ECS 289 / MAE 298, Network Theory
Spring 2014
Problem Set # 1, Solutions
Problem 1: Power Law Degree Distributions (Continuum approach)
Consider the power law distribution p(k) = Ak −γ , with support (i.e., defined from) k = 1 to
k → ∞.
a) Show that we must have γ > 1 for this to be a properly defined probability distribution
function (pdf ). Recall a pdf must have two properties: 1) p(k) ≥ 0 for all k, and 2) it must
be normalized.
∞
Z
Ak
1
−γ
∞
i
−A h
−A (−γ+1) −γ+1
k
( lim k
)−1
=
dk =
(γ − 1)
(γ − 1) k→∞
k=1
1) If γ < 1, the limit (limk→∞ k −γ+1 ) → ∞.
2) If γ = 1, the prefactor
−A
(γ−1)
→ ∞.
3) If γ > 1, then the integral is finite:
only converges if γ > 1.
R∞
1
Ak −γ dk =
A
.
(γ−1)
In other words, the integral
b) Solve for the normalization constant A.
R∞
A
From (3) above, we find 1 Ak −γ = (γ−1)
for γ > 1. Setting the integral equal to 1,
implies:
A = (γ − 1).
c) Show that if 1 < γ ≤ 2, the average value hki diverges.
Z
hki =
∞
(γ − 1)k
−γ+1
1
∞
i
(γ − 1) (−γ+2) (γ − 1) h
−γ+2
dk = −
k
=
−
(
lim
k
)
−
1
(γ − 2)
(γ − 2) k→∞
k=1
1) If γ < 2, the limit (limk→∞ k −γ+2 ) → ∞.
2) If γ = 2, the prefactor
(γ−1)
(γ−2)
→ ∞.
3) If γ > 2, then the integral is finite:
integral only converges if γ > 2.
R∞
1
1
(γ − 1)k −γ+1 dk =
(γ−1)
.
(γ−2)
In other words, the
d) Show that if 2 < γ ≤ 3, the average is finite, but the standard deviation diverges.
2
k =
Z
∞
(γ − 1)k
−γ+2
1
∞
i
(γ − 1) (−γ+3) (γ − 1) h
−γ+3
dk = −
k
=
−
(
lim
k
)
−
1
(γ − 3)
(γ − 3) k→∞
k=1
1) If γ < 3, the limit (limk→∞ k −γ+3 ) → ∞.
2) If γ = 3, the prefactor
(γ−1)
(γ−3)
→ ∞.
3) If γ > 3, then the integral is finite:
integral only converges if γ > 3.
R∞
1
(γ − 1)k −γ+2 dk =
(γ−1)
.
(γ−3)
In other words, the
Thus σ 2 = hk 2 i − hki2 → ∞ for γ ≤ 3.
In general, for a power law of the form p(k) ∝ k −γ , the moments hk n i diverge
for all n ≥ (γ − 1).
e) Plot p(k) = Ak −γ , for k = 1 to k = 100, 000 for γ = 3 (and properly normalize). Use
matlab, R, or pen and paper, etc (but make sure to label axes clearly with values).
The simple approach suggests that p(k) = 2k −3 since γ = 3.
But the solution p(k) = 2k −3 is clearly problematic since it gives probability p(k = 1) > 1.
R∞
P
To properly normalize consider that this requires ∞
k=1 p(k) = 1, not k=1 p(k)dk = 1 as used
in part (b). We can solve for the normalization constant numerically using a simple program
P
such as this C program which implements S = 100000
k −3 :
k=1
2
#i n c l u d e <s t d i o . h>
#i n c l u d e <math . h>
main ( ) {
do ub le x , S = 0 . ;
long i ;
f o r ( i =1; i $\ l e $ 1 0 0 0 0 0 ; ++i )
S+=(pow ( i , − 3 . ) ) ;
p r i n t f ( ‘ % 0. 8 l f ’ , S )
}
The inverse normalization constant, A−1 = S = 1.20206. (Note the “discrete” treatment
– in detail next – gives this value). The correctly normalized plot is:
1
f(x)
0.01
0.0001
1e-06
1e-08
1e-10
1e-12
1e-14
1e-16
1
10
100
1000
10000
100000
f ) In a finite network with N nodes, what is the largest possible value of degree, kmax , that
can ever be observed? So can we ever have hki → ∞ in a finite network?
