We construct an estimator of the unknown drift parameter θ∈R in the linear model
Xt=θt+σ1BH1(t)+σ2BH2(t),t∈[0,T],
where BH1 and BH2 are two independent fractional Brownian motions with Hurst indices H1 and H2 satisfying the condition 12≤H1<H2<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.
Consider the continuous-time linear model
X(t)=θt+σ1BH1(t)+σ2BH2(t),t∈[0,T],
where BH1 and BH2 are two independent fractional Brownian motions with different Hurst indices H1 and H2 defined on some stochastic basis (Ω,F,(F)t,t≥0,P). We assume that the filtration is generated by these processes and completed by P-negligible sets of F0.
Recall that the fractional Brownian motion (fBm) BtH, t≥0, with Hurst index H∈(0,1) is a centered Gaussian process with the covariance function
EBH(t)BH(s)=12(t2H+s2H−|t−s|2H).
From now on we suppose that the Hurst indices in (1) satisfy the inequality
12≤H1<H2<1,
and we consider the continuous modifications of both processes, which exist due to the Kolmogorov theorem. Assuming that the Hurst indices H1, H2 and parameters σ1≥0, σ2≥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>1/21/2$]]>, we call our model the model with double long-range dependence.
In the case where H1=12, the problem of drift parameter estimation in the model (1) was solved in [3], and in the case where 12<H1<H2<1 and H2−H1>1/41/4$]]>, the estimator was constructed in [6]. The goal of the present paper is to generalize the results from [6] to arbitrary 12≤H1<H2<1. The problem, more technical than principal, is that in the case where H2−H1>1/41/4$]]> and H1>1/21/2$]]>, 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 L2[0,T] exists and is unique, but for H2−H1≤1/4, the kernel does not belong to L2[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, H1 and H2, 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.
Preliminaries. How to reduce the original problem to the integral equation
Since we suppose that the Hurst parameters H1,H2 and scale parameters σ1,σ2 are known, for technical simplicity, we consider the case where σ1=σ2=1 and, as it was mentioned before, 12≤H1<H2<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 {B˜H1(t),BH2(t),t≥0}, i=1,2, defined on the space (Ω,F,(F)t) and let Pθ be a probability measure under which B˜H1 and BH2 are independent, BH2 is a fractional Brownian motion with Hurst parameter H2, and B˜H1 is a fractional Brownian motion with Hurst parameter H1 and drift θ, that is,
B˜H1(t)=θt+BH1(t).
The probability measure P0 corresponds to the case θ=0. Our main problem is the construction of maximum likelihood estimator for θ∈R by the observations of the process Z(t)=θt+BH1(t)+BH2(t)=B˜H1(t)+BH2(t),t∈[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 lH(t,s)=(t−s)1/2−Hs1/2−H and construct the integral
Y(t)=∫0tlH1(t,s)dZ(s)=θB(32−H1,32−H1)t2−2H1+MH1(t)+∫0tlH1(t,s)dBH2(s),
where B(α,β)=∫01xα−1(1−x)β−1dx is the beta function, and MH1 is a Gaussian martingale (Molchan martingale), admitting the representations
MH(t)=∫0tlH(t,s)dBH(s)=γH∫0ts1/2−HdW(s)
with γH=(2H(32−H)Γ(3/2−H)3Γ(H+12)Γ(3−2H)−1)12 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 θ∈R basing on the linear transformation Y. For simplicity, denote BH1:=B(32−H1,32−H1). Now the main problem can be formulated as follows. Let 12≤H1<H2<1,
{X˜1(t)=M˜H1(t),X2(t):=∫0tlH1(t,s)dBH2(s),t≥0},i=1,2, be a couple of processes defined on the space (Ω,F), and Pθ be a probability measure under which X˜1 and X2 are independent, BH2 is a fractional Brownian motion with Hurst parameter H2, and X˜1 is a martingale with square characteristics ⟨X˜1⟩(t)=γH122−2H1t2−2H1 and drift θBH1t2−2H1, that is,
X˜1(t)=M˜H1(t)=θBH1t2−2H1+MH1(t).
