Temporal Ontology Shervin Daneshpajouh ce.sharif.edu/~daneshpajouh.
Static Random Network Models - ce.sharif.edu
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Static Random Network Models
Social and Economic Networks
MohammadAmin Fazli
Social and Economic Networks 1
ToC
β’ Static Random Network Modelsβ’ Different static random network models
β’ Some basic mathematics
β’ Properties of random networks
β’ Readings:β’ Chapter 4 from the Jackson book
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Static Random Network Models
β’ Static random network models:β’ Static random model refers to a typed model in which all nodes are
established at the same time and then links are drawn between them according to some probabilistic rule
β’ Important to capture features of SENs which are produced by random processes
β’ Different random graph models:β’ Poisson networks & related models
β’ Watts-Strogatz model for small-world networks
β’ Markov graphs & p* networks
β’ The Configuration model
β’ The Expected Degree model
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Poisson Networks
β’ πΊ π, π :β’ Consider a set of nodes π = 1,2,β¦ , π
β’ Connect each pair π, π of nodes with probability π
β’ The expected number of edges: π2
π
β’ The expected degree of nodes: π β 1 π
β’ πΊ(π,π):
β’ Choose π edges out of all π2
pair of nodes: π2π
choices
β’ Number of edges: π
β’ The expected degree of nodes: ππ2
Γ π β 1 =2π
π
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Poisson Networks
β’ Binomal Distribution:β’ Consider a sequence of Bernoulli trials. What is the probability of m heads out
of n flips?
β’ Expected number of heads: np
β’ The variance: npq = np(1-p)
β’ Standard deviation: ππ 1 β π
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Poisson Networks
β’ Degree distribution: Binomal distributionβ’ The probability of having d neighboring edges is equal to:
π π =π β 1
πππ 1 β π πβ1βπ
β’ Binomal distribution can be approximated by π = (π β 1)π for large π
π π βπβπππ
π!
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Watts-Strogatz Model
β’ Consider a cycle and connect each node to its 2πnearest nodes
β’ Diameter: π
4
β’ Clustering Coefficient:
1
2
β’ Diameter is high, while the clustering coefficient is also high
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Watts-Strogatz Model
β’ Watts & Strogatz show that with a few random rewiring the diameter will be decreased a lot.
β’ We will speak about small-world models deeply later
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Watts-Strogatz Model
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Markov Graphs & P* Networks
β’ Think about building a random graph in which the formation of the link ij is correlated with formation of the links jk and ik?
β’ Frank & Strauss method using Clifford & Hammersley theorem:β’ Build a graph D whose nodes is the potential links in G
β’ If ij and jk are linked in D, it means that there exist some sort of dependency between them
β’ C(D) is the set of Dβs cliques
β’ πΌπ΄ πΊ = 1 for π΄ β πΆ(π·), π΄ β πΊ (consider G as a set of edges) and else πΌπ΄ πΊ = 0
β’ The probability of a given network G depends only on which cliques of D it contains:
log Pr πΊ =
π΄βπΆ(π·)
πΌπ΄πΌπ΄(πΊ) β π
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Markov Graphs & P* Networks
β’ An example: a symmetric caseβ’ Build a random graph with controllability on the number of its edges (π1(πΊ))
and its triads (π3(πΊ))
β’ C(D) consists of π3(πΊ) triads and π1(πΊ) edges. So if we weight them equally we have:
log Pr πΊ = πΌ1π1 πΊ + πΌ3π3 πΊ β π
β’ We can calibrate with different parameters to have different random networks with different number of triangles and edges. β’ πΌ3 = 0 is the Poisson networks case
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The Configuration Model
β’ A sequence of degrees is given (π1, π2, π3, β¦ , ππ) and we want to build a random graph having these degrees
β’ We generate the following sequence of numbers
β’ Randomly pick two number of elements and connect corresponding nodes
β’ The result is a multigraph
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An Expected Degree Model
β’ Form a link between node i and node j with probability ππππ
π ππ
β’ Self links are allowed
β’ The expected degree of node i will be ππ
β’ Maximum of ππ2 < π ππ
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Configuration Model vs Expected Degree Modelβ’ Consider the degree sequence < π, π, β¦ , π >
β’ In configuration model:β’ The probabilities of self links and multi links is negligible
β’ The probability of a node to have degree k will converge to 1
β’ In expected degree model:β’ The probability of a node to have degree k will converge to
whose maximum value is 1/2.
β’ Why? See the blackboard.
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Distribution of the Degree of Neighboring Nodesβ’ Consider a given graph with degree distribution P(d)β’ A related calculation π(π): the probability that a randomly chosen edge
has a (randomly chosen) neighbor with degree dβ’ π π = π(π)?
β’ π 1 = π 2 =1
2
β’ π 1 =2
3Γ
1
2=
1
3
β’ π 2 =2
3Γ
1
2+
1
3Γ
1
1=
2
3
β’ We can formulate π π
π π =π π π
< π >See the blackboard
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Distribution of the Degree of Neighboring Nodesβ’ Consider the degree sequence <1,1,2,2,1,1,2,2,β¦>. Compare two
casesβ’ In random models such as the configuration model: The distribution of the
neighboring nodes have the same distribution as π(π) for all nodes.
β’ In networks with correlation properties: The graph is highly segregated by degrees
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Multiple & Self-Links
β’ In the configuration model the probability of self links and multiple links is negligible β’ The same is true for the expected degree model
β’ Theorem: Let πππ denote the number of self or duplicate links & ππ
π
= Pr πππ > 0 . If a degree sequence (π1, π2, β¦ , ππ) is such that
maxπβ€π
ππ
π13
β 0 then maxπβ€π
πππ β 0.
