]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
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
(16)
\[I_{1}(t,s):={s}^{H_{2}-H_{1}}{\int _{0}^{1}}{z}^{\alpha _{2}-1}{(1-z)}^{-\alpha _{1}}{\bigg(1-\frac{s-t}{s}z\bigg)}^{H_{2}-H_{1}}dz,\]
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I_{1}(t,s)& \displaystyle =\mathrm{B}(1-\alpha _{1},\alpha _{2}){s}^{H_{2}-H_{1}}F\bigg(H_{1}-H_{2},\alpha _{2},1-H_{1}+H_{2};\frac{s-t}{s}\bigg)\\{} & \displaystyle =\mathrm{B}(1\hspace{0.1667em}-\hspace{0.1667em}\alpha _{1},\alpha _{2}){s}^{H_{2}-H_{1}}{\bigg(\frac{t}{s}\bigg)}^{H_{2}-H_{1}}\hspace{-0.1667em}F\bigg(\hspace{-0.1667em}H_{1}\hspace{0.1667em}-\hspace{0.1667em}H_{2},1\hspace{0.1667em}-\hspace{0.1667em}\alpha _{1},1\hspace{0.1667em}-\hspace{0.1667em}H_{1}\hspace{0.1667em}+\hspace{0.1667em}H_{2};\frac{t\hspace{0.1667em}-\hspace{0.1667em}s}{t}\hspace{-0.1667em}\bigg)\\{} & \displaystyle =\mathrm{B}(1-\alpha _{1},\alpha _{2}){t}^{H_{2}-H_{1}}F\bigg(H_{1}-H_{2},1-\alpha _{1},1-H_{1}+H_{2};\frac{t-s}{t}\bigg).\end{array}\]
Similarly, for the second term, that is, for
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
(17)
\[I_{2}(t,s):=(t-s){s}^{H_{2}-H_{1}-1}{\int _{0}^{1}}{z}^{\alpha _{2}}{(1-z)}^{-\alpha _{1}}{\bigg(1-\frac{s-t}{s}z\bigg)}^{H_{2}-H_{1}-1}dz,\]
\[\begin{array}{r@{\hskip0pt}l}\displaystyle I_{2}(t,s)& \displaystyle =(t-s){s}^{H_{2}-H_{1}-1}\mathrm{B}(1-\alpha _{1},\alpha _{2}+1)\\{} & \displaystyle \hspace{1em}\times F\bigg(H_{1}-H_{2}+1,\alpha _{2}+1,H_{2}-H_{1}+2;\frac{s-t}{s}\bigg)\\{} & \displaystyle =(t-s){s}^{H_{2}-H_{1}-1}{\bigg(\frac{t}{s}\bigg)}^{H_{2}-H_{1}-1}\\{} & \displaystyle \hspace{1em}\times B(1-\alpha _{1},\alpha _{2}+1)F\bigg(H_{1}-H_{2}+1,1-\alpha _{1},2-H_{1}+H_{2};\frac{t-s}{t}\bigg)\\{} & \displaystyle =(t-s){t}^{H_{2}-H_{1}-1}\mathrm{B}(1-\alpha _{1},\alpha _{2}+1)\\{} & \displaystyle \hspace{1em}\times F\bigg(H_{1}-H_{2}+1,1-\alpha _{1},2-H_{1}+H_{2};\frac{t-s}{t}\bigg).\end{array}\]
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:
Introduce the notation
and note that $\varPhi \in C({[0,T]_{0}^{2}})$ is bounded and homogeneous:
In terms of notation (18), the representation (15) for $\partial _{t}K_{H_{1},H_{2}}(t,s)$ can be rewritten as
(18)
\[\varPhi (t,s)=I_{1}(t,s)+I_{2}(t,s)={t}^{H_{2}-H_{1}}\varPsi _{1}(t,s)+{(t-s)}^{H_{2}-H_{1}}\varPsi _{2}(t,s)\]In turn, the kernel $k(s,u)$ from (9) can be rewritten as
Consider the kernel $k(s,u)$ for $s>u$. Then it evidently equals
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
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
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 $\kappa _{0}(s,s)=C_{H}$, $s>0$.
(25)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \kappa _{0}(s,s)& \displaystyle ={\big(\beta _{H_{2}}(H_{2}-H_{1})\big)}^{2}\varPhi (s,s)\varPhi (1,1){\int _{0}^{\infty }}{(1+st)}^{H_{2}-H_{1}-1}{t}^{H_{2}-H_{1}-1}dt\\{} & \displaystyle ={\big(\beta _{H_{2}}(H_{2}-H_{1})\big)}^{2}{s}^{H_{2}-H_{1}}\varPhi {(1,1)}^{2}{\int _{0}^{\infty }}{(1+st)}^{H_{2}-H_{1}-1}{t}^{H_{2}-H_{1}-1}dt.\end{array}\]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
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
It follows from (19) that, for $s\ne 0$,
Denote $r=\frac{s}{s-u}$ and put $t=\frac{1-z}{1-(1-r)z}$. Then
and the right-hand side of (26) can be rewritten as
(26)
\[\begin{array}{r@{\hskip0pt}l}& \displaystyle {\int _{0}^{\frac{1}{s-u}}}{\big(1-(s-u)t\big)}^{1-2H_{2}}{(1+ut)}^{H_{2}-H_{1}-1}{t}^{H_{2}-H_{1}-1}\varPhi \big(s,u-tu(s-u)\big)\\{} & \displaystyle \hspace{1em}\times \varPhi \big(1,1-t(s-u)\big)dt=\frac{1}{{(s-u)}^{H_{2}-H_{1}}}{\int _{0}^{1}}{(1-z)}^{1-2H_{2}}\\{} & \displaystyle \hspace{1em}\times {\bigg(1+\frac{u}{s-u}z\bigg)}^{H_{2}-H_{1}-1}{z}^{H_{2}-H_{1}-1}\varPhi \big(s,u(1-z)\big))\varPhi (1,1-z)dz=:I_{3}(s,u).\end{array}\](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 $s>u$,
so $\kappa _{0}$ is bounded, and the lemma is proved. □
(28)
\[\begin{array}{r@{\hskip0pt}l}\displaystyle \kappa _{0}(s,u)& \displaystyle ={\big(\beta _{H_{2}}(H_{2}-H_{1})\big)}^{2}{\int _{0}^{1}}{(1-y)}^{1-2H_{2}}{\bigg(1-\frac{u}{s}y\bigg)}^{2H_{1}-1}{y}^{H_{2}-H_{1}-1}\\{} & \displaystyle \hspace{1em}\times \varPhi \bigg(1,\frac{u(1-y)}{s-uy}\bigg)\varPhi \bigg(1,\frac{s(1-y)}{s-uy}\bigg)dy\\{} & \displaystyle \hspace{1em}\le {C_{H}^{1}}{\int _{0}^{1}}{(1-y)}^{1-2H_{2}}{y}^{H_{2}-H_{1}-1}dy,\end{array}\]Now, consider the properties of the function
participating in the kernel representation (13).
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. □Proof.
According to [2], it suffices to prove that the kernel κ defined by (13) satisfies the following two conditions:
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
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
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
The corresponding homogeneous equation is of the form
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:
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
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
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