VMSTA Modern Stochastics: Theory and Applications 2351-6054 2351-6046 2351-6046 VTeXMokslininkų g. 2A, 08412 Vilnius, Lithuania VMSTA156 10.15559/20-VMSTA156 Research Article Distance from fractional Brownian motion with associated Hurst index 0<H<1/2 to the subspaces of Gaussian martingales involving power integrands with an arbitrary positive exponent https://orcid.org/0000-0002-9730-4654 BannaOksanaokskot@ukr.neta BuryakFilippfilippburyak2000@gmail.comb https://orcid.org/0000-0002-6877-1800 MishuraYuliyamyus@univ.kiev.uab Kyiv National Taras Shevchenko University, Faculty of Economics, Volodymyrska 64, 01601 Kyiv, Ukraine Kyiv National Taras Shevchenko University, Faculty of Mechanics and Mathematics, Volodymyrska 64, 01601 Kyiv, Ukraine Corresponding author. 2020 236202072191202 2242020 662020 662020 © 2020 The Author(s). Published by VTeX2020 Open access article under the CC BY license.

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>0 0$]]>. Fractional Brownian motion martingale approximation 60G22 60G44 Introduction The subject of the present paper is a fractional Brownian motion (fBm) BH={BtH,t0} with the Hurst index H0,12 . 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|ts|2H). Its properties are rather different for H0,12 and H12,1 . In particular, H0,12 implies short-term dependence. In contrast, H12,1 implies long-term dependence. Moreover, technically it is easier to deal with fBms having H12,1 . Due to this and many other reasons, fBm with H12,1 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 H0,12 , see e.g. . 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  and . In the papers  approximation of a fractional Brownian motion with Gaussian martingales was studied and summarized in the monograph , but most of problems were considered only for H12,1 , 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 H0,12 . Let (Ω,F,P) be a complete probability space with a filtration F={Ft}t0 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  that the fBm {BtH,Ft,t0} can be represented as BtH= 0tz(t,s)dWs, where {Wt,t[0,T]} is a Wiener process, z(t,s)=cH(tH1/2s1/2H(ts)H1/2(H1/2)s1/2H stuH3/2(us)H1/2du), is the Molchan kernel, cH=2H·Γ(32H)Γ(H+12)·Γ(22H)1/2, and Γ(x) , x>0 0$]]>, 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(BtHMt)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 Mt=0ta2(s)ds . Hence we can write, see , E(BtHMt)2=E0t(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 .

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 infaL2[0,T]sup0tTEBtH 0ta(s)dWs2=infaL2[0,T]sup0tT 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]EBtH0ta(s)dWs2 could be reached, depending on γ>0 0$]]>. We also provide remarks after the theorem. Distance from fBm with <inline-formula id="j_vmsta156_ineq_037"><alternatives> <mml:math><mml:mi mathvariant="italic">H</mml:mi><mml:mo stretchy="false">∈</mml:mo><mml:mo mathvariant="normal" fence="true" stretchy="false">(</mml:mo><mml:mn>0</mml:mn><mml:mo mathvariant="normal">,</mml:mo><mml:mn>1</mml:mn><mml:mo mathvariant="normal" stretchy="false">/</mml:mo><mml:mn>2</mml:mn><mml:mo mathvariant="normal" fence="true" stretchy="false">)</mml:mo></mml:math> <tex-math><![CDATA[$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 . Let a=a(s) be a function of the form a(s)=sγ , γ>0 0$]]>, H(0,1/2) . Then:

For all γ>0 0$]]> the maximum maxt[0,1]EBtH0tsγdWs2 is reached at one of the following points: t=1 or t=t1 , where t1=(cHBγH+32,H+12(γ+1)cH2BγH+32,H+12(γ+1)22H)1γH+12. For any H(0,1/2) there exists γ0=γ0(H)>0 0$]]> such that for γ>γ0 {\gamma _{0}}$]]> the maximum maxt[0,1]EBtH0tsγdWs2=t12H2t1γ+12+HcHBγH+32,H+12γ+1γ+12+H+12γ+1t12γ+1 and is reached at the point t1 . Here B(x,y) , x,y>0 0$]]>, is a beta function.