Also, denote X1(t)=MH1(t). Our main problem is the construction of maximum likelihood estimator for θ∈R by the observations of the process
Y(t)=θBH1t2−2H1+X1(t)+X2(t)=X˜1(t)+X2(t).
Note that, under the measure Pθ, the process
W˜(t):=W(t)+θ(2−2H1)BH1γH1(32−H1)t32−H1
is a Wiener process with drift. Denote δH1=(2−2H1)BH1γH1.
By Girsanov’s theorem and independence of X1 and X2,
dPθdP0=exp{θδH1∫0Ts12−H1dW˜(s)−θ2δH124(1−H1)T2−2H1}=exp{θδH1X˜1(T)−θ2δH124(1−H1)T2−2H1}.
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
FTY:=σ{Y(t),t∈[0,T]}=FTX:=σ{X(t),t∈[0,T]},
where X(t)=X1(t)+X2(t).
We shall proceed as in [3]. Let μθ be the probability measure induced by Y on the space of continuous functions with the supremum topology under probability Pθ. Then for any measurable set A, μθ(A)=∫AΦ(x)μ0(dx), where Φ(x) is a measurable functional such that Φ(X)=E0(dPθdP0|FTX). This means that μθ≪μ0 for any θ∈R. Taking into account that X˜1=X1 under P0 and the fact that the vector process (X1,X) is Gaussian, we get that the corresponding likelihood function is given by
LT(X,θ)=E0(dPθdP0|FTX)=E0(exp{θδH1X1(T)−θ2δH124(1−H1)T2−2H1}|FTX)=exp{θδH1E0(X1(T)|FTX)+θ2δH122(V(T)−T2−2H12−2H1)},
where V(t)=E0(X1(t)−E0(X1(t)|FtX))2,t∈[0,T].
The next reasonings repeat the corresponding part of [6]. We have to solve the following problem: to find the projection PXX1(T) of X1(T) onto
X(t)=X1(t)+X2(t),t∈[0,T].
According to [4], the transformation formula for converting fBm into a Wiener process is of the form
Wi(t)=∫0t((KHi∗)−11[0,t])(s)dBHi(s),i=1,2,
where
(KH∗f)(s)=∫sTf(t)∂tKH(t,s)dt=βHs1/2−H∫sTf(t)tH−1/2(t−s)H−3/2dt,βH=(H(2H−1)B(H−1/2,2−2H))12, and the square-integrable kernel KH(t,s) is of the form
KH(t,s)=βHs1/2−H∫st(u−s)H−3/2uH−1/2du.
We have that Wi,i=1,2, are standard Wiener processes, which are obviously independent. Also, we have
X1(t)=γH1∫0ts1/2−H1dW1(s),BH2(t)=∫0tKH2(t,s)dW2(s).
Then
X2(t)=∫0tKH1,H2(t,s)dW2(s),
where
KH1,H2(t,s)=βH2s1/2−H2∫st(t−u)1/2−H1uH2−H1(u−s)H2−3/2du.
For an interval [0,T], denote by LH2[0,T] the completion of the space of simple functions f:[0,T]→R with respect to the scalar product
f,gH2:=αH∫0T∫0Tf(t)g(s)|t−s|2H−2dsdt,
where αH=H(2H−1). Note that this space contains both functions and distributions. For functions from LH22[0,T], we have that
∫0Tf(s)dX2(s)=∫0T(KH1,H2∗f)(s)dW2(s),
where
(KH1,H2∗f)(s)=∫sTf(t)∂tKH1,H2(t,s)dt.
The projection of X1(T) onto {X(t),t∈[0,T]} is a centered X-measurable Gaussian random variable and, therefore, is of the form
PXX1(T)=∫0ThT(t)dX(t)
with hT∈LH12[0,T]. Note that hT still may be a distribution. However, as we will further see, it is a continuous function. The projection for all u∈[0,T] must satisfy
EX(u)PXX1(T)=EX(u)X1(T).
