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Let \begin{equation*} X_t=xe^{-\lambda t}+\sigma e^{-\lambda t}\int_0^t e^{\lambda s} dB_s \end{equation*}

be the solution of Ornstein-Uhlenbeck equation where $B$ is Brownian motion, and $x,\sigma,\lambda$ are all constants. Compute \begin{equation*} \liminf_{t\rightarrow \infty}\frac{X_t}{\sqrt{\log t}}\quad \text{ and } \quad\limsup_{t\rightarrow \infty}\frac{X_t}{\sqrt{\log t}} \end{equation*} Furthermore, What can be said of $\liminf_{t\rightarrow \infty}X_t$ and $\limsup_{t\rightarrow \infty}X_t$?

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By vertical and horizontal rescaling, without loss of generality (wlog) $\la=\si=1$. It is then known (see e.g. formula (1.5)) that wlog \begin{equation} X_t=\frac{B_{e^t}}{e^{t/2}}. \end{equation} So, \begin{equation} \limsup_{t\to\infty}\frac{X_t}{\sqrt{\ln t}} =\lim_{t\to\infty}\sup_{s\ge t}\frac{B_{e^s}}{e^{s/2}\sqrt{\ln s}} =\lim_{t\to\infty}\sup_{u\ge e^t}\frac{B_u}{\sqrt{u\ln\ln u}}=\sqrt2 \end{equation} by the law of the iterated logarithm for the Brownian motion. Similarly, \begin{equation} \liminf_{t\to\infty}\frac{X_t}{\sqrt{\ln t}} =-\sqrt2. \end{equation} For arbitrary positive real $\la$ and $\si$ we then have \begin{equation} \limsup_{t\to\infty}\frac{X_t}{\sqrt{\ln t}}=\si\sqrt2,\quad \liminf_{t\to\infty}\frac{X_t}{\sqrt{\ln t}}=-\si\sqrt2; \end{equation} the latter limits do not depend on $\la$.

It then also follows that \begin{equation} \limsup_{t\to\infty}X_t=\si\sqrt2\,\lim_{t\to\infty}\sqrt{\ln t}=\infty,\quad \liminf_{t\to\infty}X_t=-\infty. \end{equation}

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