be an integer, then the following integral can be rewritten as $$\begin{aligned}
\int^{(i+1)T}_{iT}
p(s)g(s)\,ds
=
g(T)^i\int^T_0p(s)g(s)\,ds\end{aligned}$$ Let the time be given by $NT+t$ where $N$ is an integer number of periods. This means the following integral can be written as $$\begin{aligned}
\int^{NT+t}_0
p(s)g(s)\,ds
=
g(T)^N
\int^t_0
p(s)g(s)\,ds
+
\int^T_0p(s)g(s)\,ds
\sum_{i=0}^{N-1}
g(T)^i\end{aligned}$$ So the state probability is equal to $$\begin{aligned}
\nu_-(NT+t)
=
\frac{\nu_-(0)}{g(T)^Ng(t)}
+
\frac{\int^t_0p(s)g(s)\,ds}{g(t)}
+
\frac{\int^T_0p(s)g(s)\,ds}{g(t)}
\frac{1}{g(T)-1}
\left(
1-\frac{1}{g(T)^N}
\right)\end{aligned}$$ Letting $N\longrightarrow\infty$ gives the required result.

### Continuous Time Markov Chain - Synchronised Saddles $p=q$

If the forcing is such that the height of the barrier stays the same for both pathways the same then this corresponds to the case $p=q$

\[chap\_4\_thm:not:equal\_contin\] Let $p=q$ and $t\geq0$. The state probabilities are given by $$\begin{aligned}
\nu_-(t)&=\frac{1}{2}-\frac{\nu_+(0)-\nu_-(0)}{2}\exp\left\{-2\int_0^tp(s)\,ds\right\}\\[0.5em]
\nu_+(t)&=\frac{1}{2}+\frac{\nu_+(0)-\nu_-(0)}{2}\exp\left\{-2\int_0^tp(s)\,ds\right\}\end{aligned}$$

\[cor\_half\_continuous\] For the state probabilities in Theorem \[chap\_4\_thm:not:equal\_contin\] the invariant measures are $$\begin{aligned}
\overline{\nu}_-(t)=\overline{\nu}_+(t)=\frac{1}{2}\end{aligned}$$

Probability Density Function of Escape Times
--------------------------------------------

The escape rates from the left to right are denoted by $R_{-1+1}(\cdot)$ and right to left escape rates are denoted by $R_{+1-1}(\cdot)$. The PDFs for the escape times are given by the Theorem below.

\[chap\_4\_pdf\_thm\] Let $u$ be the time coordinate of entry into a well, then the PDFs for the escape occurring at 