Large deviations and long-time behavior of stochastic fluid queues with generalized fractional Brownian motion input

Abstract

We study the large deviation behaviors of a stochastic fluid queue with an input being a generalized Riemann–Liouville (R–L) fractional Brownian motion (FBM), referred to as GFBM. The GFBM is a continuous mean-zero Gaussian process with non-stationary increments, extending the standard FBM with stationary increments. We first derive the large deviation principle for the GFBM by using the weak convergence approach. We then obtain the large deviation principle for the stochastic fluid queue with the GFBM as the input process as well as for an associated running maximum process. Finally, we study the long-time behavior of these two processes; in particular, we show that a steady-state distribution exists and derives the exact tail asymptotics using the aforementioned large deviation principle together with some maximal inequality and modulus of continuity estimates for the GFBM.

Appendix A: Auxiliary results

In the following, we present a few results that used in the Sect. 5. Define a following continuous map from $\mathcal {C}_T$ to $\mathcal {C}_1$:

$$\begin{aligned} J_T : \xi (\cdot )\mapsto \xi (T \cdot ) . \end{aligned}$$

(A.1)

Lemma A.1

For the GFBM X in (2.1), and for any $T>0$,

$$\begin{aligned} \mathcal {P}_X\circ J_T^{-1}(\cdot )= \mathcal {P}_X (T^{-H} \cdot ), \end{aligned}$$

where $\mathcal {P}_X$ is the law of X.

Proof

For a given $n\in \mathbb {N}$, consider $0<u_1<u_2<u_3<\cdots<u_n<1$. Now from the self-similarity of X [27, Proposition 5.1], it is clear that for $n=1$,

$$\begin{aligned} Z(T u_1){\mathop {=}\limits ^{\textrm{d}}}T ^{H}Z(u_1). \end{aligned}$$

In other words,

$$\begin{aligned} \mathcal {P}_X\circ J_T^{-1}(\xi (u_1)\in \cdot )= \mathcal {P}_X (\xi (u_1)\in T^{-H} \cdot ). \end{aligned}$$

For $n>1$, assume that

$$\begin{aligned} \left( (Z(T u_1),Z(T u_2),\ldots , Z(T u_{n-1}))\in A\right) {\mathop {=}\limits ^{\textrm{d}}}\left( (Z( u_1),Z( u_2),\ldots , Z(u_{n-1}))\in T ^{-H}A\right) , \end{aligned}$$

(A.2)

for any $A\subset \mathcal {B}(\mathbb {R}^{n-1})$. Now consider the Borel set $B\in \mathcal {B}(\mathbb {R})$. Then, we have

$$\begin{aligned}&\left( (Z(T u_1),Z(T u_2),\ldots , Z(T u_{n},Z(T u_n)))\in A\times B\right) \\&\quad {\mathop {=}\limits ^{\textrm{d}}}\left( (Z( u_1),Z( u_2),\ldots , Z(u_{n-1}))\in T ^{-H}A\times T ^{-H}B\right) . \end{aligned}$$

Since the sets of the form $A\times B$ generate all the Borel sets of $\mathbb {R}^{n}$, the self-similarity property (A.2) holds for n and therefore by induction, all finite dimensional distributions. It is trivial to see that the finite dimensional distributions of $\mathcal {P}_X\circ J_T^{-1}$ and $\mathcal {P}_X$ are consistent families of measures. Therefore, using the Kolmogorov consistency theorem, we get the desired result. $\square$

Remark A.1

The above statement and proof can be generalized to processes with RCLL (right continuous with left limits) paths. Indeed, we construct a similar map $J_T$ on $\mathcal {D}_T$ to $\mathcal {D}_1$ (here, $\mathcal {D}_T$ is space of functions that right continuous with left limits equipped with the Skorohod topology). We can then proceed exactly as above.

