Foundations of Data Science

and thus k k ≤ n. Since k! ≤ k k ≤ n, the probability that a vertex has degree k =
log n/ log log n is at least 1 k! e−1 ≥ 1 . If the degrees of vertices were independent random
en
variables, then this would be enough to argue that there would be a vertex of degree
log n/ log log n with probability at least 1 − ( 1 − 1
en) n
= 1 − e
− 1 e ∼ = 0.31. But the degrees
are not quite independent since when an edge is added to the graph it affects the degree
of two vertices. This is a minor technical point, which one can get around.
4.1.2 Existence of Triangles in G(n, d/n)
What is the expected number of triangles in G ( n, d n)
, when d is a constant? As the
number of vertices increases one might expect the number of triangles to increase, but this
is not the case. Although the number of triples of vertices grows as n 3 , the probability
of an edge between two specific vertices decreases linearly with n. Thus, the probability
of all three edges between the pairs of vertices in a triple of vertices being present goes
down as n −3 , exactly canceling the rate of growth of triples.
A random graph with n vertices and edge probability d/n, has an expected number
of triangles that is independent of n, namely d 3 /6. There are ( n
3)
triples of vertices.
Each triple has probability ( d 3
n)
of being a triangle. Let ∆ijk be the indicator variable
for the triangle with vertices i, j, and k being present. That is, all three edges (i, j),
(j, k), and (i, k) being present. Then the number of triangles is x = ∑ ijk ∆ ijk. Even
though the existence of the triangles are not statistically independent events, by linearity
of expectation, which does not assume independence of the variables, the expected value
of a sum of random variables is the sum of the expected values. Thus, the expected
number of triangles is
( ∑ )
E(x) = E ∆ ijk = ∑ ( ) ( 3 n d
E(∆ ijk ) =
≈
3 n) d3
6 .
ijk
ijk
Even though on average there are d3 triangles per graph, this does not mean that with
6
high probability a graph has a triangle. Maybe half of the graphs have d3 triangles and
3
the other half have none for an average of d3 triangles. Then, with probability 1/2, a
6
graph selected at random would have no triangle. If 1/n of the graphs had d3 n triangles
6
and the remaining graphs had no triangles, then as n goes to infinity, the probability that
a graph selected at random would have a triangle would go to zero.
We wish to assert that with some nonzero probability there is at least one triangle
in G(n, p) when p = d . If all the triangles were on a small number of graphs, then the
n
number of triangles in those graphs would far exceed the expected value and hence the
variance would be high. A second moment argument rules out this scenario where a small
fraction of graphs have a large number of triangles and the remaining graphs have none.
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