# Parametric Estimation in the Vasicek-Type Model Driven by Sub-Fractional Brownian Motion
## 3 The Consistency of the Least Squares Estimator
\begin{split}&\frac{1}{T}e^{-\theta T}\left(\int_{0}^{T}X_{t}^{2} dt-\frac{1}{2}X_{T}\int_{0}^{T}X_{t}dt\right)\\ &\quad=\frac{1}{T}e^{-\theta T}\left(\int_{0}^{T}\left(f(t)+e^{ \theta t}Y_{t}\right)^{2}dt-\frac{1}{2}\left(f(T)+e^{\theta T}Y_{T}\right) \int_{0}^{T}\left(f(t)+e^{\theta t}Y_{t}\right)dt\right)\\ &\quad=\frac{1}{T}e^{-\theta T}\left(\int_{0}^{T}f(t)^{2}dt- \frac{1}{2}f(T)\int_{0}^{T}f(t)dt\right)\\ &\quad\quad\quad+\frac{1}{T}e^{-\theta T}\left(\int_{0}^{T}e^{2 \theta t}Y_{t}^{2}dt-\frac{1}{2}e^{\theta T}Y_{T}\int_{0}^{T}e^{\theta t}Y_{t} dt\right)\\ &\quad\quad+\frac{1}{T}e^{-\theta T}\left(2\int_{0}^{T}f(t)e^{ \theta t}Y_{t}dt-\frac{1}{2}e^{\theta T}Y_{T}\int_{0}^{T}f(t)dt-\frac{1}{2}f( T)\int_{0}^{T}e^{\theta t}Y_{t}dt\right)\\ &\quad\equiv\Lambda_{1}(T)+\Lambda_{2}(T)+\Lambda_{3}(T)\end{split}  \tag{25}\]for all \(T>0\). Clearly, an elementary calculus can show that:  
\[\Lambda_{1}(T) =\frac{1}{T}e^{-\theta T}\left(\int_{0}^{T}f(t)^{2}dt-\frac{1}{2}f( T)\int_{0}^{T}f(t)dt\right)\] \[=\frac{\mu^{2}}{\theta^{2}T}e^{-\theta T}\left(\int_{0}^{T}(e^{ \theta t}-1)^{2}dt-\frac{1}{2}(e^{\theta T}-1)\int_{0}^{T}(e^{\theta t}-1)dt\right)\] \[=\frac{\mu^{2}}{\theta^{2}T}e^{-\theta T}\left(\frac{1}{\theta}- \frac{1}{\theta}e^{\theta T}+\frac{1}{2}T+\frac{1}{2}Te^{\theta T}\right) \longrightarrow\frac{\mu^{2}}{2\theta^{2}},\]  
as \(T\) tends to infinity. For \(\Lambda_{2}(T)\), we have:  
\[\int_{0}^{T}e^{\theta t}Y_{t}dt =\frac{1}{\theta}\int_{0}^{T}Y_{t}d\theta^{t}\] \[=\frac{1}{\theta}\left(Y_{T}e^{\theta T}-\int_{0}^{T}e^{\theta t }dY_{t}\right)=\frac{1}{\theta}\left(Y_{T}e^{\theta T}-S_{T}^{H}\right)\]  
by integration by parts, which gives:  
\[\int_{0}^{T}e^{2\theta t}Y_{t}^{2}dt =\int_{0}^{T}e^{\theta t}Y_{t}d\left(\int_{0}^{t}e^{\theta s}Y_{s} ds\right)=\frac{1}{\theta}\int_{0}^{T}e^{\theta t}Y_{t}d\left(Y_{t}e^{\theta t }-S_{t}^{H}\right)\] \[=\frac{1}{\theta}\int_{0}^{T}e^{\theta t}Y_{t}d\left(Y_{t}e^{ \theta t}\right)-\int_{0}^{T}e^{\theta t}Y_{t}dS_{t}^{H}\] \[=\frac{1}{2\theta}\left(e^{\theta T}Y_{T}\right)^{2}-e^{\theta T} Y_{T}S_{T}^{H}+\int_{0}^{T}S_{t}^{H}d\left(e^{\theta t}Y_{t}\right)\] \[=\frac{1}{2\theta}\left(e^{\theta T}Y_{T}\right)^{2}-e^{\theta T} Y_{T}S_{T}^{H}+\theta\int_{0}^{T}S_{t}^{H}Y_{t}e^{\theta t}dt+\int_{0}^{T}S_{t}^{ H}dS_{t}^{H}\] \[=\frac{1}{2\theta}\left(e^{\theta T}Y_{T}\right)^{2}-e^{\theta T }Y_{T}S_{T}^{H}+\theta\int_{0}^{T}S_{t}^{H}Y_{t}e^{\theta t}dt+\frac{1}{2}(S_{ T}^{H})^{2}\]  
for all \(T>0\) by integration by parts. It follows from Lemma 1 and Lemma 5 that:  
\[=\frac{\mu^{2}}{\theta^{2}T}e^{-\theta T}\left(\int_{0}^{T}(e^{ \theta t}-1)^{2}dt-\frac{1}{2}(e^{\theta T}-1)\int_{0}^{T}(e^{\theta t}-1)dt\right)\]0 \[=\frac{\mu^{2}}{\theta^{2}T}e^{-\theta T}\left(\int_{0}^{T}(e^{ \theta t}-1)^{2}dt-\frac{1}{2}(e^{\theta T}-1)\int_{0}^{T}(e^{\theta t}-1)dt\right)\]1 \[=\frac{\mu^{2}}{\theta^{2}T}e^{-\theta T}\left(\int_{0}^{T}(e^{ \theta t}-1)^{2}dt-\frac{1}{2}(e^{\theta T}-1)\int_{0}^{T}(e^{\theta t}-1)dt\right)\]2  
almost surely, as \(T\) tends to infinity.