Let $dX_t=dB_t + a \cot(X_t)dt$, with $X_0=x \in (0,\pi)$, where $a$ is a specific constant so that the lifetime of the process is infinite almost surely. The process has a transition density which can be extended to $0$ and $\pi$, so the process can be started at $0$ or $\pi$ as well. Suppose $x < \pi/2$, and consider the stopping time $T=\inf\{t: X_t \geq \pi/2\}$.
The question I am trying to answer is to show that $\mathbb{P}[T \geq t]$ decays exponentially. That is, show that there is some positive constant $c$ so that $\mathbb{P}[T \geq t] \leq e^{-c t}$.
My idea is that the process has to keep trying to reach over $\pi/2$. Observe that $\mathbb{P}[T \geq N]=\mathbb{P}[\{\sup_{0 \leq t\leq 1}X_t \leq \pi/2\}, \dots, \{\sup_{N-1 \leq t \leq N}X_t \leq \pi/2\}]$. During each time interval $k-1 \leq t \leq k$, the process lies above a coupling of the same diffusion but started at $0$, so if $p=\mathbb{P}^0[T \geq 1]$, then we have $\mathbb{P}[\sup_{k-1 \leq t \leq k}X_t \leq \pi/2] \leq p$. By the Markov property, these couplings are independent of the path before each time, so the above probability can be bounded by $\mathbb{P}[T \geq N] \leq p^N$.
My advisor seems to imply that this is a standard argument, but I'm not sure if I have the rigorous details correct, or that I am completely understanding him.
Thank you.