Foundations of Data Science

√
Thus, the probability for all such u is (1 − 2p 2 ) n−3 ln n
. Substituting c
n
(
1 − 2c2 ln n
n
) n−3
∼= e
−2c 2 ln n = n −2c2 ,
for p yields
which is an upper bound on E(I ij I kl ) for one i, j, k, and l with a = 3. Summing over all
distinct triples yields n 3−2c2 for the second summation in (4.2).
For the third summation, since the value of I ij is zero or one, E ( I 2 ij)
= E (Iij ). Thus,
∑
E ( Iij) 2 = E (x) .
ij
Hence, E (x 2 ) ≤ n 4−2c2 + n 3−2c2 + n 2−c2 and E (x) ∼ = n 2−c2 , from which it follows that
for c < √ 2, E (x 2 ) ≤ E 2 (x) (1 + o(1)). By a second moment argument, Corollary 4.4, a
graph almost surely has at least one bad pair of vertices and thus has diameter greater
than two. Therefore, the property that the diameter of G(n, p) is less than or equal to
two has a sharp threshold at p = √ √
2
Disappearance of Isolated Vertices
ln n
n
The disappearance of isolated vertices in G (n, p) has a sharp threshold at ln n.
At n
this point the giant component has absorbed all the small components and with the
disappearance of isolated vertices, the graph becomes connected.
Theorem 4.6 The disappearance of isolated vertices in G (n, p) has a sharp threshold of
ln n
n .
Proof: Let x be the number of isolated vertices in G (n, p). Then,
E (x) = n (1 − p) n−1 .
Since we believe the threshold to be ln n
ln n
, consider p = c . Then,
n n
lim E (x) = lim n ( )
1 − c ln n n
n→∞ n→∞ n = lim ne −c ln n = lim n 1−c .
n→∞ n→∞
If c >1, the expected number of isolated vertices, goes to zero. If c < 1, the expected
number of isolated vertices goes to infinity. If the expected number of isolated vertices
goes to zero, it follows that almost all graphs have no isolated vertices. On the other
hand, if the expected number of isolated vertices goes to infinity, a second moment argument
is needed to show that almost all graphs have an isolated vertex and that the
isolated vertices are not concentrated on some vanishingly small set of graphs with almost
all graphs not having isolated vertices.
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