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Unformatted text preview: nvar(I1 ) + n (n − 1)cov(I1 , I2 )
�
�
nq (1 − q ) + n (n − 1) E[I1 I2 ] − E[I1 ]E[I2 ] , i j � =i i = where the second and third equalities follow since the Ii are identically distributed Bernoulli random variables with parameter q (dependent). We have E[I1 I2 ] = P(I1 = 1, I2 = 1) = P(both 1 and 2 are isolated) = (1 − p )2n−3 = q2
.
(1 − p ) Combining the preceding two relations, we obtain
�
var(X ) = nq (1 − q ) + n (n − 1) = nq (1 − q ) + n (n − 1) �
q2
− q2
(1 − p ) q2 p
.
1−p 19 Networks: Lecture 3 Introduction Proof (Continued)
For large n, we have q → 0 [cf. Eq. (3)], or 1 − q → 1. Also p → 0. Hence,
var(X ) ∼ nq + n2 q 2 p
∼ nq + n2 q 2 p
1−p = nn−λ + λn log(n)n−2λ ∼ nn−λ = E[X ], a (n ) where a(n ) ∼ b (n ) denotes b (n) → 1 as n → ∞. This implies that E[X ] ∼ var(X ) ≥ (0 − E[X ])2 P(X = 0), and therefore, P(X = 0) ≤ E[X ]
1
=
→ 0. E[X ]
E[X ]2 It follows that P(at least one isolated node) → 1 and therefore,
P(disconnected) → 1 as n → ∞, completing the proof.
20 Networks: Lecture 3 Introduction Converse
log(n) We next show claim (2), i.e., if p (n ) = λ n with λ > 1, then
P(connectivity) → 1, or equivalently P(disconnectivity) → 0.
From Eq. (4), we have E[X ] = n · n−λ → 0 for λ > 1. This implies probability of isolated nodes goes to 0. However, we need more
to establish connectivity.
The event “graph is disconnected” is equivalent to the existence of k nodes
without an edge to the remaining nodes, for some k ≤ n /2.
We have
P({1, . . . , k } not connected to the rest) = (1 − p )k (n−k ) ,
and therefore,
P(∃ k nodes not connected to the rest) = ��
n
( 1 − p ) k (n −k ) .
k
21 Networks: Lecture 3 Introduction Converse (Continued)
Using the union bound [i.e. P(∪i Ai ) ≤ ∑i P(Ai )], we obtain
P(disconnected graph) ≤
Using Stirling’s formula k ! ∼ � �k
k
e n /2 ��
n
∑ k ( 1 − p ) k (n −k ) .
k =1 k
n
, which implies (k ) ≤ n k in the
k (e) preceding relation and some (ugly) algebra, we obtain
P(disconnected graph) → 0,
completing the proof. 22...
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 Fall '09
 Acemoglu

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