Generating Functions for Random Networks

Transcription

1 for Random Networks Complex Networks CSYS/MATH 303, Spring, 2011 Prof. Peter Dodds Department of Mathematics & Statistics Center for Complex Systems Vermont Advanced Computing Center University of Vermont 1 of 35

2 Outline 2 of 35

3 functions Idea: Given a sequence a 0, a 1, a 2,..., associate each element with a distinct function or other mathematical object. Well-chosen functions allow us to manipulate sequences and retrieve sequence elements. Definition: The generating function (g.f.) for a sequence {a n } is F (x) = a n x n. n=0 Roughly: transforms a vector in R into a function defined on R 1. Related to Fourier, Laplace, Mellin,... 4 of 35

4 Simple example Rolling dice: p ( ) k = Pr(throwing a k) = 1/6 where k = 1, 2,..., 6. F ( ) (x) = 6 p k x k = 1 6 (x + x 2 + x 3 + x 4 + x 5 + x 6 ). k=1 We ll come back to this simple example as we derive various delicious properties of generating functions. 5 of 35

5 Example Take a degree distribution with exponential decay: P k = ce λk where c = 1 e λ. The generating function for this distribution is F(x) = P k x k = k=0 ce λk x k = k=0 Notice that F (1) = c/(1 e λ ) = 1. c 1 xe λ. For probability distributions, we must always have F(1) = 1 since F (1) = P k 1 k = k=0 P k = 1. k=0 6 of 35

6 Properties of generating functions Average degree: k = kp k = kp k x k 1 k=0 k=0 x=1 = d dx F (x) = F (1) x=1 In general, many calculations become simple, if a little abstract. For our exponential example: F (x) = (1 e λ )e λ (1 xe λ ) 2. So: k = F (1) = e λ (1 e λ ). 8 of 35

7 Properties of generating functions Useful pieces for probability distributions: Normalization: F (1) = 1 First moment: k = F (1) Higher moments: k n = ( x d dx ) n F (x) k th element of sequence (general): P k = 1 d k k! dx k F (x) x=1 x=0 9 of 35

8 Edge-degree distribution Recall our condition for a giant component: k R = k 2 k k > 1. Let s rëexpress our condition in terms of generating functions. We first need the g.f. for R k. We ll now use this notation: F P (x) is the g.f. for P k. F R (x) is the g.f. for R k. Condition in terms of g.f. is: k R = F R (1) > 1. Now find how F R is related to F P of 35

9 Edge-degree distribution We have F R (x) = R k x k = k=0 k=0 (k + 1)P k+1 x k. k Shift index to j = k + 1 and pull out 1 k : F R (x) = 1 k j=1 jp j x j 1 = 1 k j=1 P j d dx x j = 1 d k dx P j x j = 1 d k dx (F P(x) P 0 ) = 1 k F P (x). j=1 Finally, since k = F P (1), F R (x) = F P (x) F P (1) 12 of 35

10 Edge-degree distribution Recall giant component condition is k R = F R (1) > 1. Since we have F R (x) = F P (x)/f P (1), F R (x) = F P (x) F P (1). Setting x = 1, our condition becomes F P (1) F P (1) > 1 13 of 35

11 Size distributions To figure out the size of the largest component (S 1 ), we need more resolution on component sizes. : π n = probability that a random node belongs to a finite component of size n <. ρ n = probability a random link leads to a finite subcomponent of size n <. Local-global connection: P k, R k π n, ρ n neighbors components 15 of 35

12 Size distributions G.f. s for component size distributions: F π (x) = π n x n and F ρ (x) = ρ n x n n=0 n=0 The largest component: Subtle key: F π (1) is the probability that a node belongs to a finite component. Therefore: S 1 = 1 F π (1). Our mission, which we accept: Find the four generating functions F P, F R, F π, and F ρ. 16 of 35

13 we ll need for g.f. s Sneaky Result 1: Consider two random variables U and V whose values may be 0, 1, 2,... Write probability distributions as U k and V k and g.f. s as F U and F V. SR1: If a third random variable is defined as W = U V (i) with each V (i) d = V i=1 then F W (x) = F U (F V (x)) 18 of 35

14 Proof of SR1: Write probability that variable W has value k as W k. W k = U j Pr(sum of j draws of variable V = k) j=0 = j=0 U j {i 1,i 2,...,i j } i 1 +i i j =k V i1 V i2 V ij F W (x) = W k x k = k=0 k=0 j=0 U j {i 1,i 2,...,i j } i 1 +i i j =k V i1 V i2 V ij x k = U j j=0 k=0 {i 1,i 2,...,i j } i 1 +i i j =k V i1 x i 1 V i2 x i2 V ij x i j 19 of 35

15 Proof of SR1: With some concentration, observe: F W (x) = U j j=0 k=0 {i 1,i 2,...,i j } i 1 +i i j =k V i1 x i 1 V i2 x i2 V ij x i j }{{} ( x k piece of i =0 V i xi ) j }{{} ( i =0 V i xi ) j = (F V (x)) j = U j (F V (x)) j j=0 = F U (F V (x)) 20 of 35

