assumptions, the normalized waiting time distribution density of the activation with the time dependent rate from is given by the exponential function $$\label{eq:w0}
  w_{0,i}(t) = \gamma[f_i(t)] \,\exp\left(-\int_0^t\,\gamma[f_i(t')] \,
  \mathrm{d} t' \right).$$

Let $P_{i}(0,t)$ denote the occupation probability of unit $i$ to be in state $s_i(t)=0$ at time $t$. Analogously $P_{i}(1, t)$ for state $s_i(t)=1$. Then the balance of probability flows yields the generalized master equations $$\begin{aligned}
        \dot{P}_i(0,t)&=-J_{0\rightarrow
                  1,i}(t)+J_{1\rightarrow
                  0,i}(t),\\ \dot{P}_i(1,t)&=-J_{1\rightarrow
                  0,i}(t)+J_{0\rightarrow 1,i}(t),
    \end{aligned}
    \label{eq:flows}$$ for all $i\in\{ 1, \dots, N \}$, where $J_{0\rightarrow 1,i}(t)$ gives the probability flow from state $0$ to $1$ of unit $i$ at time $t$. Since the transition $0\rightarrow 1$ is a rate process, its probability flow is simply given by $$J_{0\rightarrow1,i}(t) = \gamma[f_i(t)]P_i(0,t).$$ The second probability flow is given by all the probability that has flown into state $1$ up to time $t$ and stayed there for a time, which is given by the waiting time distribution $w_1(t)$. Thus it is the convolution of $J_{0\rightarrow1,i}(t)$ and $w_1(t')$, $$J_{1\rightarrow0,i}(t) = \int_0^{\infty}\gamma[f_i(t-t')]P_i(0,t-t')w_1(t')\mathrm{d} t'.$$ Using the normalization condition at a given node $$P_i(0,t)=1-P_i(1,t),
    \label{eq:norm_i}$$ the temporal evolution of the occupation probabilities $P_i(1,t)$ is then given by (cf. ) $$\begin{aligned}
    \nonumber &\dot{P}_i(1,t) = \gamma[f_i(t)](1-P_i(1,t))- \\
    \nonumber &-\int_0^{\infty}\gamma[f_i(t-t')](1-P_i(1,t-t'))w_1(t') \mathrm{d} t', 