the probability of transiting from $Y_t=+1$ to $Y_t=-1$ in a small time interval $[t,t+d t]$ is $p_{+1-1}\left([t,t+d t]\right)\sim q(t)dt$. The probability of $Y_t$ staying at $-1$ in the small time interval $[t,t+d t]$ is $p_{-1-1}\left([t,t+d t]\right)$. Similarly the probability of $Y_t$ staying at $+1$ in the small time interval $[t,t+d t]$ is $p_{+1+1}\left([t,t+d t]\right)$. These probabilities satisfy $$\begin{aligned}
p_{-1-1}+p_{-1+1}=1
\quad \text{and} \quad 
p_{+1-1}+p_{+1+1}=1\end{aligned}$$ Let $p:\mathbb{R}\rightarrow \mathbb{R}$ and $q:\mathbb{R}\rightarrow \mathbb{R}$ be cyclic functions on the interval $[0,T]$ where $T$ is the period. The behaviour of $\nu(t)=
\left(
\begin{array}{cc}
\nu_-(t)&\nu_+(t)
\end{array}
\right)^\dagger$ is described by $$\begin{aligned}
\frac{d\nu}{dt}&=\label{markov_chain_main_eqn}
Q^\dagger
\nu
\quad 
\text{where}
\quad 
Q=
\left(
\begin{array}{cc}
-p(t) & p(t)\\[0.5em]
q(t) & -q(t)
\end{array}
\right)\end{aligned}$$ and $Q$ is the infinitesimal generator. The aim now is to derive the state probability by solving this differential equation for various forms of $p$ and $q$. The extra conditions we use are $$\begin{aligned}
\nu_-(t)+\nu_+(t)=1\label{markov_chain_main_condition_1}
\quad \text{and} \quad 
\nu'_-(t)+\nu'_+(t)=0\end{aligned}$$ for all times $t$ and the initial conditions at $t=0$ are $\nu_-(0)$ and $\nu_+(0)$.

After a very long time the state probabilities $\nu_\pm(\cdot)$ should not depend on the initial state probabilities $\nu_\pm(0)$. At time infinity $\nu_\pm(\cdot)$ should also be cyclic on $[0,T]$. Let the time be given by $t=NT+n$ where 