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Mixed Sub-fractional Brownian Motion and Drift Estimation of Related Ornstein–Uhlenbeck Process

Abstract

In this paper, we will first give the numerical simulation of the sub-fractional Brownian motion through the relation of fractional Brownian motion instead of its representation of random walk. In order to verify the rationality of this simulation, we propose a practical estimator associated with the LSE of the drift parameter of mixed sub-fractional Ornstein–Uhlenbeck process, and illustrate the asymptotical properties according to our method of simulation when the Hurst parameter \(H>1/2\).

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Correspondence to Qinghua Wang.

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Funding

Chunhao Cai is supported by the Fundamental Research Funds for the SUFE No. 2020110294. Weilin Xiao is supported by the National Natural Science Foundation of China, Grant No. 71871202.

Appendix

Appendix

Proof of Main Theorem

In this part, we will prove the main results of Theorem 3.3 and Theorem 3.4. First of all, let us introduce the following lemma.

Lemma 6.1

Let \(S_t^H\) be a sub-fractional Brownian motion. Then, we have

$$\begin{aligned} {\mathbf {E}}\left[ \int _s^T e^{-\vartheta (\xi -s)}\mathrm{d}S_{\xi }^H\int _t^Te^{-\vartheta (\eta -t)}\mathrm{d}S_{\eta }^H\right] \le C_{\vartheta , H}|t-s|^{2H-2} \end{aligned}$$
(6.1)

and

$$\begin{aligned} {\mathbf {E}}\left[ \int _0^t e^{-\vartheta (t-u)}\mathrm{d}S_u^H \int _0^s e^{-\vartheta (s-v)}\mathrm{d}S_v^H\right] \le C_{\vartheta , H}|t-s|^{2H-2}. \end{aligned}$$
(6.2)

Proof

In fact

$$\begin{aligned}&{\mathbf {E}}\left[ \int _s^T e^{-\vartheta (\xi -s)}\mathrm{d}S_{\xi }^H\int _t^Te^{-\vartheta (\eta -t)} \mathrm{d}S_{\eta }^H\right] \\&\quad =\alpha _H\int _t^T\int _s^T e^{-\vartheta (\xi -s)}e^{-\vartheta (\eta -t)}\left( |\xi -\eta |^{2H-2}-|\eta +\xi |^{2H-2} \right) \mathrm{d}\xi d\eta \end{aligned}$$

when for fixed real number \(t,s\ge 0\), we have \(|t+s|^{2H-2}\le |t-s|^{2H-2}\) we have

$$\begin{aligned}&0<{\mathbf {E}}\left[ \int _s^T e^{-\vartheta (\xi -s)}\mathrm{d}S_{\xi }^H\int _t^Te^{-\vartheta (\eta -t)}\mathrm{d}S_{\eta }^H\right] \\&\quad \le 2 \alpha _H\int _t^T\int _s^T e^{-\vartheta (\xi -s)}e^{-\vartheta (\eta -t)}|\xi -\eta |^{2H-2}\mathrm{d}\xi d\eta . \end{aligned}$$

From the web only Lemma 5.4 of [14] we know

$$\begin{aligned} \alpha _H\int _t^T\int _s^T e^{-\vartheta (\xi -s)}e^{-\vartheta (\eta -t)}|\xi -\eta |^{2H-2}\mathrm{d}\xi \mathrm{d}\eta \le C_{\vartheta , H}|t-s|^{2H-2}, \end{aligned}$$

which achieves the proof of (6.1) and the same for (6.2). \(\square \)

The following lemma plays a key role in the proof of Theorem 3.4 when in the norm \(\Vert \cdot \Vert _{{\mathcal {H}}}\) we have to calculate the inner product with respect to the standard Brownian motion.