Using (5) together with independency of X1 and X2, we arrive at the equation
EX1(u)∫0ThT(t)dX1(t)+X2(u)∫0ThT(t)dX2(t)=EX1(u)X1(T)=εH1u2−2H1,
where εH=γH2/(2−2H). Finally, from (3)–(6) we get the prototype of a Fredholm integral equation
εH1u2−2H1=γH12∫0uhT(s)s1−2H1ds+∫0ThT(s)rH1,H2(s,u)ds,u∈[0,T],
where
rH1,H2(s,u)=∫0s∧u∂sKH1,H2(s,v)KH1,H2(u,v)dv.
Differentiating (7), we get the Fredholm integral equation of the 2nd kind,
γH12hT(u)u1−2H1+∫0ThT(s)k(s,u)ds=γH12u1−2H1,u∈(0,T],
where
k(s,u)=∫0s∧u∂sKH1,H2(s,v)∂uKH1,H2(u,v)dv
with the function KH1,H2 defined by (4).
We will establish in Remark 2 that for the case H1=12, Eq. (8) can be reduced to the corresponding equation from [3]:
hT(u)+H2(2H2−1)∫0ThT(s)|s−u|2H2−2ds=1,u∈[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 H2−H1>14\frac{1}{4}$]]> and H1>1/21/2$]]> are satisfied, then Eq. (8) has a unique solution hTn with hTn(t)t12−H1∈L2[0,Tn] on any sequence of intervals [0,Tn] except, possibly, a countable number of Tn 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 H2 while H1=12. 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 L2-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 H2−H1>14\frac{1}{4}$]]> and can include the case H1=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
hT(u)+1γH12∫0ThT(s)κ(s,u)ds=1,u∈[0,T],
with continuous right-hand side, where
κ(s,u)=u2H1−1k(s,u),s,u∈[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).
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)=∫0TK(s,u)x(s)ds,x∈C[0,T].
Now we are in position to establish the properties of the kernel κ(s,u) defined by (12) and (9). Introduce the notation [0,T]02=[0,T]2∖{(0,0)}.
Up to a set of Lebesgue measure zero, the kernelκ(s,u),s,u∈[0,T], admits the following representation on[0,T]:
κ(s,u)=κ0(s,u)φ(s,u),s≠u,0,s=u,whereφ(s,u)=(s∧u)1−2H1u2H1−1|s−u|2H2−2H1−1, and the functionκ0is bounded and belongs toC([0,T]02).
We take (9) and first present the derivative of KH1,H2(t,s), defined by (4), in an appropriate form. To start, put u=s+(t−s)z. This allows us to rewrite KH1,H2(t,s) as
KH1,H2(t,s)=βH2s12−H2(t−s)H2−H1×∫01(1−z)12−H1(s+(t−s)z)H2−H1zH2−32dz.
Differentiating (14) w.r.t. t for 0<s<t≤T, we get
∂tKH1,H2(t,s)=(H2−H1)βH2s12−H2(t−s)H2−H1−1×∫01(1−z)12−H1(s+(t−s)z)H2−H1zH2−32dz+(H2−H1)βH2s12−H2×(t−s)H2−H1∫01(1−z)12−H1(s+(t−s)z)H2−H1−1zH2−12dz=(H2−H1)βH2s12−H2(t−s)H2−H1−1×(∫01(1−z)12−H1(s+(t−s)z)H2−H1zH2−32dz+(t−s)∫01(1−z)12−H1(s+(t−s)z)H2−H1−1zH2−12dz)=(H2−H1)βH2s12−H2(t−s)H2−H1−1×(sH2−H1∫01zH2−32(1−z)12−H1(1−s−tsz)H2−H1dz+(t−s)sH2−H1−1∫01(1−z)12−H1(1−s−tsz)H2−H1−1zH2−12dz).
Denote for technical simplicity αi=Hi−12,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
I1(t,s):=sH2−H1∫01zα2−1(1−z)−α1(1−s−tsz)H2−H1dz,
the values of parameters for the underlying integral equal a=H1−H2, b=α2, c=H2−H1+1, and x=s−ts<1, respectively; therefore, xx−1=t−st, c−b=1−α1, and
I1(t,s)=B(1−α1,α2)sH2−H1F(H1−H2,α2,1−H1+H2;s−ts)=B(1−α1,α2)sH2−H1(ts)H2−H1F(H1−H2,1−α1,1−H1+H2;t−st)=B(1−α1,α2)tH2−H1F(H1−H2,1−α1,1−H1+H2;t−st).
