Networks: Lectures 9 & 10 Random graphs

Transcription

1 Networks: Lectures 9 & 10 Random graphs Heather A Harrington Mathematical Institute University of Oxford HT 2017

2 What you re in for Week 1: Introduction and basic concepts Week 2: Small worlds Week 3: Toy models of network formation Week 4: Additional summary statistics and other concepts Week 5: Random graphs Week 6: Community structure and other mesocopic structures Week 7: Dynamical systems on networks Week 8: Other topics TBD

3 Contents 1 Motivation 2 Erdős-Rényi random graphs 3 Generating functions 4 The configuration model 5 Random graphs with degree-degree correlations 6 Random graphs with clustering

4 Motivation Generative models of graphs are very useful for understanding the properties of real networks. They are interesting to study for their own sake (and can be made progressively more complicated to incorporate features of empirical networks) They are substrate networks on which to run dynamical systems, They serve as null models with which to compare real networks in many situations (e.g., for studies of community structure ). They are useful for material in weeks 6 (null model for community detection) and 7 (dynamics on networks).

5 Random graphs Random graphs are model networks with fixed number of parameters. Very useful for understanding the properties of real networks.

6 The G(n, m) model The simplest random graph model is the G(n, m) model: n nodes. m edges. One chooses m node pairs uniformly at random out of n(n 1)/2 possible pairs.

7 The G(n, m) model Alternatively, one can think of choosing a graph uniformly at random over the set of simple graphs with n nodes and m edges. So we do not talk about a generative model but a distribution over a set of Ω graphs: P (G) = 1 Ω.

8 Why we study random graphs When we talk about random graphs we want to know about its properties and whether or not they re reasonable models of real-world networks. For example, we would calculate their: Diameter. Degree distribution. Clustering coefficients. Path lengths, etc. Also, random graphs are mathematically fascinating.

9 The G(n, p) model The G(n, m) model is simple but doing calculations with it can be messy. The G(n, p) random graph model is similar, but instead of prescribing m edges, we assign a probability p to each node pair to be connected, p [0, 1]. In other words, every edge is a Bernoulli random variable that is 1 with probability p and 0 with probability 1 p. The G(n, p) random graph model is known as the Erdős-Rényi (ER) random graph.

10 Erdős-Rényi random graphs ER random graphs can also be thought of a probability distribution over the space of graphs: P (G) = p m (1 p) n(n 1)/2 m. G(n, p) and G(n, m) are related (note they are not the same) for an appropriate choice of p P (m) = ( n(n 1) ) 2 p m (1 p) n(n 1)/2 m m

11 Erdős-Rényi random graphs G(n, p) and G(n, m) are related (note they are not the same) for an appropriate choice of p P (m) = ( n(n 1) ) 2 p m (1 p) n(n 1)/2 m m The number of edges in a ER random graph follows a binomial distribution, with expected value: n(n 1)/2 m=0 mp (m) = The mean degree is k = (n 1)p. n(n 1) p. 2

12 G(n, m) and G(n, p) The mean degree is usually referred to as c, so p = c n 1. which clarifies the relationship between G(n, m) and G(n, p).

13 Degree distribution of ER graphs The probability that a node is connected to a specific subset of k nodes is just the probability of k Bernoulli trials with probability p: p k (1 p) n 1 k. Now, the probability that a node is connected to any k nodes is binomially distributed: ( ) n 1 p k = p k (1 p) n 1 k. k What happens in the case of large n?

14 Degree distribution of ER graphs A binomial random variable becomes Poisson in the limit of large n. To see this consider (1 p) n 1 k : log (1 p) n 1 k = (n 1 k) log(1 p), = (n 1 k) log(1 c n 1 ), remember that log(1 x) = n=1 xn n the order 1 approximation: (n 1 k) log(1 c [ ) = (n k 1) n 1, when x < 1. So taking c as n. c n 1 ].

15 Degree distribution of ER graphs So now we have log (1 p) n 1 k = c, take exponentials to get (1 p) n 1 k = e c. Also, when n is large: ( ) n 1 = k (n 1)! k!(n 1 k)! We put everything together to get: ( ) n 1 p k = p k (1 p) n 1 k, k = (n 1)k p k e c (n 1)k = k! k! (n 1)k. k! ( ) k c e c = ck n 1 k! e c. So in the large-graph limit the degree distribution of an ER graph is Poisson.

16 Degree distribution of ER graphs This is a very common trick of dealing with ER graphs and probabilities in combinatorics: Take logs, Expand Taylor, Take order 1 approximation (n large), Exponentiate.

