Foundations of Data Science

same or are disjoint,
⎛( n∑
) ⎞ 2
E(x 2 ) = E ⎝ x i
⎠ = ∑
i=1
i,j
E(x i x j ) = ∑ i
E(x 2 i ) + ∑ i≠j
E(x i x j )
= E(x) + ∑ ∑
Prob ( cc(i) = cc(j) = S ) + ∑ ∑
Prob ( cc(i) = S; cc(j) = T )
i≠j S
i≠j S,T
disjoint
= E(x) + ∑ ∑
Prob(S is a c.c. ) + ∑ ∑
Prob(S, T each a cc )
i≠j S:i,j∈S i≠j S,T :i∈S,j∈T
disjoint
= E(x) + ∑ ∑
Prob ( S is a c.c. )
i≠j S:i,j∈S
+ ∑ ∑
Prob ( S is a c.c ) Prob ( T is a c.c ) (1 − p) −|S||T |
i≠j
S,T :i∈S,j∈T
disjoint
( ∑
≤ O(n) + (1 − p) −|S||T | Prob ( cc(i) = S )) ( ∑
Prob ( cc(j) = T ))
S
T
≤ O(n) + ( 1 + o(1) ) E(x)E(x).
In the next to last line, if S containing i and T containing j are disjoint sets, then the two
events, S is a connected component and T is a connected component, depend on disjoint
sets of edges except for the |S||T | edges between S vertices and T vertices. Let c 4 be a
constant in the interval (c 3 , 1). Then, by Chebyshev inequality,
Prob(x > c 4 n) ≤
Var(x)
(c 4 − c 3 ) 2 n 2 ≤ O(n) + o(1)c2 3n 2
(c 4 − c 3 ) 2 n 2 = o(1).
For the proof of (3) suppose a pair of vertices u and v belong to two different connected
components, each of size at least n 2/3 . We show that with high probability, they should
have merged into one component producing a contradiction. First, run the breadth first
search process starting at v for 1 2 n2/3 steps. Since v is in a connected component of size
n 2/3 , there are Ω(n 2/3 ) frontier vertices. The expected size of the frontier continues to grow
until some constant times n and the actual size of the frontier does not differ significantly
from the expected size. The size of the component also grows linearly with n. Thus,
the frontier is of size n 2 3 . See Exercise 4.22. By the assumption, u does not belong to
this connected component. Now, temporarily stop the breadth first search tree of v and
begin a breadth first search tree starting at u, again for 1 2 n2/3 steps. It is important to
understand that this change of order of building G(n, p) does not change the resulting
graph. We can choose edges in any order since the order does not affect independence or
conditioning. The breadth first search tree from u also will have Ω(n 2/3 ) frontier vertices
with high probability . Now grow the u tree further. The probability that none of the
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