Similarly, for the second term, that is, for
I2(t,s):=(t−s)sH2−H1−1∫01zα2(1−z)−α1(1−s−tsz)H2−H1−1dz,
the values of parameters for the underlying integral equal a=H1−H2+1, b=α2+1, c=H2−H1+2, and x=s−ts, respectively; therefore, xx−1=t−st, c−b=1−α1, and
I2(t,s)=(t−s)sH2−H1−1B(1−α1,α2+1)×F(H1−H2+1,α2+1,H2−H1+2;s−ts)=(t−s)sH2−H1−1(ts)H2−H1−1×B(1−α1,α2+1)F(H1−H2+1,1−α1,2−H1+H2;t−st)=(t−s)tH2−H1−1B(1−α1,α2+1)×F(H1−H2+1,1−α1,2−H1+H2;t−st).
It is easy to see from the initial representations (16) and (17) that I1(t,s) and I2(t,s) are continuous on the set 0<s≤t≤T.
Now, introduce the notations
Ψ1(t,s)=B(1−α1,α2)F(H1−H2,1−α1,1−H1+H2;t−st)
and
Ψ2(t,s)=(t−st)1−H2+H1B(1−α1,α2+1)×F(H1−H2+1,1−α1,2−H1+H2;t−st),
so that I1(t,s)=tH2−H1Ψ1(t,s) and I2(t,s)=(t−s)H2−H1Ψ2(t,s). Note that t−st∈[0,1); therefore,
F(H1−H2,1−α1,1−H1+H2;t−st)=1B(1−α1,α2)×∫01z−α1(1−z)α2−1(1−t−stz)H2−H1dz≤1B(1−α1,α2)∫01z−α1(1−z)α2−1dz=1,
whence the function Ψ1(t,s) is bounded by B(1−α1,α2). In order to establish that Ψ2(t,s) is bounded, we use Proposition 1. Its conditions are satisfied: a=H1−H2+1∈(0,1), b=1−α1>00$]]>, c−b=α2+1>11$]]>, and x=t−st∈[0,1). Therefore,
x1−H2+H1F(H1−H2+1,1−α1,2−H1+H2;x)≤x1−H2+H1×(1−1−α11−H1+H2x)−1−H1+H2=(1x−1−α11−H1+H2)−1−H1+H2≤(1−1−α11−H1+H2)−1−H1+H2=(1−H1+H2α2)H1−H2+1,
whence Ψ2(t,s)≤B(1−α1,α2+1)(1−H1+H2α2)H1−H2+1. Additionally, both functions are homogeneous:
Ψi(at,as)=Ψi(t,s)fora>0,i=1,2.0,\hspace{2.5pt}i=1,2.\]]]>
Introduce the notation
Φ(t,s)=I1(t,s)+I2(t,s)=tH2−H1Ψ1(t,s)+(t−s)H2−H1Ψ2(t,s)
and note that Φ∈C([0,T]02) is bounded and homogeneous:
Φ(at,as)=aH2−H1Φ(t,s),a>0.0.\]]]>
In terms of notation (18), the representation (15) for ∂tKH1,H2(t,s) can be rewritten as
∂tKH1,H2(t,s)=βH2(H2−H1)s12−H2(t−s)H2−H1−1Φ(t,s).
In turn, the kernel k(s,u) from (9) can be rewritten as
k(s,u)=(βH2(H2−H1))2×∫0s∧uv1−2H2(s−v)H2−H1−1(u−v)H2−H1−1Φ(s,v)Φ(u,v)dv.
Consider the kernel k(s,u) for s>uu$]]>. Then it evidently equals
k(s,u)=(βH2(H2−H1))2×∫0uv1−2H2(s−v)H2−H1−1(u−v)H2−H1−1Φ(s,v)Φ(u,v)dv.