Lemma 6.2

For \(H>1/2\), we have

$$\begin{aligned} \lim _{T\rightarrow \infty }\frac{1}{T}\int _0^T {\mathbf {E}}\left( \int _0^s e^{-\vartheta (s-u)}\mathrm{d}B_u^H\int _s^T e^{-\vartheta (v-s)}\mathrm{d}B_v^H \right) ds=\vartheta ^{-2H}\Gamma (2H).\quad \end{aligned}$$
(6.3)

Proof

$$\begin{aligned}&\int _0^T {\mathbf {E}}\left( \int _0^s e^{-\vartheta (s-u)}\mathrm{d}B_u^H\int _s^T e^{-\vartheta (v-s)}\mathrm{d}B_v^H \right) \mathrm{d}s\\&\quad = \int _0^T\int _s^T\int _0^s e^{-\vartheta (y-x)}(y-x)^{2H-2}\mathrm{d}x\mathrm{d}y\mathrm{d}s\\&\quad =\int _0^T\int _0^y\int _x^y e^{-\vartheta (y-x)}(y-x)^{2H-2}\mathrm{d}s\mathrm{d}x\mathrm{d}y\\&\quad =\int _0^T\int _0^ye^{-\vartheta (y-x)}(y-x)^{2H-1}\mathrm{d}x\mathrm{d}y\\&\quad =\int _0^T\int _0^ye^{-\vartheta x}x^{2H-1}\mathrm{d}x\mathrm{d}y=\int _0^T\int _x^T e^{-\vartheta x}x^{2H-1}\mathrm{d}y\mathrm{d}x\\&\quad =\int _0^Te^{-\vartheta x}x^{2H-1}(T-x)\mathrm{d}x\\&\quad =T\int _0^T e^{-\vartheta x} x^{2H-1}\mathrm{d}x-\int _0^Te^{-\vartheta x}x^{2H}\mathrm{d}x. \end{aligned}$$

The two limits

$$\begin{aligned} \int _0^{\infty }e^{-\vartheta x}x^{2H-1}\mathrm{d}x=\vartheta ^{-2H}\Gamma (2H),\,\,\,\,\,\, \lim _{T\rightarrow \infty }\frac{1}{T}\int _0^T e^{-\vartheta x}x^{2H}\mathrm{d}x=0 \end{aligned}$$

complete the proof.\(\square \)

Proof of Theorem 3.3

The result in [14] gives the strong consistency for the LSE from the ergodicity. Since the increment of the sfBm is not stationary, we cannot use the ergodicity to prove the consistency of \({\bar{\vartheta }}_T\). Now, a standard calculation yields

$$\begin{aligned}&\lim _{T\rightarrow \infty } \frac{H(2H-1)}{T}\int _0^T\int _0^t\exp (-\vartheta (t-s))(t-s)^{2H-2}\mathrm{d}s\mathrm{d}t \\&\quad =\lim _{T\rightarrow \infty }\frac{H(2H-1)}{T}\int _0^T\int _0^t u^{2H-2}e^{-\vartheta u}\mathrm{d}u\mathrm{d}t\\&\quad =\vartheta ^{1-2H}H\Gamma (2H). \end{aligned}$$

On the other hand, a straightforward calculation shows that

$$\begin{aligned}&\int _0^T\int _0^t \exp (-\vartheta (t-s))(t+s)^{2H-2}\mathrm{d}s\\&\quad =\int _0^T\int _0^t \exp (\vartheta (s+t)-2\vartheta t)(t+s)^{2H-2}\mathrm{d}s\mathrm{d}t\\&\quad =\int _0^T\exp (-2\vartheta t)\int _t^{2t}\exp (\vartheta u)u^{2H-2}\mathrm{d}u\mathrm{d}t. \end{aligned}$$

With the L’Hôspital’s rule, we have

$$\begin{aligned} \lim _{T\rightarrow \infty }\frac{1}{T}\int _0^T\exp (-2\vartheta t)\int _t^{2t}\exp (\vartheta u)u^{2H-2}\mathrm{d}u\mathrm{d}t=0. \end{aligned}$$

