Hi Tamas,

Thank you for explaining this for me. But I am still confused about why
multiply

m at last using  Ki/2m * Kj/2m * 2. Since I do not know much about random
graph

theory, could you provide me more information in detail on that?

Thanks!



Isaiah

On Wed, Feb 13, 2013 at 7:28 AM, Tamas Nepusz <[email protected]> wrote:

> > So my question is that why Ki*Kj/2m is "the probability of an edge
> existing
> > between vertices i and j if connections
> >
> > are made at random but respecting vertex degrees".
> In the Molloy-Reed model (i.e. a random graph generation model that
> "respects the vertex degrees"), edges are drawn as follows. Vertex i
> initially has Ki "stubs". Edges are drawn by selecting two stubs from all
> the stubs of all the vertices randomly and then connecting them. The total
> number of stubs is of course 2m because we are going to have m edges in the
> end and each edge connects two stubs. Now, since vertex i has Ki stubs, the
> probability of selecting the first stub from vertex i is Ki/2m, and the
> probability of selecting the second stub from vertex j is Kj/2m. Thus, the
> probability of drawing an edge between vertex i and vertex j in a single
> step of the graph generation process is Ki/2m * Kj/2m. However, since we
> are
> actually drawing m edges, and edges between i and j are the same as edges
> between j and i, the probability of drawing an edge between vertex i to
> vertex j is Ki/2m * Kj/2m * 2m, so we get Ki*Kj / 2m.
>
> --
> T.
>
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