We find the best approximation of the fractional Brownian motion with the Hurst index H∈(0,1/2) by Gaussian martingales of the form ∫0tsγdWs, where W is a Wiener process, γ>00$]]>.

The subject of the present paper is a fractional Brownian motion (fBm) BH={BtH,t≥0} with the Hurst index H∈0,12. Generally speaking, a fBm with the Hurst index H∈(0,1) is a Gaussian process with zero mean and the covariance function of the form
EBtHBsH=12(t2H+s2H−|t−s|2H).
Its properties are rather different for H∈0,12 and H∈12,1. In particular, H∈0,12 implies short-term dependence. In contrast, H∈12,1 implies long-term dependence. Moreover, technically it is easier to deal with fBms having H∈12,1. Due to this and many other reasons, fBm with H∈12,1 has been much more intensively studied in the recent years. However, the financial markets in which trading takes place quite often, demonstrate the presence of a short memory, and therefore the volatility in such markets (so called rough volatility) is well modeled by fBm with H∈0,12, see e.g. [7]. Thus interest to fBm with small Hurst indices has substantially increased recently. Furthermore, it is well known that a fractional Brownian motion is neither a Markov process nor semimartingale, and especially it is neither martingale nor a process with independent increments unless H=12. That is why it is naturally to search the possibility of the approximation of fBm in a certain metric by simpler processes, such as Markov processes, martingales, semimartingales or a processes of bounded variation. As for the processes of bounded variation and semimartingales, corresponding results are presented in [1, 2] and [12]. In the papers [4, 5, 8, 10] approximation of a fractional Brownian motion with Gaussian martingales was studied and summarized in the monograph [3], but most of problems were considered only for H∈12,1, for the reasons stated above.

In the present paper we continue to consider the approximation of a fractional Brownian motion by Gaussian martingales but concentrate on the case H∈0,12.

Let (Ω,F,P) be a complete probability space with a filtration F={Ft}t≥0 satisfying the standard assumptions. We start with the Molchan representation of fBm via the Wiener process on a finite interval. Namely, it was established in [9] that the fBm {BtH,Ft,t≥0} can be represented as
BtH=∫0tz(t,s)dWs,
where {Wt,t∈[0,T]} is a Wiener process,
z(t,s)=cH(tH−1/2s1/2−H(t−s)H−1/2−(H−1/2)s1/2−H∫stuH−3/2(u−s)H−1/2du),
is the Molchan kernel,
cH=2H·Γ(32−H)Γ(H+12)·Γ(2−2H)1/2,
and Γ(x), x>00$]]>, is the Gamma function.

Let us consider a problem of the distance between a fractional Brownian motion and the space of square integrable martingales (initially not obligatory Gaussian), adapted to the same filtration. So, we are looking for a square integrable F-martingale M with the bracket that is absolutely continuous w.r.t. (with respect to) the Lebesgue measure such that it minimizes the value
ρH(M)2:=supt∈[0,T]E(BtH−Mt)2.

We observe first that BH and W generate the same filtration, so any square integrable F-martingale M with the bracket that is absolutely continuous w.r.t. the Lebesgue measure, admits a representation
Mt=∫0ta(s)dWs,
where a is an F-adapted square integrable process such that ⟨M⟩t=∫0ta2(s)ds. Hence we can write, see [10],
E(BtH−Mt)2=E∫0t(z(t,s)−a(s))dWs2=∫0tE(z(t,s)−a(s))2ds=∫0t(z(t,s)−Ea(s))2ds+∫0tVara(s)ds.
Consequently, it is enough to minimize ρH(M) over continuous Gaussian martingales. Such martingales have orthogonal and therefore independent increments. Then the fact that they have representation (3) with a non-random a follows, e.g., from [11].

Now let a:[0,T]→R be a nonrandom measurable function of the class L2[0,T]; that is, a is such that the stochastic integral ∫0ta(s)dWs, t∈[0,T], is well defined w.r.t. the Wiener process {Wt,t∈[0,T]} (this integral is usually called the Wiener integral if the integrand is a nonrandom function). So, the problem is to find
infa∈L2[0,T]sup0≤t≤TEBtH−∫0ta(s)dWs2=infa∈L2[0,T]sup0≤t≤T∫0t(z(t,s)−a(s))2ds.

Note that the expression to be minimized does not involve neither the fractional Brownian motion nor the Wiener process, so the problem becomes purely analytic. Moreover, since the problem posed in a general form is not observable and accessible for solution, we restrict ourselves to one particular subclass of functions. We study the class
{a(s)=sγ,γ>0}.0\}.\]]]>
Our main result is Theorem 1, which shows where maxt∈[0,1]EBtH−∫0ta(s)dWs2 could be reached, depending on γ>00$]]>. We also provide remarks after the theorem.