Now we only need to prove that

$$\begin{aligned} \frac{1}{T}\int _0^TX_t^2\mathrm{d}t {\mathop {\longrightarrow }\limits ^{\textit{a.s.}}}\frac{1}{2\vartheta }+H\vartheta ^{-2H}\Gamma (2H). \end{aligned}$$
(6.4)

For \(0\le t\le T\), let \(W_t\) be a standard Brownian motion and \(X_t=X_t^{(1)}+X_t^{(2)}+X_t^{(3)}\). Then, we have

$$\begin{aligned} dX_t^{(1)}=-\vartheta X_t^{(1)}dt+dW_t,\, 0\le t \le T, \end{aligned}$$

and

$$\begin{aligned} dX_t^{(2)}= & {} -\vartheta X_t^{(2)}dt+\frac{1}{\sqrt{2}}dB_t^H,\, 0\le t\le T,\\ dX_t^{(3)}= & {} -\vartheta X_t^{(3)}dt+\frac{1}{\sqrt{2}}dB_{-t}^H,\, 0\le t \le T. \end{aligned}$$

With the ergodic property of these three processes presented in [17], we have the following results:

$$\begin{aligned}&\frac{1}{T}\int _0^T\left( X_t^{(1)}\right) ^2\mathrm{d}t {\mathop {\longrightarrow }\limits ^{a.s.}}\frac{1}{2\vartheta },\\&\frac{1}{T}\int _0^T\left( X_t^{(2)}\right) ^2\mathrm{d}t {\mathop {\longrightarrow }\limits ^{a.s.}}\frac{H}{2}\vartheta ^{-2H}\Gamma (2H) \end{aligned}$$

and

$$\begin{aligned} \frac{1}{T}\int _0^T\left( X_t^{(3)}\right) ^2\mathrm{d}t {\mathop {\longrightarrow }\limits ^{a.s.}}\frac{H}{2}\vartheta ^{-2H}\Gamma (2H) \end{aligned}$$

so (6.4) needs this result:

$$\begin{aligned} \frac{1}{T}\int _0^T X_t^{(2)} X_t^{(3)}\mathrm{d}t {\mathop {\longrightarrow }\limits ^{a.s.}} 0. \end{aligned}$$
(6.5)

First of all

$$\begin{aligned} {\mathbf {E}}X_1^{(1)}X_t^{(2)}= \int _0^t \int _0^t e^{-\vartheta (t-s)}e^{-\vartheta (t-u)}|u+s|^{2H-2}\mathrm{d}u\mathrm{d}s \end{aligned}$$
(6.6)

when \(|u+s|^{2H-2}\le (us)^{H-1}\) for \(u,s\ge 0\), then with L’Hôpital rule we have the following inequality:

$$\begin{aligned} {\mathbf {E}}X_1^{(1)}X_t^{(2)}\le C_{\vartheta , H}\left( 1 \wedge t^{2H-2} \right) . \end{aligned}$$
(6.7)

With this inequality, we can easily obtain the convergence in probability

$$\begin{aligned} \frac{1}{T}\int _0^T X_t^{(2)} X_t^{(3)}\mathrm{d}t {\mathop {\longrightarrow }\limits ^{{\mathbf {P}}}} 0. \end{aligned}$$

From (6.7), Borel–Cantelli lemma, we can easily obtain the convergence almost surely in (6.5) for \(1/2<H<3/4\). For the case \(3/4\le H<1\), we will apply the method of The Theorem 2.1 in [18]. In fact with the convergence in probability means there exists a sub-sequence which converges almost surely to 0; on the other hand, the equation (6.7) verifies the condition of second-order Winer-Ito chaos in Proposition 3.4 of [18], together with GRR inequality (see Theorem 2.1 in [19]) (6.5) will be achieved.

Proof of Theorem 3.4

Step 1: We shall use Malliavin calculus and the fourth moment theorem (see, for example, Theorem 4 in [20]) to prove (3.4).

