neither well. Only when it enters the other well would $Y^\epsilon_t$ change sign.

The escape time from the left to right well $\tau_{-1+1}$ and from the right to left well $\tau_{+1-1}$ are defined in the following way $$\begin{aligned}
\tau_{-1+1}&=l\left\{t:Y^\epsilon_t=-1\right\}\quad \text{where} \quad \left\{t:Y^\epsilon_t=-1\right\} \quad \text{is an interval}\\
\tau_{+1-1}&=l\left\{t:Y^\epsilon_t=+1\right\}\quad \text{where} \quad \left\{t:Y^\epsilon_t=+1\right\} \quad \text{is an interval}\end{aligned}$$ where $l$ denotes the Lebesgue measure. In other words the time spent being in the state $Y^\epsilon_t=-1$ is $\tau_{-1+1}$ and the time spent being in the state $Y^\epsilon_t=+1$ is $\tau_{+1-1}$. These intervals will always be closed intervals. The process $Y^\epsilon_t$ has two states, hence each sample is a piecewise constant function. The length of each piece is the escape time $\tau_{-1+1}$ or $\tau_{+1-1}$. At every point in time it is possible to define a state probability, that is the probability of the trajectory being $-1$ or $+1$ $$\begin{aligned}
P\left(Y_t=-1\right)=\nu_-(t)
\quad \text{and} \quad 
P\left(Y_t=+1\right)=\nu_+(t)\end{aligned}$$ A continuous time Markov Chain model for Equation \[chap\_4:markov\] is studied next by studying its state probabilities.

Continuous Time Markov Chain
----------------------------

Consider a two state continuous time Markov Chain given by $Y_t=\pm1$. The probability of transiting from $Y_t=-1$ to $Y_t=+1$ in a small time interval $[t,t+d t]$ is $p_{-1+1}\left([t,t+d t]\right)\sim p(t)dt$. Similarly 