Theorem A.1

[20, Theorem 9.25] For a standard Brownian motion B on [0, 1],

$$\begin{aligned} \mathbb {P}\left( \limsup _{\delta \downarrow 0} \frac{1}{\sqrt{2\delta \log (\frac{1}{\delta })}}\max _{{\mathop {t-s\le \delta }\limits ^{0\le s\le t\le 1}}}|B(t)-B(s)|=1\right) =1. \end{aligned}$$

Remark A.2

Clearly, for every $\rho >0$, there is $1>\delta _0>0$ such that for every $\delta <\delta _0$, we have

$$\begin{aligned} \max _{{\mathop {r-s\le \delta }\limits ^{0\le s\le r\le 1}}}|B(r)-B(s)|\le (1+\rho )\sqrt{2\delta \log \Big (\frac{1}{\delta }\Big )}, \; \quad \mathbb {P}-\text {a.s}. \end{aligned}$$

Corollary A.1

For $\rho >0$, there is $1>\delta _0=\delta _0(\rho )>0$ such that whenever $t>\delta _0$,

$$\begin{aligned} \sup _{0<s\le t} \frac{B(s)}{\sqrt{s\log \Big (\frac{1}{s}\Big )}}\le \max \left\{ \sqrt{2}(1+\rho ),\frac{1}{\sqrt{\delta _0\log \Big (\frac{1}{\delta _0}\Big )}} \sup _{0\le s\le t} B(s)\right\} ,\quad \mathbb {P}-\text {a.s.} \end{aligned}$$

Otherwise,

$$\begin{aligned} \sup _{0<s\le t} \frac{B(s)}{\sqrt{s\log \Big (\frac{1}{s}\Big )}}\le \sqrt{2}(1+\rho ), \quad \mathbb {P}-\text {a.s.} \end{aligned}$$

Proof

From Theorem A.1, as seen already for every $\rho >0$, there is a $1>\delta _0>0$, such that for every $\delta <\delta _0$, we have

$$\begin{aligned} \max _{{\mathop {r-s\le \delta }\limits ^{0\le s\le r\le 1}}}|B(r)-B(s)|\le (1+\rho )\sqrt{2\delta \log \Big (\frac{1}{\delta }\Big )}, \;\quad \mathbb {P}-\text {a.s}. \end{aligned}$$

In particular,

$$\begin{aligned} B(r)-B(0)=B(r)\le (1+\rho )\sqrt{2\delta \log \Big (\frac{1}{\delta }\Big )}, \text { for every }r\le \delta , \quad \mathbb {P}-\text {a.s}. \end{aligned}$$

This implies that we have

$$\begin{aligned} B(r)\le (1+\rho )\sqrt{2\delta \log \Big (\frac{1}{\delta }\Big )}, \text { for every }r= \delta \le \delta _0, \quad \mathbb {P}-\text {a.s} \end{aligned}$$

Assuming that $t>\delta _0$, we have

$$\begin{aligned} \sup _{0<s\le t} \frac{B(s)}{\sqrt{s\log \Big (\frac{1}{s}\Big )}}&\le \max \left\{ \sqrt{2}(1+\rho ),\frac{1}{\sqrt{\delta _0\log \Big (\frac{1}{\delta _0}\Big )}} \sup _{\delta _0\le s\le t} B(s)\right\} \\&\quad \le \max \left\{ \sqrt{2}(1+\rho ),\frac{1}{\sqrt{\delta _0\log \Big (\frac{1}{\delta _0}\Big )}} \sup _{0\le s\le t} B(s)\right\} , \quad \mathbb {P}- \text {a.s.} \end{aligned}$$

It is easy to see that for $t\le \delta _0$,

$$\begin{aligned} \sup _{0<s\le t} \frac{B(s)}{\sqrt{s\log \Big (\frac{1}{s}\Big )}}\le \sqrt{2}(1+\rho ), \quad \mathbb {P}- \text {a.s.} \end{aligned}$$

Hence, the proof is complete. $\square$

Using the technique similar to Theorem 5.4, we have the following.

Alternate proof of Theorem 5.2

We follow the argument almost exactly as in Theorem 5.4. We have already seen from Lemma 5.5 that for $t>0$,

$$\begin{aligned} M(t){\mathop {=}\limits ^{\textrm{d}}}t\max _{0\le v\le 1} \left( t^{H-1}Z(v) - kv\right) \doteq {\bar{M}}(t). \end{aligned}$$

And from Lemma 5.6, we know that M(t) $\mathbb {P}-$ a.s. converges to $M^*$ as $t\rightarrow \infty$.

