# Parametric Estimation in the Vasicek-Type Model Driven by Sub-Fractional Brownian Motion
## 4 Asymptotic Distribution of the Least Squares Estimator  
In this section, we consider the asymptotic normality of the LSE \(\hat{\mu}\) and \(\hat{\theta}\). We start with some preliminaries and let \(H>\frac{1}{2}\). **Lemma 6** ([PERSON] et al [14]). : _Let \(F\) be any \(\mathscr{F}^{H}=\sigma(\{S_{t}^{H},t\geq 0\})\)-measurable random variable such that \(P(F<\infty)=1\). Then, we have:_  
\[\left(F,e^{-\theta T}\int_{0}^{T}e^{\theta s}dS_{s}^{H}\right)\stackrel{{ law}}{{\longrightarrow}}(F,\theta^{-2H}\lambda_{H}\xi),\]  
_as \(T\rightarrow\infty\), where \(\xi\sim\mathcal{N}(0,1)\) is independent of \(S^{H}\) and \(\lambda_{H}=H\Gamma(2H)\)._  
Proof of Lemma 6.: The lemma is introduced in [PERSON] et al. [14]. In fact, we need to check that  
\[E\left(e^{-\theta T}\int_{0}^{T}e^{\theta s}dS_{s}^{H}\right)^{2}\longrightarrow \ell(H)=\theta^{-2H}\lambda_{H}\]  
and:  
\[E\left(e^{-\theta T}S_{T}^{H}\int_{0}^{T}e^{\theta s}dS_{s}^{H}\right)^{2} \longrightarrow 0\]  
for all fixed \(s\geq 0\), as \(T\) tends to infinity. However, the proof of the first convergence given by them is incomplete. In order to introduce the first convergence, by Lemma 3, we have that:  
\[\ell(H) =H(2H-1)\lim_{T\rightarrow\infty}\int_{0}^{T}\int_{0}^{T}e^{- \theta(T-s)}e^{-\theta(T-r)}\left(|s-r|^{2H-2}-|s+r|^{2H-2}\right)dsdr\] \[=H(2H-1)\lim_{T\rightarrow\infty}\int_{0}^{T}\int_{0}^{T}e^{- \theta(T-s)}e^{-\theta(T-r)}|s-r|^{2H-2}dsdr\] \[\quad-H(2H-1)\lim_{T\rightarrow\infty}\int_{0}^{T}\int_{0}^{T}e^{ -\theta(T-s)}e^{-\theta(T-r)}|s+r|^{2H-2}dsdr\] \[=H\theta^{-2H}\Gamma(2H)-H(2H-1)\lim_{T\rightarrow\infty}\int_{0 }^{T}\int_{0}^{T}e^{-\theta(T-s)}e^{-\theta(T-r)}|s+r|^{2H-2}dsdr.\]Notice that:  
\[\int_{0}^{T}\int_{0}^{T}e^{-\theta(T-s)} e^{-\theta(T-r)}|s+r|^{2H-2}dsdr=\int_{0}^{T}\int_{0}^{T}e^{- \theta x}e^{-\theta y}|2T-x-y|^{2H-2}dxdy\] \[\leq\int_{0}^{T}\int_{0}^{T}e^{-\theta x}e^{-\theta y}(T-x)^{2H-2}dxdy\] \[=\frac{1}{\theta}\int_{0}^{T}e^{-\theta x}(T-x)^{2H-2}dx\left(1-e ^{-\theta T}\right)\leq\frac{1}{\theta}e^{-\theta T}\int_{0}^{T}e^{\theta s}s ^{2H-2}ds\] \[E\left(e^{-\theta T}\int_{0}^{T}e^{\theta s}dS_{s}^{H}\right)^{2}\longrightarrow \ell(H)=\theta^{-2H}\lambda_{H}\]0  
as \(T\) tends to infinity. We get \(\ell(H)=H\theta^{-2H}\Gamma(2H)=\theta^{-2H}\lambda_{H}\), and the lemma follows. **Lemma 7** ([PERSON] [13]). : _Suppose that \(H>\frac{1}{2}\). Then, as \(t\rightarrow\infty\),_  
\[E\left(e^{-\theta T}\int_{0}^{T}e^{\theta s}dS_{s}^{H}\right)^{2}\longrightarrow \ell(H)=\theta^{-2H}\lambda_{H}\]1  
_in \(L^{2}(\Omega)\) and:_  
\[E\left(e^{-\theta T}\int_{0}^{T}e^{\theta s}dS_{s}^{H}\right)^{2}\longrightarrow \ell(H)=\theta^{-2H}\lambda_{H}\]2  
_as \(T\rightarrow\infty\)._  
**Theorem 2**. : _For \(\frac{1}{2}<H<1\), the convergence:_  
\[E\left(e^{-\theta T}\int_{0}^{T}e^{\theta s}dS_{s}^{H}\right)^{2}\longrightarrow \ell(H)=\theta^{-2H}\lambda_{H}\]3  
\[E\left(e^{-\theta T}\int_{0}^{T}e^{\theta s}dS_{s}^{H}\right)^{2}\longrightarrow \ell(H)=\theta^{-2H}\lambda_{H}\]4  
_and:_  
\[E\left(e^{-\theta T}\int_{0}^{T}e^{\theta s}dS_{s}^{H}\right)^{2}\longrightarrow \ell(H)=\theta^{-2H}\lambda_{H}\]5  
_hold in distribution, as \(T\) tends to infinity, where \(\xi,\eta\sim N(0,1)\) are mutually independent, \(\zeta\sim N(0,2-2^{H-1})\), and:_  
\[E\left(e^{-\theta T}\int_{0}^{T}e^{\theta s}dS_{s}^{H}\right)^{2}\longrightarrow \ell(H)=\theta^{-2H}\lambda_{H}\]6  
**Remark 1**. : It is not difficult to show that the density of \(\hat{\vartheta}=\frac{\xi}{\eta+\alpha}\) is:  
\[E\left(e^{-\theta T}\int_{0}^{T}e^{\theta s}dS_{s}^{H}\right)^{2}\longrightarrow \ell(H)=\theta^{-2H}\lambda_{H}\]7  
where \(\xi,\eta\sim N(0,1)\) are mutually independent and \(\alpha\in\mathbb{R}\).