# Parametric Estimation in the Vasicek-Type Model Driven by Sub-Fractional Brownian Motion
## 3 The Consistency of the Least Squares Estimator  
In this section, our main objective is to expound and to prove the next theorem, which gives the consistency of the estimators given by Equations (5) and (6). **Theorem 1**. : _For \(H\in(\frac{1}{2},1)\), we have:_  
1. \(\hat{\theta}_{T}\to\theta\)_, as_ \(T\) _tends to infinity, almost surely.__._
2. \(\hat{\mu}_{T}\to\mu\)_, as_ \(T\) _tends to infinity, almost surely._  
From Equation (2), one can easily get:  
\[X_{t}=\frac{\mu}{\theta}(e^{\theta t}-1)+e^{\theta t}\int_{0}^{t}e^{-\theta s} dS_{s}^{H}=\frac{\mu}{\theta}(e^{\theta t}-1)+S_{t}^{H}+\theta e^{\theta t}Z_{t} \tag{18}\]  
for all \(t\geq 0\), where \(Z_{t}=\int_{0}^{t}e^{-\theta s}S_{s}^{H}ds\). For convenience, we denote:  
\[f(t)=\frac{\mu}{\theta}(e^{\theta t}-1)\text{ and }Y_{t}=\int_{0}^{t}e^{-\theta s }dS_{s}^{H}.\]  
Then, Equation (18) can be rewritten as below:  
\[X_{t}=f(t)+e^{\theta t}Y_{t}=f(t)+S_{t}^{H}+\theta e^{\theta t}Z_{t}.\]  
It follows from the above equation that:  
\[Y_{t}=e^{-\theta t}S_{t}^{H}+\theta Z_{t}, \tag{19}\]  
for all \(t\geq 0\). **Lemma 2** (Lemma 2.1 in [PERSON] et al. [14]). : _Let \(H\in(\frac{1}{2},1)\). Then, the sub-fractional OUprocess is \(\gamma\)-Holder continuous for all \(\gamma<H\), and the Young integral:_  
\[\int_{0}^{t}u_{s}dX_{s}=u_{t}X_{t}-u_{0}X_{0}-\int_{0}^{t}X_{s}du_{s}\]  
_is well-defined for all \(t\geq 0\) if \(u\) is an adapted continuous process with bounded \(p\)-variation with \(1\leq p<\frac{1}{1-H+\theta}\) for any sufficiently small \(\epsilon\in(0,H)\). Moreover,_  
\[Z_{T}\longrightarrow Z_{\infty}=\int_{0}^{\infty}e^{-\theta r}S_{r}^{H}d\tau\]  
_almost surely and in \(L^{2}(\Omega)\), as \(T\) tends to infinity. Thus, as \(T\to\infty\),_  
\[Y_{T}\longrightarrow Y_{\infty}=\theta Z_{\infty}\]  
_almost surely and in \(L^{2}(\Omega)\)._  
**Lemma 3** (Hu-Nualart [10]). : _For all \(\frac{1}{2}<H<1\), we have:_  
\[\int_{0}^{\infty}\int_{0}^{\infty}e^{-\theta(u+v)}|u-v|^{2H-2}dudv=\frac{ \theta^{-2H}}{(2H-1)}\Gamma(2H). \tag{20}\]  
**Lemma 4**. : _Let \(H\in(\frac{1}{2},1)\). We then have that:_  
\[\lim_{T\to\infty}e^{-2\theta T}\int_{0}^{T}f^{2}(s)ds=\frac{\mu^{2}}{2\theta^ {3}} \tag{21}\]  
Proof of Lemma 4.: This is a simple calculus exercise. **Corollary 3**. : _Let \(H\in(\frac{1}{2},1)\). We then have that:_  
\[e^{-\theta T}\int_{0}^{T}X_{s}ds \longrightarrow\frac{\mu}{\theta^{2}}+\frac{1}{\theta}Y_{\infty} \tag{22}\] \[f(t)=\frac{\mu}{\theta}(e^{\theta t}-1)\text{ and }Y_{t}=\int_{0}^{t}e^{-\theta s }dS_{s}^{H}.\]0  
_almost surely, and in \(L^{2}(\Omega)\), as \(T\) tends to infinity._  
Proof of Corollary 3.: By Lemma 2, Equation (21), and L'Hopital's rule, we get that:  
\[f(t)=\frac{\mu}{\theta}(e^{\theta t}-1)\text{ and }Y_{t}=\int_{0}^{t}e^{-\theta s }dS_{s}^{H}.\]1 \[f(t)=\frac{\mu}{\theta}(e^{\theta t}-1)\text{ and }Y_{t}=\int_{0}^{t}e^{-\theta s }dS_{s}^{H}.\]2 \[f(t)=\frac{\mu}{\theta}(e^{\theta t}-1)\text{ and }Y_{t}=\int_{0}^{t}e^{-\theta s }dS_{s}^{H}.\]3  
almost surely, as \(T\) tends to infinity. Thus, the lemma follows from Equation (18). **Lemma 5**. : _Let \(H\in(\frac{1}{2},1)\). Then, the convergence:_  
\[f(t)=\frac{\mu}{\theta}(e^{\theta t}-1)\text{ and }Y_{t}=\int_{0}^{t}e^{-\theta s }dS_{s}^{H}.\]4  
_hold almost surely and in \(L^{2}\), as \(T\) tends to infinity._  
**Proposition 1**. : _Let \(H\in(\frac{1}{2},1)\). We have that:_  
\[f(t)=\frac{\mu}{\theta}(e^{\theta t}-1)\text{ and }Y_{t}=\int_{0}^{t}e^{-\theta s }dS_{s}^{H}.\]5  
_almost surely, as \(T\) tends to infinity._  
Proof of Proposition 1.: By Equation (18) and Lemma 1, we have:  
\[