17 Clustering coefficient The clustering coefficient in an ER random graph is just C = c n 1. The one thing to note here is that C 0 as n, so ER graphs are not transitive.

18 Giant component Consider an ER graph in the two following cases: p = 0: All nodes are disconnected. p = 1: Everyone is connected. What happens when 0 < p < 1?

19 Giant component Consider an ER graph in the two following cases: p = 0: All nodes are disconnected. p = 1: Everyone is connected. What happens when 0 < p < 1?

20 Giant component What is the size of the largest component? A giant component (GC) of a network is a component whose size grows linearly with n.

21 Giant component Let u be the fraction of edges not in the GC. Take a node i not in the GC, for any other node j: 1) (i, j) does not exist with prob 1 p. 2) (i, j) exists but j is not in the GC with probability pu So the probability that i is not in the GC via j is and over all nodes: an implicit function of u. 1 p + pu u = (1 p + pu) n 1

22 Giant component Taking the log expand trick we get: log(u) = (n 1) log(1 p pu),. c (n 1) (1 u) = c(1 u) n 1 So u e c(1 u), and the proportion of nodes in the GC is S = 1 u then S = 1 e cs Which again we have to solve implicitly.

23 Giant component To solve for the size of the GC we look at y = 1 e cs and y = S When c > 1 there are nontrivial solutions to S = 1 e cs, which means that when the expected degree is greater than 1 we can expect a GC to appear. The actual proof is a bit more technical but this is the general idea.

24 Giant component When c < 1 the size of the giant component is 0 (S = 0). When c > 1 the GC appears, its size is determined by S = 1 + W ( ce c ) c, where W is the Lambert W-function:

25 Small components In ER graphs the small components are those whose size does not grow with n. Let the proportion of nodes in the small component s be π s, then the proportion of nodes in a small component is π s = 1 S. s=0

26 Small components Small components are (mostly) trees (no cycles). Suppose you have a small component with m nodes and m 1 edges (i.e., a tree). The probability of having an additional edge is (( ) ) ( ) m m(m 1) p (m 1) = p (m 1), 2 2 ( ) 1 = p (m 1)(m 2), 2 = c ( ) 1 (m 1)(m 2). n 1 2 When n this probability goes to 0.

27 Small components The expected size of the small components is m = 1 1 c + cs. The denominator vanishes when c = 1 (remember S = 1 e cs ), that is when the GC emerges. This is the average size to which a randomly chosen node belongs. The average size is: 2 2 c + cs.

28 Limitations of ER graphs Useful as they are, ER random graphs are not an adequate model for many situations. For example, a social graph has a degree distribution (usually fat-tailed, like a power-law) that is not Poisson, and a clustering coefficient that does not go to 0 as n. We would like a more general random graph model that we can use to compare to real-world networks.

29 Generating functions Tool from probability that we need this week is the notion of a generating function (GF). Also known as the z-transform and discrete Laplace transform. It will be useful to examine both univariate and multivariate GFs

30 Generating functions Useful methodology for studying certain properties in random graphs (and for dynamical systems on random graphs). To use them in networks, we will need to assume the network is locally tree-like (i.e., measure of loops approaches 0 and N ) although the results of the conclusions seem to be good even for networks that are not tree-like.

31 Univariate generating functions Consider probability distribution for non-negative integer variable such that separate instances of variable are independent and have value k with probability p k (e.g., degree distribution of a network). The generating function for {p k } is: g(z) = p k z k. k=0 Given a generating function g(z), we can find p k by differentiation p k = 1 d k g k! dz k, z=0 so that g(z) gives complete information about {p k }.

32 Univariate GF: examples 1 p k = 1 k = g(z) = 1 1 z = 1 + z + z2 + 2 p k = ( n k) for fixed n = g(z) = (1 + z) n. 3 rolling a fair 20-side die (p 1 = p 20 = 1 20, p k = 0 k 21) = g(z) = Given the Poisson distribution: p k = e c c k k!, the generating function is g(z) = e c (cz) k k=0 k! = e c(z 1). 5 Power-law distribution: p 0 = 0, p k = ck α, k 1 & α > 0 normalize such that k=0 p k = 1 = c k=1 k α = 1 = c = 1 ζ(α), where ζ(α) is the Riemann-zeta function. g(z) = 1 ζ(α) k=1 k=1 where Li α (z) is the polylogarithm of z. z k. k α z k = 1 ζ(α) Li α(z),

33 Univariate GF: useful properties 1 g(1) = k=0 p k. (For p k to be a probability, it often must be normalized so that k=0 = 1, although this is not always the case). 2 Taking the derivative g (z) = k=0 kp kz k 1 = g (1) = k=0 kp k = k. 3 Higher moments: k m = dm g d(ln z) m z=1 (m 1).

