Download Probability Distributions, Node Degree in G(n,p)

CS 485 Lecture 3
Scribes: Myle Ott, Alex Tsiatas
Date: 1/26/07
1
1.1
Review of Probability (cont’d)
Distributions
Bernoulli: X ∼ Ber(p)
S = {0, 1}
P (X = 1) = p, P (X = 0) = 1 − p
E(X) = 0 · (1 − p) + 1 · p = p
var(Y ) = E(X 2 ) − (E(X))2 = p − p2 = p(1 − p)
Binomial: Y ∼ Bin(n, p)
S = {0, . . . , n}
P (Y = k) = nk pk (1 − p)n−k
P
Y = ni=1 X
∼ Ber(p), independent)
Pi (where XiP
n
E(Y ) = E( ni=1
X
)
=
i=1
Pn i
PnE(Xi ) = np
var(Y ) = var( i=1 Xi ) = i=1 var(Xi ) = np(1 − p)
Poisson: Z ∼ P oiss(λ)
1
0
...
n→∞
p→0
np → λ; λ > 0 constant
Bin(n, p) −→ P oiss(λ)
S = 0, 1, . . .
k
P (Z = k) = λk! e−k
1
Why? Fix k:
n k
nk k
λ
λk −λ
p (1 − p)n−k = n→∞
lim
lim
P (Z = k) = n→∞
lim P (Y = k) = n→∞
p (1 − )n =
e
k
k!
n
k!
p→0
p→0
p→0
np→λ
np→λ
np→λ
E(Z) = limn→∞ np = λ
var(Z) = limn→∞ np(1 − p) = λ
Normal: W ∼ N (µ, σ 2 )
S=R
Probability density function: f (x) = α(σ)e−
(x−µ)2
2σ 2
Note: P (W ∈ (µ − 3σ, µ + 3σ)) > 0.99
Note: The normal distribution is continuous, whereas the other distributions are
discrete.
2
Central Limit Theorem
“any sum is normally distributed”
If {Xi }ni=1 is independent and identically distributed* (i.i.d.) with expected value µ and
variance σ 2 , then:
n
X
Xi ∼ N (nµ, nσ 2 )
i=1
*meaning all Xi must come from the same type of distribution
3
Random Graphs (cont’d. . . )
(Back to G(n, p))
• Take the # of edges to be the random variable X
1 if edge e is present
∀ edges e, Ye =
0 otherwise
Ye ∼ Ber(p)
P
X = e Ye ∼ Bin
n+1
2
,p
2
Note: a sum of entirely Bernoulli distributions is exactly a binomial distribution. If
there are some non-Bernoulli distributions in the sum, then a normal distribution is
a reasonable estimate of the true distribution.
Q.
A.
expected
# of edges
p n+1
2
Q.
P [randomly
chosen G from G(n, p) has m edges]
n+1
n+1
2
pm (1 − p)( 2 )−m
m
A.
3.1
Degree Distribution
Let G(V, E) ∈ G(n, p).
Xv : degree of vertex v
Yv,w : edge between v and w is present ({v, w} ∈ E)
Yv,w ∼P
Ber(p)
Xv = w∈V Yv,w ∼ Bin(n, p)
Q.
A.
P [v has degree k]
P [Xv = k] = nk pk (1 − p)(n−k)
E(Xv ) = np
var(Xv ) = np(1 − p)
• vertex degree in G(n, nd )
Bin(n, nd ) −→ n→∞ P oiss(d)
p(n)→0
np(n)→d
E(Xv ) = d
var(Xv ) = d
Note: each vertex in G(n, 21 ) is ofdegree close to
√
√
i.e. P Xv ∈ n2 − 23 n, n2 + 23 n ≥ 0.99
3
n
2.