Put z=u−vs−u and transform k(s,u) to
k(s,u)=(βH2(H2−H1))2(s−u)2H2−2H1−1∫0us−uzH2−H1−1(1+z)H2−H1−1×(u−z(s−u))1−2H2Φ(s,u−z(s−u))Φ(u,u−z(s−u))dz=:k0(s,u)(s−u)1−2H2+2H1,
where
k0(s,u)=(βH2(H2−H1))2∫0us−uzH2−H1−1(1+z)H2−H1−1×(u−z(s−u))1−2H2Φ(s,u−z(s−u))Φ(u,u−z(s−u))dz.
In turn, transform k0(s,u) with the change of variables tu=z and apply (19):
k0(s,u)=(βH2(H2−H1))2∫01s−u(tu)H2−H1−1(1+tu)H2−H1−1×(u−tu(s−u))1−2H2Φ(s,u−tu(s−u))Φ(u,u−tu(s−u))udt=(βH2(H2−H1))2u1−2H1∫01s−u(1−t(s−u))1−2H2(1+tu)H2−H1−1×tH2−H1−1Φ(s,u−tu(s−u))Φ(1,1−t(s−u))dt.
Introducing the kernel κ0(s,u)=k0(s,u)u2H1−1, we can present k(s,u) as
k(s,u)=κ0(s,u)(s−u)1−2H2+2H1u2H1−1,
where, for s>u>0u>0$]]>,
κ0(s,u)=(βH2(H2−H1))2∫01s−u(1−(s−u)t)1−2H2(1+ut)H2−H1−1×tH2−H1−1Φ(s,u−tu(s−u))Φ(1,1−t(s−u))dt=(βH2(H2−H1))2∫0∞1t≤1s−u(1−(s−u)t)1−2H2(1+ut)H2−H1−1×tH2−H1−1Φ(s,u−tu(s−u))Φ(1,1−t(s−u))dt.
For the case u>s>0s>0$]]>, we can replace s and u in formulas (23) and (24). Substituting formally u=s into (24), for s>00$]]>, we get
κ0(s,s)=(βH2(H2−H1))2Φ(s,s)Φ(1,1)∫0∞(1+st)H2−H1−1tH2−H1−1dt=(βH2(H2−H1))2sH2−H1Φ(1,1)2∫0∞(1+st)H2−H1−1tH2−H1−1dt.
Note that Φ(1,1)=B(1−α1,α2) and ∫0∞(1+st)H2−H1−1tH2−H1−1dt=sH1−H2B(H2−H1,1−2H2+H1). The former equation holds due to (34). We get that κ0(s,s) does not depend on s and equals some constant CH:=(βH2(H2−H1)B(1−α1,α2))2B(H2−H1,1−2H2+H1). Therefore, we define κ0(s,s)=CH, s>00$]]>.
Now the continuity of κ0 on (0,T]2 follows from the Lebesgue dominated convergence theorem supplied by representation (24), Eq. (25), and its consequence κ0(s,s)=CH, s>00$]]>, together with the facts that Φ∈C([0,T]02) and is bounded. Consider κ0(s,u) for u↓0 and let s>00$]]> be fixed. Then
limu↓0κ0(s,u)=CH1:=(βH2(H2−H1))2Φ(1,0)×∫01(1−y)1−2H2yH2−H1−1Φ(1,1−y)dy<∞,
and we can put κ0(s,0)=κ0(0,u)=CH1, s>00$]]>, u>00$]]>, thus extending the continuity of κ0 to [0,T]02.
It is easy to see that the values κ0(s,s) and κ0(s,0) do not depend on s>00$]]> and do not coincide: CH≠CH1. Consequently, the limit
lim(s,u)→(0,0)κ0(s,u)
does not exist and depends on the way the variables s and u tend to zero. We can equate κ0(0,0) to any constant; for example, let κ0(0,0)=0.
