I am trying to calculate the remaining degree distribution of an undirected graph.
Let $q_{j,k}$ be defined as the joint probability distribution of the remaining degrees of the two nodes at either end of a randomly chosen edge. Let $G=(V,E)$ be a graph with three nodes $V=(v_1,v_2,v_3)$, and two edges $E=(e_1,e_2)$ where $e_1=(v_1,v_2)$ and $e_2=(v_2,v_3)$.
In this paper it says that the remaining degree distribution is symmetric in its indices ($q_{j,k} = q_{k,j}$) for an undirected graph. Graph $G$ has two edges both connecting a node with remaining degrees 0 and 1. So according to that paper, the probabilities of finding such an edge would be $q_{0,1} = q_{1,0} = 1/2$. But this would surely imply that there is an equal chance of finding a directed edge connecting nodes with remaining degrees 0 and 1, and another directed in the opposite direction. Shouldn't either $q_{1,0} = 1$ or $q_{0,1} = 1$?
Can anyone explain why this isn't the case?
Its important to me because I'm calculating the Mutual Information of this distribution and you get a very different result depending on if the distribution is symmetric in its indices or erm triangular(?).