Foundations of Data Science

Now breaking the j + k − j into two sums gives
∞∑ ∑k−1
∞∑ ∑k−1
δx j 2 a j x j−1 (k − j)a k−j x k−j + δx ja j x j (k − j) 2 a k−j x k−j−1 .
k=1 j=1
k=1 j=1
Notice that the second sum is obtained from the first by substituting k − j for j and that
both terms are δxg ′ g. Thus,
Hence,
−x + g(x) + 2δxg ′ (x) = 2δxg ′ (x)g(x).
g ′ = 1 2δ
1 − g x
1 − g .
Phase transition for nonfinite components
The generating function g(x) contains information about the finite components of the
graph. A finite component is a component of size 1, 2, . . . which does not depend on t.
∑
Observe that g(1) = ∞ ka k and hence g(1) is the probability that a randomly chosen
k=0
vertex will belong to a component of finite size. If g(1) = 1 there are no nonfinite components.
When g(1) ≠ 1, then 1 − g(1) is the expected fraction of the vertices that are in
nonfinite components. Potentially, there could be many such nonfinite components. But
an argument similar to Part 3 of Theorem 4.8 concludes that two fairly large components
would merge into one. Suppose there are two connected components at time t, each of
size at least t 4/5 . Consider the earliest created 1 2 t4/5 vertices in each part. These vertices
must have lived for at least 1 2 t4/5 time after creation. At each time, the probability of an
edge forming between two such vertices, one in each component, is at least δΩ(t −2/5 ) and
so the probability that no such edge formed is at most ( 1 − δt −2/5) t 4/5 /2
≤ e
−Ω(δt 2/5) → 0.
So with high probability, such components would have merged into one. But this still
leaves open the possibility of many components of size t ε , (ln t) 2 , or some other slowly
growing function of t.
We now calculate the value of δ at which the phase transition for a nonfinite component
occurs. Recall that the generating function for g (x) satisfies
g ′ (x) = 1 2δ
1 − g(x)
x
1 − g (x) .
If δ is greater than some δ critical , then g(1) ≠ 1. In this case the above formula at x = 1
simplifies with 1 − g(1) canceling from the numerator and denominator, leaving just 1 . 2δ
Since ka k is the probability that a randomly chosen vertex is in a component of size k,
∑
the average size of the finite components is g ′ (1) = ∞ k 2 a k . Now, g ′ (1) is given by
g ′ (1) = 1 2δ
120
k=1
(4.6)