Intro
Average degree
Singletons
The partial duplication random graph
Peter Pfaffelhuber
with Felix Hermann
Munich, July 2014
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Intro
Average degree
Singletons
Yeast protein-protein interaction network
from Barabsi & Oltvai, Network biology: understanding the cell’s functional organization, 2004
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Intro
Average degree
Singletons
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The model from Pastor-Sartorras, Chung, Bebek,...
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Protein-protein interactions:
key for biological functions of cells
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Evolution: proteins are copied, their function partially
retained, new interactions can emerge
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Model: copy vertex with all edges, keep each edge with
probability p, insert additional r edges at random
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With the right selection of parameters q and r , the general
duplication model well approximates the degree distribution of
the yeast proteome network. (Bebek et al 2006)
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In the following, we analyse the model for q = 0
Intro
Average degree
Singletons
The partial duplication random graph
Start with some graph
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Intro
Average degree
Singletons
The partial duplication random graph
v
Pick a vertex purely at random
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Intro
Average degree
Singletons
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The partial duplication random graph
v0
v
Copy the vertex together with all edges
Intro
Average degree
Singletons
The partial duplication random graph
v0
v
Keep every edge with probability p
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Intro
Average degree
Singletons
The partial duplication random graph
v0
v
This was one step in the partial
duplication random graph
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Intro
Average degree
Singletons
Example with 50 vertices and p = 0.6
9 ●
39 ●
49
8 ●
20 ●
23 ●
31 ●
38 ●
27 ●
5 ●
15 ●
25 ●
45 ●
1 ●
11 ●
12 ●
14 ●
21 ●
41 ●
42 ●
44 ●
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3
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19 ●
7 ●
10
17 ●
6 ●
2 ●
4 ●
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36
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46
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13
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16
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18 ●
50
22
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26
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30
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40 ●
33
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28
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29
32 ●
37 ●
47 ●
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35
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43
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48
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24
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34
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Intro
Average degree
Singletons
Theorem 1
Let E ◦ (n) be the average degree when there are n vertices. Then,
(
∞, p > 1/2,
n→∞
E ◦ (n) −−−→
0, p < 1/2.
Theorem 2
Let F0◦ (n) be the frequency of singletons and p ∗ + log p ∗ = 0.
(Note p ∗ ≈ 0.56.) Then,
n→∞
F0◦ (n) −−−→ 1
iff
p ≤ p∗.
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Intro
Average degree
Singletons
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The average degree
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Let E (n) be the number of edges when there are n vertices
2p E[E (n + 1)|Fn ] = E (n) 1 +
,
n
i
h
n0 · · · (n − 1)
n0 · · · n
Fn = E (n)
E E (n + 1)
(n0 + 2p) · · · (n + 2p)
(n0 + 2p) · · · (n − 1 + 2p)
|
{z
}
{z
}
|
n→∞ Γ(n0 +2p) −2p
n
Γ(n0 )
non-negative martingale
−−−−→
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n→∞
This implies
n−2p E (n) −−−→a.s. E (∞).
E
(n)
E ◦ (n) := n is the average degree of a vertex. So,
(
∞, p > 1/2,
n→∞
◦
E (n) −−−→
0, p < 1/2.
Intro
Average degree
Singletons
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The frequency of singletons
Let F0◦ (n) be the frequency of singletons
n→∞
E ◦ (n) −−−→ 0
●
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0.8
1.0
p < 1/2 :
n→∞
F0◦ (n) −−−→ 0
=⇒
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0.6
●
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0.4
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0.2
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at time n=1000
0.0
frequency of singletons
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0.0
0.2
●
●
0.4
0.6
p
0.8
1.0
Intro
Average degree
Singletons
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The frequency of singletons
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Fk (n): number of vertices of degree k, Fk◦ (n) =
Hx◦ (n) :=
∞
X
Fk◦ (n)(1 − x)k
=⇒
Fk (n)
n ,
F0◦ (n) = H1◦ (n).
k=0
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Using recursions,
E[Hx◦ (n + 1) − Hx◦ (n)|Fn ]
d
1 ◦
(n) − Hx◦ (n)
=
px(1 − x) Hx◦ (n) + Hpx
n+1
dx
Intro
Average degree
Singletons
The frequency of singletons
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Time continuous partial duplication graph G:
Every vertex splits at rate 1 + |G1s | . Then,
d
d
◦
E[Hx◦ (s)|Fs ] = px(1 − x) Hx◦ (s) + Hpx
(s) − Hx◦ (s)
ds
dx
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Let X be a Markov process which jumps from x to px at
rate 1, and increases at rate x(1 − x) between jumps
⇒
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d
E[f (Xs )|Fs ] = pXs (1 − Xs )f 0 (Xs ) + f (pXs ) − f (Xs )
ds
Let G, X be independent. Then,
d
E[HX◦ t−s (s)] = 0
ds
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Intro
Average degree
Singletons
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The frequency of singletons
d
E[HX◦ t−s (s)] = 0
ds
E[HX◦ 0 (t)] = E[HX◦ t (0)]
=⇒
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The process X satisfies, for p ∗ + log p ∗ = 0
(
0,
p ≤ p∗
t→∞
Xt ===⇒
X∞ 6= 1, p > p ∗
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Hence,
lim E[F0◦ (n)] = lim E[F0◦ (t)]
t→∞
n→∞
=
lim E[H1◦ (t)]
t→∞
=
E[HX◦ ∞ (0)]
(
= 1, p ≤ p ∗
< 1, p > p ∗
Intro
Average degree
Singletons
What next?
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Study connected component (power law?)
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Include additional r edges as in the original model.
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Parameter estimation using protein-protein-interaction
network.
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