Distance from fBm with <inline-formula id="j_vmsta156_ineq_037"><alternatives>
<mml:math><mml:mi mathvariant="italic">H</mml:mi><mml:mo stretchy="false">∈</mml:mo><mml:mo mathvariant="normal" fence="true" stretchy="false">(</mml:mo><mml:mn>0</mml:mn><mml:mo mathvariant="normal">,</mml:mo><mml:mn>1</mml:mn><mml:mo mathvariant="normal" stretchy="false">/</mml:mo><mml:mn>2</mml:mn><mml:mo mathvariant="normal" fence="true" stretchy="false">)</mml:mo></mml:math>
<tex-math><![CDATA[$H\in (0,1/2)$]]></tex-math></alternatives></inline-formula> to the subspaces of Gaussian martingales involving power integrands

Consider a class of power functions with an arbitrary positive exponent. Thus, we now introduce the class
{a(s)=sγ,γ>0}.0\}.\]]]>
For the sake of simplicity, let T=1.

Leta=a(s)be a function of the forma(s)=sγ,γ>00$]]>,H∈(0,1/2). Then:

For allγ>00$]]>the maximummaxt∈[0,1]EBtH−∫0tsγdWs2is reached at one of the following points:t=1ort=t1, wheret1=(cHBγ−H+32,H+12(γ+1)−cH2Bγ−H+32,H+12(γ+1)2−2H)1γ−H+12.

For anyH∈(0,1/2)there existsγ0=γ0(H)>00$]]>such that forγ>γ0{\gamma _{0}}$]]>the maximummaxt∈[0,1]EBtH−∫0tsγdWs2=t12H−2t1γ+12+HcHBγ−H+32,H+12γ+1γ+12+H+12γ+1t12γ+1and is reached at the pointt1. HereB(x,y),x,y>00$]]>, is a beta function.

According to Lemma 2.20 [3], the distance between the fractional Brownian motion and the integral ∫0tsγdWs w.r.t. Wiener process t∈[0,1] equals
EBtH−∫0tsγdWs2=EBtH2−2E∫0tz(t,s)dWs∫0tsγdWs+E∫0tsγdWs2=t2H−2∫0tz(t,s)sγds+∫0ts2γds=t2H−2tγ+H+12cHBγ−H+32,H+12γ+1γ+H+12+t2γ+12γ+1:=h(t,γ),
where cH is taken from (2).

Let us calculate the partial derivative of h(t,γ) in t:
∂h(t,γ)∂t=t2H−1(2H−2tγ−H+12cH·Bγ−H+32,H+12(γ+1)+t2(γ−H+12)).

Let us verify whether there is t∈[0,1] such that ∂h(t,γ)∂t=0, i.e.
t2(γ−H+12)−2tγ−H+12cHBγ−H+32,H+12(γ+1)+2H=0.

Changing the variable tγ−H+12=:x, we obtain the following quadratic equation:
x2−2xcHBγ−H+32,H+12(γ+1)+2H=0.

The discriminant D=D(γ) of the quadratic equation (5) equals
D(γ)=4cH2Bγ−H+32,H+12(γ+1)2−8H=8HBγ−H+32,H+12(γ+1)2Γ(32−H)Γ(H+12)Γ(2−2H)−1=8HΓ(H+12)Γ(32−H)Γ(2−2H)Γ(γ−H+32)Γ(γ+1)2−1.

Now we are going to show that D(0)>00$]]> and D(γ) is increasing in γ>00$]]>. For this we study separately the function f(H):=Γ(H+12)(Γ(32−H))3Γ(2−2H) for H∈(0,12).

The behavior of f(H) for H∈(0,12)

The behavior of D(0) as a function of H∈(0,12)

Let us calculate the values of this function at the following points: f(0)=π28>11$]]>, f(12)=1. To establish that f(H) is decreasing in H, consider Lemma 3 and the following calculation:
(lnf(H))H′=lnΓ(H+12)(Γ(32−H))3Γ(2−2H)H′=lnΓH+12+3lnΓ32−H−lnΓ2−2HH′=Γ′(H+12)Γ(H+12)−3Γ′(32−H)Γ(32−H)+2Γ′(2−2H)Γ(2−2H)=∫011−tH−121−tdt−C−3∫011−t12−H1−tdt−C+2∫011−t1−2H1−tdt−C=∫011−tH−12−3+3t12−H+2−2t1−2H1−tdt=∫013t12−H−2t1−2H−tH−121−tdt=∫01tH−12(3t1−2H−2t32−3H−1)1−tdt.

