Foundations of Data Science

problem contributes a minuscule error. Restricting k to be a fixed constant and letting
t → ∞ in this section avoids these problems.
Assume that the above equations are exactly valid. Clearly, d 0 (1) = 1 and d 1 (1) =
d 2 (1) = · · · = 0. By induction on t, there is a unique solution to (4.4), since given d k (t)
for all k, the equation determines d k (t + 1) for all k. There is a solution of the form
d k (t) = p k t, where p k depends only on k and not on t, provided k is fixed and t → ∞.
Again, this is not precisely true since d 1 (1) = 0 and d 1 (2) > 0 clearly contradict the
existence of a solution of the form d 1 (t) = p 1 t.
and
Set d k (t) = p k t. Then,
(t + 1) p 0 = p 0 t + 1 − 2δ p 0t
t
p 0 = 1 − 2δp 0
p 0 = 1
1 + 2δ
(t + 1) p k = p k t + 2δ p k−1t
− 2δ p kt
t t
p k = 2δp k−1 − 2δp k
p k =
2δ
1 + 2δ p k−1
( ) k 2δ
=
p 0
1 + 2δ
= 1 ( ) k 2δ
. (4.5)
1 + 2δ 1 + 2δ
Thus, the model gives rise to a graph with a degree distribution that falls off exponentially
fast with degree.
The generating function for component size
Let n k (t) be the expected number of components of size k at time t. Then n k (t) is
proportional to the probability that a randomly picked component is of size k. This is
not the same as picking the component containing a randomly selected vertex (see Figure
4.14). Indeed, the probability that the size of the component containing a randomly selected
vertex is k is proportional to kn k (t). We will show that there is a solution for n k (t)
of the form a k t where a k is a constant independent of t. After showing this, we focus on
the generating function g(x) for the numbers ka k (t) and use g(x) to find the threshold
for giant components.
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