34 Multivariate generating functions Consider the bivariate case: given random variables X and Y with joint distribution p kl = P (X = k, Y = l), k, l {0, 1, 2,...} The bivariate generating function is: g(z 1, z 2 ) = k,l p kl z k 1 z l 2.

35 Multivariate GF: useful properties 1 The GFs of the marginal distributions P (X = k) & P (Y = l) are { a(z) = P (z, 1) b(z) = P (1, z) 2 GFs of X + Y is P (z, z). 3 The random variables X and Y are independent if and only if g(z 1, z 2 ) = a(z 1 )b(z 2 ) for all z 1 and z 2. Note: bivariate GF useful for directed networks, which have both in-degree and out-degree.

36 The Configuration model The Configuration model (CM) is a more flexible model which creates a graph with a given degree sequence {k 1, k 2,... k n }. Assume that k 1 k 2 k n > 0. Assign half-edges (or stubs) to n nodes: To create a graph for the simplest CM version: Take any two stubs chosen uniformly at random and join them. Repeat until there are no more stubs left.

37 Note that: Number of stubs must be even. The Configuration model The CM allows multi-edges and self-edges. Since each matching of a pair of stubs is equally likely, this property allows analytical tractability. There is more than one way to obtain the same network. a b b a a b b a c e c e d e d e d f d f c f c f a b b a a b b a c f c f d f d f d e d e c e c e

38 The Configuration model a b b a a b b a c e c e d e d e d f d f c f c f a b b a a b b a c f c f d f d f d e d e c e c e Given a degree sequence {k i } the number of possible ways to obtain the same network is n N({k i }) = k i!. In this case n = 3 and k i = 2 so N({k i }) = 8. If there are Ω({k i }) possible networks, then each gets chosen with probability N/Ω. i=1

39 The Configuration model In the ER random graph model each edge has the same probability p of existing. In the CM this changes, and each possible edge (i, j) will have its own probability p ij (that an edge exists between i and j), which depends on k i and k j (where k i k j > 0). To obtain p ij, take into account that k i = 2m. Consider one stub of node i. Of the remaining 2m 1, exactly k j belong to j. The probability of choosing one of the k j stubs is k j /(2m 1). Since there are k i stubs of i we have when m is large. p ij = k ik j 2m 1 k ik j 2m,

40 The Configuration model Since the CM does not exclude multi-edges, we would like to know how probable it is to get one. Suppose i and j are already connected, the probability of getting yet another edge between them is (k i 1)(k j 1). 2m The probability of obtaining at least two edges between i and j is: k i k j (k i 1)(k j 1) (2m) 2.

41 The Configuration model The expected number of multiedges is: 1 P (A ij 2) = 2(2m) 2 k i k j (k i 1)(k j 1), i 1 = 2 k 2 n 2 k i (k i 1) i j ( k 2 ) 2 k. j k j (k j 1), = 1 2 k The number of multiedges stays constant as long as k 2 constant and finite.

42 The Configuration model For self-edges we must amend the calculations so that: p ii = k i(k i 1) 4m. The expected number of self edges will simply be p ii = i i k i (k i 1) 4m = k2 k. 2 k

43 The Configuration model We want to calculate the expected number of neighbors that i and j have in common. For node l we know p il and p jl. But if l is already connected to i, the probability it is also connected to j now is: k j (k l 1) 2m. Then the expected number of neighbors that i and j share: = l = k ik j 2m k i k l 2m k j (k l 1) 2m, l k l(k l 1), n k k 2 k = p ij. k

44 The Configuration model Let p k be the degree distribution of a CM graph (i.e., the probability of choosing a node of degree k uniformly at random). Now suppose we choose a node i and we travel along one of its edges to node j. What is the probability that the k j = k + 1?. This is called the excess-degree distribution, denoted by q k.

45 The Configuration model Given that there are np k nodes of degree k, the probability of arriving to any node of degree k is k 2m np k = kp k k. Remember that the average degree of a node is k. What is the average degree of a neighbour? Take the expected value of the expression above: k k kp k k = k2 k.