Therefore, we have

$$\begin{aligned} M^*= \lim _{t\rightarrow \infty }{\bar{M}} (t)= \lim _{t\rightarrow \infty } t\max _{0\le v\le 1} \left( t^{H-1}Z(v) - kv\right) ,\quad \mathbb {P}-\text {a.s}. \end{aligned}$$

Now we replace t in ${\bar{M}}(t)$ by $\varepsilon {^{-1}}$ and treat $t\rightarrow \infty$ as $\varepsilon \rightarrow 0$. In other words, we have

$$\begin{aligned} M^*= \lim _{\varepsilon \rightarrow 0}{\bar{M}}( \varepsilon ^{-1})= \lim _{\varepsilon \rightarrow 0} \varepsilon ^{-1}\max _{0\le v\le 1} \left( \varepsilon ^{1-H} Z(v) - kv\right) ,\; \mathbb {P}-\text {a.s}. \end{aligned}$$

(A.3)

$$\begin{aligned} {\bar{M}}( \varepsilon ^{-1})= \varepsilon ^{-1}\max _{0\le v\le 1} \left( \varepsilon ^{1-H}Z(v) - kv\right) . \end{aligned}$$

From Lemma 4.1, we know that $\varepsilon {\bar{M}}( \varepsilon ^{-1})$ satisfies an LDP. From (A.3), we also know that

$$\begin{aligned} |M^*- {\bar{M}}( \varepsilon ^{-1})|=g(\varepsilon ), \end{aligned}$$

where g is a deterministic positive function such that $g(x)\rightarrow 0$ as $x \rightarrow 0$, $\mathbb {P}-$ a.s. Then, we have $|\varepsilon M^*- \varepsilon {\bar{M}}( \varepsilon ^{-1})|=\varepsilon g(\varepsilon ).$

Now we are in a position to derive the tail behavior of $M^*$:

$$\begin{aligned} \limsup _{\varepsilon \rightarrow 0} \varepsilon ^{2(1-H)}\log \mathbb {P}\left( \varepsilon M^*>1 \right) \le \limsup _{\varepsilon \rightarrow 0} \varepsilon ^{2(1-H)}\log \mathbb {P}\left( \varepsilon {\bar{M}}(\varepsilon ^{-1})> 1 -\varepsilon g(\varepsilon ) \right) . \end{aligned}$$

Similarly,

$$\begin{aligned} \liminf _{\varepsilon \rightarrow 0} \varepsilon ^{2(1-H)}\log \mathbb {P}\left( \varepsilon {\bar{M}}(\varepsilon ^{-1})>1 \right) \le \liminf _{\varepsilon \rightarrow 0} \varepsilon ^{2(1-H)}\log \mathbb {P}\left( \varepsilon M^*> 1 -\varepsilon g(\varepsilon ) \right) . \end{aligned}$$

From Lemma 4.1 with $T=1$, we have

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} \varepsilon ^{2(1-H)}\log \mathbb {P}\left( \varepsilon M^*>1 \right)&= \lim _{\varepsilon \rightarrow 0} \varepsilon ^{2(1-H)}\log \mathbb {P}\left( \varepsilon {\bar{M}}(\varepsilon ^{-1})> 1 \right) . \end{aligned}$$

We now notice that

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} \varepsilon ^{2(1-H)}\log \mathbb {P}\left( \varepsilon M^*>1 \right) = \lambda ^{2(H-1)}\lim _{{\bar{\varepsilon }}\rightarrow 0} {\bar{\varepsilon }}^{2(1-H)}\log \mathbb {P}\left( {\bar{\varepsilon }} M^*>\lambda \right) , \end{aligned}$$

(A.4)

by changing $\varepsilon$ to $\lambda ^{-1}{\bar{\varepsilon }}$. With the same argument as above, for $\lambda >0$, we have the following

$$\begin{aligned} \lambda ^{2(H-1)}\lim _{{\bar{\varepsilon }}\rightarrow 0} {\bar{\varepsilon }}^{2(1-H)}\log \mathbb {P}\left( {\bar{\varepsilon }} M^*>\lambda \right)&= \lambda ^{2(H-1)} \lim _{\varepsilon \rightarrow 0} \varepsilon ^{2(1-H)}\log \mathbb {P}\left( \varepsilon {\bar{M}}( \varepsilon ^{-1})> \lambda \right) \\&= {\left\{ \begin{array}{ll} -\frac{(\lambda +k)^2}{2}, &{} 1< \frac{\lambda H}{k(1-H)},\\ -\frac{k^{2H}}{2H^{2H} (1-H)^{2(1-H)}}\lambda ^{2(1-H)}, &{} \text { otherwise.} \end{array}\right. } \end{aligned}$$