Let t∈(0,1). Obviously, in this case tH−12>00$]]> and 1−t>00$]]>. Changing the variables in (7) as z:=t12−H, we get
3z2−2z3−1=−(1−z)2(2z+1),
and this function is negative for all z∈(0,1). Hence, (lnf(H))′<0 and it means that f(H) is decreasing. Furthermore, f(H)>11$]]> for every H∈(0,12). The behavior of f(H) is presented in Figure 1.

So, we proved that D(0)>00$]]> (the behavior of D(0) as a function of H is presented in Figure 2), and it follows from Lemma 4 that D(γ) is increasing in γ>00$]]> for any H∈(0,1/2). Therefore, for every H∈(0,12) and γ>00$]]> we have that the quadratic equation (5) has two roots.

More precisely, if you use standard notations for the coefficients of the quadratic equation, then coefficient a at x2 in (5) is strongly positive, a=1, coefficient at x equals b=−2cHB(γ−H+32,H+12)(γ+1) and is negative, and c=2H>00$]]>. We conclude that our quadratic equation has two positive roots x1=−b−b2−4ac2a and x2=−b+b2−4ac2a, x1≤x2.

According to our notations, we let ti:=xi1γ−H+12, i=1,2. Since x=tγ−H+12∈[0,1] for t∈[0,1] and the left-hand side of (5) is negative for x∈(x1,x2), we get the following cases:

Let x1<1 and x2<1. Then maxt∈[0,1]h(t,γ) can be achieved at one of two points: t=t1 or t=1.

Let x1<1 and x2≥1. Then maxt∈[0,1]h(t,γ) is achieved at point t=t1.

Let x1≥1 (and consequently x2>11$]]>). Then maxt∈[0,1]h(t,γ) is achieved at point t=1.

Now, we rewrite the discriminant (6) in the following form:
D=4cH2Bγ−H+32,H+12(γ+1)2−2H=:4(dH2(γ)−2H)>0,0,\end{aligned}\]]]>
where dH(γ)=cHB(γ−H+32,H+12)(γ+1). From Lemma 5, x1<2H<1, so case (iii) never occurs.

According to Lemma 5, the biggest of two roots, x2=dH(γ)+dH2(γ)−2H, is increasing in γ>00$]]> and x2>1⇔dH>12+H1\Leftrightarrow {d_{H}}>\frac{1}{2}+H$]]>. Moreover, it follows from Lemma 5, (iv) that dH(γ)→+∞ as γ→+∞. Therefore, for all H∈(0,12) there exists γ0(H)>00$]]> such that for all γ>γ0{\gamma _{0}}$]]> we have x2>11$]]>.

It means that for γ>γ0{\gamma _{0}}$]]> our maximum is reached at the point t1=(x1)1γ+12−H. Finally,
maxt∈[0,1]EBtH−∫0tsγdWs2=t12H−2t1γ+12+HcHBγ−H+32,H+12γ+1γ+12+H+12γ+1t12γ+1,
where t1=(x1)1γ+12−H. □

The implicit equation dH(γ)=12+H considered as the equation for γ0 as a function of H gives us the relation between H∈0,12 and respective γ0>00$]]> which, by virtue of the foregoing, is determined unambiguously. The form of the algebraic curve γ0=γ0(H) is presented in Figure 3.

The algebraic curve γ0=γ0(H)

Consider one of the coefficients that are present in (4), namely,
cH,γ1=2cHBγ−H+32,H+12γ+1γ+H+12=22HΓ32−HΓH+12Γ2−2H1/2Γγ−H+32ΓH+12Γγ+1γ+H+12.
Let γ=0. Then
cH,01=22HΓ32−HΓH+12Γ2−2H1/2Γ32−HΓH+12H+12=2H+122HΓ32−HΓH+12Γ2−2H1/2π12−Hsinπ12−H.
For H=12, one has c1/2,01=2. Obviously, cH,01→0 as H↓0. It means that cH,01 is small in some neighborhood of zero. Having this in mind, we establish some sufficient condition for h(t,0) to get its maximum at point 1.

LetcH,01<1. Thenh(t1,0)<h(1,0).

Indeed, h(t1,0)=t12H−cH,01t11/2+H+t1, while h(1,0)=2−cH,01. Then the inequality h(t1,0)<h(1,0) is equivalent to the following one:
cH,011−t11/2+H<2−t1−t12H.
If cH,01<1, then cH,01(1−t11/2+H)<1−t11/2+H<1−t1<2−t1−t12H. □

As it was mentioned before, c1/2,01=2, and so for H=12 the condition of Lemma 1 is not satisfied. In this case d1/2(0)=1, and t1=t2=1, so that we have the equality h(t1,0)=h(1,0).