In fact, using (3.2), we have

$$\begin{aligned} \sqrt{T}\left( {\bar{\vartheta }}_T-\vartheta \right) =-\frac{\frac{1}{\sqrt{T}}\int _0^T\left( \int _0^t e^{-\vartheta (t-s)}\mathrm{d}\xi _s^H\right) \mathrm{d}\xi _t^H}{\frac{1}{T}\int _0^T X_t^2\mathrm{d}t}=\frac{-F_T}{\frac{1}{T}\int _0^T X_t^2\mathrm{d}t}, \end{aligned}$$
(6.8)

where \(F_T\) is the double stochastic integral

$$\begin{aligned} F_T=\frac{1}{2\sqrt{T}}I_2\left( e^{-\vartheta |t-s|}\right) =\frac{1}{2\sqrt{T}}\int _0^T\int _0^T e^{-\vartheta |t-s|}\mathrm{d}\xi _s\xi _t. \end{aligned}$$
(6.9)

By (6.4), we know that \(\frac{1}{T}\int _0^T X_t^2dt\) converges in probability and in \(L^2\) as T tends to infinity to \(\frac{1}{2\vartheta }+H\vartheta ^{-2H}\Gamma (2H)\). From Theorem 4 of [20], we have to check the following two conditions:

  • (i). \({\mathbf {E}}(F_T^2)\) converges to a constant as T tends to infinity

    $$\begin{aligned} \lim _{T\rightarrow \infty } {\mathbf {E}}F_T^2= & {} \vartheta ^{1-4H}H^2(4H-1)\left( \Gamma (2H)^2 +\frac{\Gamma (2H)\Gamma (3-4H)\Gamma (4H-1)}{\Gamma (2-2H)}\right) \\&+\frac{1}{2\vartheta }. \end{aligned}$$
  • (ii). \(\Vert DF_T\Vert _{{\mathcal {H}}}^2\) converges in \(L^2\) to a constant as T tends to infinity.

We first check the condition (i). When \(W_t\) and \(S_t^H\) are independent, we have \({\mathbf {E}}F_T^2={\mathbf {E}}\left( F_{1,T}^2+F_{T,2}^2\right) \), with \(F_{1,T}=\frac{1}{2\sqrt{T}}\int _0^T\int _0^T e^{-\vartheta |t-s|}\mathrm{d}W_t\mathrm{d}W_s,\,\, F_{2,T}=\frac{1}{2\sqrt{T}}\int _0^T\int _0^T e^{-\vartheta |t-s|}\mathrm{d}S_t^H\mathrm{d}S_s^H\).

A standard calculation together with (2.5) yields

$$\begin{aligned} {\mathbf {E}}F_{2,T}^2= & {} \frac{\alpha _H^2}{2T}\int _{[0,T]^4}\exp \left( -\vartheta |u_2-s_2|-\vartheta |u_1-s_1|\right) \\&\quad |u_2-u_1|^{2H-2}|s_2-s_1|^{2H-2}\mathrm{d}s_1\mathrm{d}s_2\mathrm{d}u_1\mathrm{d}u_2\\&-\frac{\alpha _H^2}{2T}\int _{[0,T]^4}\exp \left( -\vartheta |u_2-s_2|-\vartheta |u_1-s_1|\right) \\&\quad |u_2-u_1|^{2H-2}|s_2+s_1|^{2H-2}\mathrm{d}s_1\mathrm{d}s_2\mathrm{d}u_1\mathrm{d}u_2\\&-\frac{\alpha _H^2}{2T}\int _{[0,T]^4}\exp \left( -\vartheta |u_2-s_2| -\vartheta |u_1-s_1|\right) \\&\quad |u_2+u_1|^{2H-2}|s_2-s_1|^{2H-2}\mathrm{d}s_1\mathrm{d}s_2\mathrm{d}u_1\mathrm{d}u_2\\&+\frac{\alpha _H^2}{2T}\int _{[0,T]^4}\exp \left( -\vartheta |u_2-s_2|-\vartheta |u_1-s_1|\right) \\&\quad |u_2+u_1|^{2H-2}|s_2+s_1|^{2H-2}\mathrm{d}s_1\mathrm{d}s_2\mathrm{d}u_1\mathrm{d}u_2. \end{aligned}$$