46 The Configuration model Note that k 2 k k = 1 ( k 2 k 2), k = σ2 k k. where σk 2 is the variance of the degree distribution. σ k > 0 = k2 k > k. Does this seem strange? The mean degree of a neighbor of a node is larger than the mean degree of a node!

47 The Configuration model Now, to get the excess degree distribution from the average degree of a neighbour q k = (k + 1)p k+1. k This distribution allows us to calculate the clustering coefficient (the average probability that two neighbors of a node are neighbors with each other) of the CM. Consider a node v with neighbours i and j: C = q ki q kj p ij = k i k j q ki q kj 2m, k i=0 k j=0 k i=0 k j=0 ( ) 2 = 1 kq k, 2m.. k = 1 ( k 2 k ) 2 n k 3.

48 The Configuration model In this graph: n = 77. m = 254. k = 6.6. C = 0.7 In the CM the clustering coefficient would be C CM = 1 ( k 2 k ) 2 n k 3, = , =

49 Recall GF: useful properties The generating function for {p k } is: g(z) = k=0 p kz k. Given a generating function g(z), we can find p k by differentiation p k = 1 d k g k! so that g(z) gives complete information about {p dz z=0, k k }. 1 g(1) = k=0 p k. (For p k to be a probability, it often must be normalized so that k=0 = 1, although this is not always the case). 2 Taking the derivative g (z) = k=0 kp kz k 1 = g (1) = k=0 kp k = k. 3 Higher moments: k m = dm g d(ln z) m z=1 (m 1). 4 Powers: h(z) = s=0 π sz s = [ k=0 p kz k] m = [g(z)] m, where π s = k 1=0 k δ(s, m=0 i, k i)π m i=1 p k i and h(z) is the GF for π s.

50 Let g 0 (z) = k=0 p kz k be the GF for {p k }. CM: components g 1 (z) = k=0 q kz k be the GF for {q k } the excess degree distribution. Recall mean degree of a neighbor is: k k kp k k = k2 k, so q k = (k+1)p k+1 k. = g 1 (z) = 1 k = 1 k = 1 k (k + 1)p k+1 z k k=0 kp k z k 1 k=0 dg 0 dz = g 0(z) g 0 (1). If we can find g 0 (z) then we can find g 1 (z) and don t need to calculate excess degree distribution explicitly!

51 CM: components Remember that: A component of a graph ensemble is called a giant component (GC) if its size grows linearly with n (scales linearly, called extensive ). Let π s be the probability that a node chosen uniformly at random belongs to some size s component that is not the giant component (does not scale linearly, intensive ). Then the GF is h 0 (z) = s=1 π sz s. As n, a small (non-giant) component of CM becomes a tree (with degree distribution held constant). Remember, we re in for locally tree-like to apply GFs.

52 s with general degree distributions CM: components If node i is in a small component, then assuming we have a tree (which we do asymptotically) then the set of nodes reachable along each of its edges are not connected other than via i. oint to notice, however, is that the neighbors n1, rtex i are, by definition, reached by following an e, as we have discussed, these are not typical rtices, being more likely to have high degree ex. the ons - o ave. te by Figure 13.5: The size of one of the small components in the configuration model. (a) The size of the component to which a vertex i belongs is the sum of the number of vertices in each of the subcomponents (shaded regions) reachable via i's neighbors n 1,n 2,n 3, plus one for i itself. (b) If vertex i is removed the subcomponents become components in their own right. Remove i and all its edges. The small component breaks up into smaller, disconnected components. i s neighbors j 1, j 2,... are reached by following an edge from i. The distribution of the sizes of the new components is ρ s. It has GF h 1 (z) = s=0 ρ sz s.

53 CM: components Suppose i has degree k and let P (s k) be the probability that after i is removed, its k neighbors belong to small components of sizes summing to exactly s. i.e., P (s k) is the probability that i itself belongs to a small component of size s given that its degree is k. The total probability π s that i belongs to a small component of size s is this probability averaged over k. π s = = h 0 (z) = = z p k P (s 1 k). k=0 p k P (s 1 k)z s, s=1 k=0 p k k=0 s=1 P (s 1 k)z s 1, = z p k P (s 1 k)z s, k=0 with the final sum being the GF for the probability that the k neighbors belong to the small components whose size sums to s.

