I am trying to learn about SDEs and I don't seem to understand this claim:
Assume $X_t$ is the strong solution to the following SDE: \begin{equation} dX_t=c\tanh X_t \, dt +dB_t \end{equation} Where $B_t$ is a standard brownian motion and $c>0$ is a constant.
The claim is to prove that with probability $1$, \begin{equation} \limsup_{t\rightarrow \infty}\frac{|X_t|}{t}=c \end{equation} **What I know so far:
Since $|\tanh x|\leq 1$ for any $x$ and using the fact that $\lim_{t\rightarrow \infty} \frac{B_t}{t}=0$ w.p. $1$, we can see that
\begin{equation} \limsup_{t\rightarrow \infty}\frac{|X_t|}{t}\leq c \end{equation}
Also, for every $n\in\mathbb{N}$ define $Y_n=X_n-X_{n-1}$. Therefore:
\begin{equation} |Y_n|\geq |B_n-B_{n-1}|-c \end{equation} So for any $M>0$, there exist infinitely many $n$ such that $|Y_n|>2M$ and consequently, there exist infinitely many $t$ such that $|X_t|>M$.
Now, let $M>0$ and define $t_+$ to be the amount of time up to time $t$ that $X_t>M$ and $t_-$ be the amount of time up to time $t$ that $X_t<-M$. I think what I need to show is that either $\frac{t_+}{t}\rightarrow 1$ or $\frac{t_-}{t}\rightarrow 1$. But I don't seem to know how.