A little twist in the problem of a configuration model for an exponential distribution of degrees. Imagine that the probability pk of a node of degree k in this network is proportional to kak (obviously with the constant a being a number smaller than one, otherwise that expression would diverge for large k). (1) Calculate the normalization of the degrees. In other words obtain an expression for the Pk- (2) Calculate the average degree of the network. (3) Calculate the average number of second neighbors. (4) Find out for which values of a does this neighbor have a giant component. These three expressions might be useful kak k=0 a (1-a)² Σk² ak k=0 = a + a² (1 - a)³ Σk²³ ak= ª +4a² + a² (1-a)4 k=0
A little twist in the problem of a configuration model for an exponential distribution of degrees. Imagine that the probability pk of a node of degree k in this network is proportional to kak (obviously with the constant a being a number smaller than one, otherwise that expression would diverge for large k). (1) Calculate the normalization of the degrees. In other words obtain an expression for the Pk- (2) Calculate the average degree of the network. (3) Calculate the average number of second neighbors. (4) Find out for which values of a does this neighbor have a giant component. These three expressions might be useful kak k=0 a (1-a)² Σk² ak k=0 = a + a² (1 - a)³ Σk²³ ak= ª +4a² + a² (1-a)4 k=0
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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