In a finite network of N nodes, kmax = (N − 1) (or N if we allow self-loops). Regardless, we
can never observe hki → ∞ in a finite network.
3
Problem 1: Power Law Degree Distributions (Discrete approach)
In most situations in networks, the k’s take on only discrete values. We can do a similar
treatment, where now we consider k discrete:
∞
X
p(k) =
k=1
∞
X
k=1
0 −γ
Ak
1
1
1
= A 1 + γ + γ + γ + ··· .
2
3
4
0
The sum in the brackets is known as the Riemann Zeta Function, RZ(γ). The value of
RZ(γ), for many values of γ can be found in standard references (e.g., Mathworld, Wikipedia,
etc). Commonly tabulated values are:
RZ(0) = −1/2
RZ(1/2) ≈ −1.460
RZ(1) = ∞
RZ(3/2) ≈ 2.612
RZ(2) = π 2 /6 ≈ 1.645
RZ(3) ≈ 1.202
RZ(4) = π 4 /90 ≈ 1.082
Clearly, for γ < 0, RZ(γ) → ∞. Though the proof is beyond the scope of this assignment,
the relevant information is that RZ(γ) < 0 for 0 ≤ γ < 1, RZ(1) → ∞, and RZ(γ) converges
to a finite positive number for all γ > 1.
P∞
P
0 −γ
= A0 · RZ(γ). This is negative for 0 ≤ γ < 1, and it diverges
(a) ∞
k=1 A k
k=1 p(k) =
P∞
for γ = 1. Thus for k=1 p(k) to converge to a finite positive number requires γ > 1.
(b) Normalization constant, A0 = 1/RZ(γ).
P
2
3
3
0
(c) hki = ∞
1 kp(k) = A 1 + 2γ + 3γ + 4γ + · · ·
1
1
1
= A0 1 + 2γ−1
+ 3γ−1
+ 4γ−1
+ · · · = A0 RZ(γ − 1) = RZ(γ − 1)/RZ(γ),
which is negative or diverges for γ − 1 ≤ 1 or equivalently, γ ≤ 2. Thus hki only converges
to a positive finite number for γ > 2.
(d) Likewise, hk 2 i = A0 RZ(γ − 2) = RZ(γ − 2)/RZ(γ), which is negative for 2 ≤ γ < 3 and
diverges for γ = 3. Thus σ 2 = hk 2 i − hki2 = RZ(γ − 2)/RZ(γ) − [RZ(γ − 1)/RZ(γ)]2 only
converges to a positive finite number for γ > 3.
(e) For plot, see page 3.
(f) In a finite network of N nodes, kmax = (N − 1) (or N if we allow self-loops). Regardless,
we can never observe hki → ∞ in a finite network.
4
Problem 2: Random walk on a network
(a) Let Aij = 1 in the following matrix denote the presence of a link from node j to node i
and Aij = 0 denote the absence of a link. The adjacency matrix is given by


0 1 0 0 1


0 0 1 0 0 



(1)
A=
1
0
0
1
1




0 1 1 0 0 
0 0 0 0 0
(b) In order to obtain the steady state distribution we normalize the adjacency matrix given
above to obtain the transition matrix T . Entry Tij denotes the probability of going from
node j to node i in the next time step.


0 0.5 0 0 0.5


0 0 0.5 0 0 


T =
(2)
0 1 0.5
1 0



0 0.5 0.5 0 0 
0 0
0 0 0
The steady state probability corresponds to the eigenvector of T with eigenvalue 1. Using a
standard software package we find that the eigenvector that corresponds to the eigenvalue
of 1 is


0.182574


0.365148



π=
(3)
0.730296


0.547722
0
5
Note that the above eigenvector is normalized such that the sum of the square of the elements
is equal to 1. Renormalizing such that the sum of the terms is 1 we obtain
 
0.1
 
0.2
 

π=
(4)
0.4
 
0.3
0
(c) For any undirected graph, the steady state probability of finding a random walker is
directly proportional to the degree of the node. Therefore,
 
3
 
3

1 
4
(5)
π=

14 
 
2
2
6