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Threshold Effects
β’ We define the function t(n) as the threshold for the parameter p(n) for the property A(N) if
and
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Example
β’ π΄ π = π π1 π β₯ 1}
β’ In Poisson model the only model parameter is p(n)
β’ Pr π΄ π π π = 1 β 1 β π ππβ1
β’ The threshold is π‘ π =1
πβ1, on the blackboard see why?
β’ Hint: write π π =π π
πβ1
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A Summary of Thresholds for Poisson Random Network Modelβ’ π‘ π = πβ2: The first edge emerges
β’ π‘ π = nβ1.5: The first component with three nodes emerges
β’ π‘ π = nβ1: Cycles emerge as does the giant component
β’ π‘ π =log(π)
π: The network becomes connected
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Threshold for Connectedness
β’ For Poisson Random Network Modelβ’ Erdos & Renyi Theorem: A threshold function for the connectedness of the
Poisson random network is π‘ π =log(π)
πβ’ Proof: See the blackboard
β’ First we show that if π π
π‘ πβ 0 then the probability that there exist isolated nodes tends
to 1 and thus the network is clearly unconnected.
β’ Second we show that if π π
π‘ πβ β then the chance of having any component of size less
than n/2 tends to zero and thus the network is clearly connected
β’ For other static random models it is very hard to find well defined thresholds
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Threshold for Connectedness
β’ For the expected degree model: β’ ππππ = π=1
π ππ denote the total expected degree of nodes
β’ The probability of a link between nodes i and j is ππππ
ππππ
β’ The probability that node i is isolated
π
(1 βππππ
ππππ) β πβππ
β’ The probability that no node is isolated
π
(1 β πβππ) β πβ π πβππ
β’ If πβππ β β isolated nodes will occurβ’ If πβππ β 0 there will be no isolated node
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Size of the Giant Component
β’ In the Poisson Random Network Model:β’ Assume that there exist a giant component and it consists q fraction of nodes
β’ Each node is in the giant component if it has no neighbor in the giant component. Thus approximately we have
1 β π =
π
1 β π ππ π =
π
πβ πβ1 π π β 1 ππ
π!1 β π ππ π
β’ Thus since
π
π β 1 π 1 β ππ
π!= π πβ1 π(1βπ)
β’ Thus π = 1 β πβπ πβ1 π
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Threshold for the Emergence of the Giant Componentβ’ The probability that the link connects two nodes that were already
connected in a component with s nodes is the probability that both of its
end nodes lie in that component, which is proportional to π
π
2
π
π ππ
2
β€
π
π ππ
π2β€
π
π
β’ The portion of links that lie on cycles is of an order no more than π
π
β’ Before the threshold for the emergence of the giant component, the components are essentially tree.
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Threshold for the Emergence of the Giant Componentβ’ Define π as the number of nodes which can be reached by the random
processβ’ Under some constraints we have
β’ The boundries where π is NOT well defined then the giant component emerges (the mentioned constraints come here). That is where
π2 > 2 π
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Threshold for the Emergence of the Giant Componentβ’ For Poisson random network model
π 2 > 2 π
π2 = π + π 2 βΉ π 2 > π βΉ π > 1 βΉ π‘ π =1
π
β’ For k-Regular network modelπ2 = 2π βΉ π = 2
β’ For Scale-Free network model β’ π π = ππβπΎ
β’ For π2 diverges for πΎ < 3
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Diameter Estimation
β’ Estimating Diameter for Random Networksβ’ Choose a random node
β’ The expected number of reachable nodes with 1 edge: π
β’ The expected number of reachable nodes with 2 edges:
π
π
(π β 1) π(π) = ππ2 β π
π
β’ The expected number of reachable nodes with k edges:
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Diameter Estimation
β’ Estimating Diameter for Random Networksβ’ The total number of nodes is approximately:
β’ Thus
β’ Or
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Diameter Estimation
β’ Estimating Diameter for Random Networksβ’ In cases π2 is much larger than π
β’ In Poisson random network models where π2 β π = π 2
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Generating Functions
β’ Let π(. ) is a probability distribution over the set 0,1,2, β¦
β’ Define πΊπ as the generating function related to π as follows:
β’ πΊπ 1 = 1
β’ Taking a derivative from πΊπ(π₯)
β’ Thus
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Generating Functions
β’ More generally,
β’ Assume that π2(π) is the probability that the sum of two draws from π is k. Thus
π2 π =
π=0
π
π π π(π β π)
β’ Thus
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Generating Functions
β’ The relation between πΊπ2and πΊπ:
β’ More generally,
β’ Consider a distribution π that is derived by first randomly selecting a distribution from a series of distributions π1, π2, π3, β¦ picking each with probability πΎπ.
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Generating Functions
β’ Define π π = π π β 1
πΊ π π₯ =
π=1
β
π π β 1 π₯π = π₯πΊπ(π₯)
β’ Letβs define the generating function for π π =ππ(π)
π
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Distribution of Component Sizes
β’ Randomly choose an edge and then one of its nodes. What is the distribution of the number nodes of the component including this edge (Q)?β’ There is probability π π that the node at the end of the randomly selected
edge has degree d
β’ It has d β 1 edges emanating from this node
β’ The number of vertices found in the owner component is the sum of nodes found by these edges, plus one. Thus, the generating function of additional nodes found through the node if it happens to have degree d is
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Distribution of Component Sizes
β’ Randomly choose an edge and then one of its nodes. What is the distribution of the number nodes of the component including this edge (Q)?β’ With some calculations (see the blackboard) we have the following:
β’ and
β’ For 1 we have
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Distribution of Component Sizes
β’ What is the distribution of the component including a randomly chosen vertex (H)?β’ Using πΊπ we have
πΊπ» π₯ = π₯
π
π π πΊπ π₯π
= π₯πΊπ πΊπ π₯
β’ Thus
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