16 we ll need for g.f. s Sneaky Result 2: Start with a random variable U with distribution U k (k = 0, 1, 2,... ) SR2: If a second random variable is defined as V = U + 1 then F V (x) = xf U (x) Reason: V k = U k 1 for k 1 and V 0 = 0. F V (x) = V k x k = U k 1 x k k=0 k=1 = x U j x j = xf U (x). j=0 21 of 35

17 we ll need for g.f. s Generalization of SR2: (1) If V = U + i then F V (x) = x i F U (x). (2) If V = U i then F V (x) = x i F U (x) = x i U k x k k=0 22 of 35

18 Connecting generating functions Goal: figure out forms of the component generating functions, F π and F ρ. π n = probability that a random node belongs to a finite component of size n = k=0 ( sum of sizes of subcomponents P k Pr at end of k random links = n 1 ) Therefore: F π (x) = }{{} x F P (F ρ (x)) }{{} SR2 SR1 Extra factor of x accounts for random node itself. 24 of 35

19 Connecting generating functions ρ n = probability that a random link leads to a finite subcomponent of size n. Invoke one step of recursion: ρ n = probability that in following a random edge, the outgoing edges of the node reached lead to finite subcomponents of combined size n 1, = k=0 ( sum of sizes of subcomponents R k Pr at end of k random links = n 1 ) Therefore: F ρ (x) = }{{} x F R (F ρ (x)) }{{} SR2 SR1 Again, extra factor of x accounts for random node itself. 25 of 35

20 Connecting generating functions We now have two functional equations connecting our generating functions: F π (x) = xf P (F ρ (x)) and F ρ (x) = xf R (F ρ (x)) Taking stock: We know F P (x) and F R (x) = F P (x)/f P (1). We first untangle the second equation to find F ρ We can do this because it only involves F ρ and F R. The first equation then immediately gives us F π in terms of F ρ and F R. 26 of 35

21 Remembering vaguely what we are doing: Finding F π to obtain the fractional size of the largest component S 1 = 1 F π (1). Set x = 1 in our two equations: F π (1) = F P (F ρ (1)) and F ρ (1) = F R (F ρ (1)) Solve second equation numerically for F ρ (1). Plug F ρ (1) into first equation to obtain F π (1). 27 of 35

22 Example: Standard random graphs. We can show F P (x) = e k (1 x) F R (x) = F P (x)/f P (1) = e k (1 x) /e k (1 x ) x =1 = e k (1 x) = F P (x)...aha! RHS s of our two equations are the same. So F π (x) = F ρ (x) = xf R (F ρ (x)) = xf R (F π (x)) Why our dirty (but wrong) trick worked earlier of 35

23 We are down to F π (x) = xf R (F π (x)) and F R (x) = e k (1 x). F π (x) = xe k (1 Fπ(x)) We re first after S 1 = 1 F π (1) so set x = 1 and replace F π (1) by 1 S 1 : 1 S 1 = e k S 1 Or: k = 1 1 ln S 1 1 S 1 1 S k Just as we found with our dirty trick... Again, we (usually) have to resort to numerics of 35

24 Average component size Next: find average size of finite components n. Using standard G.F. result: n = F π(1). Try to avoid finding F π (x)... Starting from F π (x) = xf P (F ρ (x)), we differentiate: F π(x) = F P (F ρ (x)) + xf ρ(x)f P (F ρ(x)) While F ρ (x) = xf R (F ρ (x)) gives F ρ(x) = F R (F ρ (x)) + xf ρ(x)f R (F ρ(x)) Now set x = 1 in both equations. We solve the second equation for F ρ(1) (we must already have F ρ (1)). Plug F ρ(1) and F ρ (1) into first equation to find F π(1). 31 of 35

25 Average component size Example: Standard random graphs. Use fact that F P = F R and F π = F ρ. Two differentiated equations reduce to only one: F π(x) = F P (F π (x)) + xf π(x)f P (F π(x)) Rearrange: F π(x) = F P (F π (x)) 1 xf P (F π(x)) Simplify denominator using F P (x) = k F P(x) Replace F P (F π (x)) using F π (x) = xf P (F π (x)). Set x = 1 and replace F π (1) with 1 S 1. End result: n = F π(1) = (1 S 1 ) 1 k (1 S 1 ) 32 of 35

26 Average component size Our result for standard random networks: n = F π(1) = (1 S 1 ) 1 k (1 S 1 ) Recall that k = 1 is the critical value of average degree for standard random networks. Look at what happens when we increase k to 1 from below. We have S 1 = 0 for all k < 1 so n = 1 1 k This blows up as k 1. Reason: we have a power law distribution of component sizes at k = 1. Typical critical point behavior of 35

27 Average component size Limits of k = 0 and make sense for n = F π(1) = As k 0, S 1 = 0, and n 1. (1 S 1 ) 1 k (1 S 1 ) All nodes are isolated. As k, S 1 1 and n 0. No nodes are outside of the giant component. Extra on largest component size: For k = 1, S 1 N 2/3. For k < 1, S 1 log N. 34 of 35

28 I 35 of 35