According to Lemma 2.20 , the distance between the fractional Brownian motion and the integral 0tsγdWs w.r.t. Wiener process t[0,1] equals EBtH0tsγdWs2=EBtH22E0tz(t,s)dWs0tsγdWs+E0tsγdWs2=t2H20tz(t,s)sγds+0ts2γds=t2H2tγ+H+12cHBγH+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=t2H1(2H2tγH+12cH·BγH+32,H+12(γ+1)+t2(γH+12)).

Let us verify whether there is t[0,1] such that h(t,γ)t=0 , i.e. t2(γH+12)2tγH+12cHBγH+32,H+12(γ+1)+2H=0.

Changing the variable tγH+12=:x , we obtain the following quadratic equation: x22xcHBγH+32,H+12(γ+1)+2H=0.

The discriminant D=D(γ) of the quadratic equation (5) equals D(γ)=4cH2BγH+32,H+12(γ+1)28H=8HBγH+32,H+12(γ+1)2Γ(32H)Γ(H+12)Γ(22H)1=8HΓ(H+12)Γ(32H)Γ(22H)Γ(γH+32)Γ(γ+1)21.

Now we are going to show that D(0)>0 0$]]> and D(γ) is increasing in γ>0 0$]]>. For this we study separately the function f(H):=Γ(H+12)(Γ(32H))3Γ(22H) 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>1 1$]]>, f(12)=1 . To establish that f(H) is decreasing in H, consider Lemma 3 and the following calculation: (lnf(H))H=lnΓ(H+12)(Γ(32H))3Γ(22H)H=lnΓH+12+3lnΓ32HlnΓ22HH=Γ(H+12)Γ(H+12)3Γ(32H)Γ(32H)+2Γ(22H)Γ(22H)=011tH121tdtC3011t12H1tdtC+2011t12H1tdtC=011tH123+3t12H+22t12H1tdt=013t12H2t12HtH121tdt=01tH12(3t12H2t323H1)1tdt. Let t(0,1) . Obviously, in this case tH12>0 0$]]> and 1t>0 0$]]>. Changing the variables in (7) as z:=t12H , we get 3z22z31=(1z)2(2z+1), and this function is negative for all z(0,1) . Hence, (lnf(H))<0 and it means that f(H) is decreasing. Furthermore, f(H)>1 1$]]> for every H(0,12) . The behavior of f(H) is presented in Figure 1.

So, we proved that D(0)>0 0$]]> (the behavior of D(0) as a function of H is presented in Figure 2), and it follows from Lemma 4 that D(γ) is increasing in γ>0 0$]]> for any H(0,1/2) . Therefore, for every H(0,12) and γ>0 0$]]> we have that the quadratic equation (5) has two roots. More precisely, if you use standard notations for the coefficients of the quadratic equation, then coefficient a at x2 in (5) is strongly positive, a=1 , coefficient at x equals b=2cHB(γH+32,H+12)(γ+1) and is negative, and c=2H>0 0$]]>. We conclude that our quadratic equation has two positive roots x1=bb24ac2a and x2=b+b24ac2a , x1x2 .

According to our notations, we let ti:=xi1γH+12 , i=1,2 . Since x=tγH+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 x21 . Then maxt[0,1]h(t,γ) is achieved at point t=t1 .

Let x11 (and consequently x2>1 1]]>). Then maxt[0,1]h(t,γ) is achieved at point t=1 . Now, we rewrite the discriminant (6) in the following form: D=4cH2BγH+32,H+12(γ+1)22H=:4(dH2(γ)2H)>0, 0,\end{aligned}\]]]> where dH(γ)=cHB(γH+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>0 0]]> and x2>1dH>12+H 1\Leftrightarrow {d_{H}}>\frac{1}{2}+H$]]>. Moreover, it follows from Lemma 5, (iv) that dH(γ)+ as γ+ . Therefore, for all H(0,12) there exists γ0(H)>0 0$]]> such that for all γ>γ0 {\gamma _{0}}$]]> we have x2>1 1$]]>.

It means that for γ>γ0 {\gamma _{0}}$]]> our maximum is reached at the point t1=(x1)1γ+12H . Finally, maxt[0,1]EBtH0tsγdWs2=t12H2t1γ+12+HcHBγH+32,H+12γ+1γ+12+H+12γ+1t12γ+1, where t1=(x1)1γ+12H . □ The implicit equation dH(γ)=12+H considered as the equation for γ0 as a function of H gives us the relation between H0,12 and respective γ0>0 0$]]> 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=2cHBγH+32,H+12γ+1γ+H+12=22HΓ32HΓH+12Γ22H1/2ΓγH+32ΓH+12Γγ+1γ+H+12. Let γ=0 . Then cH,01=22HΓ32HΓH+12Γ22H1/2Γ32HΓH+12H+12=2H+122HΓ32HΓH+12Γ22H1/2π12Hsinπ12H. For H=12 , one has c1/2,01=2 . Obviously, cH,010 as H0 . 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.