In order to prove that κ0 is bounded, we consider the case s>uu$]]> (the opposite case is treated similarly) and put z=(s−u)t. Then
∫01s−u(1−(s−u)t)1−2H2(1+ut)H2−H1−1tH2−H1−1Φ(s,u−tu(s−u))×Φ(1,1−t(s−u))dt=1(s−u)H2−H1∫01(1−z)1−2H2×(1+us−uz)H2−H1−1zH2−H1−1Φ(s,u(1−z)))Φ(1,1−z)dz=:I3(s,u).
It follows from (19) that, for s≠0,
Φ(s,u(1−z))=sH2−H1Φ(1,us(1−z)).
Denote r=ss−u and put t=1−z1−(1−r)z. Then
us−u=r−1,t<1,z=1−t1−t(1−r)∈(0,1),
and the right-hand side of (26) can be rewritten as
I3(s,u)=rH2−H1∫01(1−z)1−2H2(1−(1−r)z)H2−H1−1zH2−H1−1×Φ(1,us(1−z))Φ(1,1−z)dz=r1−2H1∫01t1−2H2(1−t)H2−H1−1×(1−(1−r)t)2H1−1Φ(1,usrt1−(1−r)t)Φ(1,rt1−(1−r)t)dt.
Finally, put y=1−t. Then the right-hand side of (27) is transformed to
I3(s,u)=r1−2H1r2H1−1∫01(1−y)1−2H2yH2−H1−1(1−yr−1r)2H1−1×Φ(1,usr(1−y)r−y(r−1))Φ(1,r(1−y)r−y(r−1))dy.
Recall that r=ss−u. Then it follows from the boundedness of Φ that there exists a constant CH1 such that, for s>uu$]]>,
κ0(s,u)=(βH2(H2−H1))2∫01(1−y)1−2H2(1−usy)2H1−1yH2−H1−1×Φ(1,u(1−y)s−uy)Φ(1,s(1−y)s−uy)dy≤CH1∫01(1−y)1−2H2yH2−H1−1dy,
so κ0 is bounded, and the lemma is proved. □
Figure 1 demonstrates the graph of κ0(s,u) for H1=0.7 and H2=0.9.
Function κ0(s,u)
Now, consider the properties of the function
φ(s,u)=(s∧u)1−2H1u2H1−1|s−u|2H2−2H1−1
participating in the kernel representation (13).
The function φ has the following properties:
for anyu∈[0,T],φ(·,u)∈L1[0,T]andsupu∈[0,T]‖φ(·,u)‖L1<∞
for anyu1∈[0,T],∫0T|φ(s,u)−φ(s,u1)|ds→0asu→u1.
(i) It follows from the evident calculations that
∫0T|φ(s,u)|ds=∫0Tφ(s,u)ds=∫0uu2H1−1dss2H1−1(u−s)1+2H1−2H2+∫uTds(s−u)1+2H1−2H2=u2H2−2H1B(2−2H1,2H2−2H1)+(T−u)2H2−2H12H2−2H1≤CH1,H2T2H2−2H1<∞for allu∈[0,T].
(ii) First, let u1=0 and u↓0. Note that φ(s,0)=s2H2−2H1−1. Therefore,
∫0T|φ(s,u)−1s1+2H1−2H2|ds=∫0u|u2H1−1s2H1−1(u−s)1+2H1−2H2−1s1+2H1−2H2|ds+∫uTds(s−u)1+2H1−2H2−∫uTdss1+2H1−2H2ds≤∫0uu2H1−1dss2H1−1(u−s)1+2H1−2H2+∫0udss1+2H1−2H2+12H2−2H1((s−u)2H2−2H1−s2H2−2H1)|s=us=T=B(2−2H1,2H2−2H1)u2H2−2H1+12H2−2H1(2u2H2−2H1+(T−u)2H2−2H1−T2H2−2H1)→0,asu→0.
From now on suppose that u1>00$]]> is fixed. Without loss of generality, suppose that u↑u1. Then
∫0T|φ(s,u)−φ(s,u1)|ds=∫0u|φ(s,u)−φ(s,u1)|ds+∫uu1|φ(s,u)−φ(s,u1)|ds+∫u1T|φ(s,u)−φ(s,u1)|ds=:I1(u,u1)+I2(u,u1)+I3(u,u1).