From [14], we have

$$\begin{aligned}&\lim _{T\rightarrow \infty }\frac{\alpha _H^2}{2T}\int _{[0,T]^4}\exp \left( -\vartheta |u_2-s_2|-\vartheta |u_1-s_1|\right) |u_2\nonumber \\&\qquad -u_1|^{2H-2}|s_2-s_1|^{2H-2}\mathrm{d}s_1\mathrm{d}s_2\mathrm{d}u_1\mathrm{d}u_2 \nonumber \\&\quad =\vartheta ^{1-4H}H^2(4H-1)\left( \Gamma (2H)^2+\frac{\Gamma (2H)\Gamma (3-4H)\Gamma (4H-1)}{\Gamma (2-2H)}\right) .\qquad \end{aligned}$$
(6.10)

A simple calculation yields

$$\begin{aligned} \lim _{T\rightarrow \infty }{\mathbf {E}}F_{1,T}^2=\lim _{T\rightarrow \infty }\frac{1}{T}\int _0^T\int _0^te^{-2\vartheta (t-s)}\mathrm{d}s\mathrm{d}t=\frac{1}{2\vartheta }. \end{aligned}$$
(6.11)

Now, if we can prove

$$\begin{aligned}&\lim _{T\rightarrow \infty }\frac{1}{T}\int _{[0,T]^4}\exp \left( -\vartheta |u_2-s_2|-\vartheta |u_1-s_1|\right) \nonumber \\&\quad |u_2-u_1|^{2H-2}|s_2+s_1|^{2H-2}\mathrm{d}s_1\mathrm{d}s_2\mathrm{d}u_1\mathrm{d}u_2=0, \end{aligned}$$
(6.12)

then the last three terms of \({\mathbf {E}}F_{2,T}^2\) will tend to zero with the fact \(|s_2+s_1|^{2H-2}\le |s_2-s_1|^{2H-2}\).

Denote

$$\begin{aligned} I_T=\frac{1}{T}\int _{[0,T]^4}e^{-\vartheta |u_2-s_2|-\vartheta |u_1-s_1|}\left( |u_2-u_1|^{2H-2}|s_2+s_1|^{2H-2}\right) \mathrm{d}s_1\mathrm{d}s_2\mathrm{d}u_1\mathrm{d}u_2. \end{aligned}$$

Using the L’Hôspital’s rule, we have

$$\begin{aligned} \frac{\hbox {d}I_T}{\hbox {d}T}=\int _{[0,T]^3}e^{-\vartheta (T-s_2)-\vartheta |s_1-u_1|}\left( (T-u_1)^{2H-2}(s_2+s_1)^{2H-2}\right) \hbox {d}s_1\hbox {d}u_1\hbox {d}u_2. \end{aligned}$$

Let \(T-s_2=x_1, T-s_1=x_2, T-u_1=x_3\). Ignoring the sign, we have

$$\begin{aligned} \frac{dI_T}{dT}= & {} \int _{[0,T]^3}e^{-\vartheta x_1-\vartheta |x_2-x_3|}\left( x_3^{2H-2}(T-x_1+T-x_2)^{2H-2}\right) \hbox {d}x_1\hbox {d}x_2\hbox {d}x_3\\= & {} e^{-\vartheta T}\int _{[0,T]^3}e^{\vartheta y_1-\vartheta |x_2-x_3|}\left( x_3^{2H-2}(y_1+T-x_2)^{2H-2}\right) \hbox {d}y_1\hbox {d}x_2\hbox {d}x_3\\\le & {} e^{-\vartheta T}\int _{[0,T]^3}e^{\vartheta y_1-\vartheta |x_2-x_3|}\left( x_3^{2H-2}y_1^{2H-2}\right) \hbox {d}y_1\hbox {d}x_2\hbox {d}x_3. \end{aligned}$$

