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lec4 - Stat 206(Spring 2003 Random Graphs and Complex...

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Unformatted text preview: Stat 206 (Spring 2003): Random Graphs and Complex Networks 01/28/03 Lecture 4: Proportional Attachement models and the Yule process Lecturer: David Aldous Scribe: Rahul Jain In this lecture, we see two preferential attachment models of random graphs treated heuristically via the Yule branching process. 4.1 Yule process - a branching process in continuous time In the Yule process, individuals live forever. And for each individual living at time t during [15, t + dt] there is a chance 1dt to have a child. So, each individual gives birth at rate 1. Define Zt : #individuals at time t, Z0 : 1. Let W1 be the waiting time for first birth, which is exponentially distributed with rate 1. Let Wk be the interarrival time between the birth of the (k — 1)th and the kth individual. Then, since there are k — 1 individuals and each gives birth with rate 1, and exponential distribution has memoryless property, Wk ~ EXP(kz). Also note that Wk is independent of Wk“. Then, we can define z, = min{k‘ : 2’; W,- > t}. Yule (1924) showed that (i) EZt : et. (ii) Zt ~ GEOM(e’t). (iii) Zt/et —> W a.s., with W ~ EXP(1). Yule set up and solved the following differential equation: dP(Zt = 2') dt : —¢P(z, = i) + (2' — 1)P(Zt = i — 1) Here, the first term on RHS is because there may be 2' individuals at time t — h and there is no birth in [t — h, t], and the second term is if there are i — 1 individuals at time t — h and there is one birth in [t — h, t] with rate 1' — 1. Kendall gave a clever bijection between the Yule process and the Poisson point process. A Poisson point process is a spatial random process of points (say in R2). We can construct the bijection in the following way: Let the first point A be at (X,Y). Then, X has distribution EXP(1). The Y-axis represents the et-axis in the Yule process. Consider only points right of point A. B considered child of A.... Let N(y) be the number of points above level y. Then,... Proposition 4.1 (I) The Process (N(et),t Z 0) is emactly same as (Zt,t Z (0) (2) For large y, N(y) : number of points in the rectangle(W,y) 2 Wy. The tree structure here is what is mimicing the Yule process. 4—1 4—2 Lecture 4: Proportional Attachement models and the Yule process 4.2 Proportional attachment model I We create a digraph by adding new vertices chosen randomly but with probabilities proportional to 1 + in — degree. Fix m 2 0. Start with an arbitrary graph Gm on m vertices. To construct the graph, consider a deck of in cards. For each vertex V, deal cards until get in distinct labels V1, ..., Vm. For vertex V, create m edges V —> V,. Create (m + 1) new cards: V1, ..., Vm. We get a random graph Gn on n vertices. Study Dn : 1 + in — degree of uniform random vertex of G7,. Claim 4.2 As n —> oo,P(Dn : i) —> P(D : i), and distribution ofD GE0M(e’T’n”—$1),T ~ EXP(1). Then, as i —> oo, P(D = i) N cmi—(2+%). mT +1 To see the last part, note that Y N GEOM(p), so P(D > i|T) : (1 — e’m )i which gives mt P(D>i) = / (1—e_m+1)ie’tdt 0 m+ 1 P(i + 1)r(1 + 1/m) m F(i+2+1/m) For m=1, P(D = i) = m ~ r3. Heuristics: Make et vertices arrive by time t. Consider vertex V arriving at time t. D(t) : 1 +in — degree : #cards V in the deck. To calculate probability that D(t) increases by 1 (new in—edges) in [t,t + 6], note that there are (m + 1)et cards in deck by time t, and (Set new vertices appear in the time interval. Each new vertex has in picks, and D(t) cards have label V, Thus, the probability is given by mdetD(t) _ m (m + 1)et _ m + 1 D(t) is Yule process, rate mi+17 i.e. D(t) has GEOM(e_mm—+tl) distribution. Also note that “age” of any arrival (i.e., how long ago a given point was born) is EXP(1). 4.3 Proportional attachment model II Now, we create a digraph by adding new vertices chosen randomly but with probabilities proportional to total degree. Fix out—degree m 2 0. For each new vertex, choose m end—vertices with probabilities proportional to total degree. For m = 1, it is same as previous model. Study D7,, the total degree of a random vertex in Gn. Claim 4.3 As n —> 00, P(D” : i) —> P(D : i), and distribution ofD 2:1 GEOM(e’ %T), T N EXP(1), and the geometries are independent. Fix In: start with arbitrary digraph Gm on m vertices. For each edge V —> W, deal cards until get m distinct labels: V1, ..., Vm. Create new vertex V and m edges V —> V,. Create 2m new cards. Heuristics: Make et vertices arrive by time t. Consider typical vertex V. Create m cards V, assign m different colors. Later, when an edge W —> V is created, make the new V and copy the color of chosen V—card. Dy(t) 2 #Yellow V cards at time t. Lecture 4: Proportional Attachement models and the Yule process 4—3 Claim 4.4 For a random vertem, DyisGEOM(e’T/2),T ~ EXP(1). The process is same for arbitrary m, as for m = 1. Example: Calculate P(D = 2‘) = W 2‘ = m,m + 1. This gives P(D = i) ~ cmi_3. For rigorous proof7 see paper by Bollobas, et a1 [BRST02] References [BRST02] B.BOLLOBAS, ORIORDAN, J.SPENCEE, G.TUSNADY), “THE DEGREE SEQUENCE OF A SCALE— FREE RANDOM GRAPH PROCESS”7 Random structures and algorithms, 18:3, 2001, PP. 279— 290. ...
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