Foundations of Data Science

Proof: Apply Theorem 4.1 with α = ε √ np to get that the probability that an individual
vertex has degree outside the range [(1 − ε)np, (1 + ε)np] is at most 3e −ε2 np/8 . By the
union bound, the probability that some vertex has degree outside this range is at most
3ne −ε2 np/8 . For this to be o(1), it suffices for p to be Ω( ln n
nε 2 ). Hence the Corollary.
Note that the assumption p is Ω( ln n ) is necessary. If p = d/n for d a constant, for
nε 2
instance, then some vertices may well have degrees outside the range [(1 − ε)d, (1 + ε)d].
Indeed, shortly we will see that it is highly likely that for p = 1 there is a vertex of degree
n
Ω(log n/ log log n).
When p is a constant, the expected degree of vertices in G (n, p) increases with n. For
example, in G ( n, 2) 1 , the expected degree of a vertex is n/2. In many real applications,
we will be concerned with G (n, p) where p = d/n, for d a constant, i.e., graphs whose
expected degree is a constant d independent of n. Holding d = np constant as n goes to
infinity, the binomial distribution
( n
Prob (k) = p
k)
k (1 − p) n−k
approaches the Poisson distribution
Prob(k) = (np)k e −np = dk
k! k! e−d .
To see this, assume k = o(n) and use the approximations n − k ∼ = n, ( )
n ∼= n k
, and
k k!
(
1 −
d n−k
∼
n)
= e −d to approximate the binomial distribution by
( ( ) k n
lim
)p k (1 − p) n−k = nk d
e −d = dk
n→∞ k
k! n k! e−d .
Note that for p = d , where d is a constant independent of n, the probability of the
n
binomial distribution falls off rapidly for k > d, and is essentially zero for all but some
finite number of values of k. This justifies the k = o(n) assumption. Thus, the Poisson
distribution is a good approximation.
Example: In G(n, 1 ) many vertices are of degree one, but not all. Some are of degree
n
zero and some are of degree greater than one. In fact, it is highly likely that there is a
vertex of degree Ω(log n/ log log n). The probability that a given vertex is of degree k is
If k = log n/ log log n,
( ) ( k ( n 1
Prob (k) =
1 −
k n) 1 ) n−k
≈ e−1
n k! .
log k k = k log k =
log n (log log n − log log log n) ≤ log n
log log n
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