Let \( J_T=\int _{[0,T]^3}e^{\vartheta y_1-\vartheta |x_2-x_3|}\left( x_3^{2H-2}y_1^{2H-2}\right) \mathrm{d}y_1\mathrm{d}x_2\mathrm{d}x_3\). Then, using the L’Hôspital’s rule, we get

$$\begin{aligned} \frac{dJ_T}{dT}=e^{-\vartheta T}\int _{[0,T]^2}e^{\vartheta y_1+\vartheta x_3}(y_1^{2H-2}x_3^{2H-2})\mathrm{d}y_1\mathrm{d}x_3. \end{aligned}$$

On the other hand, we can easily obtain \(\frac{de^{-\vartheta T}}{dT}=-\vartheta e^{-\vartheta T}\). Moreover, with the L’Hôspital’s rule, it is easy to check that

$$\begin{aligned} \lim _{T\rightarrow \infty } \frac{J_T}{e^{\vartheta T}}=0\,, \end{aligned}$$

which implies the Eq. (6.12).

Consequently, with (6.12) and the fact \(|s_2+s_1|^{2H-2}\le |s_2-s_1|^{2H-2}\), it is easy to see that

$$\begin{aligned}&\lim _{T\rightarrow \infty }\frac{1}{T}\int _{[0,T]^4}\exp \left( -\vartheta |u_2-s_2|-\vartheta |u_1-s_1|\right) \nonumber \\&\quad |u_2+u_1|^{2H-2}|s_2+s_1|^{2H-2}\mathrm{d}s_1\mathrm{d}s_2\mathrm{d}u_1\mathrm{d}u_2=0. \end{aligned}$$
(6.13)

Combining (6.10), (6.11), (6.12) with (6.13), we verify condition (i).

Now we will check the condition (ii). For \(s\le T\), we have

$$\begin{aligned} D_sF_T=\frac{X_s}{\sqrt{T}}+\frac{1}{\sqrt{T}}\int _s^T e^{-\vartheta (t-s)}\mathrm{d}\xi _t. \end{aligned}$$

From (2.5) we have

$$\begin{aligned} \Vert D_sF_T\Vert _{{\mathcal {H}}}^2= & {} \frac{1}{T}\int _0^T \left( X_s+\int _s^T e^{-\vartheta (t-s)}\mathrm{d}\xi _t\right) ^2\mathrm{d}s+\frac{H(2H-1)}{T}\\&\int _0^T\int _0^T\left( X_s+\int _s^Te^{-\vartheta (t-s)} \mathrm{d}\xi _t\right) \left( X_u+\int _u^Te^{-\vartheta (t-u)}\mathrm{d}\xi _t\right) \\&\quad \left( |u-s|^{2H-2}-|u+s|^{2H-2}\right) \mathrm{d}u\mathrm{d}s. \end{aligned}$$

We first consider the first term of the above equation. A straightforward calculation shows that

$$\begin{aligned}&\frac{1}{T}\int _0^T \left( X_s+\int _s^T e^{-\vartheta (t-s)}\mathrm{d}\xi _t\right) ^2\mathrm{d}s\\&\quad =\frac{1}{T}\int _0^T \left( X_s^2+2X_s\int _s^Te^{-\vartheta (t-s)}\mathrm{d}\xi _t+\left( \int _s^T e^{-\vartheta (t-s)}\mathrm{d}\xi _t \right) ^2 \right) \mathrm{d}s\\&\quad =A_T^{(1)}+A_T^{(2)}+A_T^{(3)}. \end{aligned}$$

From the proof of Theorem 3.3, the independence of the \(W_t\) and \(S_t^H\) in the msfBm, the convergence to 0 for the standard Brownian motion case in the proof of Theorem 3.4 of [14], the ergodicity and stationary of the fractional O-U process (see [17]) and Lemma 6.2, we can easily obtain that all these three terms converge in \(L^2\) as T tends to infinity.