However, the question what will be for γ=0 and H0<H<12, where H0 is such a value that for 0<H<H0, cH,01<1, is open. In order to fill this gap, we provide the numerical results with some comments.

Consider two fuctions h(t1,γ) and h(1,γ) as functions of γ and H. We already know that maxt∈[0,1]h(t,γ)=max{h(t1,γ),h(1,γ)}. The projection of the surface of max{h(t1,γ),h(1,γ)} on the (H,γ)-plane is presented in Figure 4. Points, where h(t,γ) reaches its maximum at t=1 are represented in green color, and points where h(t,γ) reaches its maximum at t=t1 are represented in brown color. The black curve is the algebraic curve γ0=γ0(H), which is presented in Figure 3.

The projection of the surface of maxt∈[0,1]h(t,γ)

Appendix section

In the proof of Theorem 1, we use these auxiliary results.

Letf:R→Rbe a strictly convex function of one variable. Take the functiong(x1,x2)=f(x1)−f(x2)x1−x2,x1≠x2,x1,x2∈R. Theng(x1+α,x2+α)is strictly increasing inα>00$]]>.

LetΓ(x)be the Gamma function. Then(lnΓ(x))′=Γ′(x)Γ(x)=∫011−tx−11−tdt−C,where C is a fixed constant.

Proofs of Lemma 2 and Lemma 3 could be found in [6].

IfH∈0,12, then the functionz(γ):=Γ(γ−H+32)Γ(γ+1)is increasing inγ>00$]]>.

For every γ>00$]]> we have z(γ)>00$]]>. Let us calculate
lnz(γ)=lnΓγ−H+32Γ(γ+1)==12−HlnΓγ−H+32−lnΓ(γ+1)γ−H+32−(γ+1)=:12−Hω(γ).
According to Lemma 2 and the fact that ln(Γ(x)) is strictly convex we have that ω(γ) is increasing. Since 12−H>00$]]>, it is clear that z(γ) is increasing in γ>00$]]>. □

LetdH(γ)=cHB(γ−H+32,H+12)(γ+1), andx1=dH(γ)−dH2(γ)−2H,x2=dH(γ)+dH2(γ)−2Hbe roots of the quadratic equation (5). Then for allγ>00$]]>,H∈(0,12)the following statements hold.

(i) Note that dH(γ) is increasing in γ>00$]]> since
dH(γ)=cHΓH+12Γγ−H+32Γγ+1,
where cHΓH+12>00$]]> for all H∈0,12, and according to Lemma 4, Γ(γ−H+32)Γ(γ+1) is increasing in γ>00$]]>.

(ii) Discriminant (6) satisfies the following relation:
0<D0=4dH20−2H.
Therefore, dH(γ)>2H\sqrt{2H}$]]> for all γ>00$]]>. Also, we can rewrite x1=dH(γ)−dH2(γ)−2H and x2=dH(γ)+dH2(γ)−2H. Hence x2>2H\sqrt{2H}$]]>. Transform the value x1 to the following form:
x1=dH2(γ)−(dH2(γ)−2H)dH(γ)+dH2(γ)−2H=2Hx2<2H.

(iii) Let us assume that x2>11$]]> (or, what is equivalent, x1=2Hx2<2H). In turn, this is equivalent to the relation dH(γ)+dH2(γ)−2H>11$]]>, or dH2(γ)−2H>1−dH(γ)1-{d_{H}}(\gamma )$]]>. The latter inequality can be realized in one of two cases:
dH2(γ)−2H>1−2dH(γ)+dH2(γ),1−dH(γ)≥0;1-2{d_{H}}(\gamma )+{d_{H}^{2}}(\gamma ),\hspace{1em}1-{d_{H}}(\gamma )\ge 0;\]]]>
or
1−dH(γ)<0.

The couple of inequalities (9) is equivalent to 1≥dH(γ)>12+H\frac{1}{2}+H$]]>. Therefore, inequalities (9) and (10), taken together, indicate that dH(γ)>12+H\frac{1}{2}+H$]]> if and only if x2>11$]]>.

(iv) The value dH(γ) can be presented as
dH(γ)=cHΓγ−H+32ΓH+12Γγ+2γ+1,
where cHΓH+12>00$]]> is a fixed constant, and for all H∈0,12,
Γγ−H+32Γγ+2γ+1=Γγ−H+32Γγ+1=2πγ−H+32γ−H+32eγ−H+321+O1γ−H+322πγ+1γ+1eγ+11+O1γ+1∼1e12−Hγ−H+32γ−H+1γ+1γ+12→∞,γ→∞,
which follows from the Stirling’s approximation for Gamma function. □

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