# Why is the (undirected) remaining degree distribution symmetric in its indices?

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(?).

• Can you explain how your $2\times 2$ matrices are supposed to be interpreted as distributions? Googling "remaining degree distribution" suggests that you're looking for a probability distribution, but how does your notation describe that? – Henning Makholm Jan 11 at 2:59
• I have edited the question to make things more explicit. Does it make sense now? – Jonathan Jan 11 at 9:11
• @HenningMakholm I have edited the question to make it even simpler and I reference a paper which should give some context. I hope it makes sense. – Jonathan Jan 12 at 13:23