$k_{\mathrm{min}}$ and $k_{\mathrm{max}}$ be the minimum and maximum degree occurring in the network, respectively, while $N_k$ is the number of units with degree $k\in[k_{\mathrm{min}},k_{\mathrm{max}}]$. In a random network the degrees can be treated as random numbers. Their occupation probability $p(k)$ is defined by $N_k$. Then the occupation probability reads $$\begin{aligned}
    \label{eq:degree_occ}
            p(k) =  \lim_{N\to \infty }\frac{N_k}{N}.\end{aligned}$$ Mathematically, the limit make sense if $N_k~\propto~N$. Afterwards, sorting the nodes with coinciding degrees $k$ in the network gives the joint probability that any node in the network has this degree and is, respectively, in the dynamical state $s=(0,1)$ at time $t$ $$\begin{aligned}
    \label{eq:degree_prob}
          P(s,k,t)  = \sum_{i=1}^{N}  \,\delta_{s,s_i}\, \delta_{k,k_i}\,P_i(s_i,k_i,t).\end{aligned}$$ Using Bayes’ theorem we split off the occupation probability $p(k)$ as $$\begin{aligned}
    \label{eq:Bayes}
          P(s,k,t)  =  P(s,t|k)p(k).\end{aligned}$$ Normalization reads $$\begin{aligned}
    \label{eq:norm}
          P(0,k,t) \,+\,P(1,k,t)  =  p(k).\end{aligned}$$ Hence, for the conditioned probabilities the degree is fixed and it becomes $$\begin{aligned}
    \label{eq:norm_c}
         P(0,t|k)\,+\, P(1,t|k) = 1.\end{aligned}$$

The chosen conditional probabilities $P(1,t|k)$ neglect that various nodes with the same degree $k$ might be linked to $k$ nodes with different degrees. Therefore, without a further approximation, they are so far not suitable for the description of our situation.

To proceed, the signal function which is coupling the $i$th node to the 