Now let us look at the third term of \(\Vert D_sF_T\Vert _{{\mathcal {H}}}^2\). A standard calculation yields

$$\begin{aligned} C_T= & {} \frac{H(2H-1)}{T}\int _0^T\int _0^T\left( X_s+\int _s^Te^{-\vartheta (t-s)} \mathrm{d}\xi _t\right) \left( X_u+\int _u^Te^{-\vartheta (t-u)}\mathrm{d}\xi _t\right) \\&\;\left( |u-s|^{2H-2}-|u+s|^{2H-2}\right) \mathrm{d}u\mathrm{d}s\\= & {} \frac{H(2H-1)}{T}\left( C_T^{(1)}+2C_T^{(2)}+C_T^{(3)}\right) \end{aligned}$$

where

$$\begin{aligned} C_T^{(1)}= & {} \int _0^T\int _0^TX_sX_u\left( |u-s|^{2H-2}-|u+s|^{2H-2}\right) \mathrm{d}u\mathrm{d}s,\\ C_T^{(2)}= & {} \int _0^T\int _0^T\left( X_u \int _s^Te^{-\vartheta (t-s)} \mathrm{d}\xi _t\right) \left( |u-s|^{2H-2}-|u+s|^{2H-2}\right) \mathrm{d}u\mathrm{d}s \end{aligned}$$

and

$$\begin{aligned} C_T^{(3)}= & {} \int _0^T\int _0^T\left( \int _s^Te^{-\vartheta (t-s)} \mathrm{d}\xi _t \int _u^Te^{-\vartheta (t-u)}\mathrm{d}\xi _t \right) \\&\times \left( |u-s|^{2H-2}-|u+s|^{2H-2}\right) \mathrm{d}u\mathrm{d}s \end{aligned}$$

With the same method of Theorem 3.4 in [14], Lemma 6.1 and the independence of the \(W_t\) and \(S_t^H\) in msfBm, we have

$$\begin{aligned} \lim _{T\rightarrow \infty }\frac{1}{T^2}{\mathbf {E}}\left( |C_T^{(i)}-{\mathbf {E}}C_T^{(i)}|^2\right)= & {} 0,\,\,\,\,i=2,3\,, \end{aligned}$$
(6.14)

then we have

$$\begin{aligned} \lim _{T\rightarrow \infty } {\mathbf {E}}\left[ \left( C_T-{\mathbf {E}}C_T\right) ^2\right] =0, \end{aligned}$$
(6.15)

which implies that \(\lim \limits _{T\rightarrow \infty }{\mathbf {E}} C_T\) exists. Finally, we obtain that \(C_T\) converges in \(L^2\) to a constant. Thus, condition (ii) satisfies.

Step 2: Case \(H=3/4\). From (3.2) we have

$$\begin{aligned} \frac{\sqrt{T}}{\sqrt{\log (T)}}\left( {\bar{\vartheta }}_T-\vartheta \right) =-\frac{\frac{F_T}{\sqrt{\log T}}}{\frac{1}{T}\int _0^TX_t^2\mathrm{d}t}, \end{aligned}$$

where \(F_T\) is defined by (6.9).

We still use the fourth moment theorem (see, for example, Theorem 4 in [20]) and check two conditions of Step 1. Using same calculations of Step 1, we can show that

$$\begin{aligned}&\lim _{T\rightarrow \infty }\frac{1}{T\log (T)}\int _{[0,T]^4}e^{-\vartheta |s_2-u_2|-\vartheta |s_1-u_1|}|s_2-s_1|^{2H-2}\\&\qquad (u_2+u_1)^{2H-2}\mathrm{d}u_1\mathrm{d}u_2\mathrm{d}s_1\mathrm{d}s_2=0 \end{aligned}$$

and

$$\begin{aligned}&\lim _{T\rightarrow \infty }\frac{1}{T\log (T)}\int _{[0,T]^4}e^{-\vartheta |s_2-u_2|-\vartheta |s_1-u_1|}\\&\quad (s_2+s_1)^{2H-2}(u_2+u_1)^{2H-2}\mathrm{d}u_1\mathrm{d}u_2\mathrm{d}s_1\mathrm{d}s_2=0. \end{aligned}$$