Therefore, choosing $\lambda > \frac{k(1-H)}{H}$, from (A.4), we have

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} \varepsilon ^{2(1-H)}\log \mathbb {P}\left( M^*>\varepsilon ^{-1} \right) =- \frac{k^{2H}}{2H^{2H} (1-H)^{2(1-H)}}. \end{aligned}$$

This completes the proof. $\square$

Remark A.3

The intuition for the choice $\lambda >\frac{k(1-H)}{H}$ in the end of the proof is that (A.4) suggests us a scale invariance of the tail of $M^*$. Therefore, the decay rate of tail asymptotics is always one of the two cases in (4.10) which scales in $\lambda$ as $\lambda ^{2(1-H)}$. This case happens when $\lambda >\frac{k(1-H)}{H}$.

The next lemma concerns the locally stationary property of the GFBM process and is used in the proof in Sect. 5.1. Recall the definition of local stationarity for a self-similar Gaussian process in (5.28).

Lemma A.2

The GFBM $X(\cdot )$ defined in (2.1) is locally stationary.

Proof

For $0\le s\le t$,

$$\begin{aligned} \mathbb {E}[|t^{-H}X(t)-s^{-H}X(s)|^2]&= t^{-2H}\mathbb {E}[|X(t)-X(s)|^2] - (t^{-H}-s^{-H})^2\mathbb {E}[|X(s)|^2] \nonumber \\&\quad + 2t^{-H}(t^{-H}-s^{-H})\mathbb {E}[(X(t)-X(s))X(s)] \nonumber \\&= t^{-2H}\mathbb {E}[|X(t)-X(s)|^2] \nonumber \\&\quad - (ts)^{-2H}(t^{H}-s^{H})^2\mathbb {E}[|X(s)|^2] \nonumber \\&\quad + 2t^{-2H}s^{-H}(s^{H}-t^{H})\mathbb {E}[(X(t)-X(s))X(s)] \end{aligned}$$

(A.5)

To check that

$$\begin{aligned} \lim _{{\mathop {s\rightarrow t_0}\limits ^{t\rightarrow t_0}}} \frac{\mathbb {E}[|t^{-H}X(t)-s^{-H}X(s)|^2]}{(t-s)^{2H}} \text { exists,} \end{aligned}$$

(A.6)

it suffices to prove that the corresponding limits exist for the three terms on the right-hand side of (A.5). Before we proceed to do that, using (2.2) and (2.3), we rewrite $\mathbb {E}[|X(t)-X(s)|^2]$ and $\mathbb {E}[(X(t)-X(s))X(s)]$ by making the following change of variables: $u=s-(t-s)v$ and $v=xw$ with $x=\frac{s}{t-s}$ for integrals over [0, s] and $u=s+ (t-s)v$ for integrals over [s, t]. We have

$$\begin{aligned} \mathbb {E}[|X(t)-X(s)|^2]&= c^2\int _s^t(t-u)^{2\alpha } u^{-\gamma }du \nonumber \\&\quad +c^2 \int _0^s ((t-u)^\alpha -(s-u)^\alpha )^2u^{-\gamma } du\nonumber \\&= c^2 (t-s)^{2H} \bigg (\int _0^1 (1-v)^{2\alpha } \Big (v+ \frac{s}{t-s}\Big )^{-\gamma } dv\nonumber \\&\quad + \int _0^1(1-w)^{-\gamma } \big ((1+xw)^\alpha -(xw)^\alpha \big )^2x^{1-\gamma }dw\bigg ), \end{aligned}$$

(A.7)

$$\begin{aligned} \mathbb {E}[(X(t)-X(s))X(s)]&=c^2\int _0^s \big ((t-u)^\alpha -(s-u)^\alpha \big ) (s-u)^\alpha u^{-\gamma } du\nonumber \\&= c^2 (t-s)^{2H} \int _0^1(1-w)^{-\gamma } \big ((1+xw)^\alpha \nonumber \\&\quad -(xw)^\alpha \big )(xw)^{\alpha }x^{1-\gamma }dw. \end{aligned}$$

(A.8)

