Assume you have a graph with $n$ nodes/vertices and we can assign to each node a "type" : type $0$ or type 1. The types are independent. The probability of type $0$ is $$1 - \lambda \in (0,1)$$ and the probability of type $1$ is $$\lambda \in (0,1)$$ We are interested in knowing the type because it gives the capacity of the edge. If an edge connecting node $i$ to node $j$ both of type $0$ then we denote the capacity by $\alpha_{00} \in \mathbb{R}_+ ^*$. If an edge connects node $i$ to $j$ where $i$ is of type of $0$ and $j$ of type $1$ then we denote the capacity by $\alpha_{01}$ and so on. Moreover, $$\alpha_{00} < \alpha_{01} < \alpha_{11} < \alpha_{10}$$
Let $A = \{\alpha_{00} , \alpha_{01} , \alpha_{11} , \alpha_{10}\}$. If the total number of Edges is $N > 0$ then what is the proportion $f ( \alpha_{ij})$ of the capacity $\alpha_{ij}$ among all the possible edge capacities? Is it simply $\mathbb{P}$(type of i) x $\mathbb{P}$(type of j) or $\mathbb{P}$(type of i) x $\mathbb{P}$(type of j) / N?
If we have $f(\alpha_{k})$ with $ k \in \{ 00 , 01 , 10 , 11 \}$ can we write $F(\alpha_{k} ) = \sum_{j \leq k} f(\alpha_{j})$ and then maybe use the law of large numbers an say that $\alpha$ is distributed with c.d.f $F(.)$ on A ?
Thank you
I assume here that there is no self loops on the nodes. Otherwise the property is not true. Indeed for the graph $n_1\to n_1$ we have $f(\alpha_{ii})=P(i)$ and for $i\neq j$, $f(\alpha_{ij})=0$.
We show by recurrence on $n$ the number of nodes that $f(\alpha_{ij})=P(i).P(j)$.
For $n=2$ the property is clear. (If not you can go through all case of type for the two nodes and convince yourself).
Consider now a graph $G_{n+1}$ with $n+1$ nodes and $N$ edges. Let $v_0$ be a node of this graph. Let $G_{v_0}$ be the graph composed only of the neighbour of $v_0$ and of the edges with one ends in $v_0$. Let $G_{\neg v_0}$ be the graph $G_{n+1}$ without $v_0$.
We denote $n_0$ (resp. $n_{\neg 0}$ ) the number of nodes in $G_{v_0}$ (resp. $G_{\neg v_0}$), and $N_0$ (resp. $N_{\neg 0}$) the number of edges.
Assume that $n_0<n+1$, by hypothesis recurrence we have that $f_{G_{v_0}}(\alpha_{ij})= P(i).P(j)$ and $f_{G_{\neg v_0}}(\alpha_{ij})= P(i).P(j)$. Thus we obtain that $$f_{G_{n+1}}(\alpha_{ij})=(f_{G_{\neg v_0}}(\alpha_{ij})*N_{\neg 0}+f_{G_{v_0}}(\alpha_{ij})*N_0)/(N_{\neg 0}+N_0)$$ $$=(P(i).P(j)*N_{\neg 0}+P(i).P(j)*N_0)/N =P(i).P(j) $$
If $n_0=n+1$ (i.e. if $v_0$ is connected to all the nodes) you can, with the same idea, remove one node of $G_{v_0}$ (it is connected only to $v_0$) to obtain 2 graph with less than $n+1$ nodes.
Thus it is always true that $f(\alpha_{ij})=P(i).P(j)$.