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EXERCISE 5.2 Three types of Link Formation
In the appendix of Jackson and Rogers [337] the following sort of growing network formation process is considered. Newborn nodes form links to existing nodes. An existing node gets links from a newborn in three different ways:
•  some links are formed with a probability relative to the size of the existing node (as in preferential attachment),
•  some links are formed with a probability depending on the total time that has already evolved (as in the growing variation of the purely random network), and
•  some links are formed with a constant probability.
We can think of the second and third ways of forming links as different extensions of the idea of purely random Poisson networks to a growing set of nodes. The difference between these two is only in terms of how the probability of a link scales with the size of the society. In both cases, each existing node at some time has an equal chance of getting a new link from a newborn node. The difference is in terms of what the probability of a link is. Is it that we are keeping the average degree of newborn nodes constant - which necessitates a probability of link formation that decreases with the size of the society; or is it that we are holding the probability of a link between any two nodes constant - which necessitates a growing average degree. The analysis in Section
?? worked by holding average degree constant, but the other approach is also natural in some applications.
Allowing for an arbitrary combination of all three of these different methods of
forming links leads to the following expression for the change in a node i's degree over time at a time t:
ddi (t) = adi (t) + b + c                                                (5.18)
dt             t         t
Â
Â
where a b, and c are scalars.
Solve for the degree distribution under a continuous time mean-field approximation of the process under the condition that a > 0 and either c = 0 or a /= 1, and with an initial condition of di(i) = dO. Note that in those cases, the solution to (5.18) is
Â
Â
di(t) = cpt(i) =
Â
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