Unformatted text preview: d di ( t )
since each new node at each time spreads its m new links randomly over the
t existing nodes at time t .
This diﬀerential equation has a solution
di (t ) = m + m log �t �
i . From this solution, we derive an approximation to the degree distribution.
12 Networks: Lecture 6 “Expected” Degree Distribution
We ﬁrst note that the expected degrees of nodes are increasing over time.
If we ask how many nodes have degree ≤ 100 and we know that a
node born at time τ has degree = 100 at time t , then we are
equivalently asking how many nodes were born on or after time τ .
This implies that at time t , the fraction of nodes having degree less
than or equal to 100 would be t −τ .
For any d and any time t , let i (d ) be a node such that di (d ) (t ) = d . The
i (d )
resulting cumulative distribution function then is Ft (d ) = 1 − t . Applying this technique to the uniform attachment model, we solve for i (d )
i (d )
d = m + m log
= e− m ,
i (d )
and therefore the distribution function Ft (d ) = 1 − e − d −m
m . This is an exponential distribution with support from m to inﬁnity and a
mean degree of 2m.
13 Networks: Lecture 6 Preferential Attachment Model
Nodes are born over time and indexed by their date of birth.
Assume that the system starts with a group of m nodes all connected to one
Each node upon birth forms m (undirected) edges with pre-existing nodes.
Instead of selecting m nodes uniformly at random, it attaches to nodes with
probabilities proportional to their degrees.
For example, if an existing node has 3 times as many links as some
other existing node, then it is 3 times as likely to be linked to by the
Thus, the probability that an existing node i receives a new link to the
newborn node at time t is m times i ’s degree relative to the overall degree
of all existing nodes at time t , or
d (t )
m t i
∑j =1 dj (t )
14 Networks: Lecture 6 Preferential Attachment Model
Since there are tm total links at time t in the system, it follows that
∑t=1 dj (t ) = 2tm. Therefore, the probability that node i gets a new link in
d (t ) time t is i2t .
Hence, we can write down the evolution of expected degrees in continuous
d di ( t )
d (t )
with initial condition di (i ) = m (assuming degree is a continuous variable).
This equation has a solution:
di (t ) = m � t �1/2 .
As before, expected degrees of nodes are increasing over time.
Hence to ﬁnd the fraction of nodes with...
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- Fall '09