A generalized fractional Brownian motion (gfBm) with parameters a,b , and H, is a process ZH={ZtH(a,b);t≥0}={ZtH;t≥0} defined on the probability space (Ω,F,P) by (3) ∀t∈R+,ZtH=ZtH(a,b)=aBtH+bB−tH, where (BtH)t∈R is a two-sided fractional Brownian motion with parameter H.

The gfBm (ZtH(a,b))t∈R+ satisfies the following properties: •

ZH is a centered Gaussian process.

The gfBm is a self-similar process.

This follows from the fact that, for fixed h>0 0$]]>, the processes {ZhtH(a,b);t≥0} and {hHZtH(a,b);t≥0} are Gaussian, centered, and have the same covariance function.  □

We will only prove the proposition in the case where b≠0 ; the result with b=0 is known (see [10] and the references therein). The process ZH is a centered Gaussian, and for all t>0 0$]]>, Cov(ZtH,ZtH)=(a2+b2−(22H−2)ab)t2H>0. 0.\]]]> Then, if ZH were a Markov process, according to [8], for all s<t<u , we would have (4) Cov(ZsH,ZuH)Cov(ZtH,ZtH)=Cov(ZsH,ZtH)Cov(ZtH,ZuH).

In the particular case where 1<s=t<t<u=t2 , we have [12(a+b)2(1+t−3H)−ab(1+t−3/2)2H−a2+b22(1−t−3/2)2H]×[a2+b2)−(22H−2)ab]=[12(a+b)2(1+t−H)−ab(1+t−1/2)2H−a2+b22(1−t−1/2)2H]×[12(a+b)2(1+t−2H)−ab(1+t−1)2H−a2+b22(1−t−1)2H].

So [12(a+b)2(1+t−3H)−ab(1+2Ht−3/2+H(2H−1)t−3+o(t−3))−a2+b22(1−2Ht−3/2+H(2H−1)t−3+o(t−3))]×[a2+b2−(22H−2)ab]=[12(a+b)2(1+t−H)−ab(1+2Ht−1/2+H(2H−1)t−1+o(t−1))−a2+b22(1−2Ht−1/2+H(2H−1)t−1+o(t−1))]×[12(a+b)2(1+t−2H)−ab(1+2Ht−1+H(2H−1)t−2+o(t−2))−a2+b22(1−2Ht−1+H(2H−1)t−2+o(t−2))].

First case: 0<H<12 , a+b≠0 . By Taylor’s expansion we get, as t→∞ , 12(a+b)2[a2+b2−(22H−2)ab]t−3H≈14(a+b)4t−3H, which is true if and only if (a−b)22−(22H−2)ab=0. However, it is easy to check that (a−b)22−(22H−2)ab>0 0$]]> for fixed b and every real a.

Second case: 0<H<12 and a+b=0 . By Taylor’s expansion we get, as t→∞ , [a2+b2−(22H−2)ab]t−3/2≈(−2Hab+(a2+b2)H)t−3/2, which is true if and only if a=b=0 , which is false.

Third case: 12<H<1 , a−b≠0 . By Taylor’s expansion we get, as t→∞ , H(a−b)2[a2+b2−(22H−2)ab]t−3/2≈H2(a−b)4t−3/2, which is true if and only if a2(1−H)+b2(1−H)+ab(2−22H+2H)=0. However, it is easy to check that a2(1−H)+b2(1−H)+ab(2−22H+2H)>0 0$]]> for fixed b and every real a.

Fourth case: 12<H<1 and a−b=0 . By Taylor’s expansion we get, as t→∞ , 12(a+b)2[a2+b2−(22H−2)ab]t−3H≈14(a+b)4t−3H, which is true if and only if 2−22H=0 , which is false since H≠12 . The proof of Lemma 1 is complete.  □

For all (s,t)∈R+2 such that s≤t : 1.

E(ZtH(a,b)−ZsH(a,b))2=(a2+b2)|t−s|2H−22Hab(|t|2H+|s|2H)+2ab|t+s|2H;

Moreover, the constants in the inequalities of statement 2 are the best possible.

The first statement is a direct consequence of Eqs. (1) and (3). So we will just check the second one. We will do that in the case where H>12 \frac{1}{2}$]]> and ab≥0 ; the proof in the remaining cases is similar.

Since the function x⟼x2H is convex on R+ , we have 22H−1(t2H+s2H)−(t+s)2H≥0, which yields that E(ZtH(a,b)−ZsH(a,b))2≤(a2+b2)|t−s|2H.

To get the lower bound, we consider the function fλ:x⟼22H−1((s+x)2H+s2H)−(2s+x)2H−λx2H, with λ>0 0$]]>. We have fλ(0)=0 . So to get the stated lower bound, it suffices to check that, for λ=22H−1−1 , fλ is a decreasing function. But here we show more; we look for the lower bound ν (resp. the upper bound γ) of the set of reals λ>0 0$]]> such that fλ decreases (resp.  fλ increases) for x>0 0$]]>, which will give us the maximum γ≥0 and the minimum ν>0 0$]]> such that γ(t−s)2H≤22H−1(t2H+s2H)−(t+s)2H≤ν(t−s)2H.

The function fλ is differentiable in R+⋆ , and for all x>0 0$]]>, fλ′(x)=2H[22H−1(s+x)2H−1−(2s+x)2H−1−λx2H−1]. So fλ′(x)<0⟺22H−1(s+x)2H−1−(2s+x)2H−1<λx2H−1⟺22H−1(s+x)2H−1−(2s+x)2H−1x2H−1<λ.