Consider the terms separately. First, we establish that φ(s,·) is decreasing in the second argument. Indeed, for 0<s<u<u1,
φ(s,u1)=u12H1−1s2H1−1(u1−s)1+2H1−2H2=1s2H1−1(1−su1)1+2H1−2H2u12−2H2≤1s2H1−1(1−su)1+2H1−2H2u2−2H2=φ(s,u).
Therefore,
I1(u,u1)=∫0u(φ(s,u)−φ(s,u1))ds=∫0uφ(s,u)ds−∫0u1φ(s,u1)ds+∫uu1φ(s,u1)ds≤B(2−2H1,2H2−2H1)(u2H2−2H1−u12H2−2H1)+u12H1−1(u1−u)2H2−2H12H2−2H1→0,asu↑u1.
The second integral vanishes as well:
I2(u,u1)≤∫uu1φ(s,u)ds+∫uu1φ(s,u1)ds≤(12H2−2H1+(u1u)2H1−1)(u1−u)2H2−2H1→0
as u↑u1. Finally,
I3(u,u1)=∫u1Tds(s−u1)1+2H1−2H2−∫u1Tds(s−u)1+2H1−2H2=12H2−2H1×((T−u1)2H2−2H1−(T−u)2H2−2H1+(u1−u)2H2−2H1)→0asu↑u1.
The lemma is proved. □
The kernel κ generates a compact integral operatorκ:C[0,T]→C[0,T].
According to [2], it suffices to prove that the kernel κ defined by (13) satisfies the following two conditions:
for any u∈[0,T], κ(·,u)∈L1[0,T] and supu∈[0,T]‖κ(·,u)‖L1<∞;
For any u1∈[0,T], ∫0T|κ(s,u)−κ(s,u1)|ds→0 as u→u1.
The first condition follows directly from fact that κ0(s,u) is bounded (see Lemma 1) and from Lemma 2 (i).
In order to check (iv), consider
∫0T|κ(s,u)−κ(s,u1)|ds=∫0T|κ0(s,u)φ(s,u)−κ0(s,u1)φ(s,u1)|ds≤∫0Tκ0(s,u)|φ(s,u)−φ(s,u1)|ds+∫0Tφ(s,u1)|κ0(s,u)−κ0(s,u1)|ds.
Again, Lemma 1 in the part that states that κ0(s,u) is bounded, together with Lemma 2 (ii), guarantees that the first term converges to zero as u→u1. Furthermore, Lemma 1 in the part that states that κ0∈C([0,T]02) guarantees that κ0(s,u) converges to κ0(s,u1) as u→u1 for a.e. s∈[0,T]. Since
φ(s,u1)|κ0(s,u)−κ0(s,u1)|≤Cφ(s,u1)∈L1[0,T],
the proof follows from the Lebesgue dominated convergence theorem. □
In the case where H1=12, the kernel κ(s,u) can be simplified to
κ(s,u)=H2(2H2−1)|s−u|2H2−2,
and Eq. (8) coincides with (10). Indeed, let H1=12. Then the function κ0(s,u) equals H2(2H2−1). Consider the function Φ(s,v) defined by (18):
Φ(t,s)=tH2−12(∫01(1−t−stz)H2−12(1−z)H2−32dz+t−st∫01(1−z)H2−12(1−t−stz)H2−32dz)=−tH2−12H2−12∫01((1−t−stz)H2−12(1−z)H2−12)z′dz=tH2−12H2−12.
Combining (28) and (29), we get
κ0(s,u)=(βH2(H2−H1))2∫01(1−t)1−2H2tH2−32Φ(1,u(1−t)s−ut)Φ(1,s(1−t)s−ut)dt=βH22∫01(1−t)1−2H2tH2−32dt=βH22B(H2−12,2−2H2)=H2(2H2−1).
There exists a sequenceTn→∞such that the integral equation (11) has a unique solutionhTn(u)∈C[0,Tn].
We work on the space C([0,T]). Recall that (11) is of the form
hT(u)+1γH12∫0ThT(s)κ(s,u)ds=1,u∈[0,T].