54 CM: components The size of the small components are independent of each other, so (using the power property of GFs) we get: h 0 (z) = z p k [h 1 (z)] k = zg 0 (h 1 (z)) k=1 We need an expansion for ρ s ; as n. removing node i doesn t change the degree distribution. = still have probability P (s 1 k) for a degree k node belonging to a size s component. = ρ s = k=0 q kp (s 1 k) because we traverse an edge to reach this component. = h 1 (z) = = z s=1 k=0 q k P (s 1 k)z s q k [h 1 (z)] k = zg 1 (h 1 (z)) k=0

55 CM: components = h 1 (z) = = z s=1 k=0 q k P (s 1 k)z s q k [h 1 (z)] k = zg 1 (h 1 (z)) k=0 and h 0 (z) = zg 0 (h 1 (z)). if we solve for h 1 (z), then we can use that solution to obtain h 0 (z), and from there get π s and ρ s. h 0 (1) = s=1 π s is the probability that a random node belongs to the small component. This is 1 iff the network doesn t have a giant component. The fraction of nodes in the giant component is S g := 1 h 0 (1) = 1 g 0 (h 0 (1)).

56 CM: components The fraction of nodes in the giant component is S g := 1 h 0 (1) = 1 g 0 (h 0 (1)). { S g = 1 g 0 (u) Write u := h 1 (1) = g 1 (h 1 (1)) = u = g 1 (u) to find the fixed points of these equations. We would like 1 = g 1 (1) is always a fixed point. S g = 1 g 0 (1) = 0 = no giant component. there exists a giant component iff u = g 1 (u) has a fixed point u 1.

57 le given in the last section is unusual in that we can ixed-point equation (13.91) exactly for the crucial u. In most other cases exact solutions are not ut we can nonetheless get a good idea of the CM: components Usually we can t find closed-form solutions for u, but plotting u vs g(u) can help. f u by graphical means. The derivatives of g 1 (z) are s and hence are all nonhat means that for z g 1 (z) is in general The derivative of g 1 (z) are proportional to the probability ρ s and are hence non-negative. n increasing function of its argument, and upward t also takes the value 1 when z = 1. Thus it must s atively like one of the curves in Fig The the fixed-point equation (p.464) by pt of = g e he in ady e is u ght re). e at be z 0, g 1 (z) is positive, monotonically increasing, and concave up. Figure 13.6: Graphical solution of Eq. (13.91). The solution of the equation u = g 1 (u) is given by the point at which the Three possibilities 1 Fixed point at u = 1&u = u (0, 1) 2 Fixed point at u = 1 with tangency, 1 = g i (1) 3 Fixed point at u = 1 without tangency.

58 CM: components Case (1) = g (1) > 1. Case (2), giant component is born when g (1) = 1. Using equation for g 1 (z), we know g 1(1) = k=0 kq k = 1 k k=0 k(k + 1)p k = k2 k k For u < 1 we need k2 k k > 1 k 2 2 k > 0. This is the Molly-Reed condition, which is obtained through different mathematical techniques. To think about: RG with given expected degree distribution. How to examine the small component too? How would we do similar calculations for directed-graph analog of the configuration model?

59 RGs with degree-degree correlations Real graphs can be assortative or disassortatitve, so we want an ensemble of random graphs that fixes both degree and degree-degree correlation, but otherwise connects uniformly at random. Fix degree distribution p k and joint degree-degree distribution p ik. Let excess degree distribution be q k. Let e jk be joint probability distribution for the excess degree of 2 nodes at the ends of a randomly chosen edge. e jk = e kj, j,k e jk = 1, j e jk = q k. (degree assortativity = 0 = e jk = q i q k )

60 RGs with degree-degree correlations Consider an ensemble of random graphs in which e jk takes a specified value. Consider n and take some graph from the ensemble; choose an edge uniformly at random and suppose it is attached to a degree -j node. Let G j (z) be the GF for the probability distribution of the number of other nodes reachable by following that edge. We can show that G j (z) = z k e jk[g k (z)] k k e jk and that the number of nodes reachable from a randomly chose node has a GF H(z) := zp 0 + z p k [G k 1 (z)] k k=1 The mean size of a component to which such a node belongs is s = H (1) = 1 + k kp k G k 1

61 Random graphs with clustering Real networks have nontrivial clustering. It s possible to define generalizations of the configuration model that include clustering (though their clustering properties tend to be far from those of real networks).

62 Newman s Degree-Triangle model t := triangle in which node i participates. s i := stubs from i that are not part of a triangle random-graph model with fixed joint degree sequence {s i, t i } and joint distribution {p st } degree distribution in p k = s,t=0 p stδ k,s+2t GF for p st is g(x, y) = s,t=0 p sx s y t = GF for p k is f(z) = p k z k = k=0 p st z s+2t = g(z, z 2 ) s,t=0