time coordinate $t>u$ are $$\begin{aligned}
p_{-}(t,u)&=R_{-1+1}(t)\exp\left\{-\int^t_uR_{-1+1}(s)\,ds\right\}\\[0.5em] 
p_{+}(t,u)&=R_{+1-1}(t)\exp\left\{-\int^t_uR_{+1-1}(s)\,ds\right\}\end{aligned}$$ where $p_{-}(t,u)$ is for left to right and $p_{+}(t,u)$ is for right to left.

We consider escaping from the left well. The right well is similar. Divide the time interval $[u,t]$ into many small time intervals $$\begin{aligned}
\delta t =\frac{t-u}{N}\end{aligned}$$ Similar to how we derived the invariant measures we want to derive the probability of escape in a very small time interval $[t,t+\delta t]$. This is given by $$\begin{aligned}
p_{-1+1}([t,t+\delta t])
&=p(t)\delta t \\
&=1-e^{-R_{-1+1}(t)\delta t}\\
&\approx R_{-1+1}(t)\delta t \end{aligned}$$ which is valid for small $\delta t$. Large deviations allow us to say even more about the escape time $\tau_{-1+1}$ and $\tau_{+1-1}$. Theorem 1 in [@oliveri_expoen] shows that it is an exponentially distributed random variable. The probability of staying in the left well is given by $$\begin{aligned}
p_{-1-1}([t,t+\delta t])
&=1-p_{-1+1}([t,t+\delta t])\\
&=1-p(t)\delta t \\
&=1-\left(1-e^{-R_{-1+1}(t)\delta t}\right)\\
&=e^{-R_{-1+1}(t)\delta t}\end{aligned}$$ We want to know the probability of escaping in the time interval $[t,t+\delta t]$ given that the particle has entered at $u$ and stayed up to time $t$. This is given by $$\begin{aligned}
p_{-1-1}([u,t])p_{-1+1}([t,t+\delta t])
&=\prod_{i=1}^{N}p_{-1-1}\left([u+(i-1)\delta t,u+i\delta t]\right)
p_{-1+1}([t,t+\delta t])\\
&=\prod_{i=1}^{N}\exp\left\{-R_{-1+1}\left(u+(i-1)\delta t\right)\delta t\right\}
p_{-1+1}([t,t+\delta t])\\
&=\exp\left\{\sum_{i=1}^{N}-R_{-1+1}\left(u+(i-1)\delta t\right)\delta t\right\}
p_{-1+1}([t,t+\delta t])\\
&=\exp\left\{-\int^t_uR_{-1+1}(s)\,ds\right\}
R_{-1+1}(t)\delta t\end{aligned}$$ This completes 