The corresponding homogeneous equation is of the form
∫0ThT(s)κ(s,u)ds=−γH12hT(u),u∈[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>00$]]>, the following equalities hold:
κ0(sa,ua)=κ0(s,u),φ(sa,ua)=a2H2−2H1−1φ(s,u).
Consequently, κ(sa,ua)=a2H2−2H1−1κ(s,u). We can change the variable of integration s=s′T and put u=u′T in (30). Therefore, the equation will be reduced to the equivalent form
∫01hT(Ts)κ(s,u)ds=−γH12T2H1−2H2hT(Tu),u∈[0,1].
Denote λ=−γH12T2H1−2H2. 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 Tn→∞ in such a way that
λn=−γH12Tn2H1−2H2
will be not an eigenvalue. Consequently, the homogeneous equation has only the trivial solution, whence the proof follows. □
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.
The likelihood function is of the formLTn(X,θ)=exp{θδH1N(Tn)−12θ2δH12⟨N⟩(Tn)},and the maximum likelihood estimator is of the formθˆ(Tn)=N(Tn)δH1⟨N⟩(Tn),whereN(t)=E0(X1(t)|FtX)is a square-integrable GaussianFtX-martingale,N(Tn)=∫0TnhTn(t)dX(t)withhTn(t)t12−H1∈L2[0,Tn],hTn(t)is a unique solution to (11), and⟨N⟩(Tn)=γH12∫0TnhTn(t)t1−2H1dt.
The next two results establish basic properties of the estimator; their proofs repeat the proofs of the corresponding statements from [6] and [3].
The estimatorθˆTnis strongly consistent, andlimTn→∞Tn2−2H2Eθ(θˆTn−θ)2=1∫01h0(u)u12−H1du,where the functionh0(u)is the solution of the integral equationκh(u)=γH12.
The estimatorθˆTnis unbiased, and the corresponding estimation error is normalθˆTn−θ∼N(0,1∫0TnhTn(s)s1−2H1ds).
Appendix. Some properties of the hypergeometric function
Recall the integral representation of the Gauss hypergeometric function and some of its properties.
For c>b>0b>0$]]> and x<1, the Gauss hypergeometric function is defined as the integral (see [1], formula 15.3.1)
F(a,b,c;x)=2F1(a,b,c;x)=1B(b,c−b)∫01tb−1(1−t)c−b−1(1−xt)−adt.
For the same values of parameters, the following equality holds (see [1], 15.3.4):
F(a,b,c;x)=(1−x)−aF(a,c−b,c;xx−1),
Evidently, F(a,b,c;x) at x=1 is correctly defined for c−a−b>11$]]> and in this case equals
F(a,b,c;1)=Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b).
Finally, it is easy to check with the help of (31) that
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).
(i) Forc>b>1b>1$]]>,x>00$]]>, and0<a≤1, we have the inequalityF(a,b,c;−x)<1(1+x(b−1)/(c−1))a.
(ii) For0<a≤1,b>00$]]>,c−b>11$]]>, andx∈(0,1), we have the inequalityF(a,b,c;x)<1(1−bc−1x)a.
ReferencesAbramowitz, M., Stegun, I.A.: Handbook of Mathematical Functions. Applied Mathematics Series55, 62 (1966) Berezansky, Y.M., Sheftel, Z.G., Us, G.F.: Functional Analysis, vol. 1. Birkhäuser (2012) Cai, C., Chigansky, P., Kleptsyna, M.: Mixed fractional Brownian motion: The filtering perspective. To appear in Annals of Probability (2015) Jost, C.: Transformation formulas for fractional Brownian motion. Stochastic Processes and their Applications116(10), 1341–1357 (2006) MR2260738. doi:10.1016/j.spa.2006.02.006Karp, D., Sitnik, S.: Two-sided inequalities for generalized hypergeometric function. Research report collection10(2) (2007) Mishura, Y.: Maximum likelihood drift estimation for the mixing of two fractional Brownian motions. arXiv preprint arXiv:1506.04731v1 (2015)