On the other hand, a straightforward calculation shows that

$$\begin{aligned} \lim _{T\rightarrow \infty }{\mathbf {E}}\left( \frac{1}{2\sqrt{T\log T}}\int _0^T\int _0^T e^{-\vartheta |t-s|}\mathrm{d}W_t\mathrm{d}W_s\right) ^2=0\,. \end{aligned}$$

Then, we have

$$\begin{aligned}&\lim _{T\rightarrow \infty }{\mathbf {E}}\left( \frac{F_T}{\sqrt{\log T}}\right) ^2\\&\quad =\lim _{T\rightarrow \infty }\frac{H^2(2H-1)^2}{2T\log (T)}\int _{[0,T]^4}e^{-\vartheta |s_2-u_2|-\vartheta |s_1-u_1|}|s_2-s_1|^{2H-2}\\&\qquad (u_2-u_1)^{2H-2}\mathrm{d}u_1\mathrm{d}u_2\mathrm{d}s_1\mathrm{d}s_2\\&\quad =\frac{9}{4\vartheta ^2}, \end{aligned}$$

where the equality comes from Lemma 6.6 in [15].

Thus, condition (i) and condition (ii) are obvious when we add a term of \(\frac{1}{\sqrt{\log T}}\) and \(T^{8H-6}=1\) with \(H=\frac{3}{4}\).

Step 3: In this step we will prove the theorem when \(3/4<H<1\). From (3.2), we have

$$\begin{aligned} T^{2-2H}\left( {\bar{\vartheta }}_T-\vartheta \right) =-\frac{\frac{T^{1-2H}}{2}\int _0^T\int _0^Te^{-\vartheta |t-s|}\mathrm{d}\xi _s\mathrm{d}\xi _t}{\frac{1}{T}\int _0^TX_t^2\mathrm{d}t}. \end{aligned}$$

Let us mention that the condition (ii) in Step 1 will not be satisfied when \(H>3/4\). Fortunately, we still have the following convergence:

$$\begin{aligned}&\lim _{T\rightarrow \infty }T^{3-4H}\frac{1}{T}\int _{[0,T]^4}e^{-\vartheta |s_2-u_2|-\vartheta |s_1-u_1|}(s_2-s_1)^{2H-2}\\&\quad (u_2+u_1)^{2H-2}\mathrm{d}u_1\mathrm{d}u_2\mathrm{d}s_1\mathrm{d}s_2=0 \end{aligned}$$

and

$$\begin{aligned}&\lim _{T\rightarrow \infty }T^{3-4H}\frac{1}{T}\int _{[0,T]^4}e^{-\vartheta |s_2-u_2|-\vartheta |s_1-u_1|}(s_2+s_1)^{2H-2}\\&\quad (u_2+u_1)^{2H-2}\mathrm{d}u_1\mathrm{d}u_2\mathrm{d}s_1\mathrm{d}s_2=0. \end{aligned}$$

With the similarity of the process \(\xi \) and Lemma 6.6 in [15], we have

$$\begin{aligned} {T^{1-2H}}\int _0^T\int _0^Te^{-\vartheta |t-s|}\mathrm{d}\xi _s\mathrm{d}\xi _t\xrightarrow {{\mathcal {L}}}2\vartheta ^{-1}R_1 \end{aligned}$$

which achieves the proof.

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Cai, C., Wang, Q. & Xiao, W. Mixed Sub-fractional Brownian Motion and Drift Estimation of Related Ornstein–Uhlenbeck Process. Commun. Math. Stat. (2022). https://doi.org/10.1007/s40304-021-00245-8

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  • DOI: https://doi.org/10.1007/s40304-021-00245-8

Keywords

  • Sub-fractional Brownian motion
  • Ornstein–Uhlenbeck process
  • Least square estimator
  • Malliavin calculus

Mathematics Subject Classification

  • 60G22
  • 62F10