Let cH,01<1 . Then h(t1,0)<h(1,0) .

Indeed, h(t1,0)=t12HcH,01t11/2+H+t1 , while h(1,0)=2cH,01 . Then the inequality h(t1,0)<h(1,0) is equivalent to the following one: cH,011t11/2+H<2t1t12H. If cH,01<1 , then cH,01(1t11/2+H)<1t11/2+H<1t1<2t1t12H .  □

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.

Let f:RR be a strictly convex function of one variable. Take the function g(x1,x2)=f(x1)f(x2)x1x2 , x1x2 , x1,x2R . Then g(x1+α,x2+α) is strictly increasing in α>0 0$]]>. Let Γ(x) be the Gamma function. Then (lnΓ(x))=Γ(x)Γ(x)=011tx11tdtC, where C is a fixed constant. Proofs of Lemma 2 and Lemma 3 could be found in . If H0,12 , then the function z(γ):=Γ(γH+32)Γ(γ+1) is increasing in γ>0 0$]]>.

For every γ>0 0$]]> we have z(γ)>0 0$]]>. Let us calculate lnz(γ)=lnΓγH+32Γ(γ+1)==12HlnΓγH+32lnΓ(γ+1)γH+32(γ+1)=:12Hω(γ). According to Lemma 2 and the fact that ln(Γ(x)) is strictly convex we have that ω(γ) is increasing. Since 12H>0 0$]]>, it is clear that z(γ) is increasing in γ>0 0$]]>.  □

Let dH(γ)=cHB(γH+32,H+12)(γ+1) , and x1=dH(γ)dH2(γ)2H , x2=dH(γ)+dH2(γ)2H be roots of the quadratic equation (5). Then for all γ>0 0$]]>, H(0,12) the following statements hold. dH(γ) is increasing in γ>0 0$]]>.

x1<2H and x2>2H \sqrt{2H}$]]>. dH(γ)>12+Hx2>1 \frac{1}{2}+H\Leftrightarrow {x_{2}}>1$]]> and x1<2H .

dH(γ) as γ .

(i) Note that dH(γ) is increasing in γ>0 0$]]> since dH(γ)=cHΓH+12ΓγH+32Γγ+1, where cHΓH+12>0 0$]]> for all H0,12 , and according to Lemma 4, Γ(γH+32)Γ(γ+1) is increasing in γ>0 0$]]>. (ii) Discriminant (6) satisfies the following relation: 0<D0=4dH202H. Therefore, dH(γ)>2H \sqrt{2H}$]]> for all γ>0 0$]]>. Also, we can rewrite x1=dH(γ)dH2(γ)2H and x2=dH(γ)+dH2(γ)2H . Hence x2>2H \sqrt{2H}$]]>. 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>1 1$]]> (or, what is equivalent, x1=2Hx2<2H ). In turn, this is equivalent to the relation dH(γ)+dH2(γ)2H>1 1$]]>, or dH2(γ)2H>1dH(γ) 1-{d_{H}}(\gamma )$]]>. The latter inequality can be realized in one of two cases: dH2(γ)2H>12dH(γ)+dH2(γ),1dH(γ)0; 1-2{d_{H}}(\gamma )+{d_{H}^{2}}(\gamma ),\hspace{1em}1-{d_{H}}(\gamma )\ge 0;\]]]> or 1dH(γ)<0. The couple of inequalities (9) is equivalent to 1dH(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>1 1$]]>.

(iv) The value dH(γ) can be presented as dH(γ)=cHΓγH+32ΓH+12Γγ+2γ+1, where cHΓH+12>0 0\$]]> is a fixed constant, and for all H0,12 , ΓγH+32Γγ+2γ+1=ΓγH+32Γγ+1=2πγH+32γH+32eγH+321+O1γH+322πγ+1γ+1eγ+11+O1γ+11e12HγH+32γH+1γ+1γ+12,γ, which follows from the Stirling’s approximation for Gamma function.  □

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