It is clear that

$$\begin{aligned} \int _0^1 (1-v)^{2\alpha }\Big (v+ \frac{s}{t-s}\Big )^{-\gamma } dv\le \int _0^1 (1-v)^{2\alpha } v^{-\gamma } dv= \mathbf{{B}}(1+2\alpha ,1-\gamma )<\infty . \end{aligned}$$

(A.9)

Recall that $\mathbf{{B}}(a,b)$ is the Beta function for $a,b>0$. Since $-\frac{1}{2}+\frac{\gamma }{2}<\alpha <\frac{1+\gamma }{2}$, we have

$$\begin{aligned}&\sup _{y>0}\Big \{\int _0^1(1-w)^{-\gamma } \big ((1+yw)^\alpha -(yw)^\alpha \big )^2y^{1-\gamma }dw\Big \}< K \textbf{B}(1-\gamma ,\gamma )<\infty , \nonumber \\&\quad \text { for some }K>0. \end{aligned}$$

(A.10)

Indeed, we have

$$\begin{aligned}&\sup _{y>0}\Big \{\int _0^1(1-w)^{-\gamma }\big ((1+yw)^\alpha -(yw)^\alpha \big )^2y^{1-\gamma }dw\Big \}\\&\quad \le \big ( \sup _{y>0} g(y)\big )^2 \int _0^1(1-w)^{-\gamma } w^{1-\gamma }dw\\&\quad \le \big ( \sup _{y>0} g(y)\big )^2 \textbf{B}(1-\gamma ,\gamma ). \end{aligned}$$

Here,

$$\begin{aligned} g(y)\doteq \big ((1+y)^\alpha -(y)^\alpha \big )y^{\frac{1}{2}-\frac{\gamma }{2}} \text { for }y\in (0,\infty ). \end{aligned}$$

Now showing that $\sup _{y>0} g(y)<\infty$, we are done. To that end, we show that $\lim _{y\rightarrow 0}g(y)$ and $\lim _{y\rightarrow \infty }g(y)$ both exist and are finite. Then from continuity of $g(\cdot )$ in $(0,\infty )$, we know that g(y) is finite for every $y\in (0,\infty )$ and it will then imply that $\sup _{y>0} g(y)<\infty$. Consider

$$\begin{aligned}&\lim _{y\rightarrow 0} g(y)= \lim _{y\rightarrow 0}\Big ( (1+y)^\alpha y^{\frac{1}{2}-\frac{\gamma }{2}} -y^{\alpha +\frac{1}{2}-\frac{\gamma }{2}}\Big )= 0, \end{aligned}$$

where we used $\frac{1}{2}-\frac{\gamma }{2}>0$ and $\alpha > -\frac{1}{2}+\frac{\gamma }{2}$. Now consider

$$\begin{aligned} \lim _{y\rightarrow \infty } g(y)=\lim _{y\rightarrow \infty } \frac{(1+y^{-1})^\alpha -1}{y^{-\alpha -\frac{1}{2}+\frac{\gamma }{2}}}&{\mathop {=}\limits ^{H}} \lim _{y\rightarrow \infty } \frac{-\alpha (1+y^{-1})^{\alpha -1} y^{-2}}{({-\alpha -\frac{1}{2}+\frac{\gamma }{2}})y^{{-\alpha -\frac{1}{2}+\frac{\gamma }{2}}-1}}\\&= \lim _{y\rightarrow \infty } \frac{-\alpha (1+y^{-1})^{\alpha -1} y^{\alpha -\frac{1}{2}-\frac{\gamma }{2}}}{({-\alpha -\frac{1}{2}+\frac{\gamma }{2}})}\\&= 0. \end{aligned}$$

In the above, ${\mathop {=}\limits ^{H}}$ denotes that we used L’Hôpital’s rule, as the we have a $\frac{0}{0}$ form (recall that $\alpha +\frac{1}{2}-\frac{\gamma }{2}>0$). To get the final equality, we used the fact that $\alpha <\frac{1}{2}+\frac{\gamma }{2}.$