Denote g(x)=22H−1(s+x)2H−1−(2s+x)2H−1x2H−1 , that is, g(x)=22H−1(sx+1)2H−1−(2sx+1)2H−1.

Putting X=sx , we can write g(x)=h(X) where h(X)=22H−1(X+1)2H−1−(2X+1)2H−1. The function h is continuous and strictly decreasing. Consequently, h(]0,+∞[)=]limX→+∞h;limX→0h[=]0;22H−1−1[. So, supx>0g=22H−1−1 0}g={2}^{2H-1}-1$]]> and infx>0g=0 0}g=0$]]>. We deduce that the lower bound ν>0 0$]]> and the upper bound γ>0 0$]]> such that γ(t−s)2H≤22H−1(t2H+s2H)−(t+s)2H≤ν(t−s)2H are ν=22H−1−1 and γ=0 .  □

For every H∈]0,1[∖{12} , the gfBm is not a semimartingale.

By Lemma 2.1 of [2] this result is a direct consequence of Lemma 3.  □

Let H∈(0,1) . 1.

The gfBm ZH admits a version whose sample paths are almost surely Hölder continuous of order strictly less than H.

The first statement follows by the Kolmogorov criterion from Lemma 3.

The second one is easily obtained by using the specific expression of the second moment of the increments of the gfBm (10) and by following exactly the same strategies as in the proofs of Lemma 4.1 and Lemma 4.2 in [11].  □

For every n≥1 , we have, as p→∞ , RZ(p,p+n)=(a2+b2)RB(0,n)−ab(22H−1H(2H−1))p2(H−1)(1+o(1)), and, consequently, (7) limp→∞RZ(p,p+n)=(a2+b2)RB(0,n).

An easy calculus allows us to get (8) RZ(p,p+n)=a2+b22[(n+1)2H−2n2H+(n−1)2H]−ab[(2p+n+2)2H−2(2p+n+1)2H+(2p+n)2H]=(a2+b2)RB(0,n)−abfp(n), for every n≥1 , where fp(n)=(2p+n+2)2H−2(2p+n+1)2H+(2p+n)2H .

By Taylor’s expansion we have, as p→∞ , fp(n)=(2p)2H[(1+n+22p)2H−2(1+n+12p)2H+(1+n2p)2H]=(22H−1H(2H−1))p2(H−1)(1+o(1)). Since H<1 , the last term tends to 0 as p goes to infinity.  □

If a≠0 and b≠0 , then the gfBm increments are not stationary. The meaning of the proposition is that they converge to a stationary sequence.

We say that the increments of a stochastic process X are long-range dependent if for every integer p≥1 , we have ∑n≥1RX(p,p+n)=∞, where RX(p,p+n)=E((Xp+1−Xp)(Xp+n+1−Xp+n)) .

For every (a,b)∈R2∖{(0,0)} , the increments of ZH(a,b) are long-range dependent if and only if H>12 \frac{1}{2}$]]> and a≠b .

For every integer p≥1 , by Taylor’s expansion, as n→∞ , we have RZ(p,p+n)=a2+b22n2H[(1+1n)2H−2+(1−1n)2H]−abn2H[(1+2p+2n)2H−2(1+2p+1n)2H+(1+2pn)2H]=H(2H−1)n2H−2(a−b)2−4H(2H−1)(H−1)ab(2p+1)n2H−3(1+o(1)).

If a≠b , we see that as n→∞ , RZ(p,p+n)≈H(2H−1)n2H−2(a−b)2 . Then ∑n≥1RZ(p,p+n)=∞⟺2H−2>−1⟺H>12. -1\hspace{1em}\Longleftrightarrow \hspace{1em}H>\frac{1}{2}.\]]]>

If a=b , then, as n→∞ , RZ(p,p+n)≈4H(2H−1)(1−H)a2(2p+1)n2H−3 . For every H∈]0,1[ , we have 2H−3<−1 and, consequently, ∑n≥1RZ(p,p+n)<∞.  □

We know that the subfractional Brownian motion increments ZH(12,12) are short-range dependent for every H∈]0,1[ . From the theorem we see another important motivation of the investigation of the general process ZH(a,b) with a and b not necessary equal; it allows us to exhibit models taking into account not only short-range dependence of the increments, but also phenomena of long-range dependence if it exists.

For 0<H<1 , on each (time)-interval [0,T]⊂[0,+∞[ , the process (ZtH)0≤t≤T admits a local time LH([0,T],x) , which satisfies ∫RLH([0,T],x)2dx<∞.

First, denote, for s,t∈R+ and s≠t , by φZtH−ZsH the characteristic function of the random variable ZtH−ZsH and by pH(x;t,s) its probability density function. They are expressed by (9) φZtH−ZsH(u)=E[exp(iu(ZtH−ZsH))]=exp(−u22E(ZtH−ZsH)2), and (10) pH(x;s,t)=1/2πE(ZtH−ZsH)2exp(−x2/2E(ZtH−ZsH)2). It is clear that ∫−∞∞|φZtH−ZsH(u)|du<∞, and by Lemma 3 and the fact that 0<H<1 , for every T>0 0$]]>, ∫0T∫0TpH(0;s,t)dsdt≤c∫0T∫0T|t−s|−Hdsdt<∞, where c is positive constant.

So, by Lemma 3.2 of [1] the local time of (ZtH)0≤t≤T exists almost surely, and it is square integrable as a function of u.  □