Why is the (undirected) remaining degree distribution symmetric in its indices?
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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(?).
probability probability-distributions graph-theory information-theory network
asked Jan 10 at 20:15
JonathanJonathan
1957
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add a comment |
$begingroup$
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(?).
probability probability-distributions graph-theory information-theory network
asked Jan 10 at 20:15
JonathanJonathan
1957
$endgroup$
1
$begingroup$
Can you explain how your $2times 2$ 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?
$endgroup$
– Henning Makholm
Jan 11 at 2:59
$begingroup$
I have edited the question to make things more explicit. Does it make sense now?
$endgroup$
– Jonathan
Jan 11 at 9:11
$begingroup$
@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.
$endgroup$
– Jonathan
Jan 12 at 13:23
add a comment |
$begingroup$
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(?).
probability probability-distributions graph-theory information-theory network
asked Jan 10 at 20:15
JonathanJonathan
1957
$endgroup$
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(?).
probability probability-distributions graph-theory information-theory network
probability probability-distributions graph-theory information-theory network
asked Jan 10 at 20:15
JonathanJonathan
1957
asked Jan 10 at 20:15
JonathanJonathan
1957
edited Jan 14 at 18:24
Jonathan
asked Jan 10 at 20:15
JonathanJonathan
1957
asked Jan 10 at 20:15
JonathanJonathan
1957
asked Jan 10 at 20:15
JonathanJonathan
1957
1957
1
$begingroup$
Can you explain how your $2times 2$ 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?
$endgroup$
– Henning Makholm
Jan 11 at 2:59
$begingroup$
I have edited the question to make things more explicit. Does it make sense now?
$endgroup$
– Jonathan
Jan 11 at 9:11
$begingroup$
@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.
$endgroup$
– Jonathan
Jan 12 at 13:23
add a comment |
1
$begingroup$
Can you explain how your $2times 2$ 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?
$endgroup$
– Henning Makholm
Jan 11 at 2:59
$begingroup$
I have edited the question to make things more explicit. Does it make sense now?
$endgroup$
– Jonathan
Jan 11 at 9:11
$begingroup$
@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.
$endgroup$
– Jonathan
Jan 12 at 13:23
1
1
$begingroup$
Can you explain how your $2times 2$ 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?
$endgroup$
– Henning Makholm
Jan 11 at 2:59
$begingroup$
Can you explain how your $2times 2$ 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?
$endgroup$
– Henning Makholm
Jan 11 at 2:59
$begingroup$
I have edited the question to make things more explicit. Does it make sense now?
$endgroup$
– Jonathan
Jan 11 at 9:11
$begingroup$
I have edited the question to make things more explicit. Does it make sense now?
$endgroup$
– Jonathan
Jan 11 at 9:11
$begingroup$
@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.
$endgroup$
– Jonathan
Jan 12 at 13:23
$begingroup$
@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.
$endgroup$
– Jonathan
Jan 12 at 13:23
add a comment |
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$begingroup$
Can you explain how your $2times 2$ 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?
$endgroup$
– Henning Makholm
Jan 11 at 2:59
$begingroup$
I have edited the question to make things more explicit. Does it make sense now?
$endgroup$
– Jonathan
Jan 11 at 9:11
$begingroup$
@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.
$endgroup$
– Jonathan
Jan 12 at 13:23