Now consider (observe that it is $(t-s)^H$, instead of $(t-s)^{2H}$),

$$\begin{aligned}&(t-s)^{H} \Big ( \int _0^1(1-w)^{-\gamma } \big ((1+xw)^\alpha -(xw)^\alpha \big )(xw)^{\alpha }x^{1-\gamma }dw \Big )\nonumber \\&\quad \le s^{H} \sup _{y>0} \Big \{y^{-H +\alpha -\gamma +1} \big ((1+y)^\alpha -y^\alpha \big )\Big \} \int _0^1 (1-w)^{-\gamma } w^{\gamma -1}dw \nonumber \\&\quad \le s^{H}{} \textbf{B}(1-\gamma ,\gamma ) \sup _{y>0} \Big \{y^{\frac{1}{2}-\frac{\gamma }{2}} \big ((1+y)^\alpha -y^\alpha \big )\Big \}, \text { since } H=\alpha -\frac{\gamma }{2}+\frac{1}{2},\nonumber \\&\quad \le s^{H}{} \textbf{B}(1-\gamma ,\gamma ) \sup _{y>0} \Big \{y^{\frac{1}{2}-\frac{\gamma }{2}+\alpha } \big ((1+y^{-1})^\alpha -1 \big )\Big \}<\infty . \end{aligned}$$

(A.11)

The finiteness of

$$\begin{aligned} \sup _{y>0} \Big \{y^{\frac{1}{2}-\frac{\gamma }{2}+\alpha } \big ((1+y^{-1})^\alpha -1 \big )\Big \} \end{aligned}$$

can be proved in the similar way as done for g(y). From (A.9) and (A.10) ((A.11), respectively.), we can conclude that quantity in parenthesis of (A.7) ((A.8), respectively) is continuous in (s, t) when $\delta <s\le t$, for every $\delta >0$.

We are finally in a position to prove local stationarity of $X(\cdot )$. For the first term in (A.5), from (A.7) and continuity of the term in the parenthesis, we know that

$$\begin{aligned} \lim _{{\mathop {s\rightarrow t_0}\limits ^{t\rightarrow t_0}}} t^{-2H}\frac{\mathbb {E}[|X(t)-X(s)|^2]}{(t-s)^{2H}} \end{aligned}$$

exists uniformly for $t_0>\delta$, for any $\delta >0$. To see that the corresponding limit of the second term in (A.5) exists uniformly for $t_0>\delta$, for any $\delta >0$, we write

$$\begin{aligned} \lim _{{\mathop {s\rightarrow t_0}\limits ^{t\rightarrow t_0}}} \frac{(ts)^{-2H}(t^{H}-s^{H})^2\mathbb {E}[|X(s)|^2]}{(t-s)^{2H}}=\lim _{{\mathop {s\rightarrow t_0}\limits ^{t\rightarrow t_0}}} \Big (\frac{t^{H}-s^{H}}{(t-s)^{H}}\Big )^2(ts)^{-2H}\mathbb {E}[|X(s)|^2] \end{aligned}$$

and from H-Hölder continuity of function $f(t)= t^H$, we can conclude the existence of the above limit.

Now, to see that the corresponding limit of the third term in (A.5) exists uniformly for $t_0>\delta$, for any $\delta >0$, we write

$$\begin{aligned}&\lim _{{\mathop {s\rightarrow t_0}\limits ^{t\rightarrow t_0}}} \frac{t^{-2H}s^{-H}(s^{H}-t^{H})\mathbb {E}[(X(t)-X(s))X(s)]}{(t-s)^{2H}}\\&\quad =\lim _{{\mathop {s\rightarrow t_0}\limits ^{t\rightarrow t_0}}} \frac{t^{-2H}s^{-H}(s^{H}-t^{H})}{(t-s)^H}\frac{\mathbb {E}[(X(t)-X(s))X(s)]}{(t-s)^{H}}. \end{aligned}$$

From the H-Hölder continuity of function $f(t)=t^H$, we obtain

$$\begin{aligned} \lim _{{\mathop {s\rightarrow t_0}\limits ^{t\rightarrow t_0}}} \frac{t^{-2H}s^{-H}(s^{H}-t^{H})}{(t-s)^H} \text {exists} \end{aligned}$$

uniformly for $t_0>\delta$, for any $\delta >0$ and from (A.8) and continuity of the quantity inside the parenthesis, we know that

$$\begin{aligned} \lim _{{\mathop {s\rightarrow t_0}\limits ^{t\rightarrow t_0}}}\frac{\mathbb {E}[(X(t)-X(s))X(s)]}{(t-s)^{H}} \text { exists.} \end{aligned}$$

Thus, we have proved that (A.6) holds. This proves the result. $\square$